Differentiation woes with temperature/entropy relations.

In summary, the conversation discussed the relation between entropy and temperature of a black hole, and how to calculate the temperature in terms of its mass and energy. It was noted that as the energy of a black hole increases, its temperature decreases due to the cooling effect of expansion. The resulting equation for temperature in terms of mass and energy was confirmed to be correct.
  • #1
EricVT
166
6
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.
 
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  • #2
What you did here looks right to me.
 
  • #3
EricVT said:
First of all, as the ENERGY increases the TEMPERATURE decreases?
(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.
 
  • #4
Thanks then.

Cheers.
 

1. What is the relationship between temperature and entropy?

The relationship between temperature and entropy is known as the Second Law of Thermodynamics. It states that as the temperature of a system increases, the entropy (or disorder) of the system also increases.

2. How does temperature affect the rate of differentiation?

In general, an increase in temperature leads to an increase in the rate of differentiation. This is because higher temperatures provide more energy for molecules to move and react, leading to faster and more frequent differentiation processes.

3. Can temperature changes cause differentiation woes?

Yes, temperature changes can cause differentiation woes. When the temperature of a system changes rapidly, it can disrupt the balance of chemical reactions and alter the rate of differentiation. This can result in unexpected or undesirable outcomes.

4. How can temperature be controlled to avoid differentiation woes?

Temperature can be controlled by using a thermostat or other temperature-regulating equipment. It is important to maintain a stable temperature in order to avoid differentiation woes and ensure consistent results in scientific experiments.

5. Are there any specific temperature ranges that are ideal for differentiation?

The ideal temperature range for differentiation depends on the specific system and reactions involved. However, in general, temperatures within the range of 20-37 degrees Celsius (68-98.6 degrees Fahrenheit) are commonly used for biological differentiation processes.

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