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Temp of a Blackhole

  1. Oct 13, 2015 #1
    1. The problem statement, all variables and given/known data

    The problem is to calculate the temperature of a one solar mass black hole
    2. Relevant equations

    [tex] S = \frac{8\pi^2GM^2k}{hc} [/tex]
    [tex] E = Mc^2 [/tex]
    [tex] \frac{1}{T} = \frac{\partial S}{\partial U} [/tex]
    3. The attempt at a solution

    My first solution I pulled out an [tex] Mc^2 [/tex] Which left my equation looking like [tex] S = \frac{k8\pi^2GM} {hc}*U [/tex] and did the partial derivative with respect to U of the entropy equation. I found that I was off by a factor of two (I think, I don't for sure know the right answer but some friends got answers 1/2 as much as my answer).

    I think I know why and I just want to check out my reasoning. By only factoring out one U instead of U^2 I left an M in the equation. But M and U are intricately related right? So I must take out a U^2 in order to take out all of the M's in the original equation. It is only then that I get the correct result (That being 6.14 *10^-8 K, which I believe to be right but I'm also not entirely sure.)

    Does this sound like a reasonable conclusion as to why I'm probably wrong?

    Thank you for any help you can offer!
     
  2. jcsd
  3. Oct 14, 2015 #2
    You forgot that through ##E=Mc^2## you have to use that ##\frac{\partial M}{\partial U}## also contributes to the derivative.
    Then you get the right result.

    Another approach you can use is using the chain rule.
    Then you can write ##\frac{1}{T} = \frac{\partial S}{\partial U} = \frac{\partial S}{\partial M}\frac{\partial M}{\partial U}##.
    Usually I would go for this.
    In this case the relation between E and M is simple but for harder problems (in possibly other domains) the algebraic manipulations can get ugly real quick increasing the probabilities of mistakes.
     
  4. Oct 20, 2015 #3

    This is so late I know, but, THANK YOU!!

    I didn't quite get it at the time... I mean I knew of the chain rule, I just never really recognized when to use it until JUST now.

    I've been shredding through a lot of problems tonight remembering to keep this little mathematical tool in my pocket.
     
  5. Oct 21, 2015 #4
    Your welcome. It's useful to look back at old problems whenever you learn something new.
    This helps you selecting suitable tools further on.
     
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