Solving Exact Gravitational Plane Wave Confusion

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SUMMARY

The discussion focuses on the interpretation of an exact gravitational plane wave solution derived using Brinkmann coordinates. The metric presented is defined as $$g = (y^2 - x^2) \, h(u) \, du \otimes du + du \otimes dv + dv \otimes du + dx \otimes dx + dy \otimes dy$$, with the Einstein tensor equating to zero and a non-zero Riemann tensor. The confusion arises regarding the existence of a global orthonormal space-time basis, particularly when the condition $$2+h(u)(x^2-y^2) = 0$$ is met, leading to questions about potential singularities and the physical implications of the metric. The discussion references useful resources, including notes on pp-waves, which provide insights into curvature invariants and geodesics.

PREREQUISITES
  • Understanding of general relativity and gravitational waves
  • Familiarity with Brinkmann coordinates
  • Knowledge of Riemannian geometry and tensor calculus
  • Experience with geodesic equations and Killing vectors
NEXT STEPS
  • Study the properties of Brinkmann coordinates in gravitational wave solutions
  • Explore the implications of curvature invariants in general relativity
  • Investigate the geodesic equations in the context of the discussed metric
  • Examine the role of Killing vectors in determining symmetries of spacetime
USEFUL FOR

The discussion is beneficial for theoretical physicists, mathematicians specializing in differential geometry, and researchers focused on gravitational wave solutions and general relativity. It provides insights into the complexities of spacetime metrics and their physical interpretations.

pervect
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I've been looking for a simple exact, gravitational plane wave solution. Working from Wiki's short article on Brinkmann coordinates, I have what appears to be a simple exact solution - but it's significance and interpretation is confusing me a bit.

Let's start with the metric:

$$g = (y^2 - x^2) \, h(u) \, du \otimes du + du \otimes dv + dv \otimes du + dx \otimes dx + dy \otimes dy $$

The Einstein tensor is zero.
The Riemann is non-zero:

$$R = \frac{\partial}{\partial v} \, h(u) \,(dx \, du\, dx - dx \,dx\, du - dy\, du \,dy + dy \,dy \,du ) + \frac{\partial}{\partial x} \, h(u) \,(-du \, du \,dx - du \,dx \,du) + \frac {\partial}{\partial y} \, h(u) \,(du \,du \,dy - du \,dy \,du)$$

The confusion arises when we try to find an orthonormal space-time basis.

For instance, if we take
$$e_i = \left[ \frac {\frac{\partial}{\partial v} - \frac{\partial}{\partial u} } {\sqrt{2+h(u)(x^2-y^2)}} , \quad
\frac{\partial}{\partial x}, \quad
\frac{\partial}{\partial y}, \quad
\frac {\frac{\partial}{\partial u} + \left( 1 + h(u)(x^2 - y^2 )\right) \frac{\partial}{\partial v} } {\sqrt{2+h(u) (x^2-y^2) }}
\right] $$

we find that ##e_i \cdot e_j = \delta^i{}_j##, but our basis fails to make sense when ##2+h(u) (x^2-y^2) = 0##

I'm basically not sure what to make of this physically. Can we say the metric is non-singular, and that it's just impossible to have a global orthonormal space-time split? Or are there some singularity issues with the metric (but the components of the Riemann look fine before we tried to find an orthnormal basis).

I suppose the other question I should ask is if this really is a gravitational plane wave solution. I'm pretty sure it's exact, unless there's an error in calculationg the Einstein tensor as zero.
 
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Equations (2.66) and (2.67) in the notes.
 
Thank you both! There's a lot to absorb, but (2.66) is basically the same line element I was looking at, except for minor formatting differences (writhing 2 du dv as du dv + dv du and introducing h(u)).
 
pervect said:
I'm basically not sure what to make of this physically. Can we say the metric is non-singular, and that it's just impossible to have a global orthonormal space-time split? Or are there some singularity issues with the metric (but the components of the Riemann look fine before we tried to find an orthnormal basis).

In section 2.6, there is a neat proof that all curvature invariants are zero. In section 2.7, he shows that there can be singularities but only at points where the metric components are singular (as functions). So, my guess is that your space-time is non-singular and you only have coordinate problems. It is not clear to me whether there can be or not a global orthonormal basis.
 
I suppose one productive step would be to think about the geodesics in this space-time. Which entails solving the geodesic equations - well, perhaps there is some other approach, but it's the one that comes to mind.

If we set h(u) equals one, the computer algebra spits out a fair number of Killing vectors, but it doesn't look good for ##\partial / \partial x## and ##\partial / \partial y## remaining finite :(. Which looks bad for geodesic completeness.

Without setting h(u) to one, finding the Killing vectors seems to be difficult.
 
The geodesics are considered in section 2.3. If the function ##h(u)## is defined and regular for all ##u##, then there are no incomplete geodesics.
 

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