18.10 Variation principles in space-time

中文

We conclude this chapter with a description of the variation principle approach to field equations, including an appropriate variational derivation of Einstein’s field equations [13, 19]. Recall from Chapter 17 that for integration over an orientable four-dimensional space-time $(M,g)$ we need a non-vanishing 4-form $\Omega$. Let ${\epsilon }^1,{\epsilon }^2,{\epsilon }^3,{\epsilon }^4$ be any o.n. basis ofdifferential 1-forms, $g^{-1}({\epsilon }^{\mu },{\epsilon }^n)={\eta }^{\mu \nu }$ . The volume element $\Omega ={\epsilon }^1\wedge {\epsilon }^2\wedge {\epsilon }^3\wedge {\epsilon }^4$ defined by this basis is independent of the choice of orthonormal basis, provided they are related by a proper Lorentz transformation. Following the discussion leading to Eq. (8.32), we have that the components of this 4-form in an arbitrary coordinate $(U,\varphi ;x^{\mu })$ are

$${\Omega }_{\mu \nu \rho \sigma }=\frac{\sqrt{|g|}}{4!}{ϵ}_{\mu \nu \rho \sigma }\text{ where }g=\det [g_{\mu \nu }],$$

and we can write

$$\begin{array}{l}\Omega =\frac{\sqrt{|g|}}{4!}{ϵ}_{\mu \nu \rho \sigma }\mathrm{d}{x}^{\mu }\otimes \mathrm{d}{x}^{\nu }\otimes \mathrm{d}{x}^{\rho }\otimes \mathrm{d}{x}^{\sigma }\\ =\sqrt{-g}\mathrm{d}{x}^1\wedge \mathrm{d}{x}^2\wedge \mathrm{d}{x}^3\wedge \mathrm{d}{x}^4.\end{array}$$

Every 4-form $\Lambda$ can be written $\Lambda =f\Omega$ for some scalar function $f$ on $M$, and for every regular domain $D$ contained in the coordinate domain $U_i$

$${\int }_D\Lambda ={\int }_{\varphi (D)}f\sqrt{-g}{d}^{4}x\left(d^{4}x\equiv d{x}^{1}d{x}^{2}d{x}^{3}d{x}^4\right).$$

If $f\sqrt{-g}=A_{,\mu }^{\mu }$ for some set of functions $A^{\mu }$ on $M$, then

$$\Lambda =A_{,\mu }^{\mu }\mathrm{d}{x}^1\wedge \mathrm{d}{x}^2\wedge \mathrm{d}{x}^3\wedge \mathrm{d}{x}^4$$

and a simple argument, such as suggested in Problem 17.8, leads to

$$\Lambda =\mathrm{d}\alpha \text{ where }\alpha =\frac{1}{3!}{ϵ}_{\mu \nu \rho \sigma }{A}^{\mu }\mathrm{d}{x}^{\nu }\wedge \mathrm{d}{x}^{\rho }\wedge \mathrm{d}{x}^{\sigma }$$

and by Stokes’ theorem (17.3),

$${\int }_D\Lambda ={\int }_D\mathrm{d}\alpha ={\int }_{\partial D}\alpha .\text{\hspace{1em}(18.112)}$$

Let ${\Phi }_A(x)$ be any set of fields on a neighbourhood $V$ of regular domain $x\in D\subset M.$ where the index $A=1,\dots ,N$ refers to all possible components (scalar, vector, tensor, etc.) that may arise. By a variation of these fields is meant a one-parameter family of fields ${\tilde{\Phi }}_A(x,\lambda )$ on $D$ such that

1. ${\tilde{\Phi }}_A(x,0)={\Phi }_A(x)$ for all $x\in D$

2. ${\tilde{\Phi }}_A(x,\lambda )={\Phi }_A(x)$ for all $\lambda$ and al $x\in V-D$

The second condition implies that the condition holds on the boundary $\partial$ $D$ and also all derivatives of the variation field components agree there, ${\tilde{\Phi }}_{A,\mu }(x,\lambda )={\Phi }_{A,\mu }(x)$ for all $x\in \partial D$ . We define the variational derivatives

$$\delta {\Phi }_A=\frac{\partial }{\partial \lambda }{\tilde{\Phi }}_A(x,\lambda ){|}_{\lambda =0}.$$

This vanishes on the boundary $\partial D$ since ${\tilde{\Phi }}_A$ is independent of $\lambda$ there. A Lagrangian is a function $L({\Phi }_A,{\Phi }_{A,\mu })$ dependent on the fields and their derivatives. It defines a 4-form $\Lambda =L\Omega$ and an associated action

$$I={\int }_D\Lambda ={\int }_{D\cap U}L\sqrt{-g}{\mathrm{d}}^{4}x.$$

Field equations arise by requiring that the action be stationary,

$$\delta I=\frac{\mathrm{d}}{\mathrm{d}\lambda }{\int }_D\Lambda {|}_{\lambda =0}={\int }_{D\cap U}\delta (L\sqrt{-g}){\mathrm{d}}^{4}x,$$

called a field action principle that, as for path actions, can be evaluated by

$$\begin{array}{l}0={\int }_{D\cap U}\frac{\partial L\sqrt{-g}}{\partial {\Phi }_A}\delta {\Phi }_A+\frac{\partial L\sqrt{-g}}{\partial {\Phi }_{A,\mu }}\delta {\Phi }_{A,\mu }{\mathrm{d}}^{4}x\\ ={\int }_{D\cap U}\frac{\partial L\sqrt{-g}}{\partial {\Phi }_A}\delta {\Phi }_A+\frac{\partial }{\partial x^{\mu }}\left(\frac{\partial L\sqrt{-g}}{\partial {\Phi }_{A,\mu }}\delta {\Phi }_A\right)-\frac{\partial }{\partial x^{\mu }}\left(\frac{\partial L\sqrt{-g}}{\partial {\Phi }_{A,\mu }}\right)\delta {\Phi }_A{\mathrm{d}}^{4}x.\end{array}$$

Using the version of Stokes’ theorem given in Eq. (18.112), the middle term can be converted to an integral over the boundary

$${\int }_{\partial D}\frac{1}{3!}{ϵ}_{\mu \nu \rho \sigma }\frac{\partial L}{\partial {\Phi }_{A,\mu }}\sqrt{-g}\delta {\Phi }_A\mathrm{d}{x}^{\nu }\wedge \mathrm{d}{x}^{\rho }\wedge \mathrm{d}{x}^{\sigma },$$

which vanishes since $\delta {\Phi }_A=0$ on $\partial$ $D$. Since $\delta {\Phi }_A$ are arbitrary functions (subject to this boundary constraint) on $D$, we deduce the Euler–Lagrange field equations

$$\frac{\delta L\sqrt{-g}}{\delta {\Phi }_A}\equiv \frac{\partial L\sqrt{-g}}{\partial {\Phi }_A}-\frac{\partial }{\partial x^{\mu }}\left(\frac{\partial L\sqrt{-g}}{\partial {\Phi }_{A,\mu }}\right)=0.\text{\hspace{1em}(18.113)}$$

It is best to include the term $\sqrt{-g}$ within the derivatives since, as we shall see, the fields may depend specifically on the metric tensor components $g_{\mu \nu }$

Hilbert action

Einstein’s field equations may be derived from a variation principle, by setting the Lagrangian to be the Ricci scalar, $L=R$ . This is known as the Hilbert Lagrangian. For independent variables it is possible to take either the metric tensor components $g_{\mu \nu }$ or those of the inverse tensor $g^{\mu \nu }$ . We will adopt the latter, as it is slightly more convenient (the reader may try to adapt the analysis that follows for the variables ${\Phi }_A=g_{\mu \nu }$ . We cannot use the Euler–Lagrange equations Eq. (18.113) as they stand since $R$ is dependent on $g^{\mu \nu }$ and its first and second derivatives. While the Euler–Lagrange analysis can be extended to include Lagrangians that depend on second derivatives of the fields (see Problem 18.32), this would be a prohibitively complicated calculation. Proceeding directly, we have

$$\delta I_G=\delta {\int }_{D}R\Omega ={\int }_{D\cap U}\delta \left(R_{\mu \nu }{g}^{\mu \nu }\sqrt{-g}\right){\mathrm{d}}^{4}x,$$

whence

$$\delta I_G={\int }_{D\cap U}(\delta R_{\mu \nu }{g}^{\mu \nu }\sqrt{-g}+R_{\mu \nu }\delta g^{\mu \nu }\sqrt{-g}+R\delta \sqrt{-g}){\mathrm{d}}^{4}x.\text{\hspace{1em}(18.114)}$$

We pause at this stage to analyse the last term, $\delta \sqrt{-g}$ . Forgetting temporarily that $g_{\mu \nu }$ is a symmetric tensor, and assuming that all components are independent, we see that the determinant is a homogeneous function of degree in the components,

$$\delta g=\frac{\partial g}{\partial g_{\mu \nu }}\delta g_{\mu \nu }=G^{\mu \nu }\delta g_{\mu \nu }$$

where $G^{\mu \nu }$ is the cofactor of $g_{\mu \nu }$ . We may therefore write

$$\delta g=g{g}^{\nu \mu }\delta g_{\mu \nu }=g{g}^{\mu \nu }\delta g_{\mu \nu }\text{\hspace{1em}(18.115)}$$

since $g^{\mu \nu }=g^{\nu \mu }$ . The symmetry of $g_{\mu \nu }$ may be imposed at this stage, without in any way altering this result. From $g^{\mu \nu }{g}_{\mu \nu }={\delta }_{\mu }^{\mu }=4$ it follows at once that we can write Eq. (18.115) as

$$\delta g=-g{g}_{\mu \nu }\delta g^{\mu \nu }.\text{\hspace{1em}(18.116)}$$

Hence

$$\delta \left(\sqrt{-g}\right)=\frac{1}{2\sqrt{-g}}\delta (-g)=-\frac{\sqrt{-g}}{2}{g}_{\mu \nu }\delta g^{\mu \nu }.\text{\hspace{1em}(18.117)}$$

A similar analysis gives

$${\left(\sqrt{-g}\right)}_{,\rho }=-\frac{\sqrt{-g}}{2}{g}_{\mu \nu }{g}_{,\rho }^{\mu \nu }\text{\hspace{1em}(18.118)}$$

and from the formula Eq. (18.40) for Christoffel symbols,

$${\Gamma }_{\rho \mu }^{\mu }=\frac{1}{2}{g}^{\mu \nu }{g}_{\nu \mu ,\rho }=\frac{1}{\sqrt{-g}}(\sqrt{-g}{)}_{,\rho }.\text{\hspace{1em}(18.119)}$$

This identity is particularly useful in providing an equation for the covariant divergence of a vector field:

$$A_{;\mu }^{\mu }=A_{,\mu }^{\mu }+{\Gamma }_{\rho \mu }^{\mu }{A}^{\rho }=\frac{1}{\sqrt{-g}}(A^{\mu }\sqrt{-g}{)}_{,\mu }.\text{\hspace{1em}(18.120)}$$

We are now ready to continue with our evaluation of $\delta I_G$ . Using Eq. (18.117) we can write Eq. (18.114) as

$$\delta I_G={\int }_{D\cap U}\left(R_{\mu \nu }-\frac{1}{2}R\right)\delta g^{\mu \nu }\sqrt{-g}+\delta R_{\mu \nu }{g}^{\mu \nu }\sqrt{-g}{\mathrm{d}}^{4}x.\text{\hspace{1em}(18.121)}$$

To evaluate the last term, we write out the Ricci tensor components

$$R_{\mu \nu }={\Gamma }_{\mu \nu ,\rho }^{\rho }-{\Gamma }_{\mu \rho ,\nu }^{\rho }+{\Gamma }_{\mu \nu }^{\alpha }{\Gamma }_{\alpha \rho }^{\rho }-{\Gamma }_{\mu \rho }^{\alpha }{\Gamma }_{\alpha \nu }^{\rho }.$$

Since $\delta {\Gamma }_{\mu \nu }^{\rho }$ is the limit as $\lambda 0$ of a difference of two connections, it is a tensor field and we find

$$\delta R_{\mu \nu }={\left(\delta {\Gamma }_{\mu \nu }^{\rho }\right)}_{,\rho }-{\left(\delta {\Gamma }_{\mu \rho }^{\rho }\right)}_{,\nu },$$

as may be checked either directly or, more simply, in geodesic normal coordinates. Hence, using $g^{\mu \nu }{}_{;\rho }=0$

$$\delta R_{\mu \nu }{g}^{\mu \nu }=W_{;\rho }^{\rho }\text{ where }{W}^{\rho }=\delta {\Gamma }_{\mu \nu }^{\rho }{g}^{\mu \nu }-\delta {\Gamma }_{\mu \nu }^{\nu }{g}^{\mu \rho }$$

and from Eq. (18.120) we see that

$$\delta R_{\mu \nu }{g}^{\mu \nu }\sqrt{-g}={\left(W^{\rho }\sqrt{-g}\right)}_{,\rho }.$$

Since $W^{\rho }$ depends on $\delta g^{\mu \nu }$ and $\delta g^{\mu \nu }{}_{,\sigma }$ , it vanishes on the boundary $\partial D_{:}$ , and the last term in $Eq.$ Eq. (18.121) is zero by Stokes’ theorem. Since $\delta g^{\mu \nu }$ is assumed arbitrary on $D_{:}$ , the Hilbert action gives rise to Einstein’s vacuum field equations,

$$G_{\mu \nu }=0\Rightarrow R_{\mu \nu }=0.$$

Energy–stress tensor of fields

With other fields ${\Phi }_A$ present we take the total Lagrangian to be

$$L=\frac{1}{2\kappa }+L_F({\Phi }_A,{\Phi }_{A,\mu },g^{\mu \nu })\text{\hspace{1em}(18.122)}$$

and we have

$$\begin{array}{l}0=\delta I=\delta {\int }_{D\cap U}(\frac{1}{2\kappa }R+L_F)\sqrt{-g}{\mathrm{d}}^{4}x\\ ={\int }_{D\cap U}(\frac{1}{2\kappa }{G}_{\mu \nu }\delta g^{\mu \nu }\sqrt{-g}+\frac{\partial L_F\sqrt{-g}}{\partial g^{\mu \nu }}\delta g^{\mu \nu }+\frac{\delta L_F\sqrt{-g}}{\delta {\Phi }_A}\delta {\Phi }_A){\mathrm{d}}^{4}x.\end{array}$$

Variations with respect to field variables ${\Phi }_A$ give rise to the Euler–Lagrange field equations Eq. (18.113), while the coefficients of $\delta g^{\mu \nu }$ lead to the full Einstein’s field equations

$$G_{\mu \nu }=\kappa T_{\mu \nu }\text{ where }{T}_{\mu \nu }=-\frac{2}{\sqrt{-g}}\frac{\partial L_F\sqrt{-g}}{\partial g^{\mu \nu }}.\text{\hspace{1em}(18.123)}$$

Example 18.5

An interesting example of a variation principle is the Einstein–Maxwell theory, where the field variables are taken to be components of a covector field $A_{\mu }-$ essentially the electromagnetic 4-potential given in Eq. (9.47) – and the field Lagrangian is taken to be

$${\mathcal{L}}_F=-\frac{1}{16\pi }{F}_{\mu \nu }{F}^{\mu \nu }=\frac{1}{16\pi }{F}_{\mu \nu }{F}_{\rho \sigma }{g}^{\mu \rho }{g}^{\nu \sigma }\text{\hspace{1em}(18.124)}$$

where $F_{\mu \nu }=A_{\nu ,\mu }-A_{\mu ,\nu }$ . A straightforward way to compute the electromagnetic energy– stress tensor is to consider variations of $g^{\mu \nu }$

$$\begin{array}{rl}& T_{\mu \nu }\delta g^{\mu \nu }=-\frac{2}{\sqrt{-g}}\delta \left(L_F\sqrt{-g}\right)\\ & =-\frac{2}{\sqrt{-g}}\left[\frac{-1}{8\pi }{F}_{\mu \nu }{F}_{\rho \sigma }\delta g^{\mu \rho }{g}^{\nu \sigma }\sqrt{-g}+L_F\delta \left(\sqrt{-g}\right)\right]\\ & =\frac{1}{4\pi }(F_{\mu \alpha }{F}_{\nu }^{\alpha }-\frac{1}{4}{F}_{\alpha \beta }{F}^{\alpha \beta }{g}_{\mu \nu }),\end{array}\text{\hspace{1em}(18.125)}$$

on using Eq. (18.117). This expression agrees with that proposed in Example 9.5, Eq. (9.59).

Variation of the field variables $A_{\mu }$ gives

$$\begin{array}{l}\delta {\int }_D{L}_F\Omega =\frac{-1}{4\pi }{\int }_{D\cap U}\delta A_{\nu ,\mu }{F}^{\mu \nu }\sqrt{-g}{\mathrm{d}}^{4}x\\ =\frac{-1}{4\pi }{\int }_{D\cap U}\delta A_{\nu ;\mu }{F}^{\mu \nu }\sqrt{-g}{\mathrm{d}}^{4}x\\ =\frac{-1}{4\pi }{\int }_{D\cap U}{\left(\delta A_{\nu }{F}^{\mu \nu }\right)}_{;\mu }\sqrt{-g}-\delta A_{\nu }{F}_{;\mu }^{\mu \nu }\sqrt{-g}{\mathrm{d}}^{4}x\\ =\frac{-1}{4\pi }{\int }_{D\cap U}{\left(\delta A_{\nu }{F}^{\mu \nu }\sqrt{-g}\right)}_{,\mu }-\delta A_{\nu }{F}_{;\mu }^{\mu \nu }\sqrt{-g}{\mathrm{d}}^{4}x.\end{array}$$

As the first term in the integrand is an ordinary divergence its integral vanishes, and we arrive at the charge-free covariant Maxwell equations

$$F_{;\mu }^{\mu \nu }=0.\text{\hspace{1em}(18.126)}$$

The source-free equations follow automatically from $F_{\mu \nu }=A_{\nu ,\mu }-A_{\mu ,\nu };$

$$F_{\mu \nu ,\rho }+F_{\nu \rho ,\mu }+F_{\rho \mu ,\nu }=0⟺F_{\mu \nu ;\rho }+F_{\nu \rho ;\mu }+F_{\rho \mu ;\nu }=0.\text{\hspace{1em}(18.127)}$$

Problems

Problem 18.32

If a Lagrangian depends on second and higher order derivatives of the fields, $L=\mathcal{L}({\Phi }_A,{\Phi }_{A,\mu },{\Phi }_{A,\mu \nu },...)$ derive the generalized Euler–Lagrange equations

$$\frac{\delta L\sqrt{-g}}{\delta {\Phi }_A}\equiv \frac{\partial L\sqrt{-g}}{\partial {\Phi }_A}-\frac{\partial }{\partial x^{\mu }}\left(\frac{\partial L\sqrt{-g}}{\partial {\Phi }_{A,\mu }}\right)+\frac{{\partial }^2}{\partial x^{\mu }\partial x^{\nu }}\left(\frac{\partial L\sqrt{-g}}{\partial {\Phi }_{A,\mu \nu }}\right)-\cdots =0.$$

Problem 18.33

For a skew symmetric tensor $F^{\mu \nu }$ show that

$$F_{;\nu }^{\mu \nu }=\frac{1}{\sqrt{-g}}(\sqrt{-g}{F}^{\mu \nu }{)}_{,\nu }.$$

Problem 18.34

Compute the Euler–Lagrange equations and energy–stress tensor for a scalar field Lagrangian in general relativity given by

$$L_S=-{\psi }_{,\mu }{\psi }_{,\nu }{g}^{\mu \nu }-m^2{\psi }^2.$$

Verify $T^{\mu \nu }{}_{;\nu }=0.$

Problem 18.35

Prove the implication given in Eq. (18.127). Show that this equation and Eq. (18.126) imply $T^{\mu \nu }{}_{;\nu }=0$ for the electromagnetic energy–stress tensor given in Eqn. Eq. (18.125).