kent davidge
- 931
- 56
Does General Relativity allow for transformations which are not isometries of the metric?
Well, you can also do that for Newtonian gravity. ;) General covariance is not the only issue, as Kretschmann already pointed out to Einstein.king vitamin said:Yes, the Einstein-Hilbert action is invariant under any diffeomorphism.
(Formally, in any sensible physical theory, any diffeomorphism is allowed. What is special about GR is that the action is also invariant under any diffeomorphism, which also makes the equations of motion covariant under these transformations.)
I would say they only make sense once you fix a metric (coordinates be damned!). Given a (pseudo-)Riemannian manifold ##(M,g)##, an isometry of that manifold would be a function ##f: M \to M## such that ##f^* g = g##. This is fundamental for example in the definition of Killing fields as generating fields of isometries, i.e., ##K## is a Killing field if its corresponding flow ##\gamma_K(s,p)## satisfieshaushofer said:Isometries only make sense once you choose coordinates
Orodruin said:An isometry is a map between two manifolds that preserves the metric. This has little to do with coordinate transformations and more to do with maps from a manifold (in the case of GR, spacetime) to itself.
Of course, given a coordinate system, any map from a manifold to itself is going to give you a new coordinate system by assigning the coordinates of the old point to the new point, which is a source of some general confusion.I would say they only make sense once you fix a metric (coordinates be damned!). Given a (pseudo-)Riemannian manifold ##(M,g)##, an isometry of that manifold would be a function ##f: M \to M## such that ##f^* g = g##. This is fundamental for example in the definition of Killing fields as generating fields of isometries, i.e., ##K## is a Killing field if its corresponding flow ##\gamma_K(s,p)## satisfies
$$
\mathcal L_K g = \lim_{\epsilon\to 0}\left[\frac{1}{\epsilon}(\gamma_K^* g - g)\right] = 0,
$$
which is true if ##\gamma_K^*g = g##.
I agree that you usually need coordinates to do something more specific. The only beef I have with this is when people start using coordinate dependent statements as "truths" and don't consider that their interpretation may be coordinate dependent - such as considering cosmological redshift to be fundamentally different from a Doppler shift.haushofer said:for me it works best to take a concrete example and do an actual calculation in a certain coordinate basis and see what happens (i.e. transform the formal stuff into a component example which I think I understand).
I guess what you are talking about is going from one coordinate system to another one, staying at the same point on the manifold, but what I'm asking is if you can vary the coordinates (of course in a given coordinate system), i.e., if you can go from one point to another point on the manifold, like @Orodruin describes in his post. Tensors will generally change if you do that. And I'm asking if these type of transformations where the metric changes are allowed in General Relativity.king vitamin said:Yes, the Einstein-Hilbert action is invariant under any diffeomorphism
Please, see my reply to @king vitaminhaushofer said:What do you mean by "allowed"?
Yes, I'm aware of that.haushofer said:Isometries only make sense once you choose coordinates, i.e. once you chose a gauge. If in that coordinate system the form of the metric is preserved for specific transformations, you speak of isometries.
kent davidge said:I guess what you are talking about is going from one coordinate system to another one, staying at the same point on the manifold, but what I'm asking is if you can vary the coordinates (of course in a given coordinate system), i.e., if you can go from one point to another point on the manifold, like @Orodruin describes in his post. Tensors will generally change if you do that. And I'm asking if these type of transformations where the metric changes are allowed in General Relativity.
Please, see my reply to @king vitamin
Yes, I'm aware of that.
Why wouldn't they be? I'm missing context, I guess.kent davidge said:Please, see my reply to @king vitamin
very interesting, I am reading ithaushofer said: