# Is there something like ds'^2=ds^2 in GR?

1. Dec 19, 2013

### cuallito

So in special relativity we have $ds'^{2}=ds^2$, which is another way of saying $\Lambda^{T}\eta\Lambda=\eta$. Where $\eta=diag(-1,1,1,1)$.

It seems in GR the symmetry group for transformations is GL(R,4) or Diff(M) depending on who you ask: http://physics.stackexchange.com/questions/65688/group-theory-in-general-relativity

However (I might have missed it) the answers in the above link don't mention if there is an explicit formula akin to $\Lambda^{T}\eta\Lambda=\eta$ for allowed transformations, or a formula for a conserved quantity like $ds'^{2}=ds^2$?

2. Dec 19, 2013

### WannabeNewton

$ds'^2 = g_{\mu'\nu'}dx^{\mu'}dx^{\nu'} \\= g_{\alpha\beta}\frac{\partial x^{\alpha}}{\partial x^{\mu'}}\frac{\partial x^{\beta}}{\partial x^{\nu'}}\frac{\partial x^{\mu'}}{\partial x^{\gamma}}\frac{\partial x^{\nu'}}{\partial x^{\delta}}dx^{\gamma}dx^{\delta} \\= g_{\alpha\beta}\delta^{\alpha}_{\gamma}\delta^{\beta}_{\delta}dx^{ \gamma } dx^{\delta} \\= g_{\mu\nu}dx^{\mu}dx^{\nu} = ds^2$.

In SR the symmetry group is the Poincare group (with the subgroup of Lorentz transformations usually restricted to the proper, orthochronous Lorentz transformations) and the matrix representations of the proper, orthochronous Lorentz group have an explicit formula akin to what you wrote down. However in GR we are no longer dealing with special covariance but rather general covariance; there is no explicit formula for allowed transformations in the context of GR.

3. Dec 19, 2013

### bcrowell

Staff Emeritus
GR is diffeomorphism-invariant. The only transformations not allowed would be those that were not diffeomorphisms (e.g., were not differentiable). This is actually the same as SR, at least from the modern point of view. The way people today define the distinction between SR and GR is that SR applies when there's no curvature. By that definition, there is nothing wrong with using any coordinate system you like for SR. For example, you can use Rindler coordinates, which can be interpreted as an accelerated frame of reference. It is true in both SR and GR that $ds'^2=ds^2$, and in both cases this invariance is ensured in the same way: when you change coordinates, in general the metric will also change.

In old-fashioned presentations of special relativity, the form of the metric as diag(-1,1,1,1) was taken to be holy. This is where we got the idea of allowed and forbidden sets of coordinates.

4. Dec 19, 2013

### tom.stoer

I think $ds'^{2}=ds^2$ is trivial b/c it is a coordinate-free statement for a manifold.

Is the distance between New York and Berlin the same as the distance between New York and Berlin? Yes!

5. Dec 19, 2013

### bcrowell

Staff Emeritus
Maybe it's trivial for a manifold with metric, but the assumption that spacetime has a metric is highly nontrivial. In Galilean spacetime, the distance between where New York is at a certain time and where Berlin is at some other time is frame-dependent.

6. Dec 19, 2013

### tom.stoer

The thread is about GR, so we have a manifold and a metric.

7. Dec 19, 2013

### bcrowell

Staff Emeritus
The title of the thread is this: "Is there something like ds'^2=ds^2 in GR?" It looks like we both agree that it's trivial that a "yes" answer is equivalent to the existence of a metric.

If someone asks why and whether proposition P is true, and everyone agrees that P is trivially logically equivalent to proposition Q, then that's all good, but it doesn't constitute an answer to the question unless someone also gives an explanation of why Q is true.

8. Dec 19, 2013

### tom.stoer

How would a spacetime manifold w/o metric look like and what has all this to do with GR?

9. Dec 19, 2013

### ChrisVer

I guess a manifold without metric would be a manifold where you cannot define distances angles etc. Non riemannian manifolds.

Now for the ds invariance, I would suggest also the mathematics equivalence that is the 1st Fundamental form....

10. Dec 19, 2013

### tom.stoer

If you cannot define a distance then ds is not defined

11. Dec 19, 2013

### K^2

You can have distances defined, but still not have a metric. You have to have distances follow certain axioms, such as the triangle inequality, in order for it to be a metric space. So yes, you can have a manifold with ds² defined, but which is not a metric manifold, and where ds'² ≠ ds².

12. Dec 19, 2013

### tom.stoer

Can you give an example for manifold with definition of a distance but w/o metric? What is the mathematical structure? And in which sense is this relevant for GR?

13. Dec 19, 2013

### WannabeNewton

It won't be relevant to GR at all but a topological manifold can always be endowed with a metric (and by metric I mean an actual metric, the central tool in the theory of metric spaces in real analysis); a metric is a very different object from a metric tensor but it defines a notion of distance nonetheless.

14. Dec 20, 2013

### ShayanJ

What about the speed of light?Is it still frame independent and the maximum speed?
I remember reading somewhere that it is locally like SR but globally can be different but didn't understand it clearly!

15. Dec 20, 2013

### WannabeNewton

16. Dec 20, 2013

### K^2

Sure. x,y in ℝ. d²(x,y) = Sin²((x-y)(x+y)). Works basically like a distance locally; lets you have an element ds² = 4x² dx². Fails some of the metric axioms, and is not invariant under coordinate transformation x' = x+c.

Absolutely irrelevant to GR, because GR manifolds are always pseudo-Riemannian.

17. Dec 20, 2013

### ShayanJ

So the light reaching us from a star 1000 ly away, hasn't traveled 1000 years!!!(In the Earth's frame of reference)
Interesting!

18. Dec 20, 2013

### tom.stoer

This is not a distance in the sense of a metric space. The fact that it works locally is not sufficient.

19. Dec 20, 2013

### K^2

Having a distance that is one in the sense of a metric space is the definition of having a metric space.

Not every distance is a metric. That's kind of what the whole discussion so far has been about. You asked for an example of a distance that's not a metric, and I gave you one. And now you're not happy that it's not a metric?

20. Dec 20, 2013

### tom.stoer

A distance in the sense of a metric space must fulfill the following axioms:

A) d(x, y) ≥ 0
B) d(x, y) = 0 if and only if x = y
C) d(x, y) = d(y, x)
D) d(x, z) ≤ d(x, y) + d(y, z)

Your example definitly violates (B), and I bet it violates (D)

So it's not a distance in that sense. But what else shall a distance be?

21. Dec 20, 2013

### K^2

Of course. I specifically chose a function that violates B.

If distance fulfills these criteria, you have a metric space. We are specifically talking about manifolds that are not metric spaces. Therefore, it does not have a distance that fulfills all of these axioms.

On the other hand, a binary function that fulfills these axioms locally is still a distance in some sense. After all, definition of ds² requires only the local distance. Moreover, with these axioms holding locally, I can define length of a curve in a non-ambiguous way. That length is going to be frame-dependent, but again, that was sort of the point.

By the way, condition A is violated by Minkowski space. So if you are dead set on defining distances by axioms of metric space, then there is no such thing as distance in General Relativity, and you are stepping on your own argument. If you want to talk about distances in GR, you have to relax the requirements for what you call a distance.

22. Dec 20, 2013

### tom.stoer

OK, now I see what you have in mind; some sense of extrapolating local definitions of "distance" globally.

23. Dec 20, 2013

### Staff: Mentor

So is condition B; any pair of distinct points x, y connected by a null curve has d(x, y) = 0.

And so is condition D; for the Minkowski metric the "triangle inequality" is backwards. Otherwise the twin paradox would be the other way around (the traveling twin would experience more elapsed time than the stay-at-home twin).