I have a question about doing Lorentz-like coordinate transformations in general relativity. I will try not to get into too much detail about what exactly I am trying to do to not muddy the waters.

In GR, for an affinely parametrized curve with parameter $\lambda$, the geodesic equation is given by: $$ \frac{d^2x^{\mu}}{d\lambda^2} + \Gamma^{\mu}_{\alpha\beta}\frac{dx^{\alpha}}{d\lambda}\frac{dx^{\beta}}{d\lambda} = 0$$ Where because of the affine parametrization, the right hand side is equal to zero. If one uses something like coordinate time $t$, one must use the following expression: $$ \frac{d^2x^{\mu}}{dt^2} + \Gamma^{\mu}_{\alpha\beta}\frac{dx^{\alpha}}{dt}\frac{dx^{\beta}}{dt} = f(t) \frac{dx^{\mu}}{dt}$$ where $f(t)$ is related to the affine parameter and the non-affine choice of coordinates. This equation will be more difficult to solve since it is not homogeneous.

So I see this and think, "What if I solve the equations in proper time so they are easy, and then do a coordinate transformation on the solution?" I came to find out this was naive because there do not appear to be Lorentz-like coordinate transformations in GR because the metric is more complicated, which makes sense in retrospect, but still has me wondering about whether my initial plan to solve and then transform is still possible.

For example, lets say that I solved the equations for a particle in its rest frame, and I wanted to shift to an observer that is a distance $r$ away and revolving around the particle at constant radius? I think shifting the spatial coordinates would be simple enough via the standard transformation laws of $$\Lambda^{\mu}\;_{\mu'} = \frac{\partial x^{\mu}}{\partial x^{'\mu'}},$$ but transforming the time variable is confusing to me if the metric is sufficiently complicated and non-linear. In special relativity the time transformation is known and not complicated, but how to find this for complicated geometries?

Any information would be helpful, thanks!


1 Answer 1


By themselves, neither (unparametrized) curves nor manifolds have coordinates (what we call a parametrization in the case of curves). An unparametrized curve in a manifold obeys both of these equations if it is autoparallel (if the subspace its tangent generates in the tangent space is horizontal with respect to the connection).

You can define the unparametrized curve in a few way, either as the image $\mathrm{Im}(\gamma)$ of a parametrized curve $\gamma : \mathbb{R} \to M$, or as an equivalence class of every parametrization of a curve, $\gamma / \mathrm{Diff}(\mathbb{R})$, or as a map from the line as a manifold (without specific coordinates) to $M$. If you wish to once again express it in terms of coordinates, you can then map both the curve and the manifold to their respective $\mathbb{R}$ and $\mathbb{R}^n$.

Please note that what you are transforming if you wish to go from a geodesic to a pregeodesic is the coordinates of the curve (the parametrization), and not the coordinates of the manifold. So yes, you can always solve your geodesic equation in the proper time parametrization (and also in a specific coordinate system on the manifold itself, if it helps to simplify the problem), and then convert the solution back to your original parametrization/coordinates. This is a common technique you can use in solving differential equations (substitution of variables)


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