Calculate Temperature of 1 Solar Mass Black Hole

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Homework Help Overview

The problem involves calculating the temperature of a one solar mass black hole, utilizing concepts from thermodynamics and black hole physics.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive the temperature using entropy and energy equations but questions their reasoning regarding the relationship between mass and energy. They express uncertainty about their calculations and seek validation of their thought process.
  • Some participants suggest considering the derivative of mass with respect to energy as part of the calculation, indicating a potential oversight in the original poster's approach.
  • Others mention the use of the chain rule as a method to connect the variables involved in the problem.

Discussion Status

Participants are actively engaging with the original poster's reasoning, providing insights into the mathematical relationships involved. Guidance has been offered regarding the application of the chain rule and the importance of considering all derivatives in the calculations. There is an ongoing exploration of the problem without a clear consensus on the final approach.

Contextual Notes

There is a mention of potential confusion regarding the relationship between mass and energy, as well as the implications of using the chain rule in this context. The original poster expresses uncertainty about their calculations and the correctness of their results.

Crush1986
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Homework Statement



The problem is to calculate the temperature of a one solar mass black hole

Homework Equations



S = \frac{8\pi^2GM^2k}{hc}
E = Mc^2
\frac{1}{T} = \frac{\partial S}{\partial U}

The Attempt at a Solution


[/B]
My first solution I pulled out an Mc^2 Which left my equation looking like S = \frac{k8\pi^2GM} {hc}*U 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!
 
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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.
 
JorisL said:
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.
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.
 
Crush1986 said:
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.

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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