# 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 $|\phi| << 1$ 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.

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What you're looking for are "Riemann normal coordinates," but I'm having trouble finding a decent reference on the web at the moment. Sean Carroll's notes have a short section on them which might give you some pointers. The basic idea is to make a general coordinate transformation in the metric, expand it to first order and set the derivatives to zero. This gives a set of equations that determine the coordinate transformation. This is equivalent to using affinely parameterized geodesics, which is (hopefully) what you mean by the acceleration comment. –  Michael Brown Mar 27 '13 at 2:33
Note that this is only possible at a point. You will not be able to find a global coordinate system where the christoffel symbols are zero. I'm aware that you probably know this, but I"m leaving this note for future people looking at this question. –  Jerry Schirmer Mar 27 '13 at 2:37
See Wikipedia. –  Qmechanic Mar 27 '13 at 9:37
@MichaelBrown: Thanks a lot! If it is not too much to ask, please could you show me explicitly how you can change to the Riemann normal coordinate for the metric in the question? Qmechanic, Jerry: Thanks. –  ramanujan_dirac Mar 27 '13 at 12:34
I have read the relevant sections from Misner Thorpe Wheeler, and the wikipedia page. –  ramanujan_dirac Mar 27 '13 at 12:35

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

$$\mathrm{d}\tau^{2}=g_{\mu\nu}\mathrm{d}x^{\mu}\mathrm{d}x^{\nu},\ (1)$$

and we want coordinates $\tilde{x}^\mu (x)$ such that

$$\mathrm{d}\tau^{2}=\eta_{\mu\nu} \mathrm{d}\tilde{x}^{\mu}\mathrm{d}\tilde{x}^{\nu},\ (2)$$

(to linear order) in a neighbourhood of zero. We write the coordinate transformation as

$$\tilde{x}^{\mu}=ax^{\mu}+\frac{1}{2}b_{\nu\rho}^{\mu}x^{\nu}x^{\rho}+\cdots,\ (3)$$

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)

$$\begin{array}{rcl} a^{2}\eta_{\rho\sigma}&=&g_{\rho\sigma}\left(x=0\right),\\ a\left(b_{\sigma\lambda\rho}+b_{\rho\lambda\sigma}\right)&=&g_{\rho\sigma,\lambda}\left(x=0\right). \end{array}$$

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. :)

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Thanks Michael. I really appreciate the effort. I will go through the calculations. I have one more question, I have studied the equivalence principle is it possible maybe by some physical argument, to find out what is the acceleration that this metric corresponds to? I am referring to EEP: the effects of gravity and acceleration are indistinguishable. –  ramanujan_dirac Mar 27 '13 at 14:43
+1, and accepted answer. –  ramanujan_dirac Mar 27 '13 at 14:43
@ramanujan_dirac you would use the geodesic equation to find the motion of a free particle in any given metric. –  Michael Brown Mar 27 '13 at 16:09