Confusion about Transforming Christoffel Symbols I'm trying to understand how transforming Christoffel symbols works. Specifically I'm thinking about the transformation between Schwarzschild and Eddington-Finkelstein coordinates,
$$\Gamma^v_{\;vv}=\frac{\partial v}{\partial x^m}\frac{\partial x^n}{\partial v}\frac{\partial x^p}{\partial v}\Gamma^m_{\;np}+\frac{\partial^2 x^m}{\partial v^2}\frac{\partial v}{\partial x^m}$$
With $cv=ct+r+r_{S}\ln(r-r_{S})$, $m$, $n$, and $p$ only get summed through $t$ and $r$. I'm just not sure how to deal with the $\frac{\partial t}{\partial v}$ and $\frac{\partial r}{\partial v}$. I've never done anything like that before. I'm assuming it cannot be the reciprocals of $\left(\frac{\partial v}{\partial t}\right)$ and $\left(\frac{\partial v}{\partial r}\right)$, since it is a multivariable function, but beyond that I'm not sure. Just to be clear, I could just find it normally with the new metric, but I want to understand how the Christoffel symbols transform as well.
 A: I'm not sure I really understood your question, but maybe you'll find some answers in the following summary.
Firstly, the role of Christoffel's symbols has a huge importance in describing the geodesic equation for a set of local coordinates $x^\mu$
$$
\ddot{x}^\mu + \Gamma^{\mu}_{\nu\rho}(x)\dot{x}^\nu\dot{x}^\rho =0 \ \text{.}\tag{1}\label{1}
$$
Another equivalent way of determining these coefficients can be found directly from the metric, so that:
$$
\Gamma^{\mu}_{\nu\rho}(x)=\dfrac{1}{2}g^{\mu\sigma}(\partial_{\nu}g_{\sigma\rho}+\partial_{\rho}g_{\nu\sigma}-\partial_{\sigma}g_{\nu\rho})
$$
Now, equations \eqref{1} simply state that with respect to the inertial coordinates $\zeta^\alpha=\zeta^\alpha(x)$, the free motion is described by $\ddot{\zeta}^\alpha=0$, whose solutions are called geodesics.
It's crucial to see that we can totally eliminate a general field of the form:
$$
g_{\mu\nu}(x)=\eta_{\alpha\beta}\dfrac{\partial \zeta^\alpha}{\partial x^\mu}\dfrac{\partial \zeta^\beta}{\partial x^\nu}
$$
by simply inverting the transformations $x^\mu \to \zeta^\mu$.
Since they represent the "gravitational forces" (which are of course absent in the coordinates $\zeta^\mu$) the Christoffel's symbols CANNOT transform like tensors. Passing to another set of coordinates $x'^\mu$, we have indeed:
$$
\Gamma'^{\mu}_{\nu\rho}(x')= \dfrac{\partial x'^\mu}{\partial x^\alpha}\dfrac{\partial x^\beta}{\partial x'^\nu}\dfrac{\partial x^\gamma}{\partial x'^\rho}\Gamma^\alpha_{\beta\gamma}(x) + \dfrac{\partial x'^\mu}{\partial x^\alpha}\dfrac{\partial^2 x^\alpha}{\partial x'^\nu \partial x'^\rho}
$$
This formula for sure guarantees that the equations of motion are generally covariant. Thus, if the local coordinates $x^\mu$ are inertial, we have:
$$
\Gamma'^{\mu}_{\nu\rho}(x')=\dfrac{\partial x'^\mu}{\partial x^\alpha}\dfrac{\partial^2 x^\alpha}{\partial x'^\nu \partial x'^\rho}
$$
and equations \eqref{1} reduce to $\ddot{x}^\mu=0$.
For your specific case, I've never done an explicit calculation for the transformation of Christoffel symbols, since I find it more convenient to calculate them directly from the metric once the coordinate transformation has already been implemented. Nevertheless, using the general formula reported here, you should be able to achieve the result you're looking for.
Just an important (final) reminder: please note that vanishing coefficients for the set of coordinates $x^\mu$ may be transformed into non-vanishing ones for the coordinates $x'^\mu$, so that in principle the tedious calculation must be made on all coefficients, even if in some reference frame they are equal to zero.
A: transformation between Schwarzschild and Eddington-Finkelstein coordinates.
the line element of Schwarzschild metric is:
$$ds_S^2=- \left( 1-{\frac {{\it r_s}}{r}} \right) {{\it dt}}^{2}+{{\it dr}}^{2}
 \left( 1-{\frac {{\it r_s}}{r}} \right) ^{-1}+{r}^{2}{d\Omega }^{2}
$$
and of Eddington-Finkelstein metric
$$ds_E^2=-\left(1-\frac{r_s}{r}\right)\,dv^2+2\,dr\,dv+r^2\,d\Omega^2$$
from here you can obtain the transformation (Jacobi-Matrix) of the coordinates
$$  \begin{bmatrix}
   dt \\
   dr \\
   d\Omega \\
 \end{bmatrix}=\mathbf{T}\,
 \begin{bmatrix}
                dv \\
                dr \\
                d\Omega \\
              \end{bmatrix}\quad,
 \mathbf T= \left[ \begin {array}{ccc} 1&-{\frac {r}{r-{\it r_s}}}&0
\\  0&1&0\\  0&0&1\end {array}
 \right] 
$$
or
$$
 \begin{bmatrix}
   dv \\
   dr \\
   d\Omega \\
 \end{bmatrix}=\mathbf T^{-1}\,
 \begin{bmatrix}
                dt \\
                dr \\
                d\Omega \\
              \end{bmatrix}\quad,
 \mathbf T^{-1}=\left[ \begin {array}{ccc} 1&{\frac {r}{r-{\it r_s}}}&0
\\  0&1&0\\  0&0&1\end {array}
 \right]
$$
with those equations you can obtain the transformation of the  Christoffel symbols.

to obtain the transformation matrix $~\mathbf T~$ you make this ansatz
$$\mathbf T^T\,\mathbf G_S\,\mathbf T=\mathbf G_E$$
where $~\mathbf G~$ is the metric .
thus for example:
$$dt=dv-\frac{r}{r-r_s}\,dr=\frac{\partial t}{\partial v}\,dv+
\frac{\partial t}{\partial r}\,dr$$
$$dv=dt+\frac{r}{r-r_s}\,dr=\frac{\partial v}{\partial t}\,dr+
\frac{\partial v}{\partial r}\,dr$$
