- #1

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- 908

## Main Question or Discussion Point

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.

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.