# Coordinate transformations in general relativity

Let's assume a non-rotating point mass with mass $$M$$. A non-massive object travels with constant velocity $$\mathbf{v}_t$$, with respect to the point mass, in the vicinity of the point mass. A non-massive observer, with constant velocity $$\mathbf{v}_o\neq\mathbf{v}_t$$, with respect to the point mass, is observing the target.

Without the point mass special-relativistic Lorentz transformations can be applied to perform a coordinate transformation. The question is how the coordinate transformation looks like in general-relativistic case, i.e. by considering the effect of the point mass?

In principle, the transformation should contain the Lorentz transformation as a limiting case for $$M\rightarrow 0$$.

Usually, the Schwarzschild metric is cited for a point mass potential $${\displaystyle \mathrm {d} s^{2}=-\left(1-{\frac {2M}{r}}\right)\mathrm {d} t^{2}+{\frac {1}{1-{\frac {2M}{r}}}}\mathrm {d} r^{2}+r^{2}\mathrm {d} \theta ^{2}+r^{2}\sin ^{2}(\theta )\;\mathrm {d} \phi ^{2}},$$

which for $$M\rightarrow 0$$ gives

$${\displaystyle \mathrm {d} s^{2}=-\mathrm {d} t^{2}+\mathrm {d} r^{2}+r^{2}\mathrm {d} \theta ^{2}+r^{2}\sin ^{2}(\theta )\;\mathrm {d} \phi ^{2}},$$

i.e. the classical special relativistic metric in spherical coordinates. But how to derive the transformation from the Schwarzschild metric?

• Welcome to physics.SE! Your question contains a common misconception, which is that coordinate systems correspond to frames of reference in GR. See physics.stackexchange.com/q/458854 – user4552 Feb 28 '19 at 22:54
• @Ben Crowell Thanks for your comment! Ok understood, one cannot have a global reference frame in GR. But it is also said one can define local frames. Assuming that observer and target are sufficiently close to each other, can we define local transformations in GR which go beyond the Lorentz transformations and which would go to the Lorentz transformation with $M\rightarrow 0$? – Stark Mar 1 '19 at 10:34

The Schwarzschild metric $$G$$ is:

$$G=\left[ \begin {array}{cccc} -1+2\,{\frac {M}{r}}&0&0&0 \\ 0& \left( 1-2\,{\frac {M}{r}} \right) ^{-1}&0&0 \\ 0&0&{r}^{2}&0\\ 0&0&0&{r}^{2} \left( \sin \left( \theta \right) \right) ^{2}\end {array} \right]$$

we first transformed the metric $$G$$ to $$\eta$$

$$G'=T_1\,G\,T_1=\eta= \left[ \begin {array}{cccc} {\frac {1}{\sqrt {1-2\,{\frac {M}{r}}}}}&0 &0&0\\ 0&{\frac {1}{\sqrt { \left( 1-2\,{\frac {M}{r }} \right) ^{-1}}}}&0&0\\ 0&0&{r}^{-1}&0 \\ 0&0&0&{\frac {1}{r\sin \left( \theta \right) }} \end {array} \right] \left[ \begin {array}{cccc} -1+2\,{\frac {M}{r}}&0&0&0 \\ 0& \left( 1-2\,{\frac {M}{r}} \right) ^{-1}&0&0 \\ 0&0&{r}^{2}&0\\ 0&0&0&{r}^{2} \left( \sin \left( \theta \right) \right) ^{2}\end {array} \right] \left[ \begin {array}{cccc} {\frac {1}{\sqrt {1-2\,{\frac {M}{r}}}}}&0 &0&0\\ 0&{\frac {1}{\sqrt { \left( 1-2\,{\frac {M}{r }} \right) ^{-1}}}}&0&0\\ 0&0&{r}^{-1}&0 \\ 0&0&0&{\frac {1}{r\sin \left( \theta \right) }} \end {array} \right] = \left[ \begin {array}{cccc} -1&0&0&0\\ 0&1&0&0 \\ 0&0&1&0\\ 0&0&0&1\end {array} \right]$$

Then we transformed $$G'=\eta$$ to spherical coordinates

$$T_2\,\eta\,T_2= \left[ \begin {array}{cccc} 1&0&0&0\\ 0&1&0&0 \\ 0&0&r&0\\ 0&0&0&r\sin \left( \theta \right) \end {array} \right] \left[ \begin {array}{cccc} -1&0&0&0\\ 0&1&0&0 \\ 0&0&1&0\\ 0&0&0&1\end {array} \right] \left[ \begin {array}{cccc} 1&0&0&0\\ 0&1&0&0 \\ 0&0&r&0\\ 0&0&0&r\sin \left( \theta \right) \end {array} \right]= \left[ \begin {array}{cccc} -1&0&0&0\\ 0&1&0&0 \\ 0&0&{r}^{2}&0\\ 0&0&0&{r}^{2} \left( \sin \left( \theta \right) \right) ^{2}\end {array} \right]$$

so the transformation matrix to bring the Schwarzschild metric $$G$$ to a spherical coordinates (metric $$G_s\quad$$) is:

$$T=T_2\,T_1=\left[ \begin {array}{cccc} {\frac {1}{\sqrt {1-2\,{\frac {M}{r}}}}}&0 &0&0\\ 0&{\frac {1}{\sqrt { \left( 1-2\,{\frac {M}{r }} \right) ^{-1}}}}&0&0\\ 0&0&1&0 \\ 0&0&0&1\end {array} \right]$$

$$T\,G\,T=G_s$$

• I do not see how this answers the question – magma Mar 2 '19 at 0:59
• The question was : transformed the Schwarzschild metric to spherical coordinates?. This is what I did – Eli Mar 2 '19 at 7:25
• No, the (unclear) question was , as far as I understand, how to transform the metric for an Observer moving in a certain direction in Schwarzshild space-time – magma Mar 3 '19 at 2:52
• @Eli thank you for your answer, but as magma pointed out it was not the question. The question is: How to perform (local) space-time coordinate transformation in GR? – Stark Mar 3 '19 at 21:02
• @magma I think you got the question correct, although I am not familiar with the term "metric transformation". I basically look for the Lorentz-transformation analogue in GR. – Stark Mar 3 '19 at 21:06