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Calculation on hawking radiation in schutz (pg 324)

  1. May 20, 2012 #1
    on pg 324 of Schutz's "A First Course in General Relativity", i am having a little trouble with the integral (11.100). the book says that to first order in [itex] \epsilon [/itex], the answer should be [itex] 2\sqrt{2M\epsilon} [/itex] but i keep getting [itex] \sqrt{2M\epsilon} [/itex]. i am missing that factor of 2 somehow.

    the integral in (11.100) is [itex] \int_{2M}^{2M + \epsilon} (\frac{2M}{r} - \frac{2M}{2M + \epsilon})^{\frac{-1}{2}} dr [/itex]. So i treat this integral as a function [itex] f(x) = \int_{2M}^{x} (\frac{2M}{r} - \frac{2M}{2M + \epsilon})^{\frac{-1}{2}} dr [/itex] and using the fact that taylor expanding a function gives f(a + h) = f(a) + f'(a)h. I want to find [itex] f(2M + \epsilon) [/itex] and i know that f(2M) is just 0 since the upper and lower bounds are the same. by the fundamental theorem of calculus, [itex] f'(2M) = (\frac{2M}{2M} - \frac{2M}{2M + \epsilon})^{\frac{-1}{2}} = (1 - \frac{2M}{2M + \epsilon})^{\frac{-1}{2}} [/itex], and taking the first order taylor expansion of [itex] (1 + \frac{\epsilon}{2M})^{-1} [/itex] (after dividing numerator and denominator by 2M), i get [itex] (\frac{\epsilon}{2M})^{-1/2} [/itex] and multiplying this by [itex] h = \epsilon [/itex], i get [itex] \sqrt{2M\epsilon} [/itex].

    but i can't figure out why im missing a factor of 2. can someone help me out?
     
  2. jcsd
  3. May 20, 2012 #2

    Bill_K

    User Avatar
    Science Advisor

    I get the same answer you do.
     
  4. May 20, 2012 #3
    ah that's reassuring. i was afraid i was making some mistake over and over again. i believe schutz did make a mathematical error earlier in the same chapter so its not unlikely for this to be another.

    thanks for your reply!
     
  5. May 20, 2012 #4
    Basically, you have one epsilon in the upper limit, and one in the integrand. You only accounted for the former.
     
  6. May 20, 2012 #5
    Make the sunbstitution.
    [tex]
    \left( \frac{2M}{r} - \frac{2M}{2M+\epsilon}\right)^{\frac{1}{2}} = x
    [/tex]
    When [itex]r = 2M[/itex] [itex]x = \left( \frac{\epsilon}{2M + \epsilon} \right)^{\frac{1}{2}}[/itex]. When [itex]x = 2M + \epsilon[/itex], [itex]x=0[/itex].

    [tex]
    r = \frac{2M}{\frac{2M}{2M + \epsilon} + x^{2}}
    [/tex]
    [tex]
    dr = -\frac{2M \, 2 x}{\left( \frac{2M}{2M+\epsilon} + x^{2}\right)^{2}} dx
    [/tex]
    and the integral becomes
    [tex]
    \int_{0}^{\sqrt{\frac{\epsilon}{2M+\epsilon}}}{ \frac{4M \, dx}{\left( \frac{2M}{2M+\epsilon} + x^{2}\right)^{2}}}
    [/tex]
     
  7. May 20, 2012 #6
    hm i thought i did account for the epsilon in the integrand. after taylor expanding the integral itself (taking account of the epsilon in the upper limit), i taylor expanded the expression containing epsilon in what was left over (accounting for the one in the integrand).

    what do you mean exactly? so starting from [itex] (1 - \frac{2M}{2M + \epsilon})^{\frac{-1}{2}}\epsilon [/itex], which is right after taking into account the epsilon in the upper limit, where should i have gone from there?
     
  8. May 20, 2012 #7
    The asymptotic behavior of the last integral in my previous post is
    [tex]
    4M \, \sqrt{\frac{\epsilon}{2M}}, \epsilon \ll 2M
    [/tex]
    which is what the textbook states
     
  9. May 20, 2012 #8
    so i am trying to taylor expand your integral and i get: [itex] 4M\sqrt{\frac{\epsilon}{2M}}(1 - \frac{\epsilon}{4M})(1 - \frac{\epsilon}{2M})^{-2} [/itex].

    since f(h) = f(0) + hf'(0) = hf'(0) where [itex] h = \sqrt{\frac{\epsilon}{2M + \epsilon}} [/itex] and [itex]f'(0) = \frac{4M}{(\frac{2M}{2M + \epsilon})^2}[/itex].

    i'm not sure how the last 2 terms disappear. did i do this correctly?
     
  10. May 20, 2012 #9
    What is the result to lowest order in [itex]\epsilon/(2M)[/itex]?
     
  11. May 20, 2012 #10
    well i have [itex] \frac{1 - \frac{\epsilon}{4M}}{1 - \frac{\epsilon}{M}} [/itex] and i'm just not sure where to go from there. am i allowed to do [itex] (1 - \frac{\epsilon}{4M})(1 + \frac{\epsilon}{M}) = 1 + \frac{\epsilon}{M} - \frac{\epsilon}{4M} [/itex]?

    even if i did do that, i would still have those epsilon terms that don't vanish to first order. could you show the process in a little more detail? i'm still not really used to taylor expanding expressions this complicated.
     
  12. May 20, 2012 #11
    protip : what is the value of
    [tex]
    \frac{1-\frac{t}{2}}{1 - 2t}, t \equiv \frac{\epsilon}{2M} = 0
    [/tex]
    Remember that I asked for the dominant behavior in the small t limit? BTW, this is not Taylor expansion, but asymptotic expansion, since [itex]\sqrt{t}[/itex] is nonanalytic.
     
  13. May 20, 2012 #12
    You are over calculating an aproximation. The leading term in the asymptotic expansion of the above expression is simply [itex]4M \sqrt{\frac{\epsilon}{2M}}[/itex].
     
  14. May 20, 2012 #13
    one thing i am a little confused about is when authors say things like "to lowest order" of some quantity, are they referring to a taylor expansion or an asymptotic expansion?

    as in the calculation of this integral to lowest order of [itex] \epsilon [/itex], when do you use taylor expansions and when do you use asymptotic expansion? i was under the impression that we just use a taylor expansion so i didn't know that [itex] \frac{1 - \frac{\epsilon}{4M}}{1 - \frac{\epsilon}{M}} [/itex] is just 1 by setting [itex] \epsilon = 0 [/itex].

    when are we allowed to just set epsilon = 0?

    thanks for your replies so far!
     
  15. May 20, 2012 #14
    When we refer to lowest order, we usually mean the asymptotic leading non-vanishing term in the zero limit (because higher terms are negligible in that limit). If a Taylor epansion exists, then it the asymptotic expansion. A fucntion might have an asymptotic, but no taylor expansion, for example [itex]f(x) = \sqrt{x}, x \rightarrow 0[/itex]
     
  16. May 20, 2012 #15
    is there another way to do this integral without making that substitution? many of the other "approximations to lower order" of some quantity in the text can be done simply by doing taylor expansions and ignoring terms of higher than first order.

    this integral however seems tricky and i'm confused on whether to expand the integral (taking the epsilon in the upper limit into account) or the integrand (taking the epsilon in the integrand into account). although i taylor expanded both out i get the wrong answer.

    what i don't understand is why the substitution is necessary. can this be done in an alternative method (which may be more intuitive for me) without such a substitution?
     
  17. May 20, 2012 #16
    The substitution I made is standard and allows you to calculate the integral exactly! Namely:
    [tex]
    \int{\frac{dx}{(x^2 + a^2)^2}} = -\frac{1}{2a} \, \frac{d}{d a} \int{\frac{dx}{x^2 + a^2}}
    [/tex]
    [tex]
    -\frac{1}{2a} \, \frac{d}{da} \left( \frac{\tan^{-1}(x/a)}{a} \right)
    [/tex]
    [tex]
    -\frac{1}{2a} \, \left[ \left(-\frac{1}{a^2} \right) \, \tan^{-1} \left( \frac{x}{a} \right) + \frac{1}{a} \, \frac{1}{1 + \left( \frac{x}{a} \right)^2} \left( -\frac{x}{a^2} \right) \right]
    [/tex]
    [tex]
    \frac{1}{2a^2} \, \left[ \frac{1}{a} \, \tan^{-1} \left( \frac{x}{a} \right) + \frac{x}{x^2 + a^2} \right]
    [/tex]
     
  18. May 20, 2012 #17
    ah i see. i've actually never seen that in action.

    so for the original integral, i taylor expanded both the integral and the integrand and i got the wrong answer (missing factor of 2). now using the substitution and obtaining the same integral in a new form, i once again taylor expand both the integral and the integrand and i get as above: [itex] 4M\sqrt{\frac{\epsilon}{2M}}(1 - \frac{\epsilon}{4M})(1 - \frac{\epsilon}{2M})^{-2} [/itex]. but in this case, when i take the leading term i get the right answer.

    what i'm concerned about is how come i used the same methods for both integrals but only for one of them i got the right answer?

    sorry for asking so many questions. not fully understanding this calculation has really been bugging me.
     
  19. May 20, 2012 #18
    Actually, I don't know what you did in your method, nor I have the time to go through it. If you feel up to it, perhaps you could write up all you steps in detail, and I will tell you where you had made a mistake.
     
  20. May 20, 2012 #19
    for the original integral:

    [itex] \int_{2M}^{2M + \epsilon} (\frac{2M}{r} - \frac{2M}{2M + \epsilon})^{\frac{-1}{2}} dr [/itex]

    so i consider this as a function [itex] f(x) = \int_{2M}^{x} (\frac{2M}{r} - \frac{2M}{2M + \epsilon})^{\frac{-1}{2}} dr [/itex]. the original integral was [itex] f(2M + \epsilon) [/itex]. Using taylor expansions we have [itex] f(2M + \epsilon) = f(2M) + f'(2M)\epsilon [/itex]. We have [itex] f(2M) = 0 [/itex] since [itex] \int_{2M}^{2M} (\frac{2M}{r} - \frac{2M}{2M + \epsilon})^{\frac{-1}{2}} dr = 0 [/itex]. by the fundamental theorem of calculus, [itex] f'(2M)\epsilon = (\frac{2M}{2M} - \frac{2M}{2M + \epsilon})^{\frac{-1}{2}}\epsilon = (1 - \frac{2M}{2M + \epsilon})^{\frac{-1}{2}}\epsilon [/itex].

    then we divide the top and bottom of the 2nd term by 2M to get: [itex] (1 - (1 + \frac{\epsilon}{2M})^{-1})^{\frac{-1}{2}}\epsilon [/itex]. and by taylor expanding the second term: [itex] (1 - (1 - \frac{\epsilon}{2M}))^{\frac{-1}{2}} \epsilon = (\frac{\epsilon}{2M})^{\frac{-1}{2}} \epsilon = (2M\epsilon)^{\frac{1}{2}} [/itex].

    so i basically taylor expanded the actual integral itself as a function, then i taylor expanded what was leftover from that expansion, and i seemed to get the answer off by a factor of 2.

    i'm not sure why the second integral is able to give the correct answer. you said that the problem with my solution was that i didn't take into account the epsilon in the integrand. but in the integral you posted, epsilon appears in both the integrand and the upper limit as well. so i don't see how it gives me a different answer when both situations are essentially the same.

    for the second integral:

    i use the same method, taylor expanding [itex] f(x) = \int_{0}^{x}{ \frac{4M \, dx}{( \frac{2M}{2M+\epsilon} + x^{2})^{2}}}[/itex]. The integral is given by [itex] f(\frac{\epsilon}{2M + \epsilon})^{\frac{1}{2}}) = f(0) + f'(0)(\frac{\epsilon}{2M + \epsilon})^{\frac{1}{2}} [/itex]. Then f(0) = 0 since the upper and lower limit is the same as was in the original integral i posted. Then [itex]f'(0)(\frac{\epsilon}{2M + \epsilon})^{\frac{1}{2}} = 4M(\frac{2M}{2M + \epsilon})^{-2}(\frac{\epsilon}{2M + \epsilon})^{\frac{1}{2}}[/itex]. then i divide the top and bottom by 2M of the first and second term so i get: [itex] 4M(1 + \frac{\epsilon}{2M})^{2} (\frac{\epsilon}{2M})^{\frac{1}{2}}(1 + \frac{\epsilon}{2M})^{\frac{-1}{2}} = 4M(1 + \frac{\epsilon}{2M})^{\frac{3}{2}}(\frac{\epsilon}{2M})^{\frac{1}{2}} [/itex]. then i taylor expand the [itex] (1 + \frac{\epsilon}{2M})^{\frac{3}{2}} [/itex] to get [itex](1 + \frac{3\epsilon}{4M}) [/itex]. finally we have [itex] 4M(\frac{\epsilon}{2M})^{\frac{1}{2}}(1 + \frac{3\epsilon}{3M}) [/itex] and since the second term contains a term of higher order than [itex] \epsilon [/itex], we are left with [itex] 4M(\frac{\epsilon}{2M})^{\frac{1}{2}} = 2(2M\epsilon)^{\frac{1}{2}} [/itex] which is the correct answer in this case.

    i really appreciate the effort you are making to reply to me!
     
  21. May 21, 2012 #20
    You need to redefine your function as:
    [tex]
    f(x) = \int_{2M}^{x}{\left( \frac{2M}{r} - \frac{2M}{x}\right)^{-\frac{1}{2}} \, dr}
    [/tex]
    Now, what is [itex]f'(2M)[/itex]?
     
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