In the following expression of Lagrangian in General Relativity :

$$S=\int d^{4} x \sqrt{-g}\left(\frac{R}{16 \pi G}+\mathcal{L}_{\mathrm{M}}\right)$$

I understand that we can write for example :

$$c\,dt\,dx\,dy\,dz = \text{det}(J)\,c\,dt\,dr\,d\theta\,d\phi=\text{det}(J)\,d^4x$$

with $J$ the Jacobian between $(ct,x,y,z)$ and spherical coordinates.

Now, I wonder if, in the expression of this Lagrangian above, the default system of coordinates (I mean the starting space) considered is always $ct, x, y, z$.

Indeed, if this system of coordinates is always taken by default, then, I can always have invariance by writting :

$$c\,dt\,dx\,dy\,dz = \text{det}(J)\,c\,dt\,dr\,d\theta\,d\phi=\text{det}(J)\,d^4x$$

with $\text{det}(J)=\sqrt{-g}$.

  1. So should I always consider $d^4x$ like the ending system of coordinates, and not the starting system (from a Jacobian point of view) ?

  2. Is my understanding correct by saying that $\sqrt{-g}\,\,d^4x$ is always equal to $c\,dt\,dx\,dy\,dz$ ?

  • $\begingroup$ Related post by OP: physics.stackexchange.com/q/589398/2451 $\endgroup$ – Qmechanic Oct 31 '20 at 9:48
  • $\begingroup$ @Qmechanic . This is a complementary question (I have added an EDIT 1 on physics.stackexchange.com/q/589398/2451 but without answer, that's why I post it here). $\endgroup$ – youpilat13 Oct 31 '20 at 10:06

There are no preferred or default coordinates in General Relativity. You can use any coordinates you want.

There are generally no “Cartesian” coordinates in which the metric is flat and the volume element is $d^nx$, simply because a general spacetime has curvature. The volume element is $c\,dt\,dx\,dt\,dz$ in flat Minkowski spacetime but not in a general spacetime. You cannot “start” from this volume element, because there is no coordinate transformation that can transform Minkowski spacetime into a general curved spacetime.

You should be thinking in terms of wanting a volume element that is invariant under transformations between any coordinate systems.

It turns out that $\sqrt{-g}d^nx$ is this volume element because it can be shown to be invariant under any change of coordinates on a Riemannian manifold. See eq. 2.48 in Carroll and the discussion around it.

The key concept is the invariance property


Since the Lagrangian density is constructed to be invariant, and the volume element is constructed to be invariant, the action integral is invariant, so the field equations derived from it are form-invariant.

All of this expresses the key idea that coordinates are arbitrary and have no physical significance. The important quantities in a theory are the scalars, vectors, tensors, spinors, etc., which have geometric meaning — and sometimes physical significance — independent of any coordinate system.

  • $\begingroup$ Thanks a lot, I can see clearer. $\endgroup$ – youpilat13 Oct 31 '20 at 19:39

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