# 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.

• 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 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. – user7757 Mar 27 '13 at 12:34
• I have read the relevant sections from Misner Thorpe Wheeler, and the wikipedia page. – user7757 Mar 27 '13 at 12:35

## 1 Answer

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. :)

• 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. – user7757 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 Mar 27 '13 at 16:09