Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Exact gravitational plane wave confusion

  1. Jun 21, 2016 #1

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
  2. jcsd
  3. Jun 22, 2016 #2

    haushofer

    User Avatar
    Science Advisor

  4. Jun 22, 2016 #3

    martinbn

    User Avatar
    Science Advisor

    Equations (2.66) and (2.67) in the notes.
     
  5. Jun 22, 2016 #4

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    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)).
     
  6. Jun 23, 2016 #5

    martinbn

    User Avatar
    Science Advisor

    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.
     
  7. Jun 23, 2016 #6

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    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.
     
  8. Jun 26, 2016 #7

    martinbn

    User Avatar
    Science Advisor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Exact gravitational plane wave confusion
  1. Gravitational waves (Replies: 1)

  2. Gravitational waves (Replies: 3)

  3. Gravitational waves (Replies: 2)

  4. Gravitational waves (Replies: 3)

  5. Gravitational Waves (Replies: 15)

Loading...