1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Homework Helper

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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    (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.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Differentiation woes temperature Date
Proving a wave satisfies the Helmholtz equation Monday at 12:29 AM
I need help with Pendulum Movement Feb 27, 2018
Differential equations question Dec 19, 2017
Kinematics problem with differential equations. Dec 15, 2017
Tension Woes May 2, 2014