9.5 Conservation laws and energy–stress tensors

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Conservation of charge

Consider a general four-dimensional region $\Omega$ of space-time with boundary 3-surface $\partial \Omega$. The four-dimensional Gauss theorem (see Chapter 17) asserts that for any vector field $A^{\alpha }$

$${\iiint }_{\Omega }{A}_{,\alpha }^{\alpha }\mathrm{d}{x}^1\mathrm{d}{x}^2\mathrm{d}{x}^3\mathrm{d}{x}^4={\iiint }_{\partial \Omega }{A}^{\alpha }\mathrm{d}{S}_{\alpha }.\text{\hspace{1em}(9.54)}$$

If $\partial \Omega$ has the parametric form $x^{\alpha }=x^{\alpha }({\lambda }_1,{\lambda }_2,{\lambda }_3)$ , the vector 3-volume element $\mathrm{d}{S}_{\alpha }$ is defined by

$$\mathrm{d}{S}_{\alpha }={ϵ}_{\alpha \beta \gamma \delta }\frac{\partial x^{\beta }}{\partial {\lambda }_1}\frac{\partial x^{\gamma }}{\partial {\lambda }_2}\frac{\partial x^{\delta }}{\partial {\lambda }_3}\mathrm{d}{\lambda }_1\mathrm{d}{\lambda }_2\mathrm{d}{\lambda }_3,$$

with the four-dimensional epsilon symbol ${ϵ}_{\alpha \beta \gamma \delta }$ defined by Eq. (8.21). Since the epsilon symbol transforms as a tensor with respect to basis transformations having determinant 1, it is a 4-tensor if we restrict ourselves to proper Lorentz transformations, and it follows that $d{S}_{\alpha }$ is a 4-vector. Furthermore, $\mathrm{d}{S}_{\alpha }$ is orthogonal to the 3-surface $\partial \Omega$, for any 4-vector $X^{\alpha }$ tangent to the 3-surface has a linear decomposition

$$X^{\alpha }=\sum _{i=1}^3{c}_i\frac{\partial x^{\delta }}{\partial {\lambda }_i},$$

and by the total antisymmetry of ${ϵ}_{\alpha \beta \gamma \delta }$ it follows that

$$\mathrm{d}{S}_{\alpha }{X}^{\alpha }=\sum _{i=1}^3{c}_i{ϵ}_{\alpha \beta \gamma \delta }\frac{\partial x^{\alpha }}{\partial {\lambda }_i}\frac{\partial x^{\beta }}{\partial {\lambda }_1}\frac{\partial x^{\gamma }}{\partial {\lambda }_2}\frac{\partial x^{\delta }}{\partial {\lambda }_3}=0.$$

The four-dimensional Gauss theorem is a natural generalization of the well-known threedimensional result. In Chapter 17, it will become clear that this theorem is independent of the choice of parametrization ${\lambda }_i$ on $\partial \Omega$.

A 3-surface $\mathcal{S}$ is called spacelike if its orthogonal 3-volume element $\mathrm{d}{S}_{\alpha }$ is a timelike 4-covector. The reason for this terminology is that a 4-vector orthogonal to three linearly independent spacelike 4-vectors must be timelike. The archetypal spacelike 3-surface is given by the equation $t$ const. in a given inertial frame. Parametrically the surface may be given by $x={\lambda }_1,y={\lambda }_2,z={\lambda }_3$ and its 3-volume element is

$$\mathrm{d}{S}_{\alpha }={ϵ}_{\alpha 123}\mathrm{d}x\mathrm{d}y\mathrm{d}z=(0,0,0,-\mathrm{d}x\mathrm{d}y\mathrm{d}z).$$

Given a current 4-vector $J^{\alpha }=(\mathbf{j},c\rho )$ , satisfying the divergence-free condition $J_{\mathbf{\Phi },\alpha }^{\alpha }=0$ , it is natural to define the ‘total charge’ over an arbitrary spacelike 3-surface to be

$$Q=-\frac{1}{c}{\iiint }_{\mathcal{S}}{J}^{\alpha }\mathrm{d}{S}_{\alpha },\text{\hspace{1em}(9.55)}$$

as this gives the expected $Q=\int \int \int \rho dx\mathrm{d}y\mathrm{d}z$ when $\mathcal{S}$ is a surface of type $t=\mathrm{const}.$

Let $\Omega$ be a 4-volume enclosed by two spacelike surfaces $s$ and $S^{\prime }$ having infinite extent. Using the four-dimensional Gauss theorem and the divergence-free condition $J_{,\alpha }^{\alpha }=0$ we obtain the law of conservation of charge,

$$Q^{\prime }-Q=\frac{1}{c}\left({\iiint }_S{J}^{\alpha }\mathrm{d}{S}_{\alpha }-{\iiint }_{S^{\prime }}{J}^{\alpha }\mathrm{d}{S}_{\alpha }\right)=\frac{1}{c}{\iiint }_{\Omega }{J}_{,\alpha }^{\alpha }\mathrm{d}{x}^1\mathrm{d}{x}^2\mathrm{d}{x}^3\mathrm{d}{x}^4=0$$

where the usual physical assumption is made that the 4-current $J^{\alpha }$ vanishes at spatial infinity $|\mathbf{r}|\to \infty$ . This implies that there are no contributions from the timelike ‘sides at infinity’ to the 3-surface integral over ∂ $S$. Note that in Minkowski space, $\mathrm{d}{S}^{\alpha }$ is required to be ‘inwards-pointing’ on the spacelike parts of the boundary, $s$ and $S^{\prime }$ , as opposed to the more usual outward pointing requirement in three-dimensional Euclidean space.

As seen in Example 9.3 and Section 9.4 there is a converse to this result: given a conserved quantity $Q,$ , generically called ‘charge’, then $J^{\alpha }=(\mathbf{j},c\rho )$ , where $\rho$ is the charge density and $j$ the charge flux density, form the components of a divergence-free 4-vector field, $J_{,\alpha }^{\alpha }=0$

Energy–stress tensors

Assume now that the total 4-momentum $P^{\mu }$ of a system is conserved. Treating its components as four separate conserved ‘charges’, we are led to propose the existence ofa quantity $T^{\mu \nu }$ such that

$$T_{,\nu }^{\mu \nu }=0\text{\hspace{1em}(9.56)}$$

and the total 4-momentum associated with any spacelike surface $S$ is given by

$$P^{\mu }=-\frac{1}{c}{\iiint }_S{T}^{\mu \nu }\mathrm{d}{S}_{\nu }.\text{\hspace{1em}(9.57)}$$

In order to ensure that Eq. (9.56) be a tensorial equation it is natural to postulate that $T^{\mu \nu }$ is a 4-tensor field, called the energy–stress tensor of the system. This will also guarantee that the quantity $P^{\mu }$ defined by Eq. (9.57) is a 4-vector. For a surface $t=const.$ . we have

$$P^{\mu }=\left(\mathbf{p},\frac{E}{c}\right)=\frac{1}{c}{\int }_{t=\text{ const. }}{T}^{\mu 4}{d}^{3}x$$

and the physical interpretation of the components of the energy–stress tensor $T^{\mu \nu }$ are

$$T^{44}=\text{ energy density },$$ $$T^{4i}=\frac{1}{c}\times \text{ energy flux density },$$ $$T^{i4}=c\times \text{ momentum density },$$

$T^{ij}j$th component of flux of $i$th component of momentum stress tensor.

It is usual to require that $T^{\mu \nu }$ are components of a symmetric tensor, $T^{\mu \nu }=T^{\nu \mu }$ . The argument for this centres around the concept of angular 4-momentum, which for a continuous distribution of matter is defined to be

$$M^{\mu \nu }={\iiint }_S{x}^{\mu }\mathrm{d}{P}^{\nu }-x^{\nu }\mathrm{d}{P}^{\mu }\equiv -\frac{1}{c}{\iiint }_S(x^{\mu }{T}^{\nu \rho }-x^{\nu }{T}^{\mu \rho })\mathrm{d}{S}_{\rho }=-M^{\nu \mu }.$$

Conservation of angular 4-momentum $M^{\mu \nu }$ is equivalent to

$$0=(x^{\mu }{T}^{\nu \rho }-x^{\nu }{T}^{\mu \rho }{)}_{,\rho }={\delta }_{\rho }^{\mu }{T}^{\nu \rho }-{\delta }_{\rho }^{\nu }{T}^{\mu \rho }=T^{\nu \mu }-T^{\mu \nu }.$$

Example 9.4

Consider a fluid having 4-velocity $V^{\mu }=\gamma (\mathbf{v},c)$ where $\mathbf{v}=\mathbf{v}(\mathbf{r},t)$ . Let the local rest mass density (as measured in the i.r.f.) be $\rho (\mathbf{r},t)$ . In the i.r.f. at any point of the fluid the energy density is given by $\rho c^2$ , and since there is no energy flux in the i.r.f. we may set $T^{4i}=0$ . By the symmetry of $T^{\mu \nu }$ there will also be no momentum density $T^{i4}$ and the energy–stress tensor has the form

$$T^{\mu \nu }=\left(\begin{array}{cccc}& & & 0\\ & T_{ij}& & 0\\ & & & 0\\ 0& 0& 0& \rho c^2\end{array}\right)=\left(\begin{array}{cccc}{P}_1& & & 0\\ & P_2& & 0\\ & & P_3& 0\\ 0& 0& 0& \rho c^2\end{array}\right),$$

where the diagonalization of the $3\times 3$ matrix $[T_{ij}]$ can be achieved by a rotation of axes. The $P_i$ are called the principal pressures at that point. If they are all equal, $P_1=P_2=P_3=P$ then the fluid is said to be a perfect fluid and $P$ is simply called the pressure. In that case

$$T^{\mu \nu }=\left(\rho +\frac{1}{c^2}P\right)V^{\mu }{V}^{\nu }+P{g}^{\mu \nu },\text{\hspace{1em}(9.58)}$$

as may be checked by verifying that this equation holds in the i.r.f. at any point, in which frame $V^{\mu }=(0,0,0,c)$ . Since Eq. (9.58) is a 4-tensor equation it must hold in all inertia frames.

Exercise

Verify that the conservation laws $T^{\mu \nu }{}_{,\nu }=0$ reduce for $v\ll c$ to the equation of continuity and Euler’s equation

$$\begin{array}{c}\frac{\partial \rho }{\partial t}+\nabla .(\rho \mathbf{v})=0,\\ \rho \left(\frac{\partial \mathbf{v}}{\partial t}+(\mathbf{v}\cdot \nabla )\mathbf{v}\right)=-\nabla P.\end{array}$$

Example 9.5

The energy–stress tensor of the electromagnetic field is given by

$$T^{\mu \nu }=\frac{1}{4\pi }(F_{\rho }^{\mu }{F}^{\nu \rho }-\frac{1}{4}{g}^{\mu \nu }{F}_{\rho \sigma }{F}^{\rho \sigma })=T^{\nu \mu }.\text{\hspace{1em}(9.59)}$$

The energy density of the electromagnetic field is thus

$$ϵ=T^{44}=\frac{1}{16\pi }\left(4F_i^4{F}^{4i}-g^{44}{F}_{\rho \sigma }{F}^{\rho \sigma }\right)=\frac{1}{16\pi }\left(4{\mathbf{E}}^2+2({\mathbf{B}}^2-{\mathbf{E}}^2)\right)=\frac{{\mathbf{E}}^2+{\mathbf{B}}^2}{8\pi }$$

and the energy flux density has components

$$c{T}^{4i}=\frac{c}{4\pi }{F}_j^4{F}^{ij}=\frac{c}{4\pi }{E}_j{ϵ}_{ijk}{B}_k=\frac{c}{4\pi }(\mathbf{E}\times \mathbf{B}{)}_i.$$

The vector $\mathbf{S}=\frac{c}{4\pi }(\mathbf{E}\times \mathbf{B})$ is known as the Poynting vector. The spatial components $T_{ij}$

are known as the Maxwell stress tensor

$$\begin{array}{c}{T}^{ij}=T_{ij}=\frac{1}{16\pi }\left(4(F_{ik}{F}_j^k+F_{i4}{F}_j^4)-{\delta }_{ij}2({\mathbf{B}}^2-{\mathbf{E}}^2)\right)\\ =\frac{1}{4\pi }\left(-E_i{E}_j-B_i{B}_j+\frac{1}{2}{\delta }_{ij}({\mathbf{E}}^2+{\mathbf{B}}^2)\right).\end{array}$$

The total 4-momentum of an electromagnetic field over a spacelike surface $S$ is calculated from Eq. (9.57).

Exercise

Show that the average pressure $P=\frac{1}{3}\sum T_{ii}$ of an electromagnetic field is equal to $\frac{1}{3}\times$ energy density. Show that this also follows from the fact that $T^{\mu \nu }$ is trace-free, $T^{\mu }{}_{\mu }=0$

For further developments in relativistic classical field theory the reader is referred to [4, 5].

Problems

Problem 9.24

Show that as a consequence of the Maxwell equations,

$$T_{\alpha ,\beta }^{\beta }=-\frac{1}{c}{F}_{\alpha \gamma }{J}^{\gamma }$$

where $T_{\alpha }^{\beta }$ is the electromagnetic energy–stress tensor Eq. (9.59), and when no charges and currents are present it satisfies Eq. (9.56). Show that the $\alpha =4$ component of this equation has the form

$$\frac{\partial ϵ}{\partial t}+\nabla \cdot \mathbf{S}=-\mathbf{j}\cdot \mathbf{E}$$

where $\mathcal{E}$ energy density and $S$ Poynting vector. Interpret this equation physically.

Problem 9.25

For a plane wave, Problem 9.23, show that

$$T_{\alpha \beta }=ϵn_{\alpha }{n}_{\beta }$$

where $ϵ=E^2/4\pi$ and $n^{\alpha }=(\mathbf{n},1)$ is the null vector pointing in the direction of propagation of the wave. What pressure does the wave exert on a wall placed perpendicular to the path of the wave?