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Homework Help: Differentiation woes with temperature/entropy relations.

  1. Sep 29, 2007 #1
    Alright, this is probably a really redundant question but for some reason it is giving me trouble. Let's say you are given the entropy of a black hole as:

    [tex] S=\frac{8\pi^2GM^2k}{hc} [/tex] (thanks Stephen Hawking)

    And you have the relation between temperature and entropy/energy

    [tex] \frac{1}{T}= \frac{\partial S}{\partial U} [/tex] (U is energy, S is entropy)

    Now if you want an expression for the temperature of a black hole in terms of it's mass and you are using U = Mc^2, then should you rewrite:

    [tex] M^2 = \frac{U^2}{c^4} [/tex]

    [tex] S = \frac{8\pi^2GkU^2}{hc^5} [/tex]

    And then differentiate with respect to U to get:

    [tex] \frac{1}{T} = \frac{16\pi^2GkU}{hc^5} [/tex]

    [tex] T = \frac{hc^5}{16\pi^2GkU} [/tex]

    First of all, as the ENERGY increases the TEMPERATURE decreases? Is this really possible here? I'm confused by this. Finishing the problem, though, and rewriting U = Mc^2 gives:

    [tex] T = \frac{hc^3}{16\pi^2GkM} [/tex]

    Does this seem correct? I tried working the problem a different way by writing c in terms of U as well at the start, and differentiating that expression and got a completely different answer...one that is always negative no less. So with that approach you get constantly negative temperatures...I'm very confused by what result I should be looking for.
  2. jcsd
  3. Sep 29, 2007 #2


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    Homework Helper

    What you did here looks right to me.
  4. Sep 29, 2007 #3


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    (Heuristically speaking) Remember that as a black hole gains energy, it expands, which has a cooling effect. This apparently dominates other effects. This is why we expect large black holes to be very stable, whereas tiny black holes should evaporate away very quickly.
  5. Sep 29, 2007 #4
    Thanks then.

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