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

Is Hawking's statement of Raychaudhuri's equation wrong?

  1. Nov 30, 2015 #1
    In Hawkin's lectures: http://arxiv.org/abs/hep-th/9409195v1, he states that the Raychaudhuri equation
    [tex] \dot \rho=\rho^2+\sigma^{ab}\sigma_{ab}+\frac 1 n R_{ab}l^a l^b [/tex] with n=2 for null geodesics and n=3 for timelike geodesics.

    But in most places I've seen [tex] \dot \theta=-\frac 1 n \theta^2-\sigma^{ab}\sigma_{ab}-\ R_{ab}l^a l^b [/tex] It doesn't seem they are equivalent because there is a switched sign and the n factor. Is this an errata?
     
  2. jcsd
  3. Nov 30, 2015 #2

    bcrowell

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    What page of the lectures is this on? What is the definition of ##\rho##? If you have access to Hawking and Ellis, can you see how it compares on this topic?

    Seems unlikely that Hawking would simply goof on this. This is the kind of thing that he made his career out of.
     
  4. Dec 1, 2015 #3
    It's on page 8 but he doesn't define ρ. I'll use the other equation, which is the one used in Hawking & Ellis, Carroll, etc.
     
  5. Dec 1, 2015 #4

    martinbn

    User Avatar
    Science Advisor

    He does, right below the box.
     
  6. Dec 1, 2015 #5
    I meant a rigorous definition to compare it with Hawking & Ellis.
     
  7. Dec 1, 2015 #6

    bcrowell

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    But it's only a verbal definition, not a mathematical one. It's not clear to me whether there is some difference between the quantity ##\rho## he's defining and the expansion scalar ##\theta##, possibly by a factor of ##n##. If there is some difference, it would explain why he notates it ##\rho## rather than ##\theta##.

    Note that he says:

    So we shouldn't expect it to be identical to the Raychaudhuri equation, we should expect the Raychaudhuri equation to be a special case of it. It may be that ##\rho## is a generalization of ##\theta##.

    Re the sign, note that the affine parameter ##v## is arbitrary, and if we replace it with ##-v## it's still a valid affine parameter. I guess the equation is supposed to be form-invariant under a change of affine parameter, but it's not manifestly so.
     
  8. Dec 2, 2015 #7

    martinbn

    User Avatar
    Science Advisor

    Ok, I see what you meant by no definition. He says that it is the average convergence. For him that may be as clear as any of the other terms, for example the shear. So just the name is enough. Since it is convergence and not expansion that would explain the sign difference. The average must be where the factor n comes from, but it seems that there should be a factor in front of the shear. Anyway, google shows various versions of the equation under all combinations of the names Raychaudhuri, Newman, Penrose.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Is Hawking's statement of Raychaudhuri's equation wrong?
  1. Raychaudhuri equetion (Replies: 2)

Loading...