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Schwarzschild black hole/kruskal extension

  1. Apr 9, 2007 #1
    i am trying to understand why dr/dTau must be negative for a future-directed (physical) observer in the Schwarzschild metric. it says

    "we also know (e.g. from the Kruskal picture) that the sign of dr/dTau must be negative for a "future-directed" (i.e. physical) observer"

    i am probably missing something obvious.
     
  2. jcsd
  3. Apr 9, 2007 #2
    =(

    please help me, i know this has to be simple
     
  4. Apr 15, 2007 #3

    Chris Hillman

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    You haven't given enough information. What is this "it" which is apparently the source of your information? You'll also need to tell us what r and Tau are. It you spot a line element like
    [tex] ds^2 = -(1-2m/r) \, du^2 + 2 \, du \, dr + r^2 \, d\Omega^2 [/tex]
    in the book (?) you are reading, that would be a good thing to quote. (Click on the formula to see how to display a mathematical formula in this forum.) Be careful to write it out exactly as it appears. "Future-directed" probably refers to the tangent vectors to a parameterized world line of some idealized observer, not to the observer himself, so be careful to use terminology exactly as your source does. If I were to guess, I would guess that you also omitted to say that this derivative is supposed to be negative only inside the event horizon of some black hole model.
     
    Last edited: Apr 15, 2007
  5. Apr 15, 2007 #4
    i was speaking of the kruskal extension, so i'm referring to the schwarzschild black hole model with the standard schwarzschild metric. my textbook is wald so i'm assuming everything he uses is pretty much standard. i'm not sure whether this derivative is supposed to be negative inside the event horizon because i'm not sure what they meant. that could be true. the problem this was taken from involves crossing the event horizon but i'm not sure whether this derivative condition was for r < 2m or r > 2m or for all r (which is what i initially assumed, because they didn't specify).
     
  6. Apr 15, 2007 #5

    Chris Hillman

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    Suggest an easier book

    dookie, most likely you saw some remark referring to the interior region of the Schwarzschild hole, but since you can't or won't answer my questions about the notation you used in your initial post, I really can't say.

    I suggest that you study a very elementary book by Wald's colleague Robert Geroch, General Relativity from A to B. The goal of this book is to explain how to draw revealing pictures of the Schwarzschild hole without using any calculus at all, so it directly addresses the issues which seem to be troubling you. (BTW, if my guess is right about what you might have read in Wald, the Kruskal-Szekeres maximal analytic extension is irrelevant to the issue which is confusing you; Geroch's picture book employs a variant the ingoing Eddington chart, which covers half of the Carter-Penrose diagram you might have seen.) It sounds to me as though you should find this book very helpful.
     
  7. Apr 15, 2007 #6
    [tex] ds^2 = (1-2m/r) \, dt^2 -(1-2m/r)^-1 \, dr^2 - r^2 \, d\Omega^2 [/tex]

    this is the schwarzschild metric as stated in wald
     
  8. Apr 15, 2007 #7
    in any case if it IS for just the interior region of a schwarzschild black hole why is that so? because r becomes time-like and t becomes space-like? what does that mean physically?
     
  9. Apr 15, 2007 #8

    Chris Hillman

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    Not helping?

    You should have written
    [tex] ds^2 = (1-2m/r) \, dt^2 -(1-2m/r)^{-1} \, dr^2 - r^2 \, d\Omega^2 [/tex]
    This is the standard Schwarzschild chart for the Schwarzschild vacuum solution, which is only valid in "the" exterior region, so you can't use this to discuss observers in "the" interior region, as in (apparently) the discussion which you read someplace else (in Wald?) Did you really not notice that you originally mentioned [itex]\tau[/itex], not [itex]t[/itex]?

    Again, I can only guess, but most likely [itex]\tau[/itex] refers to the proper time of some observer, whereas the Schwarzschild time coordinate[itex]t[/itex] refers to the time kept by an exterior observer, static wrt the object, "at spatial infinity".

    If, as I suspect, you read something somewhere about observers in the interior, then you need to use a chart which covers "the" interior, such as the Eddington chart, Painleve chart, or Lemaitre chart.

    dookie, I think I've helped you as much as I am able, given the fact that you are unable or unwilling to clarify exactly what you read.
     
  10. Apr 15, 2007 #9
    Dookie, "Robert Geroch, General Relativity from A to B" is a popular science book for people with basic or no knowledge of relativity. Also, contrary to what the title suggests, it is mostly about special relativity.
     
    Last edited: Apr 15, 2007
  11. Apr 15, 2007 #10
    given that i no longer really need the answer (but would still very much appreciate understanding it), i wasn't going to dig out the book but here it is:

    problem book in relativity and gravitation, lightman et al

    the problem:

    "show that once a rocket ship" (the observer) "crosses the gravitation radius (horizon)" (event horizon) "of a schwarzschild black hole, it will reach r = 0 in proper time tau =< pi*M, no matter how the 'engines are fired'" (geodesic motion or with acceleration)

    their solution (which apparently *does* use the schwarzschild metric in the interior region although you say that it cannot?) i am going to use T as tau:

    "for the rocket ship to be moving along a timelike world-line its 4-velocity must satisfy (from the schwarzschild metric)

    1 = - u(dot)u = (1-2M/r)(dt/dT)^2 - (1-2M/r)^-1(dr/dT)^2 - r^2 (dtheta/dT)^@ - r^2sin^2theta (dphi/dT)^2"

    clearly they just set ds^2 = -dT^2 and divided by -dT^2, for the std schwarzschild metric (yes in a vacuum). continuing.

    "inside the horizon all of these terms are negative, except the one in (dr/dT)^2, so

    (2M/r-1)^-1(dr/dT)^2 > 1"

    this is also clear

    "we also know (e.g. from the eddington-finkelstein or kruskal picture) that the sign of dr/dT must be negative for a "future directed" (i.e. physical) observer and thus,

    dr < -(2M/r-1)^1/2dT"

    they then go on to solve the integral for Tmax which is fine.

    so i believe here they *are* using the schwarzschild metric in the interior region. but i didn't know for sure, since i don't know if that was valid which is really why i am confused. they do not say "must be negative for a future-directed observer in region II (interior)" or specify in any way where this observer is.
     
  12. Apr 15, 2007 #11

    Chris Hillman

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    Why not consider following my advice?

    Well, my previous advice stands.

    Now that you have finally bothered to tell us where you found the problem and what it actually states, it has become clear that my suspicions were correct:

    1. the problem you were trying to work does indeed concern the "future interior" of the Schwarzschild vacuum,

    2. you entirely missed the point of the problem.

    I see that you haven't even yet grasped the geometry of the local light cones inside the event horizon. This suggests to me that you are not ready for the problem book yet, much less Wald, so I advise you to put your advanced books away until you have more geometric insight. At this point they are obviously only confusing you, and possibly allowing you to believe you know much more than you really do. Please don't take offense, and don't be put off by the fact that Geroch's book is high school level--- it is sufficiently challenging to be interesting, and I repeat, at this point in your understanding it is just what you need.

    File the following points for future reference, when you revisit this material after learning enough geometric background to try again to start learning those aspects of gtr which involve calculus:

    1. Don't confuse metric (as in the chosen metric tensor, which must be valid everywhere on the given Lorentzian manifold, by definition) with "chart",

    2. Don't omit coordinate ranges when writing out a line element in a coordinate chart. In particular note that the exterior Schwarzschild chart
    [tex]ds^2 = -(1-2m/r) \, dt^2 + \frac{dr^2}{1-2m/r} + r^2 \, d\Omega^2, \; -\infty < t < \infty, \; 2m < r < \infty [/tex]
    is valid on a domain which is disjoint from the interior Schwarzschild chart
    [tex]ds^2 = -(1-2m/r) \, dt^2 + \frac{dr^2}{1-2m/r} + r^2 \, d\Omega^2, \; -\infty < t < \infty, \; 0 < r < 2m [/tex]
    Figure out why you can and should rewrite the latter as
    [tex]ds^2 = \frac{-d\tau^2}{1+2m/\tau} + (1+2m/\tau) \, dz^2 + \tau^2 \, d\Omega^2, \; -\infty < z < \infty, \; -2m < \tau < 0[/tex]

    3. If you want to study infall into a black hole, you will need to use a chart which covers both the exterior and future interior regions, such as the Eddington, Painleve, Lemaitre, or Novikov charts. If you want to understand the global structure, you will need to paste together information from various charts without getting confused (which can be hard for novices), or else to use a global chart such as the Kruskal-Szekeres chart or the "compactified" Penrose chart.

    I've said all of these things before, maybe even to yourself under a previous handle, so repeating myself it getting a bit tedious. I added you to my ignore list and hope the moderators lock this thread before it becomes even more pointless.
     
    Last edited: Apr 15, 2007
  13. Apr 15, 2007 #12
    well, it doesn't really matter whether or not you think i'm ready to study this because i have to, since i'm taking the course (and don't tell me to withdraw, too late)

    and

    i didn't really miss the point of the problem, since i already discussed it with my professor and turned in the homework, i was just trying to understand one step of the solution stated in that book.
     
  14. Apr 19, 2007 #13
    Good luck with your studies Dookie!
    And don't worry too much, even Hilbert made mistakes with regards to the Schwarzschild solution.
     
    Last edited: Apr 19, 2007
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