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 problemsee problem