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 \[cd\tau d\tilde{x}^{1}d\tilde{x}^{2}d\tilde{x}^{3} =\sqrt{-g_{00}g_{11}g_{22}g_{33}}\; dx^{0}dx^{1}dx^{2}dx^{3} =\sqrt{-g}\;dx^{0}dx^{1}dx^{2}dx^{3}.\]

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 $u^{\mu}p_{\mu}$.

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}

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 problem

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.

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 $x^{\mu}$, then 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 zero geodesics coincides with the photon's energy up to a constant factor.

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.

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 \[\partial_{0}g_{\mu\nu}=0.\]

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 $\nabla_{\mu}K_{\nu}$.

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.

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 \[\nabla_{i}R^{i}_{j}=\tfrac{1}{2}\partial_{j}R.\]

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 \[T_{ik}^{(em)} =-\frac{1}{4\pi}\Big(g^{lm}F_{il}F_{km} -\tfrac{1}{4}g_{ik}F^{lm}F_{lm}\Big).\]

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 (see problem \ref{equ_oto6b}) 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}

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 Bianchi equality (see problem \ref{equ_oto6a}), 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.