Change of coordinates from an arbitrary frame to a locally inertial frame in General Relativity If I have the following metric:
$$ds^2=(1-2\phi)\,c^2 dt^2 - (1-2\phi)(dx^2+dy^2+dz^2)$$
$\phi$ being the gravitational potential with $|\phi| \ll 1$ everywhere.
How do I find a coordinate transformation to a locally inertial coordinate system to first order in $\phi$?
One method that I know of is writing $\phi$ as an equivalent acceleration $g\,\sqrt{x^2+y^2+z^2}$, and then make a coordinate transformation such that the particle is accelerating with this acceleration in the new frame. Is this correct?
But, more importantly, I am looking for an alternative method. E.g. Is it possible to make a coordinate transformation such that the Christoffel symbols are zero?
How can I write any general metric in a locally inertial frame?
Note: I am actually studying this in an SR course which introduces GR very very briefly. But, I know the basic definitions of manifolds, such as differential forms, and tangent spaces. So, I won't mind a technical answer, but it would be great if you could talk about any terms that I am not likely to know in a couple of lines. (or I can find it out for myself, if it is too much hardwork for you). I have studied GR uptil the geodesic equation.
 A: I'll be substantially lazy and drop all quadratic terms. Keep in mind that I'm doing everything to linear order. Also, I'll suppose we are finding normal coordinates around the point $x^\mu = 0$. It's trivial to modify the technique for use anywhere else. I'll follow your mostly minus metric convention but use $c=1$. We have the metric
$$\mathrm{d}\tau^{2}=g_{\mu\nu}\mathrm{d}x^{\mu}\mathrm{d}x^{\nu},\tag1$$
and we want coordinates $\tilde{x}^\mu (x)$ such that
$$ \mathrm{d}\tau^{2}=\eta_{\mu\nu} \mathrm{d}\tilde{x}^{\mu}\mathrm{d}\tilde{x}^{\nu},\tag2$$
(to linear order) in a neighbourhood of zero. We write the coordinate transformation as
$$ \tilde{x}^{\mu}=ax^{\mu}+\frac{1}{2}b_{\nu\rho}^{\mu}x^{\nu}x^{\rho}+\cdots,\tag3$$
where $a$ and $b^\mu_{\nu\rho}$ are constants to be determined. The higher order terms don't influence the construction. Without loss of generality we take $b^\mu_{\nu\rho}=b^\mu_{\rho\nu}$. Subbing (3) in (2) I get (exercise)
$$ \mathrm{d}\tau^{2} = \left[ a^{2}\eta_{\rho\sigma}+a\left(b_{\sigma\lambda\rho}+b_{\rho\lambda\sigma}\right)x^{\lambda}+\cdots \right] \mathrm{d}x^{\rho}\mathrm{d}x^{\sigma}. $$
Matching this onto (1) order by order gives the conditions (exercise)
$$\begin{array}{rcl}
a^{2}\eta_{\rho\sigma}&=&g_{\rho\sigma}(x=0),\\
a\left(b_{\sigma\lambda\rho}+b_{\rho\lambda\sigma}\right)&=&g_{\rho\sigma,\lambda}(x=0).
\end{array}$$
In your case this gives the conditions (exercise)
$$\begin{array}{rcl}
a^{2}&=&1-2\phi^{0},\\
ab_{t\lambda t}&=&-\phi_{,\lambda}^{0},\\
ab_{i\lambda i}&=&\phi_{,\lambda}^{0},\
\text{all the other}\ b^\mu_{\nu\rho}\ \text{vanish},
\end{array}$$
(where for shorthand $\phi^0 \equiv \phi(x=0)$ and $i=x,y,z$) which I'm sure you can solve. :) I get
$$\tilde{x}^{\mu}=\sqrt{1-2\phi^{0}}x^{\mu}-\frac{\phi_{,\lambda}^{0}\,x^{\lambda}}{2\sqrt{1-2\phi^{0}}}x^{\mu}+\cdots,$$
though you should check this yourself (cause I don't feel like it) in case I made any index/sign mistakes. :)
A: I have looked at the three-dimensional analogue of the problem Michael treats.
Their are 27 equations involved,  of the form
\begin{equation*}
g_{ \rho\sigma,\lambda }=a~(b_{\sigma\lambda\rho}+b_{\rho\lambda\sigma})
\end{equation*}
The equations split into groups of size (9, 12, 6 ). The group of six is the most difficult group to solve for. These six equations involve those $b_{  \alpha \beta \gamma}$ with $\alpha ,~\beta ,~\gamma~$ all taking on  different values. It can be shown that these $b_{  \alpha \beta \gamma}$ must all equal zero.
The nine equations involving just the
\begin{equation*}
b_{i \gamma i}~~~~;i=1,2,3~~;\gamma = 1,2,3
\end{equation*}
give a value, for a ‘b’ , most straightforwardly.
To obtain a ‘b’ value from an equation in the group of twelve, use is made of the symmetry
\begin{equation*}
b_{  \alpha \beta \gamma}=b_{  \alpha \gamma \beta }
\end{equation*}
It looks as if the 27 equations may also be solved, in the case of a more general ( not diagonal ) metric.

I have looked at the full four-dimensional problem that Michael treats.
There are 64 equations involved,  of the form
\begin{equation*}
g_{ \rho\sigma,\lambda }=a~(b_{\sigma\lambda\rho}+b_{\rho\lambda\sigma})
\end{equation*}
The equations split into groups of size (16, 24, 24 ). I think the group of twenty four equations involving those $b_{  \alpha \beta \gamma}$ with $\alpha ,~\beta ,~\gamma~$ all taking on  different values will be the most difficult to solve. I guess it can be shown that these $b_{  \alpha \beta \gamma}$ must all equal zero ( not yet done).
The sixteen equations involving the
\begin{equation*}
b_{i \gamma i}~~~~;i=1,2,3,4~~;\gamma = 1,2,3,4
\end{equation*}
will give values of a ‘b’ most straightforwardly.
To obtain a ‘b’ value from an equation in the second group of twenty four , use can be made of the symmetry
\begin{equation*}
b_{  \alpha \beta \gamma}=b_{  \alpha \gamma \beta }
\end{equation*}
So, it looks as if Michael’s answer is incorrect ( Dated 01 May 2022 ), as regards the number of non-zero coefficients
Please let me know if you spot any errors in the above.
