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While trying to understand General Relativity, I'm struggeling with disentangling coordinates and curvature.

The metric tensor contains information on both: coordinates as well as curvature.

Curvature is a physical characteristic of spacetime, while the coordinates can be chosen completely arbitrary.

Is there a method to disentagle coordinates and curvature?

(Stephan Hawking must have had one as he managed to show that the singularity at the event horizon of a black hole is only a coordinate singularity, not a physical one...)

I fistly thought that one could simply use cartesian coordinates - and then would be left with only the curvature (in this question: Determinant of metric tensor in Cartesian Coordinates constant in vacuum) However, I was taught there, that cartesian coordinates can only be used in flat spacetime. Now, I'm totally clueless again how to disentangle coordinates and curvature.

I'm looking for a tensor/measure/metric/field, where only the physical curvature is in - and no coordinate curvature. Where the coordinates are flat like in a cartesian grid.

Where the metric tensor is simply that of a flat spacetime in cartesian coordinates, if the spacetime is flat (Minkowski spacetime).

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  • $\begingroup$ There is no such thing as “physical curvature” and “coordinate curvature”. There is only curvature. You can express it in any coordinates you want, and transform it using tensor transformation rules into any other coordinates that you want. You can construct curvature invariants that have the same numerical value in any coordinates. $\endgroup$
    – Ghoster
    Oct 6, 2023 at 17:15
  • $\begingroup$ Are you trying to understand curvature singularities? $\endgroup$
    – Ghoster
    Oct 6, 2023 at 17:21
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    $\begingroup$ in flat space, in spherical coordinates, there is curvature No. The curvature tensor in flat space in spherical coordinates is zero. (Have you tried calculating it?) Flat space is called flat space because it has no curvature. You are using the word curvature improperly. $\endgroup$
    – Ghoster
    Oct 6, 2023 at 17:27
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    $\begingroup$ If you have not calculated the curvature tensor for Minkowski space, or Euclidean space, in spherical coordinates and seen that all components vanish, I recommend doing so. $\endgroup$
    – Ghoster
    Oct 6, 2023 at 17:34
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    $\begingroup$ I don't know who first realized that there's no real singularity at the event horizon, but it happened before Stephen Hawking was born – see Gullstrand–Painlevé coordinates. $\endgroup$
    – benrg
    Oct 6, 2023 at 19:39

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Neither the metric nor the curvature depend on the coordinates. They are tensors. They exist independent of coordinates and are invariant under any coordinate transformation. e.g. $\boldsymbol{\eta}$ is the metric tensor for Minkowski space no matter your choice of coordinates. However, you can project a tensor onto a particular coordinate basis to obtain its $\textit{coordinate-dependent}$ components.

In Cartesian coordinates, $\boldsymbol{\eta}$ has components $\eta_{\alpha\beta} = \text{diag}(-1, 1, 1, 1)$. In spherical coordinates, $\boldsymbol{\eta}$ has components $\eta_{\bar{\mu}\bar{\nu}} = \text{diag}(-1, 1, r^2, r^2\sin^2\theta)$. At the end of the day, these components still represent the same geometric object:

$$\begin{align} \boldsymbol{\eta} &= -1\tilde{dt}\otimes\tilde{dt}+1\tilde{dx}\otimes\tilde{dx}+1\tilde{dy}\otimes\tilde{dy}+1\tilde{dz}\otimes\tilde{dz} \;\;\;\;\text{(Cartesian)}\\ &= -1\tilde{dt}\otimes\tilde{dt}+1\tilde{dr}\otimes\tilde{dr}+r^2\tilde{d\theta}\otimes\tilde{d\theta}+r^2\sin^2\theta\tilde{d\phi}\otimes\tilde{d\phi} \;\;\;\;\text{(spherical)}\\ &=\eta_{\alpha'\beta'}\tilde{d}x^{\alpha'}\otimes\tilde{d}x^{\beta'} \;\;\;\;\text{(arbitrary coordinates)} \end{align}$$

Similarly the curvature tensor is the same in all coordinate systems. A straightforward calculation shows the curvature is the zero tensor in Minkowski space - no matter the coordinate system you choose, if any.

Nonetheless, we often subscribe to a particular coordinate system to help with calculations. Then the Riemann tensor would have coordinate-dependent components: $R_{\alpha\beta\mu\nu}$. To study the curvature without being fooled by coordinate artifacts, we can compute curvature invariants which are scalars and the same in every frame.

The Kretschmann scalar $K$ is a particular curvature invariant. In the Schwarzschild geometry,

$$K = R_{\alpha\beta\gamma\delta}R^{\alpha\beta\gamma\delta} = \frac{48M^2}{r^6}$$

where we've used Schwarzschild coordinates: $(t,r,\theta,\phi)$. So the curvature is finite at the event horizon ($r=2M$), and diverges as $r\rightarrow 0$.

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  • $\begingroup$ Thank you, this helps a lot! One related question: Is the determinant of the metric tensor independent under coordinate transformation or dependent? $\endgroup$
    – MartyMcFly
    Oct 6, 2023 at 20:07
  • $\begingroup$ @MartyMcFly The determinant of the components of the metric are coordinate-dependent. Just compare the determinants of $\text{diag}(-1, 1, 1 ,1)$ and $\text{diag}(-1, 1, r^2, r^2\sin^2\theta)$ - which both describe flat space, but in different coordinates. This is because the determinant of the metric is a tensor density and transforms as such - i.e. via the determinant of the Jacobian of the transformation. $\endgroup$
    – Aiden
    Oct 6, 2023 at 20:31
  • $\begingroup$ Could it be that you mean that the determinant is a scalar density and the metric tensor is a tensor density??? $\endgroup$
    – MartyMcFly
    Oct 6, 2023 at 20:46
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    $\begingroup$ @MartyMcFly The determinant of the metric tensor is a scalar density - A.K.A. a rank 0 tensor density. The metric tensor is not a tensor density, but a proper tensor. Otherwise you would have to multiply by factors of the determinant of the Jacobian every time you wanted to transform its components. $\endgroup$
    – Aiden
    Oct 8, 2023 at 19:49
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    $\begingroup$ @MartyMcFly Or I suppose you could classify the metric as a tensor density $\textit{of weight zero}$, but this is not typically how people think of the metric. $\endgroup$
    – Aiden
    Oct 8, 2023 at 19:55

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