Equations of General Relativity

I have thought seriously about this question,
and have come to the conclusion that
what I have to say cannot reasonably be conveyed
without a certain amount of mathematical notation
and the exploration of genuine mathematical concepts.

Roger Penrose
The Road to Reality

Problem 1: proper time and distance in diagonal metrics

Consider a spacetime with diagonal metric \[ds^2=g_{00}(dx^{0})^2 + g_{11}(dx^{1})^2 +g_{22}(dx^{2})^{2}+g_{33}(dx^{3})^{2}.\] Find the explicit expressions for the intervals of proper time and spatial length, and for the 4-volume. Show that the invariant $4$-volume is given by \[\sqrt{-g}\;d^{4}x\equiv \sqrt{-g}\;dx^{0}dx^{1}dx^{2}dx^{3},\] where $g=\det(g_{\mu\nu})$.

solution

We assume that the coordinate $x^{0}$ is timelike, so $g_{00}>0$, and that $x^{1}$, $x^2$ and $x^3$ is spacelike, so $g_{11},g_{22},g_{33}<0$. The interval along the worldline of a particle at rest ($dx^1 = dx^2 = dx^3 = 0$) takes the form $ds^{2}=g_{00}(dx^{0})^2$. On the other hand, in the reference frame of the particle $ds^{2}$ is the proper time interval: $ds^{2}=c^{2}d\tau^{2}$, so $d\tau=\sqrt{g_{00}}\;dx^{0}/c$. The change of variables $\sqrt{-g_{ii}}dx^{i}=d\tilde{x}^{i}$ (there is no summation over indices here) transforms the metric to locally Lorentz form \[ds^{2}=c^{2}d\tau^{2}-(d\tilde{x}^{1})^{2} -(d\tilde{x}^{2})^{2}-(d\tilde{x}^{3})^{2},\] and it is easy to see that the element of three-dimensional length (line element) is given by \[dl^{2}=-\big[g_{11}(dx^{1})^2 +g_{22}(dx^{2})^{2}+g_{33}(dx^{3})^{2}\big].\] The element of $4$-volume is then \[cd\tau d\tilde{x}^{1}d\tilde{x}^{2}d\tilde{x}^{3} =\sqrt{-g_{00}g_{11}g_{22}g_{33}}\; dx^{0}dx^{1}dx^{2}dx^{3} =\sqrt{-g}\;dx^{0}dx^{1}dx^{2}dx^{3}.\]

Problem 2: observable invariants

Let there be an observer with $4$-velocity $u^{\mu}$. Show that the energy of a photon with $4$-wave vector $k^\mu$ that he registers is $u^{\mu}k_{\mu}$, and the energy of a massive particle with $4$-momentum $p^{\mu}$ is $u^{\mu}p_{\mu}$.

solution

Let us consider a locally flat reference frame, in which the metric tensor at a given point equals to $\eta_{\mu\nu}=diag(1,-1,-1,-1)$. If the observer in this frame has physical velocity $\mathbf{v}$, then we add a boost with the parameter $\mathbf{v}$, so in the resulting frame both the metric is locally flat and the observer is at rest, with 4-velocity equal to $u^{\mu}=(1,0,0,0)$. Let the photon in this frame have $4$-wave vector $k_{\mu}=(\omega,-\vec{k})$. Then the observer registers its frequency $\omega$ equal to \begin{equation}\label{equ_OmegaDet} \omega=u^{\mu}k_{\mu}. \end{equation} But being scalar, the expression (\ref{equ_OmegaDet}) is valid in any other reference frame. The same way one obtains that the registered momentum of a particle is $u^{\mu}p_{\mu}$.

Problem 3: covariant derivative

The covariant derivative (or connection) $\nabla_{\mu}$ is a tensorial generalization of partial derivative of a vector field $A^{\mu}(x)$ in the curved space-time. It's action on vectors is defined as \begin{equation}\label{nabla} \nabla_{\mu}A^{\nu}=\partial_{\mu}A^{\nu}+ {\Gamma^{\nu}}_{\lambda\mu}A^{\lambda};\qquad \nabla_{\mu}A_{\nu}=\partial_{\mu}A_{\nu}- {\Gamma^{\rho}}_{\nu\mu}A_{\rho}, \end{equation} where matrices ${\Gamma^{i}}_{jk}$ are called the connection coefficients, so that $\nabla_{\mu}A^{\nu}$ and $\nabla_{\mu}A_{\nu}$ are tensors. The connection used in GR is symmetric in lower indices (${\Gamma^{\lambda}}_{\mu\nu}= {\Gamma^{\lambda}}_{\nu\mu}$) and compatible with the metric $\nabla_{\lambda}g_{\mu\nu}=0$. It is called the Levi-Civita's connection, and the corresponding coefficients ${\Gamma^{\lambda}}_{\mu\nu}$ the Christoffel symbols. The action on tensors is defined through linearity and Leibniz rule. Express the Christoffel symbols through the metric tensor.

solution

The action of covariant derivative on a tensor of rank $(0,2)$ is defined as \[\nabla_{\lambda}T_{\mu\nu}= \partial_{\lambda}T_{\mu\nu} -{\Gamma^{\rho}}_{\mu\lambda}T_{\rho\nu} -{\Gamma^{\rho}}_{\lambda\nu}T_{\mu\rho}.\] We express the metric compatibility condition $\nabla_{\lambda}g_{\mu\nu}=0$ in terms of ${\Gamma^{\lambda}}_{\mu\nu}$ and make cyclic transpositions of indices to obtain \[\begin{array}{l} 0=\partial_{\lambda}g_{\mu\nu}- \Gamma^{\rho}_{\mu\lambda}g_{\rho\nu}- \Gamma^{\rho}_{\nu\lambda}g_{\mu\rho};\\ 0=\partial_{\mu}g_{\nu\lambda}- \Gamma^{\rho}_{\nu\mu}g_{\rho\lambda}- \Gamma^{\rho}_{\lambda\mu}g_{\nu\rho};\\ 0=\partial_{\nu}g_{\lambda\mu}- \Gamma^{\rho}_{\lambda\nu}g_{\rho\mu}- \Gamma^{\rho}_{\mu\nu}g_{\lambda\rho}; \end{array}\quad\Rightarrow\quad \begin{array}{l} \partial_{\lambda}g_{\mu\nu}= \Gamma^{\rho}_{\mu\lambda}g_{\rho\nu}+ \Gamma^{\rho}_{\nu\lambda}g_{\mu\rho};\\ \partial_{\mu}g_{\nu\lambda}= \Gamma^{\rho}_{\nu\mu}g_{\rho\lambda}+ \Gamma^{\rho}_{\lambda\mu}g_{\nu\rho};\\ \partial_{\nu}g_{\lambda\mu}= \Gamma^{\rho}_{\lambda\nu}g_{\rho\mu}+ \Gamma^{\rho}_{\mu\nu}g_{\lambda\rho}\;. \end{array}.\] Adding all the three equalities with signs $-++$, and taking into account the symmetry of ${\Gamma^{\lambda}}_{\mu\nu}$, we are led to \begin{equation}\label{GammaTorsion} \partial_{\mu}g_{\lambda\nu}+ \partial_{\nu}g_{\lambda\mu} -\partial_{\lambda}g_{\mu\nu} =2g_{\lambda\rho}\Gamma^{\rho}_{\mu\nu}, \end{equation} and contracting this with $g^{\sigma\lambda}$, we finally get \begin{equation}\label{GammaChristoffel} \Gamma^{\sigma}_{\mu\nu}= {\textstyle\frac{1}{2}}\; g^{\sigma\lambda} \Big(\partial_{\mu}g_{\lambda\nu}+ \partial_{\nu}g_{\lambda\mu} -\partial_{\lambda}g_{\mu\nu}\Big). \end{equation}

Problem 4: locally inertial frame

Derive the transformation rule for matrices ${\Gamma^{\lambda}}_{\mu\nu}$ under coordinate transformations. Show that for any given point of spacetime there is a coordinate frame, in which ${\Gamma^{\lambda}}_{\mu\nu}$ are equal to zero in this point. It is called a locally inertial, or locally geodesic frame.

solution

We use here the shortened notation: transfer the prime that denotes components of a tensorial quantity in the primed frame of reference to the indices ${A'}_{\nu}\equiv A_{\nu'}$, ${x'}^{m}\equiv x^{m'}$, and denote derivatives with a comma $\partial_{\mu}{T^{\ldots}}_{\ldots}={T^{\ldots}}_{\ldots\,,\mu}$ and semicolon $\nabla_{\mu}{T^{\ldots}}_{\ldots}={T^{\ldots}}_{\ldots\,;\mu}$. The transformation rule for ${\Gamma^{\mu}}_{\nu\lambda}$ is fixed by the condition that for arbitrary vector $A_\mu$ its covariant derivative $A_{\mu;\nu}$ is a tensor: \begin{align*} A_{i';k'}=& {x^m}_{,i'}{x^n}_{,k'}A_{m;n}= {x^m}_{,i'}{x^n}_{,k'} \left({A_{m}}_{,n}- {\Gamma^{p}}_{mn}A_{p}\right)=\\ &={x^m}_{,i'}{x^n}_{,k'}{A_{m}}_{,n}- {x^m}_{,i'}{x^n}_{,k'}\; {\Gamma^{p}}_{mn}A_{p}\;;\\ A_{i';k'}=&A_{i',k'}-{\Gamma^{p'}}_{i'k'}A_{p'}= \left({x^{j}}_{,i'}A_{j}\right)_{,k'}- {\Gamma^{p'}}_{i'k'}\;{x^{q}}_{,p'}A_{q}=\\ &={x^{j}}_{,i',k'}A_{j}+{x^{j}}_{,i'}{x^{n}}_{,k'}A_{j\,,n}- {\Gamma^{p'}}_{i'k'}\;{x^{q}}_{,p'}A_{q}=\\ &={x^{p}}_{,i',k'}A_{p}+ {x^{m}}_{,i'}{{x^{n}}_{,k'}}A_{m\,,n}- {x^{p}}_{,q'}{\Gamma^{q'}}_{i'k'}\;A_{p}. \end{align*} Equating the right hand sides and taking into account that $A^\mu$ is arbitrary, we have \[{x^p}_{,q'}{\Gamma^{q'}}_{i'k'}= {x^{m}}_{,i'}{x^{n}}_{,k'}\;{\Gamma^{p}}_{mn}A_{p} +{x^{p}}_{,i',k'}.\] Contracting with ${x^{j'}}_{,p}$ and taking into account that ${x^{j'}}_{,p}{x^{p}}_{,q'}=\delta^{j'}_{q'}$, we finally get \begin{equation}\label{GammaTranform} {\Gamma^{j'}}_{i'k'}={x^{p}}_{,i',k'}\,{x^{j'}}_{,p}+ {x^{j'}}_{,p}{x^{m}}_{,i'}{x^{n}}_{,k'} \;{\Gamma^{p}}_{mn}, \end{equation} or in expanded notation \begin{equation}\label{GammaTransformFull} {{\Gamma'}^{\,j}}_{ik}= \frac{\partial^{2}x^{p}} {\partial {x}^{\,i'}\partial {x}^{\,k'}} \frac{\partial x'^j}{\partial x^p} +\frac{\partial {x}^{\,j'}}{\partial x^{p}} \frac{\partial x^{m}}{\partial {x}^{\,i'}} \frac{\partial {x}^{\,n}}{\partial {x}^{\,k'}} {\Gamma^{p}}_{mn}. \end{equation} Let the value of $\Gamma^{i}_{kl}$ at a given point $X$ be $\tilde{\Gamma}^{i}_{kl}=\tilde{\Gamma}^{i}_{lk}$. Then we can move the coordinate origin to $X$ and make the following coordinate transformation: \begin{align}\label{LocalGeodesic} x^{i'}&=x^{i}-{\textstyle\frac{1}{2}} \tilde{\Gamma}^{i}_{kl}x^{k}x^{l};\\ {x^{i'}}_{,n}&=\delta^{i}_{n}- \tilde{\Gamma}^{i}_{kn}x^{k};\nonumber\\ {x^{i'}}_{,n,m}&=-\tilde{\Gamma}^{i}_{mn}.\nonumber \end{align} At the point $x=x'=0$ then ${x^{i'}}_{,j}=\delta^{i}_{j}$, and therefore ${x^{i}}_{,j'}=\delta^{i}_{j}$, so the considered transformation preserves tensor components at the origin. Substituting the derivatives into the transformation law for $\Gamma$ (\ref{GammaTransformFull}), we obtain the connection coefficients in $X$ in the new reference frame: \[\tilde{\Gamma}^{i'}_{k'l'}= \delta^{m}_{k'}\delta^{n}_{l'}\delta^{i'}_{p} \tilde{\Gamma}^{p}_{mn}- \delta^{i'}_{p}\tilde{\Gamma}^{p}_{k'l'}= \tilde{\Gamma}^{i'}_{k'l'} -\tilde{\Gamma}^{i'}_{k'l'}= 0.\] Thus for any given point $X$ we can always bring the metric in the point to the diagonal form and then carry out the above coordinate transformation, turning to zero the connection coefficients. This frame, in which the metric tensor differs from $\eta_{\mu\nu}=diag(1,-1,-1,-1)$ only in second derivatives, is called the locally inertial frame, or locally geodesic frame. Its existence is very useful for calculations, as will be seen in the problem on Bianchi identity

Problem 5: geodesics

Free falling particles' worldlines in General Relativity are geodesics of the spacetime, i.e the curves $x^{\mu}(\lambda)$ with tangent vector $u^{\mu}=dx^{\mu}/d\lambda$, such that covariant derivative of the latter along the curve equals to zero: \[u^{\mu}\nabla_{\mu}u^{\nu}=0.\]

In a (pseudo-)Euclidean space the geodesics are straight lines. Obtain the general equation of geodesics in terms of the connection coefficients. Show that the quantity $u^{\mu}u_{\mu}$ is conserved along the geodesic.

solution

Expressing the derivative $\nabla$ through $\Gamma^{\lambda}_{\mu\nu}$, and taking into account that \[u^{\mu}\frac{\partial u^{\nu}}{\partial x^{\mu}} =\frac{dx^{\mu}}{d\lambda} \frac{\partial u^{\nu}}{\partial x^{\mu}} =\frac{du^{\mu}}{d\lambda},\] we obtain the geodesic equation in the form \begin{equation}\label{DefGeodesics2} \frac{du^{\mu}}{d\lambda}+ \Gamma^{\mu}_{\rho\sigma}u^{\rho}u^{\sigma}=0 \quad\Leftrightarrow\quad\boxed{ \frac{d^{2}x^{\mu}}{d\lambda^2}+ \Gamma^{\mu}_{\rho\sigma} \frac{dx^{\rho}}{d\lambda} \frac{dx^{\sigma}}{d\lambda}=0}. \end{equation} The rate of change of the scalar $u^{\mu}u_{\mu}$ along the geodesic is \[\frac{d u^{\mu}u_{\mu}}{d\lambda} =u^{\nu}\partial_{\nu}(u^{\mu}u_{\mu}) =u^{\nu}\nabla_{\nu}(u^{\mu}u_{\mu}) =2u^{\nu}u^{\mu}\nabla_{\nu}u_{\mu} =0.\] In the second equality sign we used the fact that the covariant derivative of a scalar coincides with its partial derivative. Thus the type of the tangent vector---whether it is timelike, spacelike or null (lightlike)---is conserved along a geodesic, and the latter can be grouped into the three corresponding types. Massive particles move along timelike geodesics, while the massless, in particular photons, move along the null geodesics.

Problem 6: principle of least action

Consider the action for a massive particle of the form \[S_{AB}=-mc\int_{A}^{B} ds,\quad\text{where}\quad ds=\sqrt{g_{\mu\nu}dx^{\mu}dx^{\nu}}\] and derive the geodesic equation from the principle of least action. Find the canonical $4$-momentum of a massive particle and the energy of a photon.

solution

We consider here only timelike curves, for which $dx^{\mu}dx_{\mu}>0$ at every point, so $ds^2$ never turns to zero along the curve. Then variation of $\int_{1}^{2}ds$ and integration of the term $\sim \delta dx^{\nu}=d\delta x^{\nu}$ by parts results in the following \begin{align*} 0&=\delta\int ds= \delta\int\sqrt{g_{\mu\nu}dx^{\mu}dx^{\nu}}=\\ &=\int\!\!{\textstyle\frac{1}{2ds}}\left[ \delta x^{\lambda}\partial_{\lambda}g_{\mu\nu} dx^{\mu}dx^{\nu}+ 2g_{\mu\nu}dx^{\mu}\delta dx^{\lambda}\right]=\\ &=\tfrac{1}{2}\int dx^{\mu}\delta x^{\nu} u^{\lambda}\partial_{\nu}g_{\mu\lambda}+ g_{\mu\nu}u^{\mu}\delta x^{\nu}\Big|_{1}^{2}- \int\delta x^{\nu} d\big(g_{\lambda\nu}u^{\lambda}\big)=\\ &=u_{\mu}\delta x^{\mu}\Big|_{1}^{2} +\int\!\! dx^{\mu}\delta x^{\nu}\left\{ {\textstyle\frac{1}{2}} u^{\lambda}\partial_{\nu}g_{\mu\lambda} -\partial_{\mu} (g_{\lambda\nu}u^{\lambda})\right\}=\\ &=u_{\mu}\delta x^{\mu}\Big|_{1}^{2} +\int\!\! dx^{\mu}\delta x^{\nu}\left\{ -g_{\lambda\nu}\partial_{\mu}u^{\lambda}+ u^{\lambda} [{\textstyle\frac{1}{2}} \partial_{\nu}g_{\mu\lambda}- \partial_{\mu}g_{\lambda\nu}]\right\}. \end{align*} Variation of the action with the ends of the interval fixed gives us the equation of motion: equating the integrand to zero, on contracting with $g^{\nu\sigma}$ we get \[\partial_{\mu}u^{\lambda}= {\textstyle\frac{1}{2}}u^{\lambda}g^{\nu\sigma} [\partial_{\nu}g_{\mu\lambda}- 2\partial_{\mu}g_{\lambda\nu}].\] Multiplying by $u^{\mu}=dx^{\mu}/ds\;$ and symmetrizing the bracket over $\mu\lambda$, we eventually obtain the geodesic equation \[\frac{du^{\sigma}}{ds}+u^{\mu}u^{\lambda} \Gamma^{\sigma}_{\mu\lambda}=0,\] where $\Gamma^{\sigma}_{\mu\lambda}$ are the Christoffel's symbols of Levi-Civita connection (\ref{GammaChristoffel}). Thus the time-like geodesics, that represent worldlines of massive particles, actually realize the extremum of the spacetime interval between the two fixed points. For real motion of a particle the variation of action with respect to variation of the end points of the curve is \[\delta S=-mcu_{\mu}\delta x^{\mu}\Big|_{1}^{2},\] so the canonical $4$-momentum of a particle is \[p_{\mu}=-\frac{\delta S}{\delta x^{\mu}} =mcu_{\mu}.\] Then in case the spacetime has the symmetry with respect to translations along $x^{\mu}$, the $\mu$-component of momentum is conserved. If it is invariant under time translations, then the conserved quantity is energy $p_{0}=mcu_{0}$. Likewise for massless particles, e.g. photons, the $4$-momentum and wave vector are proportional to the $4$-velocity $u^{\mu}=dx^{\mu}/d\lambda$. However, in this case the normalization of $u^{\mu}$ cannot be fixed since $u^{\mu}u_{\mu}=0$, so it is arbitrary. Therefore the quantity $dt/d\lambda$ along a null geodesics coincides with the photon's energy up to a constant factor.

Problem 7: symmetries and Killing vectors

A Killing vector field, or just Killing vector, is a vector field $K^{\mu}(x)$, such that infinitesimal coordinate transformation $x\to x'+\varepsilon K $ (where $\varepsilon\to 0$) leaves the metric invariant in the sense* \[g_{\mu\nu}(x)=g'_{\mu\nu}(x).\] A Killing vector defines a one-parametric symmetry group of the metric tensor, called isometry.

Show that a Killing vector obeys the equation \[\nabla_{\mu}K_{\nu}+\nabla_{\nu}K_{\mu}=0,\] called the Killing equation.

*That is, let $g_{\mu\nu}(x)$ be the components of the metric at some point $A$ in the original frame. Then $g'_{\mu\nu}(x)$ are the components of the metric in the new frame, taken at point $A'$, which has the same coordinates in the new frame as $A$ had in the old frame.

solution

Let us rewrite the Killing equation in terms of $K^\mu$ and derivatives of the metric $g_{\mu\nu}$ \begin{align*} 0&=\nabla_{\mu}K_{\nu}+\nabla_{\nu}K_{\mu} =\partial_{\mu}K_{\nu}+\partial_{\nu}K_{\mu} -\Gamma^{\lambda}_{\mu\nu}K_{\lambda} -\Gamma^{\lambda}_{\mu\nu}K_{\lambda}=\\ &=\partial_{\mu}(g_{\nu\lambda}K^\lambda) +\partial_{\nu}(g_{\mu\lambda}K^\lambda) -2\Gamma_{\lambda,\mu\nu}K^\lambda =g_{\nu\lambda}\partial_{\mu}K^{\lambda} +g_{\mu\lambda}\partial_{\nu}K^{\lambda} +K^\lambda \partial_{\lambda}g_{\mu\nu}, \end{align*} thus \begin{equation}\label{KillingEqCoords} g_{\nu\lambda}\partial_{\mu}K^{\lambda} +g_{\mu\lambda}\partial_{\nu}K^{\lambda} +K^\lambda \partial_{\lambda}g_{\mu\nu}=0. \end{equation} Using the small parameter $\varepsilon$ and the tensor tranformation law for $g_{\mu\nu}$, we find that \begin{align*} g'_{\mu\nu}(x')=&g_{\rho\sigma}(x) \frac{\partial x^\rho}{\partial x'^\mu} \frac{\partial x^\sigma}{\partial x'^\nu}= g_{\rho\sigma}(x)(\delta^{\rho}_{\mu}- \varepsilon\partial_{\mu}K^\rho) (\delta^{\sigma}_{\nu}- \varepsilon\partial_{\nu}K^\sigma)\approx\\ &\approx g_{\mu\nu}(x)-\varepsilon\big( g_{\lambda\nu}\partial_{\mu}K^{\lambda} +g_{\mu\lambda}\partial_{\nu}K^{\lambda}\big),\\ g'_{\mu\nu}(x')\approx&g'_{\mu\nu}(x) +\partial_{\lambda}g_{\mu\nu}\cdot \varepsilon K^\lambda. \end{align*} Taking the difference, and taking into account the given condition $g'_{\mu\nu}(x)=g_{\mu\nu}(x)$, we obtain exactly the Killing equation in terms of partial derivatives, written above.

Problem 8: Killing vectors and selected frames of reference

Suppose there is a coordinate frame, in which the metric does not depend on one of the coordinates $x^{k}$. Show that in this case the vectors $\partial_{k}$ constitute the Killing vector field, and that the inverse is also true: if there is a Killing vector, we can construct such a coordinate frame.

solution

Let us prove the first statement. The vector $\partial_{k}$ has coordinates $\delta_{k}^{\mu}=const$, so using the left hand part of the Killing equation in terms of partial derivatives (\ref{KillingEqCoords}), we obtain \[\nabla_{\mu}K_{\nu}+\nabla_{\nu}K_{\mu} =g_{\nu\lambda}\partial_{\mu}K^{\lambda} +g_{\mu\lambda}\partial_{\nu}K^{\lambda} +K^{\lambda}\partial_{\lambda}g_{\mu\nu} =\partial_{k}g_{\mu\nu}=0.\] Thus $K^{\mu}=\delta_{k}^{\mu}$ is a Killing vector. The second part is more geometrical. Suppose we have the Killing vector field $K^{\mu}$. As it determines an isometry, its integral curves, $x^{\mu}(\lambda)$ with $dx^{\mu}/d\lambda=K^{\mu}$, pass through every point of spacetime. We take an arbitrary 3-dimensional surface $\Sigma$, parametrized by $x^{\alpha}$, $\alpha=1,2,3$, and reconstruct the integral curve of $K^{\mu}$ through each point. Then each point $X$ in the neighborhood of $\Sigma$ lies on one of these curves and is thus determined by the four numbers -- $x^{0}$, which give the length from $\Sigma$ to $X$ along this curve (we assume $K^\mu$ is not null), and $x^{\alpha}$, which parametrize the set of the integral curves. We adopt these numbers as the coordinate frame. In this frame $K^{\mu}=\delta^{\mu}_{0}$ and from the Killing equation (\ref{KillingEqCoords}) we get what we need \[\partial_{0}g_{\mu\nu}=0.\]

Problem 9: Killing vectors and integrals of motions

Prove that if $K^{\mu}$ is a Killing vector, the quantity $K^{\mu}u_{\mu}$ is conserved along a geodesic with tangent vector $u^{\mu}$.

solution

From the Killing equation $\nabla_{\mu}K_{\nu}+\nabla_{\nu}K_{\mu}=0$ and geodesic equation $u^{\mu}\nabla_{\mu}u^{\nu}=0$ we get \[u^{\mu}\nabla_{\mu}(u^{\nu}K_{\nu})= u^{\mu}u^{\nu}\nabla_{\mu}K_{\nu}= \frac{1}{2}u^{\mu}u^{\nu} (\nabla_{\mu}K_{\nu}-\nabla_{\nu}K_{\mu})=0,\] due to symmetry of $u^{\mu}u^{\nu}$ and antisymmetry of $\nabla_{\mu}K_{\nu}$.

Problem 10: Riemann tensor

The Riemann curvature tensor ${R^{i}}_{klm}$ can be defined through the so-called Ricci identity, written for arbitrary $4$-vector $A^i$: \[\nabla_{m}\nabla_{l}A^{i}-\nabla_{l}\nabla_{m}A^{i} ={R^{i}}_{klm}A^{k}.\] Express ${R^{i}}_{klm}$ in terms of the Christoffel symbols. Show that the Ricci tensor \[R_{km}={R^{l}}_{klm}\] is symmetric.

solution

Straightforward calculation results in the following: \begin{equation}\label{RiemannTensor} {R^{i}}_{klm}=\partial_{l}{\Gamma^{i}}_{km} -\partial_{m}{\Gamma^{i}}_{kl} +{\Gamma^{n}}_{km}{\Gamma^{i}}_{nl} -{\Gamma^{n}}_{kl}{\Gamma^{i}}_{nm}. \end{equation} Contracting on indices $i$ and $l$ gives the Ricci tensor in the form \begin{equation}\label{RicciTensor} R_{km}\equiv {R^{i}}_{kim}=\partial_{i}{\Gamma^{i}}_{km} -\partial_{m}{\Gamma^{i}}_{ki} +{\Gamma^{n}}_{km}{\Gamma^{i}}_{ni} -{\Gamma^{n}}_{ki}{\Gamma^{i}}_{nm}, \end{equation} where all terms are explicitly symmetric except the second one $\partial_{m}{\Gamma^{i}}_{ik}$. Using the explicit expression for ${\Gamma^i}_{kl}$ through the metric tensor and taking into account symmetry of $g^{ij}$, we can bring it to the form \[{\Gamma^{i}}_{ik}=\tfrac{1}{2}g^{ij} \big[\partial_{i}g_{jk}+\partial_{k}g_{ij} -\partial_{j}g_{ik}\big] =\tfrac{1}{2}g^{ij}\partial_{k}g_{ij}.\] Note now that $g^{ij}$ is by definition the inverse matrix to $g_{ij}$, and its components can be presented as $g^{ij}=G^{ij}/g$, where $G^{ij}$ is the algebraic complement to $g_{ij}$, and $g=det(g_{ij})$. Therefore when calculating the differential of $g=det(g_{ij})$, we have \begin{equation}\label{detG} dg=G^{ij}dg_{ij}=gg^{ij}dg_{ij}. \end{equation} Then \[{\Gamma^{i}}_{ik}=\frac{\partial_{k}g}{2g} =\tfrac{1}{2}\partial_{k}\ln |g|,\quad \Rightarrow\quad \partial_{m}{\Gamma^{i}}_{ik} =\tfrac{1}{2}\partial_{m}\partial_{k}\ln |g| =\partial_{k}{\Gamma^{i}}_{im},\] and thus Ricci tensor is symmetric.

Problem 11: Bianchi identity

Prove the differential Bianchi identity for the curvature tensor: \begin{equation}\label{BianchiId} \nabla_{i}{R^{j}}_{klm}+\nabla_{l}{R^{j}}_{kmi} +\nabla_{m}{R^{j}}_{kil}=0, \end{equation} and show that \[\nabla_{i}R^{i}_{j}=\tfrac{1}{2}\partial_{j}R.\]

solution

Let us use the locally inertial frame, in which ${\Gamma^{i}}_{jk}=0$ and $\nabla_{i}=\partial_{i}$. Then the terms quadratic in $\Gamma$ after differentiation turn to zero and \[\nabla_{i}{R^{j}}_{klm}=\nabla_{i}\big( \partial_{l}{\Gamma^{j}}_{km} -\partial_{m}{\Gamma^{j}}_{kl}\big) =\partial_{i}\partial_{l}{\Gamma^{j}}_{km} -\partial_{i}\partial_{m}{\Gamma^{j}}_{kl}.\] Taking into account the symmetry of ${\Gamma^{i}}_{jk}$ and adding the three terms of this kind, we obtain zero in the left hand side of the Bianchi identity (\ref{BianchiId}). As this equality is tensorial, it is valid in any other reference frame. Contracting on $j$ and $l$ and taking into account the anti-symmetry of the curvature tensor over the last two indices, we get \[\nabla_{i}R_{km}-\nabla_{m}R_{ki} +\nabla_{j}R^{j}_{kmi}=0.\] Contracting one more time on $k$ and $m$, we are led to \[0=\nabla_{i}R-\nabla_{m}R^{m}_{i} +\nabla_{j}{R^{jk}}_{ki} =\nabla_{i}R-\nabla_{m}R^{m}_{i} -\nabla^{j}{R^{k}}_{jki} =\nabla_{i}R-\nabla_{m}R^{m}_{i} +\nabla_{j}R^{j}_{i}= \nabla_{i}R-2\nabla_{j}R^{j}_{i},\] and thus finally \[\nabla_{i}R^{i}_{j}=\tfrac{1}{2}\partial_{j}R.\]

Problem 12: energy-momentum tensor

The energy-momentum tensor in General Relativity is defined through the variational derivative of the action for matter \[S_{m}[g^{\mu\nu},\psi] =\frac{1}{c}\int d^{4}x\sqrt{-g} \;L_{m}(g^{\mu\nu},\psi)\] with respect to metric $g^{\mu\nu}$: \[\delta_{g}S=\tfrac{1}{2}\int d^{4}x\sqrt{-g}\; \delta g^{\mu\nu}T_{\mu\nu}.\] Here $L_{m}$ is the Lagrange function for the matter fields $\psi$. Show that for the cases of a massless scalar field and electromagnetic field the above definition reduces to the usual one.

solution

The Lagrange function for a massless scalar field in a potential $V$ is \[L_{\phi}= \tfrac{1}{2}g^{\mu\nu} \partial_{\mu}\phi\;\partial_{\nu}\phi-V.\] Here $\partial_{\mu}\phi\;\partial_{\nu}\phi\equiv (\partial_{\mu}\phi)(\partial_{\nu}\phi)$. Varying the action with respect to $g^{\mu\nu}$, and using (\ref{detG}), one gets (here $c=1$) \begin{align*} \delta_{g}S&=\tfrac{1}{c} \int d^{4}x \delta(\sqrt{-g}\;L) =\tfrac{1}{c}\int d^{4}x\Big(\tfrac{1}{2}\sqrt{-g}\; \delta g^{\mu\nu} \partial_{\mu}\phi\;\partial_{\nu}\phi -\tfrac{1}{2}\sqrt{-g}\;g_{\mu\nu}\delta g^{\mu\nu} L\Big)=\\ &=\tfrac{1}{2c}\int d^{4}x \sqrt{-g}\; \delta g^{\mu\nu} \Big\{\partial_{\mu}\phi\;\partial_{\nu}\phi -Lg_{\mu\nu}\Big\}, \end{align*} so \[T_{\mu\nu}=\partial_{\mu}\phi\;\partial_{\nu}\phi -Lg_{\mu\nu} =\tfrac{1}{2}g^{\mu\nu} \partial_{\mu}\phi\;\partial_{\nu}\phi +Vg_{\mu\nu},\] which coincides with the classical expression. Similarly for the electromagnetic field with \[L_{em}=-\tfrac{1}{16\pi}F^{\mu\nu}F_{\mu\nu} =\tfrac{1}{16\pi}g^{il}g^{km}F_{ik}F_{lm}, \quad\text{where}\quad F_{ik}=\partial_{i}A_{k}-\partial_{k}A_{i},\] after variation with respect to $g^{\mu\nu}$ one obtains the usual expression \[T_{ik}^{(em)} =-\frac{1}{4\pi}\Big(g^{lm}F_{il}F_{km} -\tfrac{1}{4}g_{ik}F^{lm}F_{lm}\Big).\]

Problem 13: Einstein equations

The full action consists of the action for matter, discussed in the previous problem, and the action for the gravitational field $S_{g}$: \[S=S_{g}+S_{m};\qquad\text{where}\quad S_{g}=-\frac{c^{3}}{16\pi G}\int d^{4}x\sqrt{-g}\;R,\] $R=R^{i}_{i}=g^{ik}R_{ik}$ is scalar curvature, $G$ is the gravitational constant. Starting from the variational principle, derive the Einstein-Hilbert equations$^*$ for the gravitational field \begin{equation}\label{EinsteinHilbetEq} R_{\mu\nu}-\tfrac{1}{2}Rg_{\mu\nu} =\frac{8\pi G}{c^4}T_{\mu\nu}. \end{equation}

solution

We vary the action with respect to the metric tensor $g^{\mu\nu}$. As the scalar curvature is the contraction of the Ricci tensor $R=g^{km}R_{km}$, this variation can be presented in the form \[\delta S_{g}= -\frac{c^3}{16\pi G}\int d^{4}x \big\{\delta (\sqrt{-g}g^{\mu\nu})R_{\mu\nu} +\sqrt{-g}g^{\mu\nu}\delta R_{\mu\nu}\big\}.\] Using relation (\ref{detG}), we get \[\delta (\sqrt{-g}g^{\mu\nu}) =\sqrt{-g}\delta g^{\mu\nu} -\tfrac{1}{2}\sqrt{-g} g_{\rho\sigma}\delta g^{\rho\sigma}g^{\mu\nu},\] and therefore \[\delta (\sqrt{-g}g^{\mu\nu})R_{\mu\nu} =\sqrt{-g}\delta g^{\mu\nu} \big(R_{\mu\nu}-\tfrac{1}{2}Rg_{\mu\nu}\big).\] In order to obtain $\delta R_{\mu \nu }$ we make use of the locally inertial reference frame, in which ${\Gamma^{\lambda}}_{\mu\nu}$ and the first derivatives of the metric tensor turn to zero at a given point. Then from (\ref{RicciTensor}) we obtain \begin{align*} g^{\mu \nu }\delta R_{\mu \nu } &= g^{\mu \nu }\left( \partial_\lambda \delta\Gamma_{\mu \nu}^\lambda - \partial_\mu \delta\Gamma_{\lambda\nu}^\lambda \right) = \partial_{\lambda} (g^{\mu\nu}\delta \Gamma_{\mu\nu}^\lambda) - \partial_{\mu} (g^{\mu \nu}\delta\Gamma_{\lambda\nu}^\lambda=\\ &=\partial_{\lambda} (g^{\mu\nu}\delta\Gamma_{\mu \nu }^\lambda) - \partial_{\lambda} (g^{\lambda\nu}\delta\Gamma_{\mu\nu}^\mu = \partial_{\lambda}( g^{\mu\nu}\delta\Gamma_{\mu \nu }^\lambda -g^{\lambda\nu}\delta\Gamma_{\mu\nu}^\mu) =\partial_{\lambda}\delta w^{\lambda}, \end{align*} where we introduced notation \[\delta w^\lambda = g^{\mu\nu}\delta\Gamma_{\mu \nu }^\lambda -g^{\lambda\nu}\delta\Gamma_{\mu\nu}^\mu.\] Then \[\int d^{4}x \sqrt{-g}\;g^{\mu\nu}\delta R_{\mu\nu} =\int d^{4}x\; \partial_{\lambda} \left(\sqrt {- g}\delta w^\lambda \right).\] According to the divergence (Gauss) theorem this integral can be transformed into one of $\delta w^\lambda$ over the hypersurface encasing the whole $4$-volume. As the variation is zero at the integration limits, this surface term vanishes. Thus the variation of the Einstein-Hilbert action takes the form: \[\delta S_g = - \frac{c^3}{16\pi G}\int d^{4}x\sqrt{-g}\; \delta g^{\mu\nu} \left( R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}\right).\] Variation of the action for matter, by the definition of the energy momentum tensor is \[\delta S_m = \frac{1}{2c}\int d^{4}x\sqrt{-g} \delta g^{\mu\nu} T_{\mu \nu }.\] Thus from the principle of least action we obtain: \[- \frac{c^3}{16\pi G}\int d^{4}x\sqrt{-g}\; \delta g^{\mu\nu} \left(R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu} -\frac{8\pi G}{c^4} T_{\mu \nu } \right)= 0.\] Due to arbitrariness of $\delta g^{\mu \nu }$ one finally obtains the Einstein equations \begin{equation}\label{EinsteinEq} R_{\mu \nu } - \frac{1}{2}Rg_{\mu \nu } = \frac{8\pi G}{c^4} T_{\mu \nu }. \end{equation} $^*$ These equations are called either Einstein-Hilbert equations, or Einstein field equations, or just Einstein equation (either in singular or plural); the action $S_{g}$ is referred to as Hilbert or Einstein-Hilbert action.

Problem 14: energy conservation

Show that the Einstein's equation can be presented in the following form \[R_{\mu\nu} = \frac{8\pi G}{c^4} \left( T_{\mu\nu}-\tfrac{1}{2}T g_{\mu\nu}\right), \quad\text{where}\quad T=T^{\mu}_{\mu}.\] Show that it leads to the energy-momentum conservation law for matter \[\nabla_{\mu}T^{\mu\nu}=0.\] Does it mean the energy and momentum of matter are actually conserved in general?

solution

On multiplying both sides of the Einstein's equations (\ref{EinsteinEq}) by $g^{\mu\nu}$, we have \[g^{\mu\nu}R_{\mu\nu} -\frac{1}{2}g^{\mu\nu}g_{\mu\nu}R = \frac{8\pi G}{c^4}g^{\mu\nu}T_{\mu\nu}.\] Taking into account that $g^{\mu\nu}g_{\mu\nu}=4$, $R = g^{\mu\nu}R_{\mu\nu}$ and $T = g^{\mu\nu}T_{\mu\nu}$, then \[-R = \frac{8\pi G}{c^4} T,\] and therefore \[R_{\mu\nu} =\frac{8\pi G}{c^4} \left(T_{\mu\nu}- \frac 1 2 g_{\mu\nu}T\right).\] On the other hand, acting with $\nabla^{\mu}$ on the equations (\ref{EinsteinEq}), one gets zero in the left hand side due to the Bianchi identity, and therefore the equation for $T_{\mu\nu}$: \[\nabla^{\mu}T_{\mu\nu}=0.\] In the flat space-time with Cartesian coordinates we have $\nabla_{\mu}=\partial_{\mu}$, so the equation reduces to differential conservation laws for the energy and momentum of matter \[\dot{T}^{0i}+\partial_{\alpha}T^{\alpha i}=0, \quad i=0,1,2,3,\quad \alpha=1,2,3.\] In the non-flat case there are additional terms $\sim\Gamma$, so energy and momentum are not generally conserved. This should not be surprising in the presence of the gravitational field.