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Carroll Problem #15 of chapter 3 - As trivial as it seems?

  1. Mar 29, 2013 #1

    WannabeNewton

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    Hey guys! Problem #15 of Chapter 3 in Carroll's text says: Use Raychaudhuri's equation to show that if a fluid is flowing on geodesics through space - time with zero shear and expansion, then space - time must have a time - like killing vector.

    As background: Raychaudhuri's equation and all the things formulated below apply to a time - like geodesic congruence. The equation itself is ##\frac{\mathrm{d} \theta}{\mathrm{d} \tau} = -\frac{1}{3}\theta^2 - \sigma_{\mu\nu}\sigma^{\mu\nu} + \omega_{\mu\nu}\omega^{\mu\nu} - R_{\mu\nu}U^{\mu}U^{\nu}## where ##\omega_{\mu\nu}## is the rotation of the 4 - velocity field ##U^{\mu}## of the congruence, ##\sigma_{\mu\nu}## its shear, and ##\theta## its expansion. Given the tensor ##B_{\mu\nu} = \triangledown _{v}U_{\mu}## (this is its definition - its physical significance is that it measures the failure of separation vectors between neighboring geodesics in the congruence to be parallel transported), the rotation,shear, and expansion are defined as ##\theta = \triangledown _{\mu}U^{\mu},\sigma_{\mu\nu} = B_{(\mu\nu)} - \frac{1}{3}\theta P_{\mu\nu}, \omega_{\mu\nu} = B_{[\mu\nu]}## where ##P_{\mu\nu}## is just the projection tensor onto the subspace orthogonal to the 4- velocity field at each point.

    Here we have a fluid traveling on geodesics which is exactly a time - like geodesic congruence so the above formulations can be used. Now the problem says that both the shear and expansion are zero which tells us immediately that ##B_{(\mu\nu)} = \triangledown _{(\nu}U_{\mu)} = 0## thus ##U^{\mu}## solves killing's equation and is time - like since it is the 4 - velocity field so it gives us a time - like killing vector field. I don't even see why Carroll mentioned explicitly using Raychaudhuri's equation because the result the problem wanted is so immediate from the definitions above when given the zero shear and expansion that it makes me question why this problem was even put in...did I over look something?
     
    Last edited: Mar 29, 2013
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  3. Mar 29, 2013 #2

    bcrowell

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    Your reasoning looks right to me, and it does seem strange to me that he would explicitly say to use Raychaudhuri's equation when it's not needed. Maybe there's something we're both missing.
     
  4. Mar 29, 2013 #3

    WannabeNewton

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    Thanks for responding! I hope that we are indeed missing something otherwise this has got to be one of the most effortless problems I've seen in a physics book in a while lol.
     
  5. Mar 30, 2013 #4

    Mentz114

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    Obviously if ##B_{\mu\nu} = \triangledown _{v}U_{\mu}=0## then the acceleration vector ##\triangledown _{v}U_{\mu}U^{\nu}=0## and U is a geodesic. Maybe Carroll wants you to show that ##\Gamma^\mu_{\alpha\beta}U^\alpha U^\beta = 0##. I doubt if that's trivial.
     
  6. Mar 30, 2013 #5

    WannabeNewton

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    Hmm? We are given that the fluid travels on a geodesic etc. already in the problem. That was so that we could actually use the formulations surrounding the Raychaudhuri equation. The question itself just asked given these conditions (zero shear and expansion), show that space -time has a time - like killing vector field, nothing else. If you'll notice the definition of shear is given by ##\sigma_{\mu\nu} = B_{(\mu\nu)} - \frac{1}{3}P_{\mu\nu}\theta##. Since ##\sigma_{\mu\nu} = 0,\theta = 0## as given in the problem, ##B_{(\mu\nu)} = \triangledown _{(\nu}U_{\mu)} = 0## (here the parenthesis around the indices are the symmetrization brackets - the shear is the symmetric part of ##B_{\mu\nu}## and it's the symmetric part that vanishes - not ##B_{\mu\nu}##). This immediately tells us the space - time has a time - like killing vector field given by the 4 - velocity field.

    There was like literally zero work involved which is why it made me question why this problem existed in the first place lol. I'm not sure why we have to show what you said - not seeing how anything beyond the above is even needed for what the problem wants.
     
  7. Mar 30, 2013 #6

    Mentz114

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    Somehow I missed this bit so my post is less than irrelevant.:redface:
     
  8. Mar 30, 2013 #7

    WannabeNewton

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    xD. Well I scrutinized really hard (in fact I spent like 1000x more time trying to see if my solution had a flaw somewhere than in actually coming up with the solution lol) and I couldn't really see anything wrong in what I did. Maybe the problem really is that insultingly trivial.There is another problem (chapter 4 problem 5 IIRC) that seems to be more challenging. That is to show that given a fluid that generates a metric solution to the EFEs, the metric can be static only if the fluid 4 - velocity is parallel to the resulting time - like killing vector field. I'll have to have a go at it :smile:
     
  9. Mar 30, 2013 #8

    Ben Niehoff

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    It sounds like you found the most efficient way to do this problem.
     
  10. Mar 30, 2013 #9

    WannabeNewton

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    Heh well maybe for kicks I can try and show it using just the Raychaudhuri equation. Thanks guys!
     
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