# Curvature polynomials vanish for plane waves?

1. Nov 17, 2015

### bcrowell

Staff Emeritus
Geroch 1968 touches on the Kundt type I and II curvature invariants. If I'm understanding correctly, then type I means curvature polynomials. Type II appears to be something else that I confess I don't understand very well. (I happen to own a copy of the book in which the Kundt paper appeared. I looked at the paper, and I can't make heads or tails of it.) Geroch says, "For example, in the plane wave solutions all the type I invariants vanish, yet the Riemann tensor is not zero."

This seems very surprising to me. Am I understanding correctly that all curvature polynomials vanish in the case of a plane wave solution? Why would this be?

Geroch, "What is a singularity in general relativity?," Ann Phys 48 (1968) 526

2. Nov 17, 2015

### fzero

The paper Hans-Jürgen Schmidt, Why do all the curvature invariants of a gravitational wave vanish?, http://arxiv.org/abs/gr-qc/9404037 gives an argument based on form-invariance of the metric under a boost along a null direction and the fact that the curvature invariants are continuous functions of their arguments.

3. Nov 18, 2015

### bcrowell

Staff Emeritus
Thanks, that's an extremely interesting paper, despite the sometimes unintelligible English. It's going to take me some time to digest. I hadn't realized that the Riemannian and semi-Riemannian cases were so different from each other.

4. Nov 18, 2015

### martinbn

Very intersting, also the paper in post 2. Is the following possible or is there an "obvious" reason for it not to work? To have two plane waves coliding and forming a singularity (with or without black hole) and thus a physically reasonable example of a singular space-time with all curvature scalors regular (in fact zero).

5. Nov 18, 2015

### bcrowell

Staff Emeritus
Hmm...but the field equations are nonlinear, and the formation of a singularity is about as spectacular a nonlinearity as one could imagine. So I would expect that the curvature scalars would not be zero in this situation.

But a single plane wave can be singular: https://www.physicsforums.com/threads/wave-of-death.93654/#post-1176988

6. Nov 18, 2015

### bcrowell

Staff Emeritus
I'm puzzled by this remark in the Schmidt paper:

He seems to give this as motivation for being interested in curvature invariants. But I don't know what "constructively" measurable means, or whether he has some precise meaning of "locally" in mind. LIGO clearly can't be designed to measure curvature scalars, since what it's designed to detect is a plane wave. I don't understand why it would be of interest to talk about measurements in the absence matter or of a frame of reference; ordinarily your measuring apparatus is material and has its own well-defined rest frame.

Geroch gives this instead as motivation for curvature invariants:

I don't understand this either. When we change frames, the change of coordinates can be represented as a diffeomorphism. If the components of a tensor are finite in one set of coordinates, then they remain finite under a diffeomorphism.

Schmidt defines curvature invariants and generalized curvature invariants. I think Kundt's type II are ratios of curvature polynomials, and these do not qualify as curvature invariants according to Schmidt's definition, since they depend in a discontinuous way on the curvature components (the denominator can go to zero).

Last edited: Nov 18, 2015
7. Nov 19, 2015

### bcrowell

Staff Emeritus
I think I understand this now. Schmidt's proof doesn't depend on signature, and is valid in 3 or more dimensions. However, the physical content is easiest to understand in 3+1 dimensions. Here we have gravitational waves, and the argument is basically that when we redshift a gravitational wave, in the limit of arbitrarily large redshifts, all of the components of the Riemann tensor, as well as their derivatives, approach zero. If the curvature invariant is a continuous function of these components, then it also approaches zero. But it's a scalar, so it has to be invariant under boosts, and therefore it vanishes identically.

As a side note, Schmidt seems to claim at the bottom of p. 3 that for a Riemannian signature, the space is flat if and only if the Kretschmann scalar vanishes. I'm assuming this is a problem with his tortured English, because it would seem to contradict the main result of his paper, in which he constructs an explicit counterexample in 3 dimensions for arbitrary signature. He says "If the metric has definite signature or if n=2...," but I think the "or" has to be wrong, doesn't it?

Last edited: Nov 19, 2015
8. Nov 19, 2015

### martinbn

The exmale is for indefinite matrics of any signature. By the way what is $I_0$? It seems to be defined as the sing of a sum of absolute values!

9. Nov 19, 2015

### bcrowell

Staff Emeritus
Which would contradict what he says at the bottom of p. 3, right?

It simply equals 0 if the Riemann tensor is zero, 1 otherwise.

10. Nov 19, 2015

### bcrowell

Staff Emeritus
Oh, now I think I get the point about the signature. I wasn't understanding his terminology ("indefinite"=not Riemannian) correctly. His results really don't apply to Riemannian signatures at all. His metric $ds^2=2dudv\pm a^2 dw^2$ has signature +-- or ++-, depending on the choice of sign on the $dw^2$ term, but either way, it's not a Riemannian signature.

11. Nov 20, 2015

### bcrowell

Staff Emeritus
After thinking about this a little more, I think the idea is that you can have badly behaved coordinates, which are not related by diffeomorphism to the well-behaved coordinates. For example, the Riemann tensor for the Schwarzschild metric, expressed in Schwarzschild coordinates, has a component $R_{\theta rr\theta}=m/(r-2m)$, which blows up at the event horizon. We can get rid of this divergence by a change of coordinates, but not by a diffeomorphism. Going the opposite way, we can introduce a blow-up in the Riemann tensor. E.g., the change of coordinates from Kruskal–Szekeres coordinates to Schwarzschild coordinates has $\partial r/\partial X=\partial r/\partial T=0$ at the horizon, so it's not a diffeomorphism.