Tensors

Tensor algebra and calculus in general coordinates.

In this chapter 6

Tick a unit to add it to your PDF basket.

Tensor algebra and calculus

Tensor algebra and calculus

Definition

Tensors and their transformation laws

Adapted

Add to PDF

Defines contravariant and covariant components through the two bases of a general coordinate system, gives the transformation law of a tensor of type (r,s), builds new tensors by tensor product and contraction, and proves the quotient rule that recognises a tensor from the way it acts on arbitrary vectors.

Prerequisites: Linear maps and matrices, Vector spaces and bases, Vector algebra and index notation, Orthogonal curvilinear coordinates, Partial derivatives and the chain rule, Implicit and inverse function theorems

Definition(General coordinates and the two bases)AdaptedAdapted from Riley, Hobson & Bence (2006), section 26.14, pp. 955-956; Arfken, Weber & Harris (2013), section 4.3, Eqs. (4.37) and (4.44), pp. 218-220

Let $D \subseteq \mathbb{R}^n$ be open and let

x : D \to \mathbb{R}^n , \qquad (u^1, \dots, u^n) \mapsto x(u^1,\dots,u^n)

be an injective $C^2$ map whose Jacobian matrix is nonsingular at every point: a general coordinate system on the image. Nothing here requires the coordinate curves to meet at right angles. Coordinate labels are written as superscripts, $u^i$ , because they will turn out to transform contravariantly.

At each point there are two natural bases of $\mathbb{R}^n$ :

the covariant basis $e_i = \dfrac{\partial x}{\partial u^i}$ , the displacement per unit increase of $u^i$ with the other coordinates held fixed, tangent to the $u^i$ -coordinate curve;
the contravariant basis $e^i = \operatorname{grad} u^i$ , normal to the surface $u^i = \text{const}$ .

They are bases because the Jacobian is nonsingular, and by the inverse function theorem (Implicit and inverse function theorems) each $u^i$ is a $C^2$ function of position, so $e^i$ is defined. The chain rule applied to $u^i\bigl(x(u)\bigr) = u^i$ gives the reciprocity relation

e^i \cdot e_j = \delta^i_{\;j} = \begin{cases} 1, & i = j,\\ 0, & i \neq j.\end{cases}

Neither basis need consist of unit vectors, and the two coincide only when the coordinates are Cartesian. For an orthogonal system they differ only by scale factors, $e^i = \hat e_i / h_i$ and $e_i = h_i \hat e_i$ (Orthogonal curvilinear coordinates).

Summation convention. From here on, an index that occurs twice in a term, once below and once above, is summed from $1$ to $n$ . A repeated pair in the same position is not summed; such an expression is a warning that something has gone wrong. In the derivative $\partial u^i / \partial \bar{u}^{\,j}$ the upper index in the denominator counts as a lower index, which makes the balance of indices in every formula below checkable at a glance.

Every vector $v$ has an expansion in each basis,

v = v^i e_i = v_i e^i , \qquad v^i = v\cdot e^i, \qquad v_i = v \cdot e_i ,

obtained by dotting with $e^i$ and with $e_i$ respectively. The numbers $v^i$ are the contravariant components of $v$ and the $v_i$ its covariant components. They are two descriptions of the one vector, and they agree in Cartesian coordinates.

Definition(Tensors of type

(r,s)

and their transformation law)AdaptedAdapted from Riley, Hobson & Bence (2006), section 26.16, Eqs. (26.67)-(26.71), pp. 960-962; Arfken, Weber & Harris (2013), section 4.1, Eqs. (4.4)-(4.6), pp. 206-207

Let $\bar{u}^1,\dots,\bar{u}^n$ be a second general coordinate system on the same region, with $\bar u = \bar u(u)$ invertible and $C^1$ . Differentiating $x$ by the chain rule gives the two bases in the new system,

\bar{e}_i = \frac{\partial u^k}{\partial \bar{u}^{\,i}}\, e_k , \qquad \bar{e}^{\,i} = \frac{\partial \bar{u}^{\,i}}{\partial u^k}\, e^k ,

the second because $\operatorname{grad}\bar u^{\,i}$ expands by the chain rule in the same way. The two matrices of partial derivatives are inverse to each other, since $\partial \bar u^{\,i}/\partial u^k \cdot \partial u^k / \partial \bar u^{\,j} = \partial \bar u^{\,i}/\partial \bar u^{\,j} = \delta^i_{\;j}$ .

Because $v = v^i e_i = \bar v^{\,i}\bar e_i$ is one and the same vector, the components must compensate for the change of basis:

\bar{v}^{\,i} = \frac{\partial \bar{u}^{\,i}}{\partial u^k}\, v^k \quad\text{(contravariant)}, \qquad \bar{v}_i = \frac{\partial u^k}{\partial \bar{u}^{\,i}}\, v_k \quad\text{(covariant)} .

These two laws are taken as the definition of what it means to be a vector's components, and generalise as follows.

A tensor of type $(r,s)$ on the region is an assignment, to each general coordinate system, of $n^{r+s}$ numbers $T^{i_1\cdots i_r}{}_{j_1\cdots j_s}$ at each point, such that under any change of coordinates

\bar{T}^{\,i_1\cdots i_r}{}_{j_1\cdots j_s} = \frac{\partial \bar{u}^{\,i_1}}{\partial u^{a_1}} \cdots \frac{\partial \bar{u}^{\,i_r}}{\partial u^{a_r}} \; \frac{\partial u^{b_1}}{\partial \bar{u}^{\,j_1}} \cdots \frac{\partial u^{b_s}}{\partial \bar{u}^{\,j_s}} \; T^{a_1\cdots a_r}{}_{b_1\cdots b_s} .

The number $r$ is its contravariant degree, $s$ its covariant degree, and $r+s$ its rank (Arfken, Weber and Harris) or order (Riley, Hobson and Bence); we say rank. Components with upper indices only are contravariant, with lower indices only covariant, and with both mixed. A type $(0,0)$ tensor is a scalar, a type $(1,0)$ tensor a vector. Arfken, Weber and Harris stress that the rank is not limited by $n$ : a tensor in $n$ dimensions may carry any number of indices, each running from $1$ to $n$ .

The one tensor is represented in the different bases by

T = T^{ij} \, e_i \otimes e_j = T^i{}_j \, e_i \otimes e^j = T_{ij}\, e^i \otimes e^j

and so on for higher rank; the three sets of numbers describe the same object against three different tensor bases. Adding tensors, and multiplying by scalars, is done componentwise, and is only defined for tensors of the same type.

A tensor of type $(0,2)$ or $(2,0)$ is symmetric if $T_{ij} = T_{ji}$ and antisymmetric if $T_{ij} = -T_{ji}$ ; writing $T_{ij} = \tfrac12 (T_{ij}+T_{ji}) + \tfrac12 (T_{ij}-T_{ji})$ splits any such tensor into a symmetric and an antisymmetric part, and both halves are tensors because the transformation law treats the two lower slots alike. Symmetry of this kind is preserved by every change of coordinates; the symmetry of a mixed pair of indices is not, and is not a meaningful notion.

Note(The same definition without coordinates)AdaptedAdapted from Hassani (2013), Definitions 26.1.4 and 26.1.5 and Theorem 26.1.6, pp. 784-785

Hassani starts from the opposite end. Given a vector space $V$ with dual $V^*$ , a tensor of type $(r,s)$ is a multilinear map

T : \underbrace{V^* \times \cdots \times V^*}_{r} \times \underbrace{V \times \cdots \times V}_{s} \longrightarrow \mathbb{R} ,

with no coordinates mentioned. Choosing a basis $\{e_i\}$ of $V$ and its dual basis $\{\epsilon^j\}$ , the products $e_{i_1}\otimes\cdots\otimes e_{i_r}\otimes \epsilon^{j_1}\otimes\cdots\otimes\epsilon^{j_s}$ form a basis of the space of such maps, and the components of $T$ in it are its values on the basis covectors and vectors. Changing the basis then forces on the components exactly the law displayed above. The two definitions therefore describe the same objects: the coordinate definition says how the numbers behave, the multilinear one says what the numbers are the components of. In this chapter $V$ is the tangent space at a point, $\{e_i\}$ the covariant basis of the coordinate system and $\{e^j\}$ its reciprocal, which the reciprocity relation identifies as the dual basis. The manifold version is Tangent and vector bundles.

Definition(Tensor product and contraction)AdaptedAdapted from Arfken, Weber & Harris (2013), section 4.1, pp. 209-210, Eq. (4.16) and Example 4.1.1; Hassani (2013), Definitions 26.1.5 and 26.1.9, pp. 784, 788

Two operations build every tensor met in practice out of vectors.

Tensor product (direct, or outer, product). Multiply all the components of a tensor $A$ of type $(r,s)$ by all those of a tensor $B$ of type $(k,l)$ , keeping every index:

(A \otimes B)^{i_1\cdots i_r\, p_1 \cdots p_k}{}_{j_1\cdots j_s\, q_1\cdots q_l} = A^{i_1\cdots i_r}{}_{j_1\cdots j_s}\; B^{p_1\cdots p_k}{}_{q_1\cdots q_l} .

The result is a tensor of type $(r+k, s+l)$ : each index of the product picks up exactly the derivative factor it carried in $A$ or in $B$ , and multiplying the two transformation laws gives the law for the product. The order in which the indices are written is ours to choose, but an index that was upper must stay upper.

Contraction. Set one upper and one lower index equal and sum. On a mixed tensor of type $(r,s)$ with $r,s \geq 1$ this produces a tensor of type $(r-1,s-1)$ : contraction lowers the rank by two. For a type $(1,1)$ tensor it gives the trace $T^i{}_i$ , a scalar. Hassani defines it without coordinates, as a linear map $T^r_s(V)\to T^{r-1}_{s-1}(V)$ , which makes its independence of the basis part of the definition.

Contracting an upper with an upper index, or a lower with a lower, is not an operation on tensors: the result depends on the coordinate system, as Hassani's Example 26.1.8 shows for a would-be "trace" of a type $(2,0)$ tensor. (With a metric available one may first lower an index and then contract; that is a different operation, and it depends on the metric.)

Two consequences used constantly. First, the Kronecker delta $\delta^i_{\;j}$ is a type $(1,1)$ tensor, and has the same components in every coordinate system: its transformation law collapses by the chain rule to $\partial\bar u^{\,i}/\partial\bar u^{\,j} = \delta^i_{\;j}$ . Second, the direct product of a covariant and a contravariant vector, contracted, is the scalar $v_i w^i$ , whose value is the same in all coordinate systems.

Definition(The quotient rule)AdaptedAdapted from Arfken, Weber & Harris (2013), section 4.1, Eqs. (4.20)-(4.24), pp. 211-212

Suppose that in every general coordinate system on the region a set of $n^2$ numbers $K_{ij}$ is given, and that

B_i = K_{ij}\, A^j

holds in every such system, where:

$A^j$ ranges over all contravariant vectors at the point (not one fixed vector), and
for each such $A$ the resulting $B_i$ are the covariant components of a vector, that is, they obey the covariant transformation law.

Then the $K_{ij}$ are the covariant components of a tensor of type $(0,2)$ .

The same conclusion holds for any other arrangement of indices: if contracting an unknown array with arbitrary tensors always yields a tensor, the unknown array is a tensor of the type its indices advertise. Arfken, Weber and Harris state the rule for the list

K^iA_i = B , \qquad K^j{}_iA_j = B_i , \qquad K^j{}_iA_{jk} = B_{ik} , \qquad K^{ijkl}A_{ij} = B_{kl} , \qquad K^{ij}A_k = B^{ijk} ,

and prove the second; the version stated here, with both of $K$ 's indices down, is the same argument and is the case needed for the metric in The metric tensor. They add one caution: the rule says nothing when $B$ vanishes identically, because the zero array transforms as a tensor of every type.

Idea of the proofAdaptedAdapted from Arfken, Weber & Harris (2013), section 4.1, p. 212

Write the defining relation once in each coordinate system. In the new system it reads $\bar B_i = \bar K_{ij}\bar A^{\,j}$ directly. But $\bar B_i$ can also be computed the long way round: transform $B$ back to the old system, use the relation there, and transform $A$ forward again. That second route produces $\bar A^{\,j}$ multiplied by exactly the combination of derivatives that a $(0,2)$ tensor law would require. Subtracting the two expressions leaves a bracket contracted with $\bar A^{\,j}$ and equal to zero; since $\bar A$ may be chosen freely, each basis choice kills one column of the bracket in turn.

ProofAdaptedAdapted from Arfken, Weber & Harris (2013), section 4.1, Eqs. (4.21)-(4.24), p. 212

Fix a point and two coordinate systems $u$ and $\bar u$ .

1. The relation in the barred system. By hypothesis 1 the relation holds there for the arbitrary vector $\bar A$ , so

\bar B_i = \bar K_{ij}\,\bar A^{\,j} . \tag{i}

2. The same quantity, computed through the unbarred system. By hypothesis 2 the $B_i$ are covariant components, so $\bar B_i = (\partial u^m/\partial\bar u^{\,i})\,B_m$ . Using the relation in the unbarred system, $B_m = K_{mn}A^n$ . Finally $A$ is a contravariant vector, and the inverse of its transformation law (obtained by exchanging the roles of the two systems) is $A^n = (\partial u^n/\partial \bar u^{\,j})\,\bar A^{\,j}$ . Chaining the three substitutions,

\bar B_i = \frac{\partial u^m}{\partial \bar u^{\,i}}\,\frac{\partial u^n}{\partial \bar u^{\,j}}\; K_{mn}\;\bar A^{\,j} . \tag{ii}

3. Subtract. Both (i) and (ii) give $\bar B_i$ , so for each fixed $i$

\left( \bar K_{ij} - \frac{\partial u^m}{\partial \bar u^{\,i}}\,\frac{\partial u^n}{\partial \bar u^{\,j}}\, K_{mn} \right) \bar A^{\,j} = 0 .

4. Use the arbitrariness of $A$ . Hypothesis 1 lets us choose $\bar A$ to be the vector with $\bar A^{\,j} = \delta^j_{\;p}$ for each $p = 1,\dots,n$ in turn. Each choice leaves a single term of the sum, so the bracket vanishes for every $i$ and every $p$ :

\bar K_{ij} = \frac{\partial u^m}{\partial \bar u^{\,i}}\,\frac{\partial u^n}{\partial \bar u^{\,j}}\, K_{mn} ,

which is the transformation law of a tensor of type $(0,2)$ . $\square$

Remark(Where the hypotheses are used)AdaptedAdapted from Arfken, Weber & Harris (2013), section 4.1, pp. 212-213

"For every $A$ " is essential, and step 4 is where it is used. One vector is not enough. In two dimensions take $A$ with components $(1,0)$ in every coordinate system. Then $B_i = K_{i1}$ : the relation constrains only the first column of $K$ , and the second column may be filled in independently in each coordinate system with numbers that obey no transformation law at all. The resulting array satisfies $B_i = K_{ij}A^j$ everywhere and is not a tensor.

"In every coordinate system" is used in steps 1 and 2. The proof compares the relation in two systems; a relation known only in one says nothing about how $K$ transforms.

$B \neq 0$ . If $B_i = 0$ identically the final bracket is contracted with vectors but the preceding steps give no information, because both routes produce zero whatever $K$ does. This is Arfken, Weber and Harris's caution that the transformation properties of zero are indeterminate.

Why the rule is worth having. It is the only practical test for tensor character in physics, where a quantity is usually met through the linear relation it imposes between two measured vectors rather than through its own transformation law. Its most important negative companion is that an array with indices need not be a tensor: the partial derivatives $\partial v^i/\partial u^j$ of a vector field pick up a second-derivative term under a change of coordinates and fail the law, which is what forces the covariant derivative of Christoffel symbols and the covariant derivative, and the Christoffel symbols there are the standard example of a three-index array that is not a tensor. Behaviour under reflections is a further refinement that the law above does not capture; see Pseudotensors and the Levi-Civita tensor.

Method(Deciding whether an indexed array is a tensor)Original

Count the indices and note their position. This fixes the type $(r,s)$ to be tested, hence the number of $\partial\bar u/\partial u$ and $\partial u/\partial\bar u$ factors in the law.
If the array is built from known tensors by addition (of the same type), tensor product, contraction, or raising and lowering with a metric, it is a tensor; no calculation is needed.
If instead the array is defined by a linear relation $B = K A$ that holds in all coordinate systems with $A$ arbitrary and $B$ a tensor, apply the quotient rule.
If the array is defined by differentiating something, be suspicious: carry out the change of coordinates explicitly and look for a term containing a second derivative $\partial^2 u / \partial \bar u\, \partial \bar u$ . Such a term cannot be cancelled and disqualifies the array.
Otherwise, verify the transformation law directly on a change of coordinates that is not linear; a linear change will not detect the failure in step 4.

Example(A contraction is the same number in two coordinate systems)Original

Work in the plane, and let the second coordinate system be

x = u^2 - v^2 , \qquad y = 2uv , \qquad u > 0 ,

which maps the right half-plane bijectively onto the plane cut along the negative $x$ -axis, with Jacobian matrix and determinant

J = \frac{\partial(x,y)}{\partial(u,v)} = \begin{pmatrix} 2u & -2v \\ 2v & 2u\end{pmatrix}, \qquad \det J = 4(u^2+v^2) > 0 ,

so it is a general coordinate system. It is not linear, which is what makes the test meaningful.

The two objects. Let $\varphi(x,y) = x^2 y$ and let $\omega_i = \partial\varphi/\partial x^i$ be its gradient, a covariant vector. Let $A$ be the rotation field with Cartesian components $A^i = (-y, x)$ .

At a point. Take $(u,v) = (2,1)$ , so $x = 2^2-1^2 = 3$ and $y = 2\cdot 2\cdot 1 = 4$ . There

\omega_i = (2xy,\; x^2) = (24,\; 9), \qquad A^i = (-y,\; x) = (-4,\; 3).

Their tensor product is the mixed tensor $T^i{}_j = A^i\omega_j$ , with matrix

[T^i{}_j] = \begin{pmatrix} -4 \\ 3\end{pmatrix}\begin{pmatrix} 24 & 9\end{pmatrix} = \begin{pmatrix} -96 & -36 \\ 72 & 27 \end{pmatrix}, \qquad T^i{}_i = -96 + 27 = -69 .

The same point in the new coordinates. At $(u,v) = (2,1)$ ,

J = \begin{pmatrix} 4 & -2 \\ 2 & 4\end{pmatrix}, \qquad \det J = 20, \qquad J^{-1} = \frac{1}{20}\begin{pmatrix} 4 & 2 \\ -2 & 4\end{pmatrix} .

Here $J$ has entries $\partial x^i/\partial u^j$ and $J^{-1}$ has entries $\partial u^i/\partial x^j$ . The contravariant law uses $J^{-1}$ :

\bar A^{\,u} = \tfrac{1}{5}(-4) + \tfrac{1}{10}(3) = -\tfrac12 , \qquad \bar A^{\,v} = -\tfrac{1}{10}(-4) + \tfrac{1}{5}(3) = 1 .

The covariant law uses $J$ itself, in the transposed pattern $\bar\omega_i = (\partial x^k/\partial u^i)\,\omega_k$ :

\bar\omega_u = 4(24) + 2(9) = 114 , \qquad \bar\omega_v = -2(24) + 4(9) = -12 .

As a check on the covariant law, $\varphi$ expressed in the new coordinates is $2uv(u^2-v^2)^2$ , whose partial derivatives at $(2,1)$ are $2v(u^2-v^2)^2 + 8u^2v(u^2-v^2) = 18 + 96 = 114$ and $2u(u^2-v^2)^2 - 8uv^2(u^2-v^2) = 36 - 48 = -12$ : the gradient computed in the new coordinates agrees with the transformed one, as it must for a covariant vector.

The contraction. In the new coordinates

[\bar T^{\,i}{}_j] = \begin{pmatrix} -\tfrac12 \\ 1 \end{pmatrix}\begin{pmatrix} 114 & -12\end{pmatrix} = \begin{pmatrix} -57 & 6 \\ 114 & -12 \end{pmatrix}, \qquad \bar T^{\,i}{}_i = -57 - 12 = -69 .

The individual components are completely different in the two systems; the contraction is

\boxed{\,T^i{}_i = \bar T^{\,i}{}_i = -69\,}

in both. Note also that the two off-diagonal entries changed sign pattern and that $-96+27$ and $-57-12$ have nothing to do with each other term by term: it is only the sum that is invariant, which is exactly what the chain rule in the proof of contraction delivers.

Application(Recognising material tensors from a measured linear response)AdaptedAdapted from Riley, Hobson & Bence (2006), section 26.12, Eqs. (26.43)-(26.44), p. 952; Arfken, Weber & Harris (2013), section 4.1, p. 212

A laboratory never measures a transformation law. What it measures is a linear response: apply a field, read off a current. In an anisotropic crystal the current density and the applied electric field are related by

j^i = \sigma^i{}_j \, E^j ,

and the nine conductivities $\sigma^i{}_j$ are whatever the experiment returns in the axes the apparatus happens to use. The quotient rule is what licenses treating them as a tensor: $E$ can be applied in any direction, so $E^j$ is arbitrary; $j$ and $E$ are both vectors; the relation holds whatever axes the next laboratory chooses. Hence $\sigma$ is a tensor of type $(1,1)$ , and only then is it legitimate to diagonalise it, to speak of principal axes of conduction, or to compare measurements made in different frames. Arfken, Weber and Harris make the same argument for the polarisability relating an induced dipole moment to an applied field, and Riley, Hobson and Bence list magnetic susceptibility, $M_i = \chi_{ij}H_j$ , alongside conductivity.

The argument is used at every rank. Hooke's law for an elastic solid relates two second-rank tensors, $p_{ij} = c_{ijkl}e_{kl}$ , and the quotient rule promotes the $81$ elastic constants $c_{ijkl}$ to a fourth-rank tensor; that case, with the strain and stress tensors themselves, is Stress and strain tensors. In relativity the same reasoning identifies the four-potential as a four-vector from the fact that a scalar wave operator applied to it equals the four-current. What the rule cannot do is manufacture information: if the measured response vanishes in all frames, nothing is learned.

Definition

The metric tensor

Adapted

Add to PDF

Defines the metric tensor as the matrix of dot products of the covariant basis vectors, proves that it is a symmetric second-rank tensor whose inverse is the contravariant metric, and shows how it produces the line element, the volume element and the raising and lowering of indices, with a worked non-orthogonal example.

Prerequisites: Tensors and their transformation laws, Orthogonal curvilinear coordinates, Inner products and norms, Determinants, Multiple integrals and change of variables

Definition(The metric tensor and its inverse)AdaptedAdapted from Riley, Hobson & Bence (2006), section 26.15, Eqs. (26.56) and (26.62), pp. 957-958; Arfken, Weber & Harris (2013), section 4.3, Eqs. (4.40), (4.41) and (4.47), pp. 219-220

Let $u^1,\dots,u^n$ be a general coordinate system with covariant basis $e_i = \partial x/\partial u^i$ and contravariant basis $e^i = \operatorname{grad} u^i$ , as in Tensors and their transformation laws. The covariant components of the metric tensor are the $n^2$ functions

g_{ij} = e_i \cdot e_j ,

and the contravariant components are

g^{ij} = e^i \cdot e^j .

Both are symmetric, because the dot product is. The mixed components are $g^i{}_j = e^i\cdot e_j = \delta^i{}_j$ by reciprocity, so the Kronecker delta is the metric with one index of each kind.

We write $G = [g_{ij}]$ for the matrix of covariant components and

g = \det G

for its determinant. The coordinate system is assumed nonsingular, so (as proved below) $g \neq 0$ and $G$ is invertible.

In an orthogonal system the matrix is diagonal, $g_{ij} = h_i^2\,\delta_{ij}$ with no sum, where the $h_i$ are the scale factors of Orthogonal curvilinear coordinates, and $g = h_1^2h_2^2h_3^2$ . The point of the general $g_{ij}$ is precisely to handle systems in which the coordinate curves are not perpendicular, where off-diagonal entries survive.

Definition(What the metric does)AdaptedAdapted from Riley, Hobson & Bence (2006), section 26.15, Eqs. (26.57)-(26.58) and (26.61)-(26.65), pp. 957-959, and p. 963; Arfken, Weber & Harris (2013), section 4.3, Eqs. (4.39), (4.41)-(4.43) and (4.48), pp. 219-221

With the hypotheses of the definition, at every point of the region:

Tensor character. The $g_{ij}$ are the covariant components of a tensor of type $(0,2)$ and the $g^{ij}$ the contravariant components of the same tensor, of type $(2,0)$ .
The two matrices are inverse. $g^{ij}g_{jk} = \delta^i{}_k$ ; equivalently $[g^{ij}] = G^{-1}$ . In particular $g \neq 0$ .
The bases are related by the metric. $e^i = g^{ij}e_j$ and $e_i = g_{ij}e^j$ .
Line element. For a displacement $dx = e_i\,du^i$ , $(ds)^2 = dx\cdot dx = g_{ij}\,du^i\,du^j ,$ with the cross terms $i \neq j$ present exactly when the system is not orthogonal. The length of a curve $t \mapsto u(t)$ , $a \le t \le b$ , is $\int_a^b \sqrt{g_{ij}\,\dot u^i\dot u^j}\;dt$ .
Volume element. In three dimensions $dV = \sqrt{g}\;du^1du^2du^3$ , and in general the Jacobian factor of the coordinate change is $\lvert\det[\,e_1\cdots e_n]\rvert = \sqrt{g}$ .
Scalar product. $v\cdot w = g_{ij}v^iw^j = g^{ij}v_iw_j = v^iw_i = v_iw^i$ .
Raising and lowering. The covariant and contravariant components of one vector are related by $v_i = g_{ij}v^j , \qquad v^i = g^{ij}v_j ,$ and the same holds index by index for a tensor of any type: contracting with $g_{ij}$ turns an upper index into a lower one, contracting with $g^{ij}$ does the reverse, and the two operations are mutually inverse.

Idea of the proofOriginal

Everything follows from two facts and no computation worth the name. The first is that $\{e_i\}$ and $\{e^i\}$ are bases of the same space that are dual to one another; so each is expressible in terms of the other, and the coefficients can only be the dot products, which are the $g$ 's. The second is that the components of a vector against a basis are obtained by dotting with the reciprocal basis; so passing between $v^i$ and $v_i$ is passing between the two bases, and the matrix that does it is the metric. Tensor character then costs one line, because $g_{ij}$ is built from two objects, $e_i$ and $e_j$ , each of which already transforms with one $\partial u/\partial\bar u$ factor.

ProofAdaptedAdapted from Riley, Hobson & Bence (2006), section 26.15, pp. 958-959 and p. 963; Arfken, Weber & Harris (2013), section 4.3, pp. 219-220

1. Tensor character. Under a change of coordinates the covariant basis transforms as $\bar e_i = (\partial u^k/\partial\bar u^{\,i})e_k$ (Tensors and their transformation laws). Hence

\bar g_{ij} = \bar e_i\cdot\bar e_j = \frac{\partial u^k}{\partial \bar u^{\,i}}\frac{\partial u^l}{\partial \bar u^{\,j}}\, e_k\cdot e_l = \frac{\partial u^k}{\partial \bar u^{\,i}}\frac{\partial u^l}{\partial \bar u^{\,j}}\, g_{kl} ,

which is the type $(0,2)$ law. The contravariant basis transforms with the inverse derivatives, so the same two lines with $e^i$ in place of $e_i$ give the type $(2,0)$ law for $g^{ij}$ . Arfken, Weber and Harris reach the first of these differently: $(ds)^2$ is a scalar and $du^i$ is an arbitrary contravariant vector, so the quotient rule applied twice to item 4 identifies $g_{ij}$ as a $(0,2)$ tensor. Both routes are available; the second needs item 4 first.

2 and 3. The two matrices are inverse. The Jacobian of the coordinate map is nonsingular, so $\{e_j\}$ is a basis of $\mathbb{R}^n$ and we may expand the reciprocal basis in it: $e^i = c^{ij}e_j$ for some numbers $c^{ij}$ . Dot this with $e^l$ and use the expansion's own reciprocity, $e_j\cdot e^l = \delta_j^{\;l}$ :

e^i\cdot e^l = c^{ij}\,(e_j\cdot e^l) = c^{ij}\delta_j^{\;l} = c^{il} .

The left-hand side is $g^{il}$ by definition, so $c^{ij} = g^{ij}$ and $e^i = g^{ij}e_j$ , which is the first half of item 3. Now dot $e^i = g^{ij}e_j$ with $e_k$ :

\delta^i{}_k = e^i\cdot e_k = g^{ij}\,(e_j\cdot e_k) = g^{ij}g_{jk} ,

which is item 2. Since $[g^{ij}]G = I$ , the determinant $g$ cannot vanish, and multiplying $e^i = g^{ij}e_j$ by $g_{ki}$ and summing gives $g_{ki}e^i = g_{ki}g^{ij}e_j = \delta_k^{\;j}e_j = e_k$ , the second half of item 3. This is the step that uses the nonsingularity of the coordinate system, and nothing else in the proof does.

4. Line element. A displacement produced by increments $du^i$ is $dx = (\partial x/\partial u^i)du^i = e_i\,du^i$ to first order. Dotting it with itself,

(ds)^2 = (e_i\,du^i)\cdot(e_j\,du^j) = (e_i\cdot e_j)\,du^i du^j = g_{ij}\,du^i du^j ,

where the middle step is bilinearity of the dot product, not orthogonality: no term is discarded. For a curve, divide by $(dt)^2$ and take the square root, which requires $g_{ij}\dot u^i\dot u^j \geq 0$ — true here because it is $\lvert dx/dt\rvert^2$ .

5. Volume element. Let $J$ be the matrix whose columns are $e_1,\dots,e_n$ , that is the Jacobian matrix of $x$ . Its Gram matrix is $J^{\mathsf T}J = [\,e_i\cdot e_j\,] = G$ , so

(\det J)^2 = \det(J^{\mathsf T}J) = \det G = g ,

giving $\lvert\det J\rvert = \sqrt{g}$ ; the change-of-variables theorem (Multiple integrals and change of variables) then supplies $dV = \sqrt{g}\,du^1\cdots du^n$ . In three dimensions $\det J$ is the scalar triple product $e_1\cdot(e_2\times e_3)$ , which is Riley, Hobson and Bence's form of the same statement.

6 and 7. Scalar products, raising and lowering. A vector has the two expansions $v = v^ie_i = v_ie^i$ , and dotting each with $e_i$ or with $e^i$ and using reciprocity gives $v_i = v\cdot e_i$ and $v^i = v\cdot e^i$ . Therefore

v_i = (v^je_j)\cdot e_i = g_{ij}v^j , \qquad v^i = (v_je^j)\cdot e^i = g^{ij}v_j ,

which is item 7 for a vector; by item 2 the two operations undo each other. Item 6 is obtained by expanding $v\cdot w$ with each of the four combinations of the two expansions and using $g_{ij}$ , $g^{ij}$ or $\delta^i{}_j$ as the case requires. For a tensor of higher type, the same contraction applied in one slot at a time converts that slot, because the transformation law treats the slots independently: for instance $T^i{}_j = g_{jk}T^{ik}$ and $T_{ij} = g_{ik}g_{jl}T^{kl}$ . $\square$

Remark(Where the hypotheses are used, and which convention we follow)AdaptedAdapted from Arfken, Weber & Harris (2013), section 4.3, Example 4.3.1, p. 221; Hassani (2013), section 26.5, Definition 26.5.2 and Proposition 26.5.3, pp. 805-806

Nonsingularity is the whole hypothesis. Only step 2-3 of the proof used anything beyond bilinearity, and what it used was that $\{e_i\}$ is a basis. Where that fails the structure collapses in a way worth seeing: at the origin of plane polar coordinates $e_\phi = \partial x/\partial\phi = 0$ , so $g_{\phi\phi}=0$ and $g = 0$ ; the covariant components of a vector there are $v_\rho = v\cdot e_\rho$ and $v_\phi = 0$ whatever $v$ is, and no amount of raising will recover $v$ . This is why the definition removes the singular points from the domain.

Hassani's version of the same hypothesis. In the coordinate-free account the metric is defined as a nondegenerate symmetric bilinear form $g$ , nondegeneracy meaning that the induced map from $V$ to $V^*$ is invertible, equivalently that $\det[g_{ij}] \neq 0$ . Lowering an index is applying that map and raising is applying its inverse; Lee calls the two flat and sharp, writing $v^\flat$ and $\omega^\sharp$ , from the musical signs, and notes the useful consequence that the gradient of a function is the sharp of its differential, $(\operatorname{grad} f)^i = g^{ij}\partial_j f$ . Our $g_{ij} = e_i\cdot e_j$ is a special case: it is nondegenerate for the reason just given, and positive definite because it comes from the Euclidean dot product.

Positive definiteness is not used in items 1, 2, 3, 6 or 7. It is used only to read $\sqrt{g}$ as a volume and $\sqrt{(ds)^2}$ as a length. This matters, because the metrics of relativity are not positive definite. Arfken, Weber and Harris give the Minkowski metric of special relativity as $\operatorname{diag}(1,-1,-1,-1)$ in their Example 4.3.1 and point out explicitly that $ds^2$ need not be positive, so that $ds$ may be imaginary. With that signature $\det G = -1 < 0$ and the volume factor must be written $\sqrt{\lvert g\rvert}$ ; every statement above except items 4 and 5 goes through unchanged. The general manifold setting, and the name pseudo-Riemannian for the indefinite case, is Riemannian manifolds.

A convention to state once. Riley, Hobson and Bence and Arfken, Weber and Harris both build $g_{ij}$ from an embedding in Euclidean $\mathbb{R}^3$ and therefore always have $g > 0$ ; Hassani and Lee start from an abstract nondegenerate form and allow either sign. We follow the embedded definition in this unit, and flag the indefinite case where it matters rather than silently merging the two. The sign convention $(+,-,-,-)$ used below for Minkowski space is Arfken, Weber and Harris's; the opposite convention $(-,+,+,+)$ is equally common in the relativity literature and changes the sign of every squared interval.

Method(Computing the metric of a coordinate system)Original

Given the coordinates $u^1,\dots,u^n$ through the Cartesian position $x(u)$ :

Differentiate: form the $n$ covariant basis vectors $e_i = \partial x/\partial u^i$ as explicit Cartesian column vectors. These are the columns of the Jacobian matrix $J$ .
Take dot products: $g_{ij} = e_i\cdot e_j$ , that is $G = J^{\mathsf T}J$ . Compute the full matrix, including off-diagonal entries; zeros there mean the system is orthogonal, and only then may scale factors be used instead.
Check nonsingularity: compute $g = \det G$ and record where it vanishes. Remove those points from the domain.
Invert: $[g^{ij}] = G^{-1}$ . For a diagonal $G$ this is just reciprocals of the diagonal entries.
Read off the line element $(ds)^2 = g_{ij}du^idu^j$ and the volume element $\sqrt{\lvert g\rvert}\,du^1\cdots du^n$ .
To lower an index, contract with $g_{ij}$ ; to raise one, contract with $g^{ij}$ . Verify by checking that $v_iv^i = g_{ij}v^iv^j$ agrees with the Euclidean $\lvert v\rvert^2$ at one point.

The differential operators that this makes available in general coordinates are Differential operators in general coordinates.

Example(A non-orthogonal coordinate system: shear coordinates in the plane)Original

Take the plane with coordinates $(u^1,u^2) = (u,v)$ defined by

x = u , \qquad y = v + u^2 ,

a shear whose amount grows with $u$ . The coordinate curves $v = \text{const}$ are the parabolas $y = x^2 + \text{const}$ and the curves $u = \text{const}$ are vertical lines; they are not perpendicular, so this system is outside the reach of Orthogonal curvilinear coordinates. The map is a bijection of the plane onto itself with inverse $u = x$ , $v = y - x^2$ .

Step 1: the basis. $e_1 = \partial x/\partial u = (1, 2u)$ and $e_2 = (0,1)$ .

Step 2: the metric.

g_{11} = 1 + 4u^2 , \qquad g_{12} = g_{21} = 2u , \qquad g_{22} = 1 , \qquad G = \begin{pmatrix} 1+4u^2 & 2u \\ 2u & 1\end{pmatrix} .

The off-diagonal entry $2u$ is the whole content of the example: it vanishes only on the line $u=0$ , where the two coordinate directions happen to be perpendicular.

Step 3: the determinant. $g = (1+4u^2)\cdot 1 - (2u)^2 = 1$ everywhere. The shear preserves area, so the area element is $\sqrt g\,du\,dv = du\,dv = dx\,dy$ , as it should be.

Step 4: the inverse. Since $g=1$ ,

[g^{ij}] = G^{-1} = \begin{pmatrix} 1 & -2u \\ -2u & 1+4u^2 \end{pmatrix} .

Step 5: the line element and a length.

(ds)^2 = (1+4u^2)\,(du)^2 + 4u\,du\,dv + (dv)^2 ,

the middle term being $2g_{12}du\,dv$ . Take the coordinate curve $v = 0$ from $u=0$ to $u=1$ : along it $dv = 0$ , so its length is

L = \int_0^1 \sqrt{g_{11}}\;du = \int_0^1\sqrt{1+4u^2}\;du = \frac{\sqrt5}{2} + \frac{1}{4}\ln\bigl(2+\sqrt5\bigr) = 1.478943\ldots

In Cartesian terms that curve is the parabola $y=x^2$ from $(0,0)$ to $(1,1)$ , whose arc length $\int_0^1\sqrt{1+4x^2}\,dx$ is the same integral — the metric has reproduced the Euclidean length from the coordinates alone.

Step 6: raising an index. Let $\varphi(x,y) = y - x^2$ , which in the new coordinates is simply $\varphi = v$ . Its gradient is a covariant vector with components

\omega_i = \frac{\partial\varphi}{\partial u^i} = (0,\,1) .

Raising with $g^{ij}$ ,

\omega^1 = g^{11}\omega_1 + g^{12}\omega_2 = -2u , \qquad \omega^2 = g^{21}\omega_1 + g^{22}\omega_2 = 1+4u^2 ,

so the covariant components $(0,1)$ and the contravariant components $(-2u,\,1+4u^2)$ of the one gradient look nothing alike — the clearest possible warning against treating the two kinds of index as interchangeable. Their contraction is

\omega_i\omega^i = 0\cdot(-2u) + 1\cdot(1+4u^2) = 1 + 4u^2 .

Checking against Euclidean geometry: $\operatorname{grad}\varphi = (-2x, 1)$ with $\lvert\operatorname{grad}\varphi\rvert^2 = 4x^2+1$ , and $x = u$ , so the two agree. At the point $u = 3/2$ ,

\boxed{\;\omega_i\omega^i = 1 + 4\left(\tfrac32\right)^2 = 10\;}

by either route. Had we forgotten the metric and summed $\omega_i\omega_i = 0^2+1^2 = 1$ , we would have been out by a factor of ten.

Application(The interval of special relativity)AdaptedAdapted from Arfken, Weber & Harris (2013), section 4.3, Example 4.3.1, p. 221

The metric earns its keep wherever "distance" is not the Euclidean one. In special relativity the coordinates are $(x^0,x^1,x^2,x^3) = (ct,x,y,z)$ and the metric is the constant Minkowski matrix

[\eta_{\mu\nu}] = [\eta^{\mu\nu}] = \operatorname{diag}(1,-1,-1,-1) ,

given in this form by Arfken, Weber and Harris, who remark that it is included precisely to show that $ds^2$ need not be positive. The line element is then the interval

(ds)^2 = \eta_{\mu\nu}\,dx^\mu dx^\nu = c^2(dt)^2 - (dx)^2-(dy)^2-(dz)^2 ,

and the entire kinematic content of the theory is that this number is the same in every inertial frame. A coordinate change here is a Lorentz transformation, and the statement that $\eta$ is a type $(0,2)$ tensor with the same components in every inertial frame is the statement that Lorentz transformations are exactly those that preserve the interval.

Three everyday consequences are pure index gymnastics with $\eta$ . Raising and lowering flips the sign of the three spatial components and leaves the time component alone, so that the four-momentum with contravariant components $p^\mu = (E/c,\, p^1, p^2, p^3)$ has covariant components $p_\mu = (E/c,\, -p^1,-p^2,-p^3)$ ; the contraction $p_\mu p^\mu = E^2/c^2 - \lvert p\rvert^2$ , where $\lvert p\rvert$ is the ordinary length of the three-momentum, is therefore a scalar, and setting it equal to its value in the rest frame gives $E^2 = \lvert p\rvert^2c^2 + m^2c^4$ with no further work. Since $\det\eta = -1$ , the volume factor is $\sqrt{\lvert\eta\rvert} = 1$ and four-dimensional integrals carry no extra weight in inertial coordinates — a simplification that disappears in general relativity, where the metric varies from point to point and $\sqrt{\lvert g\rvert}$ must be carried through every integral. Finally the sign of $(ds)^2$ classifies separations as timelike, null or spacelike, which is the indefiniteness of the metric doing physical work rather than causing trouble.

Definition

Christoffel symbols and the covariant derivative

Adapted

Add to PDF

Defines the Christoffel symbols as the components of the derivatives of the basis vectors, derives them from the metric, shows that they are not tensor components, builds the covariant derivative of a tensor of any type, and uses it to define parallel transport, worked out on the sphere.

Prerequisites: The metric tensor, Tensors and their transformation laws, Orthogonal curvilinear coordinates, Linear systems and the matrix exponential, Existence and uniqueness of solutions, Determinants

Definition(Christoffel symbols)AdaptedAdapted from Riley, Hobson & Bence (2006), section 26.18, Eqs. (26.75)-(26.77), pp. 965-966; Arfken, Weber & Harris (2013), section 4.3, Eqs. (4.53)-(4.54) and (4.59), pp. 222-223

In a general coordinate system the basis vectors $e_i = \partial x/\partial u^i$ change from point to point, so their derivatives do not vanish. Each derivative is itself a vector and may be expanded in the basis. The coefficients are the Christoffel symbols of the second kind $\Gamma^k{}_{ij}$ :

\frac{\partial e_i}{\partial u^j} = \Gamma^k{}_{ij}\, e_k , \qquad\text{equivalently}\qquad \Gamma^k{}_{ij} = e^k\cdot\frac{\partial e_i}{\partial u^j} ,

the second form obtained by dotting the first with $e^k$ and using reciprocity. There are $n^3$ of them at each point — $27$ in three dimensions — and they depend on the coordinate system, not only on the point. Differentiating the reciprocity relation $e^i\cdot e_j = \delta^i{}_j$ and substituting gives the companion rule for the other basis,

\frac{\partial e^i}{\partial u^j} = -\,\Gamma^i{}_{kj}\, e^k ,

with the opposite sign. The Christoffel symbol of the first kind is the same object with its upper index lowered,

[ij,k] = g_{mk}\,\Gamma^m{}_{ij} = e_k\cdot\frac{\partial e_i}{\partial u^j} ,

a notation due to Arfken, Weber and Harris that is convenient in the derivation below. In Cartesian coordinates the $e_i$ are constant, so every $\Gamma^k{}_{ij}$ and every $[ij,k]$ is zero.

Despite carrying three indices, the $\Gamma^k{}_{ij}$ are not the components of a tensor; the statement below says precisely how they transform and why.

Definition(The metric formula, the symmetry, and the transformation law)AdaptedAdapted from Riley, Hobson & Bence (2006), section 26.18, Eqs. (26.78) and (26.82), pp. 966-967; Arfken, Weber & Harris (2013), section 4.3, Eqs. (4.61)-(4.63), pp. 223-224

Let $u^1,\dots,u^n$ be a general coordinate system with metric $g_{ij}$ (The metric tensor), and suppose the coordinate map $x$ is $C^3$ so that the symbols are differentiable. Then:

Symmetry. $\Gamma^k{}_{ij} = \Gamma^k{}_{ji}$ , and likewise $[ij,k]=[ji,k]$ .
Derivative of the metric. $\dfrac{\partial g_{ij}}{\partial u^k} = [ik,j] + [jk,i] = \Gamma^l{}_{ik}\,g_{lj} + \Gamma^l{}_{jk}\,g_{il}$ .
The metric formula. The symbols are determined by the metric and its first derivatives alone: $\Gamma^m{}_{ij} = \tfrac12\, g^{mk}\left( \frac{\partial g_{jk}}{\partial u^i} + \frac{\partial g_{ki}}{\partial u^j} - \frac{\partial g_{ij}}{\partial u^k} \right) .$
Transformation law. Under a change of coordinates $u \to \bar u$ , $\bar\Gamma^{\,k}{}_{ij} = \frac{\partial\bar u^{\,k}}{\partial u^n}\frac{\partial u^l}{\partial\bar u^{\,i}}\frac{\partial u^m}{\partial\bar u^{\,j}}\,\Gamma^n{}_{lm} \;+\; \frac{\partial\bar u^{\,k}}{\partial u^l}\,\frac{\partial^2 u^l}{\partial\bar u^{\,i}\,\partial\bar u^{\,j}} .$ The first term is the type $(1,2)$ tensor law; the second term, containing a second derivative, is not of that form, so the $\Gamma^k{}_{ij}$ are not tensor components. The second term does not depend on $\Gamma$ , so the difference of two sets of Christoffel symbols on the same region is a tensor.

Idea of the proofOriginal

The symmetry is the equality of mixed partial derivatives in disguise: $\partial e_i/\partial u^j$ and $\partial e_j/\partial u^i$ are both $\partial^2 x/\partial u^i\partial u^j$ . The metric formula then comes from a trick that is worth remembering rather than re-deriving. Differentiating $g_{ij}=e_i\cdot e_j$ produces a sum of two Christoffel symbols; one such equation cannot be solved for one symbol. Writing the same equation three times with the indices $i,j,k$ cycled gives three sums of two, and the particular combination "second plus third minus first" is arranged so that four of the six terms cancel in pairs and the survivors are two copies of the same symbol. Lowering an index with $g^{mk}$ finishes it. The transformation law is a direct computation: express the barred basis and reciprocal basis in terms of the unbarred ones and differentiate, and the product rule emits exactly one term in which the derivative lands on the transformation coefficients rather than on a basis vector. That is the offending term.

ProofAdaptedAdapted from Riley, Hobson & Bence (2006), section 26.18, pp. 966-967; Arfken, Weber & Harris (2013), section 4.3, pp. 223-224

1. Symmetry. Because $e_i = \partial x/\partial u^i$ and $x$ is at least $C^2$ ,

\frac{\partial e_i}{\partial u^j} = \frac{\partial^2 x}{\partial u^j\,\partial u^i} = \frac{\partial^2 x}{\partial u^i\,\partial u^j} = \frac{\partial e_j}{\partial u^i} .

Dotting with $e^k$ gives $\Gamma^k{}_{ij} = \Gamma^k{}_{ji}$ , and contracting with $g_{mk}$ gives the same for the first-kind symbols. This step uses that the basis comes from coordinates; see the remark.

2. Derivative of the metric. Differentiate $g_{ij}=e_i\cdot e_j$ by the product rule and insert the definition twice:

\frac{\partial g_{ij}}{\partial u^k} = \frac{\partial e_i}{\partial u^k}\cdot e_j + e_i\cdot\frac{\partial e_j}{\partial u^k} = \Gamma^l{}_{ik}\,e_l\cdot e_j + \Gamma^l{}_{jk}\,e_i\cdot e_l = \Gamma^l{}_{ik}\,g_{lj} + \Gamma^l{}_{jk}\,g_{il} ,

which in first-kind notation is $[ik,j]+[jk,i]$ .

3. The metric formula. Write item 2 three times, cycling $i\to j\to k\to i$ :

\partial_k g_{ij} = \Gamma^l{}_{ik}g_{lj} + \Gamma^l{}_{jk}g_{il}, \qquad \partial_i g_{jk} = \Gamma^l{}_{ji}g_{lk} + \Gamma^l{}_{ki}g_{jl}, \qquad \partial_j g_{ki} = \Gamma^l{}_{kj}g_{li} + \Gamma^l{}_{ij}g_{kl} .

Add the second and third and subtract the first. Using the symmetry from item 1 and the symmetry of $g$ , the term $\Gamma^l{}_{ki}g_{jl}$ cancels against $\Gamma^l{}_{ik}g_{lj}$ , and $\Gamma^l{}_{kj}g_{li}$ cancels against $\Gamma^l{}_{jk}g_{il}$ . What remains is $\Gamma^l{}_{ji}g_{lk}+\Gamma^l{}_{ij}g_{kl} = 2\,\Gamma^l{}_{ij}g_{kl}$ , so

\frac{\partial g_{jk}}{\partial u^i} + \frac{\partial g_{ki}}{\partial u^j} - \frac{\partial g_{ij}}{\partial u^k} = 2\,\Gamma^l{}_{ij}\,g_{kl} .

Contract with $g^{mk}$ and use $g^{mk}g_{kl}=\delta^m{}_l$ (the nondegeneracy of the metric is used here, and only here) to get item 3. Arfken, Weber and Harris organise the same cancellation through the first-kind symbols, obtaining $[ij,k]$ from the three derivatives first and raising the index afterwards; the arithmetic is identical.

4. Transformation law. In the barred system $\bar\Gamma^{\,k}{}_{ij} = \bar e^{\,k}\cdot\partial\bar e_i/\partial\bar u^{\,j}$ . Substitute the two basis transformation laws of Tensors and their transformation laws, namely $\bar e^{\,k} = (\partial\bar u^{\,k}/\partial u^n)\,e^n$ and $\bar e_i = (\partial u^l/\partial \bar u^{\,i})\,e_l$ . Differentiating the second by the product rule,

\frac{\partial\bar e_i}{\partial\bar u^{\,j}} = \frac{\partial^2 u^l}{\partial\bar u^{\,j}\partial\bar u^{\,i}}\,e_l + \frac{\partial u^l}{\partial\bar u^{\,i}}\,\frac{\partial u^m}{\partial\bar u^{\,j}}\,\frac{\partial e_l}{\partial u^m} ,

where the chain rule converted $\partial/\partial\bar u^{\,j}$ acting on $e_l$ into $\partial u^m/\partial\bar u^{\,j}$ times $\partial/\partial u^m$ . Now dot with $\bar e^{\,k}$ . In the first term $e^n\cdot e_l = \delta^n{}_l$ collapses the sum and leaves $(\partial\bar u^{\,k}/\partial u^l)(\partial^2u^l/\partial\bar u^{\,i}\partial\bar u^{\,j})$ ; in the second, $e^n\cdot\partial e_l/\partial u^m = \Gamma^n{}_{lm}$ by definition. Adding the two gives item 4.

For the last sentence of item 4: if $\Gamma$ and $\tilde\Gamma$ are two such arrays on the same region, both transforming by item 4, their difference $\Gamma - \tilde\Gamma$ loses the second-derivative term, which is the same for both, and obeys the pure type $(1,2)$ law. $\square$

Definition(The covariant derivative)AdaptedAdapted from Riley, Hobson & Bence (2006), section 26.19, Eqs. (26.84)-(26.90) and the collected rules on p. 971, pp. 968-971; Arfken, Weber & Harris (2013), section 4.3, Eqs. (4.56)-(4.58), pp. 222-223

Partial differentiation destroys tensor character. Transforming $\partial v^i/\partial u^j$ and differentiating the transformation coefficients along the way produces

\frac{\partial \bar v^{\,i}}{\partial \bar u^{\,j}} = \frac{\partial \bar u^{\,i}}{\partial u^l}\frac{\partial u^k}{\partial \bar u^{\,j}}\frac{\partial v^l}{\partial u^k} \;+\; \frac{\partial u^k}{\partial\bar u^{\,j}}\frac{\partial^2\bar u^{\,i}}{\partial u^k\partial u^l}\,v^l ,

whose second term spoils the type $(1,1)$ law. The cure is to differentiate the vector, basis and all, rather than its components. From $v = v^ie_i$ ,

\frac{\partial v}{\partial u^j} = \frac{\partial v^i}{\partial u^j}\,e_i + v^i\,\frac{\partial e_i}{\partial u^j} = \frac{\partial v^i}{\partial u^j}\,e_i + v^i\,\Gamma^k{}_{ij}\,e_k = \left( \frac{\partial v^i}{\partial u^j} + \Gamma^i{}_{kj}\,v^k \right) e_i ,

the last step exchanging the two dummy labels $i$ and $k$ inside the second term so that $e_i$ factors out. The bracket is the covariant derivative of $v$ , written with a semicolon (a comma denoting the plain partial derivative):

v^i{}_{;\,j} = \frac{\partial v^i}{\partial u^j} + \Gamma^i{}_{kj}\,v^k = v^i{}_{,\,j} + \Gamma^i{}_{kj}\,v^k .

Since $\partial v/\partial u^j = v^i{}_{;j}\,e_i$ is a genuine vector for each $j$ , and $du^j$ is an arbitrary contravariant vector, the quotient rule of Tensors and their transformation laws identifies $v^i{}_{;j}$ as the mixed components of a type $(1,1)$ tensor, written $\nabla v$ . The same calculation starting from $v = v_ie^i$ , using the minus sign in the derivative of $e^i$ , gives the covariant form

v_{i\,;\,j} = \frac{\partial v_i}{\partial u^j} - \Gamma^k{}_{ij}\,v_k .

For a tensor of any type the rule is: differentiate the components, then add one Christoffel term per index — with a plus sign and the symbol's upper index for each contravariant index, with a minus sign for each covariant index — and in every term the index of differentiation sits in the last lower slot of the symbol. For second rank this reads

T^{ij}{}_{;\,k} = T^{ij}{}_{,\,k} + \Gamma^i{}_{lk}T^{lj} + \Gamma^j{}_{lk}T^{il} ,\qquad T^i{}_{j\,;\,k} = T^i{}_{j\,,\,k} + \Gamma^i{}_{lk}T^l{}_j - \Gamma^l{}_{jk}T^i{}_l ,\qquad T_{ij\,;\,k} = T_{ij\,,\,k} - \Gamma^l{}_{ik}T_{lj} - \Gamma^l{}_{jk}T_{il} ,

and these are the components of one third-rank tensor $\nabla T$ against the three tensor bases. A scalar has no basis vectors to differentiate, so

\varphi_{;\,j} = \varphi_{,\,j} = \frac{\partial\varphi}{\partial u^j} ,

which is already a covariant vector. In Cartesian coordinates every $\Gamma$ vanishes and the covariant derivative is the partial derivative.

Definition(The metric is covariantly constant)AdaptedAdapted from Riley, Hobson & Bence (2006), section 26.18, Eq. (26.79), p. 966; Stakgold & Holst (2011), section 10.4, p. 728

With the Christoffel symbols defined from the metric as above,

g_{ij\,;\,k} = 0 , \qquad g^{ij}{}_{;\,k} = 0 , \qquad \delta^i{}_{j\,;\,k} = 0 .

Consequently raising and lowering indices commutes with covariant differentiation: $(g_{il}T^{l}{}_{j})_{;k} = g_{il}\,T^l{}_{j\,;k}$ , so one may raise an index before or after differentiating and get the same tensor. A rule for differentiating that annihilates the metric in this way is called a Levi-Civita connection; Stakgold and Holst characterise it exactly so, and it is the connection used throughout this chapter.

ProofAdaptedAdapted from Riley, Hobson & Bence (2006), section 26.18, Eq. (26.79), p. 966

Apply the covariant-derivative rule for a type $(0,2)$ tensor to $g$ :

g_{ij\,;\,k} = \frac{\partial g_{ij}}{\partial u^k} - \Gamma^l{}_{ik}\,g_{lj} - \Gamma^l{}_{jk}\,g_{il} .

Item 2 of the statement above says that the first term is exactly $\Gamma^l{}_{ik}g_{lj}+\Gamma^l{}_{jk}g_{il}$ , so the right-hand side is zero. For the contravariant metric, differentiate $g^{ij}g_{jk}=\delta^i{}_k$ covariantly. The covariant derivative obeys the product rule (it is a partial derivative plus terms linear in the components, and both respect products), and $\delta^i{}_k$ has constant components with its two Christoffel terms cancelling, so $\delta^i{}_{k\,;l}=0$ ; hence $g^{ij}{}_{;\,l}\,g_{jk} = 0$ , and contracting with $g^{km}$ gives $g^{im}{}_{;\,l}=0$ . $\square$

Definition(Parallel transport)AdaptedAdapted from do Carmo (2016), section 4-4, Definitions 1, 4, 5 and 6 and Eq. (1), pp. 241-245

Let $t\mapsto u(t)$ , $t\in[a,b]$ , be a $C^1$ curve in the coordinate region, and let $v(t)$ be a vector field defined along it, with contravariant components $v^i(t)$ . Its covariant derivative along the curve is

\frac{Dv^i}{dt} = \frac{dv^i}{dt} + \Gamma^i{}_{jk}\bigl(u(t)\bigr)\,v^j(t)\,\frac{du^k}{dt} ,

obtained from $v^i{}_{;k}$ by contracting with the velocity $\dot u^k$ ; it is therefore a vector, and it depends on the curve only through the point and the velocity there. The field is parallel along the curve if $Dv^i/dt = 0$ for all $t$ , and the value $v(t_1)$ of the parallel field with prescribed initial value $v(t_0)=v_0$ is the parallel transport of $v_0$ from $u(t_0)$ to $u(t_1)$ along that curve.

Do Carmo arrives at the same equation from outside: for a surface sitting in $\mathbb{R}^3$ , the covariant derivative is the ordinary derivative $dv/dt$ followed by orthogonal projection onto the tangent plane, and a field is parallel when its ordinary derivative is purely normal to the surface — it does not turn as seen from within the surface. His Eq. (1) shows that the projection depends only on the Christoffel symbols and the velocity, which is what makes the notion intrinsic.

Definition(Parallel transport exists, is unique, and preserves the metric)AdaptedAdapted from do Carmo (2016), section 4-4, Propositions 1 and 2 and the remark on p. 245, pp. 244-245

Let $t\mapsto u(t)$ be a $C^1$ curve on $[a,b]$ lying in a region where the metric is $C^1$ , and let $t_0\in[a,b]$ .

Existence and uniqueness. For each $v_0$ there is exactly one parallel field $v(t)$ along the curve with $v(t_0)=v_0$ , defined on the whole of $[a,b]$ .
Linearity and isometry. The map $P: v_0\mapsto v(t_1)$ is linear, and if $v(t)$ and $w(t)$ are both parallel then $g_{ij}v^iw^j$ is constant along the curve. In particular parallel transport preserves lengths and the angles between vectors, so it is a linear isometry from the tangent space at $u(t_0)$ onto the tangent space at $u(t_1)$ .
Independence of the parametrisation. Reparametrising a regular curve rescales $Dv/dt$ by the nonzero factor $dt/d\sigma$ and so does not change which fields are parallel.

ProofAdaptedAdapted from do Carmo (2016), section 4-4, Proposition 1, p. 244

1. The condition $Dv^i/dt = 0$ reads

\frac{dv^i}{dt} = -\,\Gamma^i{}_{jk}\bigl(u(t)\bigr)\,\dot u^k(t)\; v^j(t) ,

a system of $n$ first-order ordinary differential equations that is linear and homogeneous in the unknowns $v^1,\dots,v^n$ , with coefficient matrix $A^i{}_j(t) = -\Gamma^i{}_{jk}(u(t))\dot u^k(t)$ continuous on $[a,b]$ . A linear system with continuous coefficients has a unique solution for each initial value, and that solution exists on the whole interval on which the coefficients are continuous, with no possibility of blow-up (Linear systems and the matrix exponential, Existence and uniqueness of solutions). Linearity of $P$ in item 2 is the linearity of the solution map of a linear system.

2. Let $v,w$ be parallel. Since $g$ is covariantly constant, the product rule for the covariant derivative along the curve gives

\frac{d}{dt}\bigl( g_{ij}v^iw^j \bigr) = \frac{Dg_{ij}}{dt}v^iw^j + g_{ij}\frac{Dv^i}{dt}w^j + g_{ij}v^i\frac{Dw^j}{dt} = 0 + 0 + 0 = 0 ,

because the left-hand side is a scalar, whose ordinary and covariant derivatives agree, and each of the three terms vanishes: the first by $g_{ij;k}=0$ , the other two by parallelism. Hence $g_{ij}v^iw^j$ is constant; taking $w=v$ gives constancy of $\lvert v\rvert$ , so $\cos\angle(v,w) = g_{ij}v^iw^j/(\lvert v\rvert\,\lvert w\rvert)$ is a ratio of constants and is itself constant. Since $P$ is a linear map between spaces of the same dimension that preserves lengths, it is injective and therefore an isomorphism, that is, an isometry onto. Do Carmo proves the same proposition geometrically for surfaces: both $dv/dt$ and $dw/dt$ are normal to the tangent plane, so the derivative of $v\cdot w$ is zero.

3. Under $t = t(\sigma)$ the chain rule multiplies every term of $Dv^i/dt$ by $dt/d\sigma$ , and $dt/d\sigma \neq 0$ for a regular reparametrisation, so the vanishing of one expression is equivalent to the vanishing of the other. $\square$

Remark(Where the hypotheses are used)AdaptedAdapted from Riley, Hobson & Bence (2006), section 26.18, p. 965; Stakgold & Holst (2011), section 10.4, p. 728

The coordinate basis, in the symmetry. $\Gamma^k{}_{ij}=\Gamma^k{}_{ji}$ was proved by exchanging the order of two partial derivatives of $x$ , which requires $x$ to be $C^2$ and, more importantly, requires $e_i$ to be $\partial x/\partial u^i$ for an actual coordinate system. Connection coefficients defined the same way for a frame that is not the coordinate frame — for instance the orthonormal frame $\hat e_i = e_i/h_i$ of an orthogonal system — are not symmetric in their lower indices, and the metric formula does not apply to them. Everything in this unit is written in the coordinate basis.

Nondegeneracy of the metric, in the metric formula. The contraction with $g^{mk}$ that solves $2\Gamma^l{}_{ij}g_{kl}$ for $\Gamma^m{}_{ij}$ needs $[g_{ij}]$ to be invertible. Where it is not — on the axis of polar coordinates, for example — the symbols are undefined, which is visible in the cylindrical case below where $\Gamma^\phi{}_{\rho\phi}=1/\rho$ blows up at $\rho = 0$ . That blow-up is a defect of the coordinates, not of the space: in Cartesian coordinates covering the same points all the symbols are zero.

Metric compatibility, in the isometry. If the connection did not annihilate the metric, transport would still exist and be unique — item 1 of the transport statement uses only linearity of the ODE — but lengths would drift, and the whole idea of moving a vector "without changing it" would be empty. Compatibility is exactly what rules that out.

What is not assumed: path independence. Nothing above says that transporting a vector from $p$ to $q$ gives the same answer along two different curves, and in general it does not. Equivalently, covariant derivatives need not commute: Stakgold and Holst note that in flat space $v^a{}_{;bc} - v^a{}_{;cb} = 0$ , whereas on a curved manifold the difference equals $R^a{}_{dbc}v^d$ , and the array $R^a{}_{dbc}$ built from the Christoffel symbols and their derivatives is what is left over. That array is the Riemann curvature tensor of The Riemann curvature tensor; the example below exhibits the failure of path independence on the sphere as a concrete rotation angle, and The Gauss-Bonnet theorem ties the angle to the enclosed curvature.

Why the $\Gamma$ cannot be a tensor, said once more. In Cartesian coordinates every $\Gamma^k{}_{ij}$ is zero, and in plane polar coordinates they are not. A tensor whose components all vanish in one coordinate system has vanishing components in all of them, because every term of the transformation law carries a factor of a component. So no tensor can have these components, and item 4 of the statement shows precisely which extra term is to blame.

Method(Christoffel symbols and covariant derivatives from a metric)Original

Write the metric $g_{ij}$ in the chosen coordinates, as a matrix, and invert it to get $g^{ij}$ . For a diagonal metric both are diagonal and the inversion is free.
List the independent symbols. Symmetry in the lower pair leaves $n^2(n+1)/2$ of them: $18$ in three dimensions, $6$ in two.
For each, use $\Gamma^m{}_{ij} = \tfrac12 g^{mk}(\partial_i g_{jk} + \partial_j g_{ki} - \partial_k g_{ij})$ . With a diagonal metric only $k=m$ contributes, which reduces the work to three derivatives per symbol.
Sanity-check: a symbol can be nonzero only if some $g_{ij}$ actually depends on a coordinate appearing among $i,j,m$ . If the metric has constant entries, all symbols vanish.
Assemble covariant derivatives by the index rule: copy the partial derivative, then add $+\Gamma^{(\text{upper})}{}_{l\,k}$ terms for upper indices and $-\Gamma^{l}{}_{(\text{lower})\,k}$ terms for lower ones, with $k$ the differentiation index always last.
Check the result on the metric itself: $g_{ij;k}$ must come out zero. This catches sign and index-order errors immediately.
For transport along a curve, substitute into $\dot v^i = -\Gamma^i{}_{jk}v^j\dot u^k$ and solve the linear system. Verify that $g_{ij}v^iv^j$ is constant along the solution.

Example(Parallel transport around a circle of latitude on the sphere)Original

Take the unit sphere with the usual colatitude and longitude $(u^1,u^2) = (\theta,\varphi)$ , for which

(ds)^2 = (d\theta)^2 + \sin^2\theta\,(d\varphi)^2 , \qquad [g_{ij}] = \begin{pmatrix} 1 & 0 \\ 0 & \sin^2\theta\end{pmatrix}, \qquad [g^{ij}] = \begin{pmatrix} 1 & 0 \\ 0 & \sin^{-2}\theta\end{pmatrix} .

Step 1-3: the symbols. The only coordinate-dependent entry is $g_{\varphi\varphi}=\sin^2\theta$ , and it depends only on $\theta$ , so a symbol survives only when it can use $\partial_\theta g_{\varphi\varphi} = 2\sin\theta\cos\theta$ . Applying the formula with the metric diagonal:

\Gamma^\theta{}_{\varphi\varphi} = \tfrac12 g^{\theta\theta}\bigl(-\partial_\theta g_{\varphi\varphi}\bigr) = -\sin\theta\cos\theta , \qquad \Gamma^\varphi{}_{\theta\varphi} = \Gamma^\varphi{}_{\varphi\theta} = \tfrac12 g^{\varphi\varphi}\,\partial_\theta g_{\varphi\varphi} = \frac{\cos\theta}{\sin\theta} = \cot\theta ,

and all other symbols vanish. Note $\Gamma^\varphi{}_{\theta\varphi}\to\infty$ at the poles: the coordinates, not the sphere, break down there.

Step 6: the check. $g_{\varphi\varphi;\theta} = \partial_\theta g_{\varphi\varphi} - 2\Gamma^l{}_{\varphi\theta}g_{l\varphi} = 2\sin\theta\cos\theta - 2\cot\theta\cdot\sin^2\theta = 0$ , as required.

Step 7: transport along a latitude circle. Fix a colatitude $\theta_0$ with $0<\theta_0<\pi/2$ and run once round the circle $\theta = \theta_0$ , using $\varphi$ itself as the parameter, so $\dot\theta = 0$ and $\dot\varphi = 1$ . The transport equations are

\frac{dv^\theta}{d\varphi} + \Gamma^\theta{}_{\varphi\varphi}\,v^\varphi = 0 , \qquad \frac{dv^\varphi}{d\varphi} + \Gamma^\varphi{}_{\theta\varphi}\,v^\theta = 0 ,

that is

\frac{dv^\theta}{d\varphi} = \sin\theta_0\cos\theta_0\; v^\varphi , \qquad \frac{dv^\varphi}{d\varphi} = -\cot\theta_0\; v^\theta .

The mixed coefficients are an artefact of the basis not being orthonormal. Change to the physical components $a = v^\theta$ and $b = \sin\theta_0\, v^\varphi$ , which are the components against the unit vectors and satisfy $\lvert v\rvert^2 = a^2+b^2$ . Then

\frac{da}{d\varphi} = \cos\theta_0\, b , \qquad \frac{db}{d\varphi} = \sin\theta_0\frac{dv^\varphi}{d\varphi} = -\cos\theta_0\, a .

Writing $z = a+ib$ turns this into $dz/d\varphi = -i\cos\theta_0\,z$ , so

z(\varphi) = e^{-i\varphi\cos\theta_0}\,z(0) .

The transported vector keeps its length, as item 2 of the transport statement promised, and rotates in the local orthonormal frame at the constant rate $\cos\theta_0$ per radian of longitude. After one full circuit, $\varphi: 0\to2\pi$ , it has turned through

\Delta = 2\pi\cos\theta_0 .

A number. Take $\theta_0 = \pi/3$ , the circle $30^\circ$ north of the equator. Then $\cos\theta_0 = 1/2$ and

\boxed{\;\Delta = 2\pi\cdot\tfrac12 = \pi \;}

so a vector carried once around that circle comes back exactly reversed: $v(2\pi) = -v(0)$ . Starting from the unit vector pointing due south, $(v^\theta,v^\varphi) = (1,0)$ , the solution is $a=\cos(\varphi/2)$ , $b=-\sin(\varphi/2)$ , which at $\varphi=2\pi$ is $(a,b) = (-1,0)$ .

The transport is therefore genuinely path-dependent: along the circle the vector is reversed, while along any path staying at a fixed point it is unchanged. The discrepancy from the identity, $2\pi - \Delta = 2\pi(1-\cos\theta_0)$ , is numerically equal to the area of the spherical cap the circle encloses, which is the local Gauss-Bonnet theorem (The Gauss-Bonnet theorem) and the reason the obstruction is called curvature (The Riemann curvature tensor). At the equator, $\theta_0=\pi/2$ , the rotation is zero: the equator is a great circle, hence a geodesic (Geodesics), and transport along it is trivial in this frame.

Application(The Foucault pendulum)Original

A Foucault pendulum swings in a vertical plane whose horizontal direction is a tangent vector to the Earth's surface. Over one swing the suspension exerts no horizontal torque about the vertical, so the only change in that direction is the change forced by the support being dragged along a circle of latitude by the Earth's rotation: the swing direction is parallel-transported along that circle, in exactly the sense defined above. Over one sidereal day the support completes one circuit of the latitude circle $\theta_0 = 90^\circ - \lambda$ , where $\lambda$ is the latitude, and the calculation just performed applies verbatim with $\cos\theta_0 = \sin\lambda$ .

The plane of swing therefore rotates relative to the ground by

\Delta = 2\pi\sin\lambda \quad\text{per sidereal day} ,

clockwise in the northern hemisphere. At $\lambda = 45^\circ$ this is $2\pi\sin 45^\circ = 4.4429$ radians, or $254.56^\circ$ per sidereal day. With the sidereal day equal to $23.9345$ hours, the precession rate is

\frac{254.56^\circ}{23.9345\ \text{h}} = 10.64^\circ\ \text{per hour} ,

so the plane takes $360/10.64 = 33.8$ hours to come back to its starting direction — more than a day, which is the observation that made the original 1851 demonstration persuasive. The two extreme cases are the ones the geometry predicts: at the pole, $\lambda = 90^\circ$ , the latitude circle degenerates to a point, the holonomy is a full turn, and the plane appears to rotate once per day as the ground turns beneath it; at the equator, $\lambda = 0$ , the path is a geodesic, the holonomy vanishes and the pendulum does not precess at all.

The precession rate itself is the classical $\Omega\sin\lambda$ of rotating-frame mechanics; what the geometry adds is the reason the answer depends on latitude through the enclosed solid angle rather than on anything local to the pendulum.

Application(The covariant derivative in general relativity)AdaptedAdapted from Stakgold & Holst (2011), section 10.4, p. 728

The same machinery, with the same Christoffel symbols, carries much heavier loads. Stakgold and Holst set out the chain in a page. Spacetime is a pseudo-Riemannian four-manifold; its metric $g_{ab}$ determines a Levi-Civita connection, characterised by annihilating the metric exactly as in the statement above, and written in components as $v^a{}_{;b} = v^a{}_{,b} + \Gamma^a{}_{cb}v^c$ with the Christoffel symbols of the surface. On a curved manifold covariant derivatives do not commute: the combination $v^a{}_{;bc} - v^a{}_{;cb}$ , which vanishes identically in flat space where partial derivatives commute, equals $R^a{}_{dbc}\,v^d$ , and

R^a{}_{dbc} = \Gamma^a{}_{bd,c} - \Gamma^a{}_{cd,b} + \Gamma^a{}_{ec}\Gamma^e{}_{bd} - \Gamma^a{}_{eb}\Gamma^e{}_{cd}

is what is left over — the Riemann tensor, built from the symbols and their first derivatives alone, and so from the metric and its first two derivatives. Contracting it gives the Ricci tensor $R_{ab}$ and the scalar curvature $R$ , and from them the Einstein tensor $G_{ab} = R_{ab} - \tfrac12 R\,g_{ab}$ , which encodes the deformation of spacetime. The field equations

G_{ab} = \kappa\,T_{ab} , \qquad \kappa = \frac{8\pi G}{c^4} ,

then set it equal to the stress-energy tensor that encodes matter and energy. Everything on the left-hand side descends from the one definition in this unit; the covariant derivative is the whole of the differential apparatus of the theory. The curvature tensor itself is The Riemann curvature tensor, and the manifold language is Riemannian manifolds.

Method

Differential operators in general coordinates

Adapted

Add to PDF

Gives the gradient, divergence, Laplacian and curl in an arbitrary coordinate system, orthogonal or not, as formulas built from the metric and the square root of its determinant, proves each from the covariant derivative, and works all three out in a skew coordinate system with numerical checks.

Prerequisites: Christoffel symbols and the covariant derivative, The metric tensor, Tensors and their transformation laws, Orthogonal curvilinear coordinates, Determinants

Definition(The data a general coordinate system supplies)AdaptedAdapted from Riley, Hobson & Bence (2006), §26.20, p. 972 and §26.14–§26.15, pp. 955–958; Arfken, Weber & Harris (2013), §4.3, pp. 218–220, Eq. (4.37)

Let $D \subseteq \mathbb{R}^3$ be open and let $u^1, u^2, u^3$ be coordinates on $D$ , that is, let the position map $r(u^1,u^2,u^3)$ be $C^2$ with a $C^2$ inverse onto its image. (Tensors and their transformation laws calls this map $x$ ; it is written $r$ here so that $x$ , $y$ , $z$ stay free for Cartesian components in the example.) Write $\partial_i = \partial/\partial u^i$ . The objects used below are the ones built in Tensors and their transformation laws, The metric tensor and Christoffel symbols and the covariant derivative:

the covariant basis $e_i = \partial r/\partial u^i$ , tangent to the $u^i$ -coordinate curve, and the contravariant basis $e^i = \nabla u^i$ , normal to the $u^i$ -coordinate surface, with $e_i \cdot e^j = \delta_i^j$ ;
the metric $g_{ij} = e_i\cdot e_j$ , its inverse $g^{ij}$ , and the number $g = \det\,[g_{ij}] ,$ the determinant of the matrix $G = [g_{ij}]$ , which is positive on $D$ because $G = (Dr)^{\mathsf T}(Dr)$ with $Dr$ the (invertible) Jacobian matrix of $r$ , so that $g = (\det Dr)^2$ ;
the components of a vector field $v$ in the two bases, $v = v^ie_i = v_ie^i$ , with $v^i = v\cdot e^i$ and $v_i = v\cdot e_i$ , related by $v_i = g_{ij}v^j$ and $v^i = g^{ij}v_j$ ;
the Christoffel symbols $\Gamma^k{}_{ij}$ defined by $\partial_j e_i = \Gamma^k{}_{ij}e_k$ , which are symmetric in $i$ and $j$ , and the covariant derivative, written with a semicolon, $v^i{}_{;j} = \partial_j v^i + \Gamma^i{}_{kj}v^k$ and $v_{i;j} = \partial_j v_i - \Gamma^k{}_{ij}v_k$ . A scalar has $\phi_{;j} = \partial_j\phi$ , since a scalar does not refer to the basis at all.

Nothing here assumes the coordinates are orthogonal. When they happen to be, $g_{ij} = h_i^2\delta_{ij}$ with the scale factors $h_i$ of Orthogonal curvilinear coordinates, so $g = h_1^2h_2^2h_3^2$ and $g^{ij} = h_i^{-2}\delta_{ij}$ . Riley, Hobson and Bence emphasise that even then the $e_i$ are not unit vectors: a physical component $\hat v^i$ , the coefficient of the unit vector $\hat e_i = e_i/h_i$ , is related to the contravariant component by $v^i = \hat v^i/h_i$ , with no sum on $i$ . Arfken, Weber and Harris attach the same warning to the covariant components $v_i$ . The formulas of this unit are stated for $v^i$ and $v_i$ , never for $\hat v^i$ .

Method(The four operators in general coordinates)AdaptedAdapted from Riley, Hobson & Bence (2006), §26.20, Eqs. (26.91), (26.97), (26.98) and the curl on p. 974, pp. 972–975; Arfken, Weber & Harris (2013), §4.3, Eqs. (4.64), (4.69), (4.70), (4.71), pp. 223–225

With the hypotheses of the definition, let $\phi$ be a $C^1$ scalar field ( $C^2$ for the Laplacian) and $v$ a $C^1$ vector field on $D$ .

Gradient. Its covariant components are the plain partial derivatives: $\nabla\phi = (\partial_i\phi)\,e^i .$
Divergence. With $v^j$ the contravariant components, $\nabla\cdot v \;=\; v^i{}_{;i} \;=\; \frac{1}{\sqrt{g}}\,\partial_j\!\left(\sqrt{g}\,v^j\right) .$
Laplacian. $\nabla^2\phi \;=\; \frac{1}{\sqrt{g}}\,\partial_j\!\left(\sqrt{g}\,g^{jk}\,\partial_k\phi\right) .$
Curl. In any dimension the antisymmetric covariant tensor $(\operatorname{curl} v)_{ij} \;=\; v_{i;j} - v_{j;i} \;=\; \partial_j v_i - \partial_i v_j$ is defined, and the Christoffel symbols have dropped out of the right-hand side. In three dimensions Riley, Hobson and Bence associate with it the vector whose contravariant components are $(\nabla\times v)^i \;=\; -\frac{1}{2\sqrt{g}}\,\epsilon^{ijk}\,(\operatorname{curl} v)_{jk} \;=\; \frac{1}{\sqrt{g}}\,\epsilon^{ijk}\,\partial_j v_k ,$ where $\epsilon^{ijk}$ is the permutation symbol, equal to $+1$ on even permutations of $123$ , to $-1$ on odd ones and to $0$ otherwise. It is $\epsilon^{ijk}/\sqrt g$ , not $\epsilon^{ijk}$ , that is a tensor (Pseudotensors and the Levi-Civita tensor), and it is a pseudotensor, so the sign of $\nabla\times v$ depends on the handedness of the labelling $u^1,u^2,u^3$ .

Setting $g_{ij} = h_i^2\delta_{ij}$ recovers the orthogonal formulas of Orthogonal curvilinear coordinates: $\sqrt g = h_1h_2h_3$ turns statement 3 into $\nabla^2\phi = (h_1h_2h_3)^{-1}\partial_j\bigl(h_1h_2h_3\,h_j^{-2}\partial_j\phi\bigr)$ , which is the formula given there once the sum over $j$ is written out.

Idea of the proofAdaptedAdapted from Riley, Hobson & Bence (2006), §26.20, pp. 972–974; Arfken, Weber & Harris (2013), §4.3, pp. 223–225

The divergence and the curl are the Cartesian expressions with every partial derivative promoted to a covariant derivative, because that is the only way to produce a tensor. The work is then to get rid of the Christoffel symbols again, and in each case something special happens. The divergence contracts $\Gamma^i{}_{ki}$ over two of its three indices; two of the three metric derivatives in that contraction cancel against each other, and what is left is exactly the combination that appears in the derivative of a determinant, so the whole Christoffel term collapses into $\partial_k\log\sqrt g$ and can be absorbed into a single derivative of $\sqrt g\,v^k$ . The curl antisymmetrises in the two lower indices of $\Gamma^k{}_{ij}$ , where the symbol is symmetric, so the Christoffel terms simply cancel. The gradient is free: a scalar has no basis to differentiate. The Laplacian is the divergence applied to the gradient, which costs one index raise.

ProofAdaptedAdapted from Riley, Hobson & Bence (2006), §26.20, Eqs. (26.92)–(26.98), pp. 972–974; Arfken, Weber & Harris (2013), §4.3, Eqs. (4.65)–(4.71), pp. 223–225

1. The gradient. The gradient of a scalar is the covariant vector field whose components are the covariant derivatives $\phi_{;i}$ , and for a scalar the covariant derivative is simply the partial derivative: the correction terms in the covariant derivative all arise from differentiating basis vectors, and a scalar refers to no basis. Hence $\phi_{;i} = \partial_i\phi$ , and $\nabla\phi = (\partial_i\phi)\,e^i$ — which is also what the chain rule gives directly, since $e^i = \nabla u^i$ . That is statement 1.

2. Contracting the Christoffel symbol. From Christoffel symbols and the covariant derivative,

\Gamma^i{}_{kj} = \tfrac12\,g^{il}\bigl(\partial_k g_{jl} + \partial_j g_{kl} - \partial_l g_{kj}\bigr) .

Put $j = i$ and sum. The last two terms cancel: in the middle one rename the dummy indices $i\leftrightarrow l$ and then use the symmetry $g^{li} = g^{il}$ , which turns $g^{il}\partial_i g_{kl}$ into $g^{il}\partial_l g_{ki}$ , the same as the third term. What survives is

\Gamma^i{}_{ki} = \tfrac12\,g^{il}\,\partial_k g_{il} . \tag{1}

3. The derivative of the determinant. Let $A = [a_{ij}]$ be an invertible matrix whose entries depend on the coordinates, let $a = \det A$ and let $\Delta^{ij}$ be the cofactor of $a_{ij}$ . Expanding the determinant along the $i$ th row, $a = \sum_j a_{ij}\Delta^{ij}$ for each fixed $i$ , and no cofactor in that sum contains $a_{ij}$ ; differentiating with respect to the single entry therefore gives $\partial a/\partial a_{ij} = \Delta^{ij}$ . The entries of the inverse are $(A^{-1})_{ij} = \Delta^{ji}/a$ (Determinants), so the chain rule gives

\partial_k a \;=\; \frac{\partial a}{\partial a_{ij}}\,\partial_k a_{ij} \;=\; \Delta^{ij}\,\partial_k a_{ij} \;=\; a\,(A^{-1})_{ji}\,\partial_k a_{ij} .

Apply this to $A = [g_{ij}]$ , whose inverse is $[g^{ij}]$ and which is symmetric, so that $(A^{-1})_{ji} = g^{ij}$ :

\partial_k g = g\,g^{ij}\,\partial_k g_{ij} . \tag{2}

Comparing (1) and (2), and using $g > 0$ to write $\partial_k\sqrt g = (2\sqrt g)^{-1}\partial_k g$ ,

\Gamma^i{}_{ki} = \frac{1}{2g}\,\partial_k g = \frac{1}{\sqrt g}\,\partial_k \sqrt g . \tag{3}

4. The divergence. The divergence is the contraction $v^i{}_{;i}$ of the covariant derivative, which is a tensor of one lower rank, hence a scalar. Writing it out and inserting (3),

v^i{}_{;i} = \partial_i v^i + \Gamma^i{}_{ki}v^k = \partial_i v^i + \frac{v^k}{\sqrt g}\,\partial_k\sqrt g .

Both terms are what the product rule produces from one derivative: multiplying through by $\sqrt g$ gives $\sqrt g\,\partial_j v^j + v^j\partial_j\sqrt g = \partial_j(\sqrt g\,v^j)$ . Dividing by $\sqrt g$ , which is legitimate because $g>0$ , gives statement 2.

5. The Laplacian. By definition $\nabla^2\phi = \nabla\cdot(\nabla\phi)$ . Step 1 gives the covariant components of $\nabla\phi$ , while step 4 wants the contravariant ones, so raise the index with the metric: $v^j = g^{jk}\partial_k\phi$ . These are $C^1$ when $\phi$ is $C^2$ and the coordinates are $C^2$ . Substituting into statement 2 gives statement 3.

6. The curl. Expand the two covariant derivatives of the covariant components:

v_{i;j} - v_{j;i} = \bigl(\partial_j v_i - \Gamma^k{}_{ij}v_k\bigr) - \bigl(\partial_i v_j - \Gamma^k{}_{ji}v_k\bigr) = \partial_j v_i - \partial_i v_j ,

the Christoffel terms cancelling because $\Gamma^k{}_{ij} = \Gamma^k{}_{ji}$ . The left-hand side is a difference of components of a tensor, hence a tensor; it is antisymmetric in $i$ and $j$ by construction. This is valid in every dimension, and it is the reason the curl of a gradient vanishes identically: for $v_i = \partial_i\phi$ the right-hand side is a difference of mixed second partials of a $C^2$ function.

In three dimensions, contract with the permutation symbol as Riley, Hobson and Bence do. Using the antisymmetry of $(\operatorname{curl} v)_{jk}$ ,

-\frac{1}{2\sqrt g}\,\epsilon^{ijk}(\operatorname{curl} v)_{jk} = -\frac{1}{2\sqrt g}\,\epsilon^{ijk}\bigl(\partial_k v_j - \partial_j v_k\bigr) = \frac{1}{\sqrt g}\,\epsilon^{ijk}\,\partial_j v_k ,

the two halves being equal after the dummy indices $j$ and $k$ are exchanged in the first of them, which costs a sign from $\epsilon^{ijk}$ . That is statement 4. $\square$

Remark(Where the hypotheses are used)Original

$g > 0$ is used twice in step 3 and once in step 4: to form $\sqrt g$ , to divide by it, and to differentiate it. In Euclidean space with a genuine coordinate system this is automatic, since $g = (\det Dr)^2$ ; but it fails on the sets where the chart degenerates, which is why the polar axis and the origin have to be deleted from the domain of spherical polars before these formulas may be used there. For an indefinite metric, where $g<0$ , the same derivation runs with $|g|$ in place of $g$ , because (2) and hence $\Gamma^i{}_{ki} = \tfrac12\partial_k\log|g|$ are unaffected by the sign; this is the form used in relativity.
The coordinates are $C^2$ so that the $g_{ij}$ are $C^1$ and the Christoffel symbols exist; step 5 differentiates the metric once more against $\phi$ , and the vanishing of $\operatorname{curl}\nabla\phi$ in step 6 needs $\phi$ itself to be $C^2$ for the mixed partials to commute.
The symmetry $\Gamma^k{}_{ij} = \Gamma^k{}_{ji}$ is the whole content of step 6. It holds because the $e_i$ come from a coordinate system, so $\partial_j e_i = \partial_j\partial_i r = \partial_i\partial_j r$ . In a frame that is not the coordinate frame of any chart — an orthonormal frame on a curved surface, say — the symbol is not symmetric, the Christoffel terms do not cancel, and $\partial_j v_i - \partial_i v_j$ is no longer a tensor.
Which components go into which formula is a hypothesis, not a notational detail. The divergence takes contravariant components $v^j$ ; the curl takes covariant components $v_k$ ; the Laplacian needs the full inverse metric $g^{jk}$ , including the off-diagonal entries that an orthogonal system does not have. Feeding physical components $\hat v^i$ into the divergence formula is the commonest way to get a wrong answer, and both of our sources warn about it.
Three dimensions and a handedness are needed only for the last part of statement 4. The tensor $(\operatorname{curl} v)_{ij}$ exists in every dimension; it has three independent entries only when $n = 3$ , which is what lets it be repackaged as a vector. Relabelling $u^1\leftrightarrow u^2$ changes $\det Dr$ from $+\sqrt g$ to $-\sqrt g$ and reverses $\nabla\times v$ , so the formula as written presupposes that the coordinates are listed in right-handed order.
Conventions differ on the sign of the curl. Riley, Hobson and Bence set $(\operatorname{curl} v)_{ij} = v_{i;j} - v_{j;i}$ , so that the first index is the one being differentiated against; the compensating minus sign in front of their $\epsilon^{ijk}$ contraction is what makes $(\nabla\times v)^i$ come out with the familiar Cartesian sign. Books that instead write $(\operatorname{curl} v)_{ij} = \partial_i v_j - \partial_j v_i$ drop that minus sign. The two routes give the same vector; mixing them gives its negative.

Note(The same two formulas on a Riemannian manifold)AdaptedAdapted from Lee (2013), Problems 16-11 and 16-13, p. 436

Statements 2 and 3 are not special to three dimensions or to a surface sitting inside $\mathbb{R}^3$ . For a Riemannian metric $g$ on an $n$ -manifold, in any smooth local coordinates $x^i$ , Lee states the divergence of a vector field $X = X^i\partial/\partial x^i$ as

\operatorname{div} X = \frac{1}{\sqrt{\det g}}\,\partial_i\!\left(X^i\sqrt{\det g}\right) ,

where $\det g$ is the determinant of the component matrix $(g_{kl})$ , and the Laplacian as

\Delta u = -\frac{1}{\sqrt{\det g}}\,\partial_i\!\left(g^{ij}\sqrt{\det g}\,\partial_j u\right) .

The minus sign is Lee's convention: he defines the geometric Laplacian as $\Delta u = -\operatorname{div}(\operatorname{grad} u)$ , the negative of the operator $\nabla^2$ used here and throughout this handbook, so that its eigenvalues are non-negative, and he warns that authors disagree about the sign. Apart from that sign the two statements are statements 2 and 3 verbatim, with $\det g$ playing the role of $g$ . See Riemannian manifolds.

Method(Applying the formulas in a given coordinate system)Original

Write down the chart and its domain. Give $r(u^1,u^2,u^3)$ explicitly, and state the open set on which it is injective with a nowhere-singular Jacobian matrix. Delete the degenerate sets; nothing below is valid on them.
Differentiate once. Compute the three columns $e_i = \partial r/\partial u^i$ in Cartesian components.
Build the metric. Form the full matrix $g_{ij} = e_i\cdot e_j$ , off-diagonal entries included. Do not assume they vanish; test them.
Determinant and inverse. Compute $g = \det G$ with $G=[g_{ij}]$ , check $g>0$ on the domain, take $\sqrt g$ , and invert the matrix to get $g^{ij}$ . Check the inverse by multiplying out: a slip here poisons every later step. As a cross-check, $\sqrt g$ must equal $|\det Dr|$ , the Jacobian factor of the chart.
Fix the orientation, if you need the curl: compute $e_1\cdot(e_2\times e_3)$ and confirm it is $+\sqrt g$ rather than $-\sqrt g$ . If it is negative, swap two coordinates.
Convert the field to the right components. From a Cartesian field $v$ , the covariant components are $v_i = v\cdot e_i$ and the contravariant ones are $v^i = v\cdot e^i$ , most easily obtained by solving the linear system $v = v^ie_i$ . Raise and lower with $g^{ij}$ , $g_{ij}$ as needed: the divergence wants $v^j$ , the curl wants $v_k$ .
Apply the formula and simplify the coefficient functions $\sqrt g$ and $\sqrt g\,g^{jk}$ before differentiating; in most useful charts they collapse.
Convert back and check. A contravariant answer $w^i$ becomes the Cartesian vector $w^ie_i$ . Then test the whole chain on a field whose answer you already know in Cartesian coordinates.

How it fails. Step 3 is the one that gets skipped: a chart that looks orthogonal at the origin usually is not orthogonal elsewhere, and the Laplacian then needs the cross terms $g^{jk}$ , $j \neq k$ . Step 6 is the second: physical components and contravariant components differ by a factor $h_i$ even in an orthogonal system, and in a skew system the physical components are not defined at all. Step 5 is the third, and it is silent: nothing in the algebra warns you that the curl came out with the wrong sign.

Example(Fan coordinates: a genuinely skew chart)Original

Take the chart

x = u, \qquad y = uv, \qquad z = w, \qquad (u,v,w)\in (0,\infty)\times\mathbb{R}\times\mathbb{R} .

Its inverse is $u = x$ , $v = y/x$ , $w = z$ , smooth on the half-space $x>0$ , so the chart is a diffeomorphism onto it. The surfaces $u = \text{const}$ are the planes $x = \text{const}$ ; the surfaces $v = \text{const}$ are the half-planes $y = vx$ hinged on the $z$ -axis, so the $v$ -coordinate fans out from that axis.

Steps 2–4. The basis vectors are

e_u = (1, v, 0), \qquad e_v = (0, u, 0), \qquad e_w = (0,0,1),

so that $e_u\cdot e_v = uv$ , which is non-zero wherever $v\neq 0$ : the chart is not orthogonal, and nothing from Orthogonal curvilinear coordinates applies. The metric, its determinant and its inverse are

[g_{ij}] = \begin{pmatrix} 1+v^2 & uv & 0\\ uv & u^2 & 0\\ 0&0&1\end{pmatrix}, \qquad g = u^2(1+v^2) - u^2v^2 = u^2 , \qquad \sqrt g = u ,

[g^{ij}] = \begin{pmatrix} 1 & -v/u & 0\\ -v/u & (1+v^2)/u^2 & 0\\ 0&0&1\end{pmatrix} .

(Check the inverse by multiplying the two upper-left $2\times2$ blocks: the $(1,1)$ entry is $(1+v^2)\cdot 1 + uv\cdot(-v/u) = 1$ , the $(1,2)$ entry is $(1+v^2)(-v/u) + uv(1+v^2)/u^2 = 0$ , and the other two go the same way.) Also $\det Dr = u$ , and $\sqrt g = u$ agrees with it, as step 4 requires.

Step 5. $e_u\cdot(e_v\times e_w) = \det[e_u\;e_v\;e_w] = u = +\sqrt g > 0$ , so $(u,v,w)$ is already right-handed.

Step 7, the Laplacian. The three coefficient vectors $\sqrt g\,g^{jk}\partial_k\phi$ are

u\phi_u - v\phi_v , \qquad -v\phi_u + \frac{1+v^2}{u}\,\phi_v , \qquad u\phi_w ,

\nabla^2\phi = \frac{1}{u}\left[ \partial_u\bigl(u\phi_u - v\phi_v\bigr) + \partial_v\!\left(-v\phi_u + \frac{1+v^2}{u}\phi_v\right) + \partial_w\bigl(u\phi_w\bigr) \right] .

Test it on $\phi = u^3v$ , which is the Cartesian field $x^2y$ . Here $\phi_u = 3u^2v$ , $\phi_v = u^3$ , $\phi_w = 0$ . The first bracket is $3u^3v - u^3v = 2u^3v$ , whose $u$ -derivative is $6u^2v$ . The second is $-3u^2v^2 + u^2(1+v^2) = u^2(1-2v^2)$ , whose $v$ -derivative is $-4u^2v$ . The third vanishes. Hence

\nabla^2\phi = \frac{6u^2v - 4u^2v}{u} = 2uv = 2y ,

which is $\nabla^2(x^2y)$ computed in Cartesian coordinates. At the point $u=2$ , $v=3$ , $w=0$ — that is, $(x,y,z) = (2,6,0)$ — the value is $2\cdot 2\cdot 3 = 12$ .

Step 7, the divergence. Let $F$ be the field with contravariant components $F^u = u^2v$ , $F^v = 0$ , $F^w = uw$ . Then

\nabla\cdot F = \frac1u\Bigl[\partial_u\bigl(u\cdot u^2v\bigr) + \partial_v(0) + \partial_w\bigl(u\cdot uw\bigr)\Bigr] = \frac{3u^2v + u^2}{u} = u(3v+1) .

To check it, convert to Cartesian: $F = F^ue_u + F^we_w = (u^2v,\, u^2v^2,\, uw) = (xy,\, y^2,\, zx)$ , whose Cartesian divergence is $y + 2y + x = 3y + x = 3uv + u$ . The two agree, and at $u=2$ , $v=3$ both give $2\cdot(9+1) = 20$ .

Step 7, the curl. Let $G$ be the Cartesian field $(0,0,xy)$ . Step 6 gives its covariant components $G_i = G\cdot e_i$ , namely $G_u = 0$ , $G_v = 0$ , $G_w = u^2v$ . Then

(\nabla\times G)^u = \frac{\partial_v G_w - \partial_w G_v}{u} = \frac{u^2}{u} = u, \qquad (\nabla\times G)^v = \frac{\partial_w G_u - \partial_u G_w}{u} = \frac{-2uv}{u} = -2v,

and $(\nabla\times G)^w = (\partial_u G_v - \partial_v G_u)/u = 0$ . Reassembling,

\nabla\times G = u\,e_u - 2v\,e_v = u(1,v,0) - 2v(0,u,0) = (u, -uv, 0) = (x, -y, 0),

which is indeed $\nabla\times(0,0,xy)$ . Note how little the contravariant answer $(u,-2v,0)$ resembles the Cartesian one: the components are coefficients of basis vectors that are neither unit nor orthogonal, and reading them as if they were Cartesian would give a vector of the wrong length and the wrong direction.

ApplicationOriginal

Terrain-following coordinates in atmosphere and ocean models. Weather and circulation codes do not use height as their vertical coordinate near the ground, because the lowest surfaces would then cut through mountains. They use a coordinate that follows the orography: in the simplest version, $x = \xi$ , $y = \eta$ , $z = h(\xi,\eta) + s$ , with $h$ the terrain height and $s \geq 0$ the height above it. This chart is a shear, so $\sqrt g = 1$ , but it is strongly non-orthogonal: $e_\xi = (1,0,h_\xi)$ and $e_s = (0,0,1)$ have $e_\xi\cdot e_s = h_\xi$ . Working through the method gives

[g^{ij}] = \begin{pmatrix} 1 & 0 & -h_\xi \\ 0 & 1 & -h_\eta \\ -h_\xi & -h_\eta & 1 + h_\xi^2 + h_\eta^2\end{pmatrix},

and statement 3 then gives the Laplacian in the model's own coordinates as

\nabla^2\phi = \phi_{\xi\xi} + \phi_{\eta\eta} + \bigl(1 + h_\xi^2 + h_\eta^2\bigr)\phi_{ss} - 2h_\xi\,\phi_{\xi s} - 2h_\eta\,\phi_{\eta s} - \bigl(h_{\xi\xi} + h_{\eta\eta}\bigr)\phi_s .

Every term after the first two is a metric term that an orthogonal-coordinate formula would not produce. Over steep topography $h_\xi$ is of order one, the cross terms $-2h_\xi\phi_{\xi s}$ are as large as the terms they correct, and a horizontal derivative taken at constant $s$ differs from one taken at constant $z$ by $h_\xi\phi_s$ . In hydrostatic models the horizontal pressure gradient is the small difference of two such large terms, and the resulting cancellation error — the long-standing "pressure-gradient error" of sigma-coordinate models, which can drive spurious currents along a continental slope in an ocean at rest — is exactly a failure to evaluate these metric terms consistently with the rest of the discretisation.

Scalar fields on a curved background. In general relativity the wave operator on a scalar field is $\Box\phi = (|g|)^{-1/2}\partial_\mu\bigl(\sqrt{|g|}\,g^{\mu\nu}\partial_\nu\phi\bigr)$ , which is statement 3 with the modulus inserted as the remark describes, and it is the form actually used in numerical work because it needs only the metric components and their first derivatives — no Christoffel symbols have to be assembled. The same formula on a Riemannian manifold is the Laplace–Beltrami operator, whose eigenvalues are the normal-mode frequencies of a drum of any shape (Riemannian manifolds).

Finite elements on distorted meshes. An isoparametric finite element is a chart: the reference cube is mapped to a curved hexahedron by the element's own shape functions. Element stiffness matrices are integrals $\int g^{jk}\,\partial_j N_a\,\partial_k N_b\,\sqrt g\;du^1du^2du^3$ over the reference cube, which is statement 3 integrated against a test function. Mesh-quality measures used by solvers — aspect ratio, skewness, Jacobian ratio — are, in this language, statements about how far $g_{jk}$ departs from a multiple of the identity and how close $\sqrt g$ comes to zero.

Definition

Pseudotensors and the Levi-Civita tensor

Adapted

Add to PDF

Defines Cartesian pseudotensors by the extra factor of $\det L$ they carry under improper rotations, proves that the permutation symbol is an isotropic third-order pseudotensor and that it pairs antisymmetric tensors with axial vectors, extends both to relative tensors of arbitrary weight in general coordinates, and separates the passive notion of a pseudovector from the active notion of an axial vector.

Prerequisites: Tensors and their transformation laws, The metric tensor, Determinants, Vector algebra and index notation

Definition(Improper rotations and Cartesian pseudotensors)AdaptedAdapted from Riley, Hobson & Bence (2006), §26.10, Eqs. (26.38)–(26.39), pp. 946–947; Arfken, Weber & Harris (2013), §4.2, Eqs. (4.25)–(4.27), pp. 215–216

Throughout this unit the transformations are passive: the physical situation is left alone and only the Cartesian axes used to describe it are changed. Let the new axes be given by $x_i' = L_{ij}x_j$ with $L$ orthogonal, so that $\det L = \pm 1$ . Following Riley, Hobson and Bence we write $|L|$ for $\det L$ and call the transformation a proper rotation when $|L| = +1$ and an improper rotation when $|L| = -1$ . Every improper rotation is the inversion $x_i' = -x_i$ , for which $L_{ij} = -\delta_{ij}$ , composed with a proper rotation; equivalently, an improper rotation is one that turns a right-handed set of axes into a left-handed one.

A set of $3^n$ numbers attached to each choice of axes is

a Cartesian tensor of order $n$ if, under every orthogonal $L$ , proper or improper, $T'_{ij\cdots k} = L_{ip}L_{jq}\cdots L_{kr}\,T_{pq\cdots r} ;$
a Cartesian pseudotensor of order $n$ if instead $T'_{ij\cdots k} = |L|\;L_{ip}L_{jq}\cdots L_{kr}\,T_{pq\cdots r} .$

The two laws agree on proper rotations and differ by a sign on improper ones. A pseudotensor of order $0$ is a pseudoscalar: invariant under rotation, sign-reversing under reflection. A pseudotensor of order $1$ is a pseudovector, for which Arfken, Weber and Harris write the rule in matrix form as $A' = \det(S)\,SA$ .

Because the extra factor is multiplicative and $|L|^2 = 1$ , the status of a product is settled by a sign rule, which Arfken, Weber and Harris state as $T\otimes T = P\otimes P = T$ and $T\otimes P = P\otimes T = P$ , with $T$ for tensor and $P$ for pseudotensor: an outer product or a contraction of two tensors or of two pseudotensors is a tensor, and of one of each is a pseudotensor.

Definition(Relative tensors and their weights in general coordinates)AdaptedAdapted from Riley, Hobson & Bence (2006), §26.17, Eq. (26.74) and the worked example on the determinant of the metric, pp. 963–965

Outside Cartesian coordinates the transformation matrices are the Jacobian matrices of Tensors and their transformation laws, whose determinants are not $\pm1$ , and the single sign $|L|$ generalises to a power of the Jacobian. For a change of general coordinates $u^i \mapsto u'^i$ , write $J = \lvert \partial u'/\partial u\rvert$ for the determinant of $[\partial u'^i/\partial u^j]$ , so that $\lvert \partial u/\partial u'\rvert = 1/J$ .

A relative tensor of weight $w$ is a set of components transforming by the usual tensor rule multiplied by $\lvert \partial u/\partial u'\rvert^{\,w}$ ; for example, for one contravariant and one covariant index,

T'^{\,i}{}_{l} \;=\; \frac{\partial u'^i}{\partial u^a}\,\frac{\partial u^d}{\partial u'^l}\;T^{a}{}_{d}\;\left\lvert \frac{\partial u}{\partial u'}\right\rvert^{\,w} .

Riley, Hobson and Bence call a relative tensor of weight $0$ an absolute (or true) tensor, one of weight $-1$ a general pseudotensor, and one of weight $+1$ a tensor density; they note that a Cartesian pseudotensor, where $J = \pm1$ , satisfies both of the last two descriptions at once. Weights add under outer products and are unchanged by contraction.

Definition(The permutation symbol, its weights, and the dual correspondence)AdaptedAdapted from Riley, Hobson & Bence (2006), §26.10, pp. 947–948; Riley, Hobson & Bence (2006), §26.11, Eqs. (26.40)–(26.41), pp. 949–950; Riley, Hobson & Bence (2006), §26.17, pp. 964–965; Arfken, Weber & Harris (2013), §4.2, Example 4.2.1 and Eqs. (4.31)–(4.33), pp. 216–217

Let $\epsilon_{ijk}$ denote the permutation symbol: $\epsilon_{123} = \epsilon_{231} = \epsilon_{312} = +1$ , $\epsilon_{132} = \epsilon_{213} = \epsilon_{321} = -1$ , and $\epsilon_{ijk} = 0$ whenever two indices are equal. It is the same array of numbers in every coordinate system, by decree.

It is an isotropic third-order Cartesian pseudotensor. For every orthogonal $L$ , $\epsilon_{ijk} \;=\; |L|\,L_{ip}L_{jq}L_{kr}\,\epsilon_{pqr} ,$ which is the pseudotensor law of the definition with $T' = T = \epsilon$ . It is not a tensor: without the factor $|L|$ the right-hand side equals $-\epsilon_{ijk}$ under every improper rotation.
Cross products are pseudovectors. If $b_j$ and $c_k$ are the components of (true) vectors, then $a_i = \epsilon_{ijk}b_jc_k$ are the components of a pseudovector, and $\epsilon_{ijk}b_ic_jd_k$ is a pseudoscalar.
Duality. To an antisymmetric second-order Cartesian tensor $A_{ij}$ associate $p_i = \tfrac12\,\epsilon_{ijk}A_{jk} , \qquad\text{so that}\qquad (p_1,p_2,p_3) = (A_{23},\,A_{31},\,A_{12}) .$ Then $p$ is a pseudovector, called the dual of $A$ , and the correspondence is invertible: $A_{ij} = \epsilon_{ijk}p_k$ . Likewise the dual of a totally antisymmetric third-order tensor $A_{ijk}$ is the pseudoscalar $s = \tfrac1{3!}\epsilon_{ijk}A_{ijk}$ , and then $A_{ijk} = s\,\epsilon_{ijk}$ .
Weights in general coordinates. $\epsilon_{ijk}$ is a covariant relative tensor of weight $-1$ ; the numerically identical array $\epsilon^{ijk}$ , carried with upper indices, is a contravariant relative tensor of weight $+1$ ; and the determinant $g = \det[g_{ij}]$ of the metric is a relative scalar of weight $2$ . Consequently $\epsilon_{ijk}$ and $\epsilon^{ijk}$ may be used to pass between antisymmetric tensors and pseudovectors in general coordinates exactly as in the Cartesian case, the weights bookkeeping the extra Jacobian factors.

Idea of the proofOriginal

Everything here comes from one identity about determinants: contracting the three indices of $\epsilon$ with three copies of the same matrix reproduces $\epsilon$ times the determinant of that matrix. For an orthogonal $L$ that determinant is $\pm 1$ , which is precisely the extra sign in the pseudotensor law; for a Jacobian matrix it is the Jacobian, which is precisely the weight. The identity itself holds because the contraction is a totally antisymmetric three-index array in three dimensions, and there is only a one-dimensional space of those, so it must be a multiple of $\epsilon$ — and evaluating at $(1,2,3)$ names the multiple as the determinant. Duality is then arithmetic: an antisymmetric $3\times3$ array and a vector both have three independent entries, and $\epsilon$ is the only isotropic object that can match them up.

ProofAdaptedAdapted from Arfken, Weber & Harris (2013), §4.2, Eqs. (4.29)–(4.30) and Exercise 4.2.7, pp. 216–218; Riley, Hobson & Bence (2006), §26.10, the worked example on p. 948; Riley, Hobson & Bence (2006), §26.11, the worked example on p. 950; Riley, Hobson & Bence (2006), §26.17, pp. 964–965

0. The determinant identity. Let $M$ be any $3\times3$ matrix and set $f_{ijk} = M_{ip}M_{jq}M_{kr}\epsilon_{pqr}$ . Interchanging $i$ and $j$ and then renaming the dummy indices $p \leftrightarrow q$ turns $f_{jik}$ into $M_{iq}M_{jp}M_{kr}\epsilon_{pqr} = M_{ip}M_{jq}M_{kr}\epsilon_{qpr} = -f_{ijk}$ , by the antisymmetry of $\epsilon$ ; the same works for the other two pairs. So $f_{ijk}$ is totally antisymmetric. In three dimensions such an array is determined by its value at $(1,2,3)$ , since every non-zero entry is reached from that one by permuting indices and every entry with a repeated index vanishes; hence $f_{ijk} = f_{123}\,\epsilon_{ijk}$ . Finally $f_{123} = \epsilon_{pqr}M_{1p}M_{2q}M_{3r}$ is the Leibniz expansion of $\det M$ (Determinants). Therefore

M_{ip}M_{jq}M_{kr}\,\epsilon_{pqr} \;=\; (\det M)\,\epsilon_{ijk} \tag{$\ast$}

for every $3\times3$ matrix $M$ .

1. The permutation symbol. Put $M = L$ in $(\ast)$ : $L_{ip}L_{jq}L_{kr}\epsilon_{pqr} = |L|\,\epsilon_{ijk}$ . Multiply both sides by $|L|$ . Orthogonality is used here, and only here: it forces $|L| = \pm1$ , so $|L|^2 = 1$ and the left-hand side acquires the factor $|L|$ while the right-hand side becomes $\epsilon_{ijk}$ . That is statement 1. If instead $\epsilon$ were asked to obey the plain tensor law, $(\ast)$ would give $\epsilon'_{ijk} = |L|\epsilon_{ijk} = -\epsilon_{ijk}$ for improper $L$ , contradicting the decree that the array is the same in every frame.

2. Cross products. In the new axes $a'_i = \epsilon_{ijk}b'_jc'_k$ , with the same numbers $\epsilon_{ijk}$ . Substitute statement 1 for $\epsilon_{ijk}$ and $b'_j = L_{jm}b_m$ , $c'_k = L_{kn}c_n$ :

a'_i = |L|\,L_{ip}L_{jq}L_{kr}\epsilon_{pqr}\,L_{jm}b_m\,L_{kn}c_n .

Now use orthogonality in its second role, $L_{jq}L_{jm} = \delta_{qm}$ and $L_{kr}L_{kn} = \delta_{rn}$ , which collapses the sums over $j$ and $k$ :

a'_i = |L|\,L_{ip}\,\epsilon_{pqr}b_qc_r = |L|\,L_{ip}\,a_p ,

the pseudovector law. Contracting once more with a true vector $d$ gives $\epsilon_{ijk}b_ic_jd_k$ , which by the sign rule (pseudotensor times tensor) is a pseudoscalar; Arfken, Weber and Harris identify it as the scalar triple product, the determinant of the three rows $b$ , $c$ , $d$ .

3. Duality. That $p_i = \tfrac12\epsilon_{ijk}A_{jk}$ is a pseudovector is again the sign rule. The explicit components follow by writing out the sum: for $i=1$ only $(j,k) = (2,3)$ and $(3,2)$ contribute, giving $\tfrac12(A_{23} - A_{32}) = A_{23}$ by antisymmetry, and cyclically. For the inverse, contract with $\epsilon_{ijk}$ and use the standard identity $\epsilon_{ijk}\epsilon_{klm} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}$ (Vector algebra and index notation):

\epsilon_{ijk}p_k = \tfrac12\,\epsilon_{ijk}\epsilon_{klm}A_{lm} = \tfrac12\bigl(A_{ij} - A_{ji}\bigr) = A_{ij} ,

the last step using the antisymmetry of $A$ . The third-order case is the same computation with $\epsilon_{ijk}\epsilon_{ijk} = 3!$ .

4. Weights. Apply $(\ast)$ with $M = [\partial u^p/\partial u'^i]$ , whose determinant is $\lvert\partial u/\partial u'\rvert$ :

\frac{\partial u^p}{\partial u'^i}\frac{\partial u^q}{\partial u'^j}\frac{\partial u^r}{\partial u'^k}\,\epsilon_{pqr} \;=\; \left\lvert \frac{\partial u}{\partial u'}\right\rvert\,\epsilon_{ijk} .

The left-hand side is the plain covariant tensor law applied to $\epsilon$ ; the array in the new coordinates is again $\epsilon_{ijk}$ , so to match we must supply a further factor $\lvert\partial u/\partial u'\rvert^{-1}$ , which is the weight- $(-1)$ law. The contravariant array is the same computation with $M = [\partial u'^i/\partial u^p]$ and gives weight $+1$ . For the metric determinant, the transformation $g'_{ij} = (\partial u^k/\partial u'^i)(\partial u^l/\partial u'^j)g_{kl}$ reads in matrix form $G' = U^{\mathsf T}GU$ with $U = [\partial u^i/\partial u'^j]$ ; taking determinants, $g' = \lvert\partial u/\partial u'\rvert^{2}g$ , which is weight $2$ . $\square$

Remark(Where the hypotheses are used)Original

Orthogonality of $L$ is used twice in step 2 and nowhere else in step 1: once as $|L|^2 = 1$ and once as $L_{jq}L_{jm} = \delta_{qm}$ . Identity $(\ast)$ itself needs no orthogonality, which is exactly why the general-coordinate statement 4 comes out of the same line of algebra with a Jacobian in place of a sign.
Three dimensions is a real hypothesis for statement 3, not a convenience. In $n$ dimensions $\epsilon$ carries $n$ indices, an antisymmetric second-order tensor has $\tfrac12 n(n-1)$ independent components and a vector has $n$ , and these agree only at $n = 3$ . Arfken, Weber and Harris set this as Exercise 4.2.2, and it is the reason the vector cross product has no $n$ -dimensional analogue. Statements 1, 2 and 4 generalise to $n$ indices without change; so does $(\ast)$ , which is the Leibniz formula.
Antisymmetry of $A_{ij}$ is used at the last equality of step 3. Drop it and the computation returns $\tfrac12(A_{ij} - A_{ji})$ : the dual sees only the antisymmetric part, and a symmetric tensor has dual zero.
$g > 0$ is needed before $\sqrt{g}$ can be used to build absolute tensors from $\epsilon$ , as in the curl formula of Differential operators in general coordinates: non-vanishing is not enough, because a real positive root is wanted. That unit records where positivity holds and what replaces it when the metric is indefinite.
Orientation and the square root. Riley, Hobson and Bence state that $g$ is a relative scalar of weight $2$ . Taking the positive square root gives $\sqrt{g'} = \bigl\lvert\det[\partial u/\partial u']\bigr\rvert\sqrt{g}$ — with a modulus, because the root is positive by definition. So $\sqrt{g}$ behaves as a relative scalar of weight $1$ for coordinate changes with positive Jacobian, and acquires an extra minus sign for those with negative Jacobian. The combinations $\varepsilon_{ijk} = \sqrt{g}\,\epsilon_{ijk}$ and $\varepsilon^{ijk} = \epsilon^{ijk}/\sqrt{g}$ , whose weights cancel to $0$ , are therefore absolute tensors under orientation-preserving changes of coordinates and change sign under orientation-reversing ones. This is the general-coordinate form of "pseudo", and it is why an orientation must be chosen before the curl, the vector area or the volume element can be written down.

Remark(Polar and axial vectors: what is physical and what is bookkeeping)AdaptedAdapted from Riley, Hobson & Bence (2006), §26.10, pp. 947–949, including the footnotes on p. 948 and p. 949

Riley, Hobson and Bence are emphatic that a pseudovector is not a geometrical object in the ordinary sense: it is not an arrow in space, because a mere relabelling of the axes reverses it, and they draw it with a broken line to make the point. Their conclusion is that real physical quantities are described by tensors — temperature and density by scalars, velocity, magnetic field strength and angular momentum by vectors — and not by pseudotensors.

The genuinely physical distinction they then draw is a different one, and it concerns active transformations, in which the system itself is inverted rather than the axes. Under an active inversion through the origin, a polar vector such as position or velocity reverses direction, while an axial vector such as the angular momentum $J = m\,(x\times\dot x)$ does not, because both factors of the cross product have reversed. The words polar and axial therefore classify physical quantities; the words tensor and pseudotensor classify transformation laws under a change of description. The two are often conflated, and this unit keeps them apart in the way the source does.

Two further points from the same pages, both reported as the source gives them. First, in a left-handed Cartesian system the array $\epsilon_{ijk}b_jc_k$ fails to represent the arrow $b\times c$ at all: each of its three entries comes out with the opposite sign. The handedness of the axes must therefore be fixed before the cross product is written in components, even though the vector relation $a = b\times c$ itself needs no coordinates. Second, a footnote records that pairing an axial with a polar vector in a dot product yields a pseudoscalar, and that parity was shown not to be conserved by detecting a dependence of the angular distribution of $\beta$ -decay electrons, of momentum $p_e$ and emitted by polarised nuclei of spin $J_N$ , on the pseudoscalar quantity $J_N\cdot p_e$ . The footnote does not say which transformation it has in mind; it is the active one — under an active inversion $p_e$ reverses and $J_N$ does not, so the product changes sign — since under the passive reading of the rest of the section both are ordinary vectors and their product would be invariant.

Example(A reflection, worked in numbers)Original

Work in a right-handed Cartesian frame $S$ with three true vectors

b = (1,2,3), \qquad c = (4,5,6), \qquad d = (1,0,1) .

The cross product. $a_i = \epsilon_{ijk}b_jc_k$ gives $a_1 = 2\cdot6 - 3\cdot5 = -3$ , $a_2 = 3\cdot4 - 1\cdot6 = 6$ , $a_3 = 1\cdot5 - 2\cdot4 = -3$ , so $a = (-3, 6, -3)$ .

Change to a left-handed frame. Take $L = \operatorname{diag}(1,1,-1)$ , the reflection in the plane $x_3 = 0$ ; it is orthogonal with $|L| = -1$ , hence improper. Since $b$ , $c$ , $d$ are true vectors their new components are $b' = (1,2,-3)$ , $c' = (4,5,-6)$ , $d' = (1,0,-1)$ .

Evaluate both candidate laws. Computing the cross product afresh from the primed components, with the same numerical array $\epsilon_{ijk}$ :

a'_1 = 2(-6) - (-3)5 = 3, \qquad a'_2 = (-3)4 - 1(-6) = -6, \qquad a'_3 = 1\cdot5 - 2\cdot4 = -3 ,

so $a' = (3,-6,-3)$ . The tensor law would predict $L_{ip}a_p = (-3, 6, 3)$ , which is wrong in all three entries. The pseudotensor law predicts $|L|\,L_{ip}a_p = -(-3,6,3) = (3,-6,-3)$ , which is right. The same test with the improper matrix $M = -R$ , where $R$ is the quarter turn about $x_3$ , gives $M b = (2,-1,-3)$ and $Mc = (5,-4,-6)$ , whose cross product is $(-6,-3,-3)$ , and $|M|\,Ma = -(6,3,3) = (-6,-3,-3)$ again.

A pseudoscalar. The triple product $s = \epsilon_{ijk}b_ic_jd_k = a\cdot d = -3 + 0 - 3 = -6$ , which is also the determinant of the matrix with rows $b$ , $c$ , $d$ . In the primed frame $s' = a'\cdot d' = 3 + 0 + 3 = +6 = -s$ : a single number that reverses sign when the axes are reflected, which no scalar does.

The dual. The antisymmetric tensor dual to $a$ is $A_{ij} = \epsilon_{ijk}a_k$ , that is

[A_{ij}] = \begin{pmatrix} 0 & -3 & -6 \\ 3 & 0 & -3 \\ 6 & 3 & 0 \end{pmatrix} ,

since $A_{12} = a_3 = -3$ , $A_{23} = a_1 = -3$ and $A_{31} = a_2 = 6$ . Reading off $(A_{23}, A_{31}, A_{12}) = (-3, 6, -3)$ returns $a$ , as statement 3 requires. Here $A$ is a true tensor, being the product of the pseudotensor $\epsilon$ with the pseudovector $a$ .

A general-coordinate check. Take the skew chart $x = u$ , $y = uv$ , $z = w$ of Differential operators in general coordinates, on which $g = u^2$ and $\sqrt g = u$ . The absolute tensor of statement 4 and the remark is $\varepsilon_{ijk} = u\,\epsilon_{ijk}$ , so $\varepsilon_{123} = u$ ; and indeed $e_u\cdot(e_v\times e_w) = u$ , the signed volume of the cell that the three basis vectors span — which is the coordinate-free quantity $\varepsilon_{123}$ is meant to record. At the point $u = 2$ the three numbers $\varepsilon_{123} = 2$ , $\varepsilon^{123} = \tfrac12$ and $\epsilon_{123} = 1$ are all different, and only the first two are components of tensors.

ApplicationOriginal

Parity, and why the distinction is not pedantic. In electromagnetism the electric field is polar and the magnetic field is axial: under an active inversion of a current loop the circulating charges reverse their velocities, and $B$ is unchanged. The product $E\cdot B$ is therefore a pseudoscalar, and a term proportional to it in a Lagrangian violates parity. That observation is the whole content of the $\beta$ -decay experiment the previous remark describes, and it is also why the axion is introduced as a pseudoscalar field: it is allowed to couple to $E\cdot B$ and so to shift the parity-violating angle of quantum chromodynamics. In optics the same arithmetic explains why a chiral molecule rotates the plane of polarisation: the rotation angle per unit length is a pseudoscalar property of the medium, so the two enantiomers of a sugar necessarily rotate it by equal and opposite amounts, and a racemic mixture does not rotate it at all.

Rotation and the hat map. The angular velocity of a rigid body is the dual of an antisymmetric tensor, and in computation it is almost always the tensor that is stored. If $R(t)$ is the rotation taking body axes to space axes, then differentiating $RR^{\mathsf T} = I$ shows that $\Omega = \dot R R^{\mathsf T}$ is antisymmetric, and the angular velocity is the pseudovector $\omega$ with $\Omega v = \omega\times v$ , that is $\Omega_{ik} = \epsilon_{ijk}\omega_j$ in components. Beware the sign: contracting that with $\tfrac12\epsilon_{lik}$ and using $\epsilon_{ilk}\epsilon_{ijk} = 2\delta_{lj}$ gives $\tfrac12\epsilon_{lik}\Omega_{ik} = -\omega_l$ , so the dual of $\Omega$ in the sense of statement 3 is $-\omega$ , and the angular velocity is recovered as $\omega_i = -\tfrac12\epsilon_{ijk}\Omega_{jk} = \tfrac12\epsilon_{ijk}\Omega_{kj}$ . Robotics and graphics libraries call the passage from $\omega$ to $\Omega$ the "hat map"; it is statement 3 read backwards, with that minus sign kept in view — transposing the two indices of $\Omega$ is exactly what hides it. The same algebra puts the inertia tensor on a sound footing: Riley, Hobson and Bence obtain $J_i = I_{il}\omega_l$ with $I_{il} = \sum_\alpha m^{(\alpha)}\bigl(r_{(\alpha)}^2\delta_{il} - x^{(\alpha)}_ix^{(\alpha)}_l\bigr)$ by contracting two permutation symbols, and since $J$ and $\omega$ are both pseudovectors the sign rule makes $I$ an ordinary tensor — which is what licenses the use of the spectral theorem to find its principal axes, as in Stress and strain tensors.

Crystal and continuum physics. Material properties are tensors whose form is constrained by the symmetry group of the material, and whether the property is a tensor or a pseudotensor decides the constraint. The arithmetic is one line: put $L = -I$ in the two laws of the definition. A true tensor of order $n$ is multiplied by $(-1)^n$ and a pseudotensor of order $n$ by $|L|(-1)^n = (-1)^{n+1}$ . A centrosymmetric crystal — one whose point group contains the inversion — requires the property to come back to itself, so in such a crystal every true tensor of odd order and every pseudotensor of even order vanishes identically, while a pseudotensor of odd order is unconstrained by the inversion.

Each half has its standard consequence. No centrosymmetric crystal is piezoelectric, because the piezoelectric moduli $d_{ijk}$ form a true third-order tensor. Nor is one optically active, because optical rotation is carried by the second-order gyration pseudotensor $g_{ij}$ : quartz, whose point group $32$ has no centre, rotates the plane of polarisation, and calcite, whose group $\bar3m$ does, does not. The surviving case is odd-order and pseudo: the spontaneous magnetisation of a ferromagnet is an axial vector, order $1$ , and the inversion leaves it alone — which is why iron can be both centrosymmetric and magnetically ordered. In continuum mechanics the same bookkeeping distinguishes the ordinary stress tensor from the couple-stress tensors of micropolar media.

In the language of forms. Everything in this unit is said more economically with differential forms, where duality is the Hodge star and the factor $\sqrt g$ appears by itself: see Differential forms and The Hodge star and Maxwell's equations. The payoff of that language is exactly the point of the remark on orientation — the pseudo-ness is isolated into the single choice of an orientation on the manifold, rather than being carried as a sign on every object.

Application

Stress and strain tensors

Adapted

Add to PDF

Builds the infinitesimal strain tensor from a displacement field and the stress tensor from the traction on a surface element, reduces the generalised Hooke's law to the two Lamé constants for an isotropic medium, relates those to Young's modulus and Poisson's ratio, and computes the principal stresses in a loaded steel block.

Prerequisites: Tensors and their transformation laws, Pseudotensors and the Levi-Civita tensor, Spectral theorem for Hermitian matrices, Eigenvalues and eigenvectors, The divergence theorem

Definition(Displacement and the strain tensor)AdaptedAdapted from Riley, Hobson & Bence (2006), §26.12, p. 953; Stakgold & Holst (2011), §0.4, 'Continuum model', p. 20

Model the body as a continuum: a material point is labelled by its position $x$ in a chosen reference configuration — whatever shape the body is taken to start from, usually its undeformed one — and is found afterwards at $x + u(x)$ . The vector field $u$ is the displacement. Stakgold and Holst note that these are Lagrangian, or material, coordinates, and that their advantage is that integrals may be taken over the reference region, whose geometry is usually the simpler of the two.

Riley, Hobson and Bence describe the local deformation at an interior point by the symmetric second-order Cartesian tensor

e_{ij} \;=\; \frac12\left(\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\right) ,

the strain tensor. Its trace $\theta = e_{kk} = \nabla\cdot u$ is called the dilatation; to first order in the displacement gradients it is the fractional change of volume, since the volume element is multiplied by $\det(I + Du) = 1 + \operatorname{tr}Du + O(|Du|^2)$ .

Only the symmetric part of $\partial u_i/\partial x_j$ appears. The antisymmetric part $\omega_{ij} = \tfrac12(\partial_j u_i - \partial_i u_j)$ is discarded because it describes a rotation rather than a deformation: for the infinitesimal rigid motion $u_i = a_i + \epsilon_{ijk}\omega_j x_k$ one has $\partial u_i/\partial x_l = \epsilon_{ijl}\omega_j$ , which is antisymmetric in $i$ and $l$ , so $e_{ij}$ vanishes identically. (The pseudovector $\omega_j$ and the tensor $\omega_{ij}$ are the dual pair of Pseudotensors and the Levi-Civita tensor, up to a sign that is worth fixing once: the same computation gives $\omega_{ij} = \tfrac12(\partial_ju_i - \partial_iu_j) = -\epsilon_{ijk}\omega_k$ , so the dual in the sense used there is $\tfrac12\epsilon_{ijk}\omega_{jk} = -\omega_i$ , not $+\omega_i$ . For $\omega = (0,0,1)$ , say, the rigid rotation is $u = (-x_2,x_1,0)$ , giving $\omega_{12} = -1$ and a dual $(0,0,-1)$ .)

Definition(Traction and the stress tensor)AdaptedAdapted from Stakgold & Holst (2011), §0.4, Eqs. (0.4.5)–(0.4.6) and the tetrahedron argument, pp. 20–21; Riley, Hobson & Bence (2006), §26.12, p. 953

The forces one part of a body exerts on a neighbouring part are short-range, so they act, to the accuracy of the continuum model, across the shared surface. Let $B$ be a surface inside the body with unit normal $n$ at a point. The traction $T(n)$ is the force per unit area transmitted across $B$ at that point, so that the total force on a region $R$ from the material outside it is $\int_{\partial R} T(n)\,dS$ .

To find how $T$ depends on $n$ , Stakgold and Holst shrink a tetrahedron whose three short faces lie in the coordinate planes and whose slanted face is normal to $n$ . Surface terms scale as the squared size of the tetrahedron and volume terms as its cube, so in the limit the tractions on the four faces must balance by themselves; since the three short faces have areas proportional to the components of $n$ , what emerges is that $T(n)$ is a linear function of those components. Linearity in $n$ is precisely what makes the coefficients a tensor, here the stress tensor:

T_j(n) \;=\; n_i\,p_{ij} .

Both sources index it the same way, the first slot recording which plane the force is transmitted across and the second recording which way the force points; Riley, Hobson and Bence phrase this as $p_{ij}$ being the component along $x_j$ of the force per unit area on a surface element whose normal points along $x_i$ . Riley, Hobson and Bence additionally take $p_{ij}$ to be symmetric, writing that into the definition, and this unit follows them; Stakgold and Holst, who get the tensor from linearity in $n$ alone, say nothing about the symmetry of $T$ on pp. 20–22. The remark below collects what that assumption buys and what it costs.

Feeding $T_j(n) = n_ip_{ij}$ into the impulse–momentum law for an arbitrary subregion and converting the surface integral with the divergence theorem (The divergence theorem), Stakgold and Holst arrive at the equation of motion

\rho\,\frac{\partial^2 u_j}{\partial t^2} \;=\; \frac{\partial p_{ij}}{\partial x_i} + f_j ,

with $\rho$ the density in the reference configuration and $f$ the body force per unit volume. This is three equations for the nine quantities $u_j$ and $p_{ij}$ ; a constitutive law relating stress to strain is what closes the system.

Application(Hooke's law for an isotropic medium)AdaptedAdapted from Riley, Hobson & Bence (2006), §26.12, Eqs. (26.46)–(26.49), pp. 953–954

The generalised Hooke's law takes the stress to be a linear function of the strain,

p_{ij} = c_{ijkl}\,e_{kl} ,

with $c_{ijkl}$ a fourth-order Cartesian tensor. Suppose the medium is isotropic, so that $c_{ijkl}$ is an isotropic tensor, and take — as Riley, Hobson and Bence do, assuming rather than proving it — the most general isotropic fourth-order tensor to be

c_{ijkl} = \lambda\,\delta_{ij}\delta_{kl} + \eta\,\delta_{ik}\delta_{jl} + \nu\,\delta_{il}\delta_{jk} .

Then, with $\theta = e_{kk}$ and $2\mu = \eta + \nu$ :

The law has two constants. $p_{ij} \;=\; \lambda\,\theta\,\delta_{ij} + 2\mu\,e_{ij} ,$ where $\lambda$ and $\mu$ are the Lamé constants. Only the combination $\eta + \nu$ survives, so an isotropic linear elastic solid has exactly two independent elastic constants.
Stress and strain share their principal axes. Every eigenvector of $e$ is an eigenvector of $p$ , and if $\varepsilon_1,\varepsilon_2,\varepsilon_3$ are the principal strains then the principal stresses are $\lambda\theta + 2\mu\varepsilon_a$ . In particular $e_{ij} = 0$ for $i \neq j$ forces $p_{ij} = 0$ for $i \neq j$ .
Young's modulus and Poisson's ratio. Apply a simple tension $p_{11} = S$ with all other $p_{ij} = 0$ , and assume $\mu \neq 0$ , $\lambda + \mu \neq 0$ and $3\lambda + 2\mu \neq 0$ . Define $E$ by $S = E\,e_{11}$ and $\sigma = -e_{22}/e_{11}$ . Then $E = \frac{\mu(3\lambda + 2\mu)}{\lambda + \mu} , \qquad \sigma = \frac{\lambda}{2(\lambda+\mu)} ,$ and solving these for $\lambda$ and $\mu$ puts the law in the form engineers use, $p_{ij} \;=\; \frac{\sigma E}{(1+\sigma)(1-2\sigma)}\,\theta\,\delta_{ij} \;+\; \frac{E}{1+\sigma}\,e_{ij} , \qquad\text{i.e.}\qquad \lambda = \frac{\sigma E}{(1+\sigma)(1-2\sigma)}, \quad \mu = \frac{E}{2(1+\sigma)} .$

A notation warning. Riley, Hobson and Bence write $p_{ij}$ for the stress and reserve $\sigma$ for Poisson's ratio, and this unit follows them. Most engineering texts do the opposite, writing $\sigma_{ij}$ for the stress and $\nu$ for Poisson's ratio — and $\nu$ is one of the three constants in the isotropic tensor above. Check which convention a formula is in before using it.

Idea of the proofOriginal

Isotropy is a very strong constraint, and the whole of statement 1 is an exercise in how little survives it. A fourth-order tensor has $81$ components; an isotropic one must be built from the only isotropic tensors available, which in three dimensions are multiples of $\delta_{ij}$ and products of them, leaving three constants; and because the strain it is contracted against is symmetric, two of those three constants are only ever seen in the combination $\eta+\nu$ . Two constants remain. Statement 2 is then immediate: the law expresses $p$ as a linear combination of $e$ and the identity, and adding a multiple of the identity to a matrix does not move its eigenvectors. Statement 3 is a change of variables: $E$ and $\sigma$ are what a tension test measures, and reading the law in a state of simple tension converts one pair of constants into the other.

ProofAdaptedAdapted from Riley, Hobson & Bence (2006), §26.12, the worked example on pp. 953–954

1. The reduction. Contract the assumed isotropic form against the strain:

c_{ijkl}e_{kl} = \lambda\,\delta_{ij}\,\delta_{kl}e_{kl} + \eta\,\delta_{ik}\delta_{jl}e_{kl} + \nu\,\delta_{il}\delta_{jk}e_{kl} = \lambda\,e_{kk}\,\delta_{ij} + \eta\,e_{ij} + \nu\,e_{ji} .

The strain tensor is symmetric, so $e_{ji} = e_{ij}$ and the last two terms merge into $(\eta+\nu)e_{ij}$ . Writing $\eta + \nu = 2\mu$ gives statement 1. The symmetry of $e$ is used exactly here, and it is why $\eta$ and $\nu$ cannot be separated by any experiment on an elastic solid.

2. Principal axes. In matrix form statement 1 reads $p = \lambda\theta\,I + 2\mu\,e$ , with $\theta = \operatorname{tr}e$ a scalar. The strain tensor is real and symmetric, so by the spectral theorem (Spectral theorem for Hermitian matrices) it has an orthonormal basis of eigenvectors $q_a$ with real eigenvalues $\varepsilon_a$ . For each of them $p\,q_a = \lambda\theta\,q_a + 2\mu\varepsilon_a q_a = (\lambda\theta + 2\mu\varepsilon_a)q_a$ , which is statement 2. Referred to those axes both matrices are diagonal, which is Riley, Hobson and Bence's observation that the principal axes of stress and strain coincide.

3. Simple tension. Suppose $p_{11} = S$ and every other $p_{ij} = 0$ . For $i\neq j$ statement 1 gives $0 = p_{ij} = 2\mu e_{ij}$ , so $e_{ij} = 0$ as well, using $\mu \neq 0$ ; both tensors are diagonal in the given axes. The three diagonal equations are

S = \lambda\theta + 2\mu e_{11}, \qquad 0 = \lambda\theta + 2\mu e_{22}, \qquad 0 = \lambda\theta + 2\mu e_{33} .

Add them. The left side is $S$ ; on the right $3\lambda\theta + 2\mu(e_{11}+e_{22}+e_{33}) = 3\lambda\theta + 2\mu\theta$ , by the definition of $\theta$ . Hence

S = (3\lambda + 2\mu)\,\theta , \tag{1}

and $\theta$ may be eliminated because $3\lambda+2\mu \neq 0$ . Substituting $\theta = S/(3\lambda+2\mu)$ into the first equation and solving for $e_{11}$ ,

e_{11} = \frac{1}{2\mu}\left(S - \frac{\lambda S}{3\lambda+2\mu}\right) = \frac{S}{2\mu}\cdot\frac{2(\lambda+\mu)}{3\lambda+2\mu} = \frac{S\,(\lambda+\mu)}{\mu(3\lambda+2\mu)} .

Since $E = S/e_{11}$ by definition, this is the stated formula for $E$ . For Poisson's ratio, the second equation gives $e_{22} = -\lambda\theta/(2\mu)$ , so

\sigma = -\frac{e_{22}}{e_{11}} = \frac{\lambda\theta}{2\mu\,e_{11}} = \frac{\lambda}{2\mu}\cdot\frac{S}{3\lambda+2\mu}\cdot\frac{\mu(3\lambda+2\mu)}{S(\lambda+\mu)} = \frac{\lambda}{2(\lambda+\mu)} ,

which uses $\lambda+\mu \neq 0$ . Finally, the two displayed relations are a pair of equations for $\lambda$ and $\mu$ in terms of $E$ and $\sigma$ ; solving them gives the engineering form of the law, and a convenient by-product of (1) is

3\lambda + 2\mu = \frac{E}{1-2\sigma} , \qquad\text{so}\qquad p_{kk} = \frac{E}{1-2\sigma}\,\theta . \tag{2}

$\square$

Remark(Where the hypotheses are used)Original

Small displacement gradients. Neither source attaches a smallness condition to $e_{ij}$ , but the definition is linear in $u$ while the actual change in the squared length of a material line element is read off from $dx' = (I + Du)\,dx$ as $\lvert dx'\rvert^2 - \lvert dx\rvert^2 = \bigl(Du + Du^{\mathsf T} + Du^{\mathsf T}Du\bigr)_{ij}dx_idx_j = 2e_{ij}\,dx_idx_j + \partial_iu_k\,\partial_ju_k\,dx_idx_j$ , the Green–Lagrange strain. Note which slot the summed index sits in: it is $(Du^{\mathsf T}Du)_{ij} = \partial_iu_k\,\partial_ju_k$ , not $(Du\,Du^{\mathsf T})_{ij} = \partial_ku_i\,\partial_ku_j$ , and the two differ — for the simple shear $u = (x_2,0,0)$ and $dx = (1,0,0)$ the exact change is $0$ , which the first reproduces and the second does not. The quadratic term has been dropped, so $e_{ij}$ measures deformation only when $|\partial u_i/\partial x_j| \ll 1$ . This is why the whole theory is called linear elasticity, and why a thin strip bent through a right angle — small strains, large rotations — needs a different treatment.
Symmetry of $e_{ij}$ is used once, at the merging of $\eta$ and $\nu$ in step 1. It is built into the definition, and it is the reason an isotropic solid has two elastic constants rather than three.
Symmetry of $p_{ij}$ is assumed, not derived. Riley, Hobson and Bence make it part of the definition; Stakgold and Holst remark that an angular impulse–momentum law accompanies the linear one but do not pursue it for a general continuum, turning instead to rods, where they do write the one-component angular momentum equation (0.4.12) down. It is this symmetry that lets the spectral theorem be applied in step 2 and that makes "principal stress" meaningful. Media in which it fails — those supporting distributed body couples — need couple-stress tensors and lie outside this unit.
Isotropy enters only through the assumed form of $c_{ijkl}$ . Arfken, Weber and Harris set as Exercise 4.2.4 the verification that three independent combinations of Kronecker deltas are isotropic — $\delta_{ij}\delta_{kl}$ , the sum $\delta_{ik}\delta_{jl} + \delta_{il}\delta_{jk}$ and the difference $\delta_{ik}\delta_{jl} - \delta_{il}\delta_{jk}$ — from which the bare products used above follow as half the sum and half the difference of the last two. That the three exhaust the isotropic fourth-order tensors is the part Riley, Hobson and Bence assume. The companion fact for second-order tensors, that an isotropic one is a multiple of $\delta_{ij}$ , is their Exercise 4.1.4. For an anisotropic crystal $c_{ijkl}$ has up to $21$ independent components and statement 2 is false: the principal axes of stress and strain need not coincide.
$\mu \neq 0$ is needed in step 3 to conclude that the strain is diagonal; a material with $\mu = 0$ supports no shear stress and is a fluid, for which the whole construction degenerates. In the output variables $\mu = 0$ is exactly $\sigma = \tfrac12$ , since $\sigma = \lambda/(2(\lambda+\mu))$ equals $\tfrac12$ precisely when $\mu$ vanishes. $3\lambda+2\mu \neq 0$ is needed to solve (1); it is equivalent to $\sigma \neq -1$ , and also to $E \neq 0$ , because $\lambda = -\tfrac23\mu$ gives both. $\lambda+\mu \neq 0$ is needed for the last step of the Poisson calculation, and on it $E$ and $\sigma$ are not even defined as finite numbers. None of the three excludes a real solid: ordinary materials have $\mu > 0$ and $-1 < \sigma < \tfrac12$ .
The incompressible limit is the one place where these conditions nearly bite. Holding $E$ fixed and letting $\sigma \to \tfrac12$ sends $\lambda \to \infty$ with $\mu \to E/3$ finite, and relation (2) shows what happens: $3\lambda+2\mu = E/(1-2\sigma)$ diverges rather than vanishes, so the bulk modulus $K = \tfrac13(3\lambda+2\mu)$ becomes infinite and the dilatation is forced to zero. Rubber, with $\sigma \approx 0.4999$ , is close enough that naive finite-element formulations in the displacement $u$ alone lock up and must be replaced by mixed formulations carrying the pressure as a separate unknown.
The reference configuration is where $x$ , $\rho$ and all the derivatives live. Mixing reference and deformed coordinates is harmless at this order and disastrous beyond it.

Method(From a displacement field to principal stresses)Original

Differentiate the displacement. Form the matrix $\partial u_i/\partial x_j$ . Split it into its symmetric part $e_{ij}$ and its antisymmetric part; only the first carries stress, and inspecting the second tells you how much of the motion is a local rotation.
Take the dilatation $\theta = e_{kk}$ . If the material is nearly incompressible this number should be small, and if it is not, suspect an error.
Convert the elastic data. Laboratory data come as $E$ and $\sigma$ ; the law needs $\mu = E/\bigl(2(1+\sigma)\bigr)$ and $\lambda = \sigma E/\bigl((1+\sigma)(1-2\sigma)\bigr)$ . Keep the units straight: moduli in GPa, strains dimensionless, stresses in MPa, so a strain of $10^{-4}$ against a modulus of $200$ GPa is $20$ MPa.
Assemble the stress, $p_{ij} = \lambda\theta\delta_{ij} + 2\mu e_{ij}$ .
Diagonalise the strain. Being real and symmetric it has an orthonormal eigenbasis (Eigenvalues and eigenvectors); the eigenvalues are the principal strains and the eigenvectors the principal directions. By statement 2 you do not need to diagonalise the stress separately: the principal stresses are $\lambda\theta + 2\mu\varepsilon_a$ on the same axes.
Read off what the design question needs. The largest principal stress governs brittle fracture; the maximum shear stress, which is half the difference between the largest and smallest principal stresses, governs yielding in the Tresca criterion.
Check the trace. By (2) the sum of the principal stresses must equal $E\theta/(1-2\sigma)$ . This is a genuine independent check: it uses the dilatation computed in step 2 and never touches the eigenvectors.

How it fails. Step 1 is where engineering strain conventions bite: the shear strain $\gamma_{12}$ quoted by strain gauges and material data sheets is $2e_{12}$ , not $e_{12}$ , and a factor of two inserted here propagates into every shear stress. Step 3 is where the notation warning matters. Step 5 tempts one to diagonalise $p$ from scratch and then wonder why the axes agree.

Example(A sheared and stretched steel block)Original

Take structural steel with Young's modulus $E = 200$ GPa and Poisson's ratio $\sigma = 0.3$ , and the displacement field

u = c\,\bigl(2x_1 + x_2,\;\; x_1 + 2x_2,\;\; 0\bigr), \qquad c = 10^{-4} .

Step 1. The gradient matrix is $c\begin{pmatrix}2&1&0\\1&2&0\\0&0&0\end{pmatrix}$ , already symmetric, so there is no rotational part and

[e_{ij}] = 10^{-4}\begin{pmatrix}2&1&0\\1&2&0\\0&0&0\end{pmatrix} .

The displacement gradients are of order $10^{-4}$ , comfortably within the linear theory.

Step 2. $\theta = (2 + 2 + 0)\times 10^{-4} = 4\times10^{-4}$ .

Step 3. $\mu = 200/(2\times1.3) = 1000/13 \approx 76.923$ GPa and $\lambda = (0.3\times200)/(1.3\times0.4) = 1500/13 \approx 115.385$ GPa.

Step 4. The hydrostatic part is $\lambda\theta = (1500/13)\times 4\times10^{-4}$ GPa $= 600/13 \approx 46.154$ MPa, and $2\mu = 2000/13$ GPa, so

p_{11} = p_{22} = \frac{600}{13} + \frac{2000}{13}\times2\times10^{-4}\times10^{3}\ \text{MPa} = \frac{600 + 400}{13} = \frac{1000}{13} \approx 76.923\ \text{MPa},

p_{12} = \frac{2000}{13}\times10^{-4}\times10^{3} = \frac{200}{13} \approx 15.385\ \text{MPa}, \qquad p_{33} = \frac{600}{13} \approx 46.154\ \text{MPa},

with $p_{13} = p_{23} = 0$ . Note that $p_{33}$ is not zero even though $e_{33}$ is: stretching the block in the $x_1x_2$ -plane while forbidding it to contract in $x_3$ requires a stress in $x_3$ . This is the plane-strain condition.

Step 5. The upper-left $2\times2$ block of $[e_{ij}]/c$ , namely $\left(\begin{smallmatrix}2&1\\1&2\end{smallmatrix}\right)$ , has eigenvalues $3$ and $1$ with eigenvectors $(1,1,0)/\sqrt2$ and $(1,-1,0)/\sqrt2$ , and the third principal direction is $(0,0,1)$ with eigenvalue $0$ . So the principal strains are

\varepsilon_1 = 3\times10^{-4}, \qquad \varepsilon_2 = 1\times10^{-4}, \qquad \varepsilon_3 = 0 ,

and the principal stresses $\lambda\theta + 2\mu\varepsilon_a$ are

\frac{600}{13} + \frac{600}{13} = \frac{1200}{13} \approx 92.308\ \text{MPa}, \qquad \frac{600}{13} + \frac{200}{13} = \frac{800}{13} \approx 61.538\ \text{MPa}, \qquad \frac{600}{13} \approx 46.154\ \text{MPa} .

Step 6. The maximum principal stress is $1200/13 \approx 92.3$ MPa, and the maximum shear stress is

\tfrac12\left(\frac{1200}{13} - \frac{600}{13}\right) = \frac{300}{13} \approx 23.1\ \text{MPa} ,

both far below the roughly $250$ MPa yield stress of mild steel, so the block stays elastic.

Step 7, the check. The principal stresses sum to $(1200 + 800 + 600)/13 = 2600/13 = 200$ MPa exactly. Independently, $E\theta/(1-2\sigma) = 200\ \text{GPa}\times4\times10^{-4}/0.4 = 0.2$ GPa $= 200$ MPa. The two agree.

ApplicationOriginal

The field equations of elasticity. Substituting Hooke's law into the equation of motion closes the system. With $p_{ij} = \lambda u_{k,k}\delta_{ij} + \mu(u_{i,j} + u_{j,i})$ ,

\frac{\partial p_{ij}}{\partial x_i} = \lambda\,\partial_j(\nabla\cdot u) + \mu\,\nabla^2 u_j + \mu\,\partial_j(\nabla\cdot u) = \mu\,\nabla^2u_j + (\lambda+\mu)\,\partial_j(\nabla\cdot u) ,

so the motion of an isotropic elastic body with no body force obeys Navier's equation $\rho\,u_{tt} = \mu\nabla^2u + (\lambda+\mu)\nabla(\nabla\cdot u)$ . With $\rho$ scaled to one these are exactly the evolution equations of linear elasticity that Evans lists among the basic linear systems, and dropping the time derivative gives his equilibrium equations. They are a coupled elliptic system, not three copies of Laplace's equation: the coupling term $(\lambda+\mu)\nabla(\nabla\cdot u)$ is what makes elasticity harder than electrostatics.

Two wave speeds, and how seismologists use them. Evans sets as a problem that $w = \nabla\cdot u$ and $\mathbf w = \nabla\times u$ each satisfy wave equations with different speeds. Taking the divergence of Navier's equation and using $\nabla\cdot\nabla(\nabla\cdot u) = \nabla^2(\nabla\cdot u)$ gives $\rho\,w_{tt} = (\lambda+2\mu)\nabla^2w$ ; taking the curl and using $\nabla\times\nabla\phi = 0$ gives $\rho\,\mathbf w_{tt} = \mu\nabla^2\mathbf w$ . So compressional (P) disturbances travel at $c_P = \sqrt{(\lambda+2\mu)/\rho}$ and shear (S) disturbances at $c_S = \sqrt{\mu/\rho}$ , and the ratio $c_P/c_S = \sqrt{(\lambda+2\mu)/\mu}$ depends only on Poisson's ratio. For the steel of the example, $\lambda + 2\mu = 3500/13 \approx 269.2$ GPa and $\rho = 7850$ kg m $^{-3}$ give $c_P \approx 5.86$ km s $^{-1}$ and $c_S \approx 3.13$ km s $^{-1}$ , which is what an ultrasonic thickness gauge measures. For a "Poisson solid" with $\sigma = \tfrac14$ one has $\lambda = \mu$ and the ratio is exactly $\sqrt3$ ; most of the Earth's crust is close to this, which is why the delay between the P and S arrivals at a seismometer converts so simply into a distance to the epicentre, and why three stations suffice to locate an earthquake.

Measuring the tensor. A strain-gauge rosette measures the normal strain along three directions in a surface, which is three linear equations for the three independent components of the in-plane strain tensor; the gauge rotation formula it is read with is just the transformation law of a second-order Cartesian tensor (Tensors and their transformation laws). Photoelasticity does the same job optically: a transparent loaded model placed between crossed polarisers shows fringes along which the difference of the two in-plane principal stresses is constant, so the fringe pattern is a contour map of the maximum shear stress computed in step 6 of the method. Both techniques deliver principal directions, which is why the spectral theorem is the working tool of experimental stress analysis and not an ornament.

Thermal and residual stress. Adding a thermal term, $p_{ij} = \lambda\theta\delta_{ij} + 2\mu e_{ij} - \beta\,\Delta T\,\delta_{ij}$ , is enough to explain why a constrained rail buckles on a hot day and why a thick weld retains stress after it has cooled: the body is in equilibrium, with $\partial_ip_{ij} = 0$ , yet the stress is not zero, because the strain field that would relieve it is not compatible with the constraints. The same arithmetic, with the dilatation driven by swelling instead of heating, governs the cracking of drying mud and of lithium-ion battery electrodes.