# A Difference Between Covariance and Diffeomorphism Invarience

1. Jan 14, 2018

### Staff: Mentor

Hi

I have studied GR form a number of sources - my favorite being Wald - he uses Diffeomorphism Invarience, the rest General Covarience/Invarience. Wald is more mathematically sophisticated but at rock bottom is there really any difference. My suspicion is Wald does it because there is a debate about if General Covarience is correct - it should really be Genenal Invarience - the difference being, for invarience, it also applies to physical content rather than just form.

Thanks
Bill

2. Jan 14, 2018

### stevendaryl

Staff Emeritus
I've struggled with understanding exactly the content of general covariance, or whatever you call it, and I've come to the conclusion that there are three related but slightly different notions.
1. Coordinate independence. The laws of physics have the same form in any coordinate system. This is not actually a constraint on the laws of physics, but is a constraint on how we formulate those laws. Any laws can be written in a coordinate-independent way. Some people try to salvage some physical content by revising it to say that the laws have their simplest form in coordinate-free language, but simplicity is in the eye of the beholder, so I'm not sure how useful that is.
2. Background independence: The laws of physics when written in covariant have no non-dynamic scalar, vector or tensor fields. This is in some sense philosophically a generalization of Newton's third law about equal and opposite forces. If there is some field that affects the motion of particles, then those particles should affect the field, in return. In other words, the field should be dynamic. In contrast, Newtonian physics has a scalar field, universal time, that is non-dynamic, and Special Relativity has a tensor field, the metric, that is non-dynamic. Somebody argued, I think, that this principle doesn't really have any physical content, either, because you can always put some unobservable dynamics into your theory to make it satisfy this principle.
3. Active diffeomorphism invariance. (I'm not sure about the distinction between "invariance" and "invarience")
Some people use "diffeomorphism invari(a/e)nce" to mean the same thing as coordinate independence. However, conceptually, there is a subtle distinction to be made (or maybe it's a false distinction) which was illustrated by Einstein's "hole argument".

Physics (or at least, most physics) takes place on some manifold, $\mathcal{M}$. A coordinate system is a map $c$ from that manifold to the manifold $R^n$ (or rather, it's a map from some open set of the first to some open set of the second). So if you have a smooth transformation on $R^n$:

$m: R^n \rightarrow R^n$,

then you can use it to transform a coordinate system on $\mathcal{M}$. Let $c'(\mathcal{P}) = m(c(\mathcal{P}))$ or more compactly, $c' = m \circ c$ where $\circ$ is function composition. (I've seen different mathematicians use the opposite convention as to whether to write this as $m \circ c$ or $c \circ m$)

Now, let me introduce something I'm going to call a "situation", which is just a collection of scalar fields, vector fields, tensor fields, etc., on the manifold. Given a situation $S$ and a coordinate system $c$, we can come up with a coordinate-dependent description of that situation: $\mathcal{D}(S,c)$.

The same situation $S$ can be described using different coordinate systems, and the descriptions may look very different: $\mathcal{D}(S,c) \neq \mathcal{D}(S,c)$.

But what about the opposite case? Can there be two different situations, $S$ and $S'$ and two different coordinate systems, $c$ and $c'$, such that $\mathcal{D}(S,c) = \mathcal{D}(S', c')$. Different situations, described by different coordinate systems, but the coordinate-dependent descriptions are the same. Is that possible?

Here's a toy example: Situation 1 has Jill holding a ball. Situation 2 has Jack holding a kitten. If we just started calling Jill "John" and started calling balls "kittens", then we would have a description of Situation 1 in our new language that sounds the same as the description of Situation 2 in our old language. But they aren't the same situation. So different situations can have the same description.

However, if a "situation" includes absolutely everything: fields, scalars, particles, metric, etc., then maybe two situations with the same descriptions are actually the same situation.

I think that that's the intuitive idea behind diffeomorphism invari(a/e)nce. If you have two points $\mathcal{A}$ and $\mathcal{B}$ on your manifold $\mathcal{M}$, then you could imagine two different situations: One in which there is a black hole located at point $\mathcal{A}$, and another where there is a black hole at point $\mathcal{B}$. Those are different situations, in the sense that the metric tensor is a different function on $\mathcal{M}$ in the two cases. But they have the same description: There is a black hole at one point. Diffeomorphism invariance implies that they really aren't different situations, at all.

I think that what it really amounts to, mathematically, is that the arena for physics isn't actually manifolds, but are manifolds modulo diffeomorphisms. If you move points around on a manifold, and correspondingly adjust the metric and all other scalar, vector, and tensor fields, then you haven't done anything. Another way of putting it is that points on the manifold have no identity other than the values of the scalar, vector and tensor fields on those points. There is no difference between: "It's the same manifold, but the fields have different values" and "It's a different manifold."

3. Jan 14, 2018

### Staff: Mentor

Thanks Steven.

In practice it's not actually hard. You want to construct a Lagrangian where the geometry is dynamical. From books on differential geometry, not even the pseudo Riemanian geometry of GR, but simple two dimensional curved spaces you see the simplest curvature invariant is the Ricci Scalar. So going over to the 4 dimensional Pseudo-Riemannian geometry of GR its simply a generalization of that.

You write down the most reasonable Lagrangian that contains the Ricci scalar then vary it eg:
http://www.tapir.caltech.edu/~chirata/ph236/2011-12/lec33.pdf

That's the way I have ended up looking at it.

I have read Ohanian, MTW, Wald, Sean Carroll, Dirac plus others I cant recall at the moment and like you struggle hard with what they are saying, like for example the only scalar that can be constructed from the Ricki Curvature tensor is the Ricki Scalar etc:
https://arxiv.org/pdf/gr-qc/0401099.pdf

Ok I get it - that's how you get the scalar - but really where are you invoking all this principle of general invariance, covariance stuff etc, except you just want, on simplicity grounds, the measure of curvature you use in the Lagrangian to be invariant. This is the same as SR - the equations should not vary between coordinate systems otherwise you can tell what coordinate system you are in. For example the only invariant Lagrangian of a free particle in SR is ∫αdt where t is the proper time because its invariant. I don't know if you can come up with another, but its the only one I can think of. You define -α as the mass in units c=1. Basically from what I can see this is all you are doing in GR at a much more sophisticated level. There is a whole heap of words written to justify it and I have read them, and individually understand them, but for me the way it hooks together never really gelled.

If I was asked for more than that, even though Wald is my favorite book, I would go to Ohanian. Here he develops GR by first, after a careful analysis of EM, writing down the equations of linearised gravity. Then by a simple argument invoking the principle of invariance you get full GR. He comments, and it has always struck me as really weird, this is the only theory where the linearised equations implies the full equations.

So for me - yes I have read all the books and yes they make sense. But overall the only thing that gels for me is what I wrote at the start - you are simply in the most reasonable way extending the ideas of SR to curved space time.

And yes - its invariance - not invarience - its just my crappy spelling and ignoring my spell checker.

Thanks
Bill

Last edited: Jan 14, 2018
4. Jan 14, 2018

### Staff: Mentor

More precisely, the Ricci scalar plus a constant is the only scalar that can be constructed from the Riemann tensor if we require that it contains no more than two derivatives--i.e., it must be no more than linear in second derivatives of the metric and no more than quadratic in first derivatives of the metric. There are other scalars you can construct from the Riemann tensor or the Ricci tensor--for example, $R^{\mu \nu} R_{\mu \nu}$, which doesn't have a short name that I'm aware of, or $R^{\mu \nu \lambda \rho} R_{\mu \nu \lambda \rho}$, the Kretschmann scalar. But these scalars do not meet the "no more than two derivatives" restriction.

5. Jan 15, 2018

### martinbn

I've always thought that these are historical remnants from the time the theory was developed because the mathematical concepts were not there yet. Even the modern definition of a manifold was given around that time. By the way since the metric is important shouldn't one look at isometries rather than just diffeomorphisms.

6. Jan 15, 2018

### stevendaryl

Staff Emeritus
That paragraph seems to be along the lines of the principle that the laws of physics are as simple as possible, when expressed in coordinate-free language.

As I said, though, it seems to me that being coordinate-free is not the whole content of being diffeomorphism-invariant. It seems that it is an additional claim about the world to say the following:

Let $\mathcal{M}$ be a manifold, and let $A, B, C, ...$ be a bunch of fields of various types (scalar, vector, tensor, etc.) on $\mathcal{M}$. We start with that "situation". Then we perform the following operation:
1. Stretch, compress, twist, etc., $\mathcal{M}$ to get a new manifold, $\mathcal{M}'$ related to the first through some diffeomorphism.
2. Perform the appropriate transformations on the fields to get new fields $A', B', C', ...$
3. Then this new situation is equivalent to the original one.
This seems at first blush to be a different claim than "the laws of physics are invariant under coordinate transformations", because it's not talking about coordinates at all. But maybe it is the same as general covariance, in the sense that every theory that is invariant under all coordinate transformations, and has no non-dynamic scalar, vector or tensor fields will also be diffeomorphism-invariant.

7. Jan 15, 2018

### stevendaryl

Staff Emeritus
I don't know what the terminology is, but as I said in the above post, the real symmetry of GR is that if you perform a diffeomorphism on the manifold, and simultaneously transform all scalar, vector and tensor fields, then the result is a situation equivalent to what you started with.

I don't know to what extent the metric is special in this sense.

8. Jan 15, 2018

### martinbn

If that's what the principle is, then the metric wouldn't need to be mention explicitly. But I don't get the point. It seems like a trivial observation not a principle.

9. Jan 15, 2018

### Staff: Mentor

Of course - thanks for clarifying what I said. The other thing about the Ricci scalar is its, as explained in the second paper I linked to, just the generalization of the good old Gaussian curvature scalar (yep it was really Gauss not Riemann that laid the foundation for this stuff) which is a very natural measure of curvature and because of its definition (ie a ratio of infinitesimal areas - the area of a circle divided by one of the same infinitesimal radius, if it was in the euclidean space). That always seemed a really natural measure of it. So if you generalize that its quire reasonable to get GR. You don't invoke the Riemann tensor etc - it's all very intuitive and IMHO simpler to understand. You can explain it easily to undergrads then go into greater detail with a more usual treatment.

Thanks
Bill

10. Jan 15, 2018

### Staff: Mentor

You cant get away from the metric - otherwise what do you vary against in the Lagrangian?

From the days when I was into this stuff if you consider velocities a lot less than light you get the motion of a particle the same as if it was in a gravitational field where the metric (in the approximation) is the field - so its reasonable to take the 'field' of GR as the metric - see 87 of Landau - Classical Theory Of Fields - it moves as if g00 = 1 +2Φ where Φ is the gravitational potential. This leads to the metric having the role of the field in the full relativistic' theory.

Hence that's what you vary against i the Lagrangian.

Thanks
Bill

11. Jan 15, 2018

### Staff: Mentor

Yes one can use isometries - see page 438 - Wald - but I haven't seem them used much - maybe historical, and I am not sure exactly how applicable it would be using that language.

Thanks
Bill

12. Jan 16, 2018

### Staff: Mentor

I just want to add was digging around my library last night and found the book that explains exactly how I look on GR - Soper - Classical Field Theory:
https://www.amazon.com/Classical-Field-Theory-Dover-Physics/dp/0486462609

When I started reading the chapter on GR I thought - yes, yes - that's exactly whats going on. When I looked up - it was 6 in the morning - didn't wake up until 2.00 pm. Whoops.

Excellent first exposure IMHO - you can move onto more 'advanced' material like Wald later.

Thanks
Bill