# Diffeomorphism and constant curvature

1. Mar 14, 2011

### jfy4

I have a question...

Since solutions to Einstein's field equations are diffeomorphism invariant, does that mean that solutions are metrics of constant curvature?

2. Mar 14, 2011

### atyy

No. Any metric is a solution to Einstein's equations. However, each metric solution specifies the distribution of the energy of matter in spacetime. And the question is whether there is any metric solution that has a distribution of energy compatible with what we observe (the answer seems to be yes).

The diffeomorphism invariance simply means that if two metrics are related by diffeomorphisms, then they represent the same physical reality.

3. Mar 14, 2011

### jfy4

is this true? The worm hole metric is not a physical solution... in the sense that it is not real, as far as we know... but it is a metric...

If you put a metric through a diffeomorphism, and it is unchanged, then the curvature tensor would be the same also, correct? Then any solution to the Einstein equation under a diffeomorphism has the same curvature, which means they are surfaces of constant curvature no?

4. Mar 14, 2011

### bcrowell

Staff Emeritus
It is a solution to the field equations, because for the reasons Atyy gave in #2, any metric is a solution. Being a solution doesn't mean it really happens in our universe.

No, both the metric and the curvature will change according to the tensor transformation law.

5. Mar 14, 2011

### jfy4

Let me make sure I have this straight... We have field equation for GR that when solved, give solutions which are not physically real... this does not sound correct, I am skeptical.

OK, can we go a bit deeper? solutions to the GR field equations are diffeomorphism invariant, then under a diffeomorphism, the physics of the metric should remain unchanged, and hence the physics of the curvature tensor should remain unchanged also. Thus one would arrive with the same curvature. Is this reasoning not sound? Please show me where it deviates.

6. Mar 14, 2011

### bcrowell

Staff Emeritus
For comparison, Newton's laws admit a solution where a pig drops out of an airplane and falls in my swimming pool. That doesn't mean it really happens.

Many spacetimes also violate various energy conditions, but that isn't a violation of the GR field equations, it's a violation of the way we think matter normally behaves.

You haven't defined what you mean by "the physics of the metric." The metric does have some properties that are diff-invariant (e.g., the Kretchmann invariant), but that doesn't mean that, e.g., the components of the tensor are diff-invariant -- they aren't.

7. Mar 14, 2011

### jfy4

I don't mean to get pedantic, and this may be a matter of definition, but your situation (with the pig) could happen, whereas a worm hole, could not... This is my point.

Fair enough, and I will continue to be a bit vauge Can we agree that under a diffeomorphism, the gravitation field before is the same as the gravitational field after, in so far as it represents the same physical gravitational field?

If we can agree on that, then I believe the curvature tensor with that metric will also be the same, in the same sense that the gravitational field was the same after the diffeomorphism.

8. Mar 14, 2011

### atyy

Ok, make it 7 pigs exactly.

9. Mar 14, 2011

### bcrowell

Staff Emeritus
How do you know the wormhole could not? If it's because it violates an energy condition, then that's a separate issue than the field equations.

No. The equivalence principle tells us that the gravitational field can be whatever you like, based on your frame of reference. This is exactly why there is no really useful analog in GR of the Newtonian gravitational field.

No. Tensors obey the tensor transformation law, which requires them to change when you change coordinates (assuming it's not a rank-0 tensor, i.e., a scalar).

10. Mar 15, 2011

### Sam Gralla

Just to clarify the point atty is making:

Einstein's equation says $G_{\mu \nu}[g]=T_{\mu \nu}$. If you take any metric g, compute its Einstein tensor, and simply call that the stress-energy, then you have a "solution". This is sometimes called "Synge's method" for solving Einstein's equation (I guess becuase Synge called attention to the falacy), and if somebody says you are using it, it is not a compliment.

The point is that for something to be an interesting solution you want to check that the stress-energy has reasonable properties, like timelike observers measure positive energy density. Otherwise you've just found something made of some crazy type of matter that probably doesn't exist.

11. Mar 15, 2011

### jfy4

The non-free-field equations depend on energy density. Classically, and quantum-ly, we have yet to observe any energy violations, I think that is an important condition to take into account.

This was the point I was making. I don't believe you can have a solution to the field equations just by inserting a metric into the Einstein curvature. It must be physical.

I really agree with this post.

We are talking about diffeomorphisms, and as I understand it's definition, After a diffeomorphism, you have the same gravitational field as seen by a new set of coordinates.

I thought the point of tensors was their coordinate invariance... you are really confusing me here.

12. Mar 16, 2011

### atyy

It is possible to consider the metric tensor to be the same "physical object" under a change of coordinates. Given a coordinate system on a manifold, the metric tensor has a coordinate representation eg. as a specific line element. When the coordinate system is changed, the metric tensor will have a different coordinate representation.

It is the same with a vector (which is a sort of tensor, anyway), eg. the tangent vector to a worldline. It is the same "physical object" in any coordinate system, but it has a different coordinate representation in different coordinates.

A tensor is basically something that eats one or more vectors and spits out a number. The point of the coordinate transformation rules is that under a coordinate change, the tensor and the vectors that it eats change their coordinate representation, but the number ("physical observable") remains unchanged - it is in this sense that the tensor and vectors are the same "physical objects" under arbitrary changes in coordinates.

Last edited: Mar 16, 2011
13. Mar 16, 2011

### tom.stoer

Let's make an axample: you can introduce coordinates (t, x), (t', x'), (t'', x''), ... on spacetime. You are labelling the same event on spacetime (a supernova explodes) with different coordinates. But we all agree that it happens "here and then", at a specific point To be more precise, the event itself defines this point P, the "here and then". In an empty spacetime the "here and then" cannot be defined reasonably b/c there is nothing "at" a specific point.

So if different observers introduce different frames of reference (t, x), (t', x'), (t'', x''), ... they do not agree "where w.r.t. these reference frames" the supernova explodes, but they agree on the fact that the explosion defines a point P on spacetime uniquely (observer-independently). In addition once they have two different events these events define two points P and Q. Again they do not agree on the coordinates, but they agree on "distance" which is calculated using the line element ds² = gabdxadxb.

What GR says is that the essential physics is encoded via invariants like ds². They do not change under coordinate changes. The "same" spacetime geometry means that you have one "geometry" which is described via one metric tensor "g" which has different coordinate representations gab, g'ab, g''ab, ... with different values. But that any invariant derived from these different representations has the same value for all representations.

The curvature is a complicated object; the curvature tensor Rabcd depends on the choice of coordinates, but different scalars derived from it like the curvature scalar R and the Kretschmann scalar K are indeed invariants.

14. Mar 16, 2011

### atyy

tom.stoer makes a very important point. We must have some event marked by matter - such as an explosion - to have a physical observable. If we just say eg. R(x), we don't know what x is, so different experimentalists cannot know what quantity is being measured. We have to say R at some event.

15. Mar 17, 2011

### jfy4

Thank you guys very much for giving me a hand with this. Let me restate my question then from the beginning of this post.

With real-physical solutions to Einstein's equation, the solutions are diffeomorphism invariant, in that, when a solution is put through a diffeomorphism it represents the same gravitational field in a new set of coordinates. The curvature scalar created from the curvature tensor for both the gravitational field from before the diffeo, and after, will be the same, and hence, the curvature of the surface must be of constant curvature.

Is that statement ok?

16. Mar 17, 2011

### Sam Gralla

What you say is all correct until you say something about constant curvature. I think you're missing a basic point (which is actually pretty confusing if not taught correctly). Let's think physically--hopefully you can understand how the math relates once you have the basic picture down. Imagine you have a blob of stuff that is really dense on one end (call it the dense end) and not so dense elsewhere, and this blob is sitting in 3D space. If you rotate the blob, the dense end moves (or its coordinate value changes), but the value of the density stays the same there. That is the statement that the density is a "scalar". Now suppose you attached a little arrow (pointing outward, say) to the dense end. Now when you rotate the whole blob the place where the arrow is moves AND the arrow points in a new direction. That's why you have to use the tensor transformation law on directied quantities: not only do the "little arrows" move (or change coordinate position), but the values of their coordinate components change too.

In your argument, you'd rotate the blob and then demand that the density at the dense end equal the density at whatever part of the blob is sitting where the dense end used to be. Of course, you'd conclude that the density is uniform. But this is the statement of an isometry, not of diffeomorphism invariance. Diffeomorphism invariance is much more trivial!

Last edited: Mar 17, 2011
17. Mar 17, 2011

### jfy4

With a diffeomorphism, are we not transplanting the gravitational field across a manifold, only to end up with the same gravitational field, in a new set of coordinates? Then the curvature scalar evaluated for both of these gravitational fields would be the same, thus the manifold is a surface of constant curvature right?

18. Mar 17, 2011

### Sam Gralla

I guess my explanation didn't help. I'll try more a mathematical one. A diffeo maps a points on a manifold to other points. Let's say we have points P1, P2, and P3, and our diffeo is set up to map P1 to P2, P2 to P3, and P3 to P1. By definition, the action of the diffeo on a scalar field (such as the Ricci scalar) is to make the value at P2 be *what it used to be* at P1, and so on.

So if we have the Ricci scalar being 5 at P1, 10 at P2, and 15 at P3, then after the diffeo it is 15 at P1, 5 at P2, and 10 at P3. That's all there is to a diffeomorphism. Being "diffeomorphism invariant" just means this property holds for your scalar field. Diffeomorphism invariance is pretty trivial.

My interpretation of your confusion is that you think a diffeomorphism means just to compare the value at P2 with P1, and that diffeomoprhism invariance would then mean that the values must be equal. But diffeomorphisms tell you to compare the value at P2 with what used to be the value at P1.

It's tricky to explain, but extremely simple.

19. Mar 18, 2011

### tom.stoer

Think about a surface like an ellipsoid with non-constant curvature. You can introduce coordinates with origin at the north pole; then you can shift the origin slightly which means you introduce a second coordinate system. Such coordinate systems can be related to each other using diffeomorphisms, but there is no reason why this should force the ellispoid to have constant curvature i.e. to "become" a sphere.

20. Mar 18, 2011

### atyy

Perhaps jfy4 is using the (correct) terminology that a diffeomorphism moves points on the manifold only. The physics is not invariant under such an operation. Physically, this would mean that one can stand at a point on the blank manifold. However, this isn't possible, because a person is a collection of physical fields that include the metric, and once he is on the manifold, it is no longer blank.

To be very correct, the physics is invariant under a diffeomorphism on the manifold and the corresponding pullback of the metric. Physicists, perhaps sloppily, call this "diffeomorphism invariance" with the pullback implied.