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Black Holes in String Theory - calculation help needed!

  1. Mar 31, 2013 #1
    I have a question about the paper:

    C. G. . Callan, R. C. Myers and M. J. Perry, “Black Holes In String Theory,” Nucl. Phys. B
    311, 673 (1989).

    I have attached the relevant section.

    I am having trouble using equations (2.1) and (2.4) to derive (2.5) and (2.6). When I do the calculation, I do not get any [itex]\phi'[/itex] terms (i.e. first derivatives --- I do get all the second-derivatives of phi and all other terms).

    For example, I don't understand how the (tt) and (θθ) equations can have [itex]\phi'[/itex] terms. Doesn't (2.1a) contain
    [itex]\nabla_{a}\nabla_{b}\phi = \partial_a \partial_b \varphi(r) = 0[/itex] unless [itex]a = b = r [/itex]?
    Thus shouldn't only the (rr) equation contain a [itex]\phi[/itex]-term... and shouldn't it just be a second derivative?

    For another example, since [itex]\nabla \phi[/itex] is first-order in [itex]\lambda[/itex], isn't [itex](\nabla \phi)^2[/itex] second-order in [itex]\lambda[/itex] and thus irrelevant for equation (2.6)? This would lead me to conclude that there should be no first derivative of [itex]\phi[/itex] in (2.6). Where does it come from?

    I would greatly appreciate some insight. Tell me where I am going wrong!

    Thanks in advance,

    Attached Files:

  2. jcsd
  3. Mar 31, 2013 #2
    Maybe I should clarify my question, in order not to scare people away. What I am asking should not take too much time for someone familiar with general relativity.

    I am only asking that you look at equation (2.1a), mentally insert (2.4) into it, and let me know how a phi-prime appeared in all three equations (2.5).

    I am very grateful to anyone who tries to help.

  4. Apr 1, 2013 #3
    Does anyone have any ideas?

    I might need to reproduce this calculation in front of my professor tomorrow evening, so I will be extremely grateful for any help soon.

  5. Apr 1, 2013 #4
    I have no idea how to understand these equations and have very little experience but from a computer programming perspective I can usually look through foreign "code" and get a general idea.

    From the excerpt provided I looked at figure (2.5) and looked very closely at lines (tt) and (θθ) I don't see the phi-prime symbol ϕ′ but I do see φ' several times.

    Sorry if this is no help at all, I'm trying to learn these things coming from a different perspective.
  6. Apr 1, 2013 #5
    You are right - I have accidentally been using [itex]\phi'[/itex] and [itex]\varphi'[/itex] interchangeably, since they differ only by a multiplicative factor. You can see this from equation (2.4) since [itex]\phi_0[/itex] is an arbitrary constant.

    But I still don't understand why [itex]\varphi'[/itex] should appear anywhere in any equation of (2.5).

    I can rephrase the question as:

    Using (2.4), what is [itex]\nabla_a \nabla_b \phi[/itex] ?

    It seems like a very easy question to me, but my answer to that questions seems to be in contradiction with what is in (2.5).

    Please keep asking questions so we can figure this out together.

    Thank you,
  7. Apr 1, 2013 #6


    Staff: Mentor

    This equality is not correct in curved spacetime, and it looks like the text is using curved spacetime (i.e., gravity is present). In curved spacetime there are extra terms in the covariant derivative [itex]\nabla_a[/itex] involving the connection coefficients.
  8. Apr 1, 2013 #7
    PeterDonis -

    Thank you very much! You figured it out. Indeed [itex]\nabla_a \phi = \partial_a \phi[/itex] since [itex]\phi[/itex] is a scalar field, but the second covariant derivative must incorporate the connection coefficients and the [itex]\partial_a \phi[/itex]'s. I feel silly for overlooking this - my mind was stuck for very long.

    Anyway, thanks again for the help. I am very grateful.

  9. Apr 1, 2013 #8
    Just to check:

    I also need to use [itex]\square \phi = g^{ab} \nabla_a \nabla_b \phi[/itex] - correct?
  10. Apr 1, 2013 #9
    Yes, it is. Alright - everything works out perfectly now.

    Thanks again.
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