Stumped on mathematical proof

[laser1]

Homework Statement

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Relevant Equations

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From Blundell and Blundell Chapter 20 Problem 20.3.

I have proved that $$1-e^{-\beta \omega}=2\sinh\left(\frac{\beta \omega}{2}\right)$$ with no problem, but I am stuck on the $\coth$ term. I have tried to solve this but it gets messy and I'd rather not include them here. Thanks!

 

[PeroK]

If that $\coth$ identity holds, then:
$$e^{\beta \omega} + 1 = 2\beta$$

 

[laser1]

If that $\coth$ identity holds, then:
$$e^{\beta \omega} + 1 = 2\beta$$

Okay, assuming that is true, surely that equation can't be true for all $\beta$ and $\omega$?

 

[PeroK]

Okay, assuming that is true, surely that equation can't be true for all $\beta$ and $\omega$?

That is clearly not an identity. Why do you think the textbook is infallible?

 

[laser1]

That is clearly not an identity. Why do you think the textbook is infallible?

Alr fair enough. As a mere undergrad, I always assume I am wrong first rather than the textbook! Of course, if I can't reason with the textbook I will ask here/my lecturer.

 

[PeroK]

Alr fair enough. As a mere undergrad, I always assume I am wrong first rather than the textbook! Of course, if I can't reason with the textbook I will ask here/my lecturer.

It's not a question of assumptions. Are $\beta$ and $\omega$ related in that way?

 

[laser1]

It's not a question of assumptions. Are $\beta$ and $\omega$ related in that way?

I don't think so.

 

[PeroK]

I don't think so.

Perhaps check that first identity!

 

[PeroK]

I have proved that $$1-e^{-\beta \omega}=2\sinh\left(\frac{\beta \omega}{2}\right)$$ with no problem,

That's the mistake!

 

[laser1]

That's the mistake!

Edit: oh yeah I see the denominator now yikes

 

[PeroK]

View attachment 352694


Edit: oh yeah I see the denominator now yikes

Yikes, indeed!

 

[PeroK]

It's not a question of assumptions. Are $\beta$ and $\omega$ related in that way?

The expression just looked wrong to me. There is a factor of $\beta$ missing in (20.51). It should be:
$$\frac{\beta \omega}{2}\coth\big (\frac{\beta \omega}{2}\big )$$

 

[laser1]

The expression just looked wrong to me. There is a factor of $\beta$ missing in (20.51). It should be:
$$\frac{\beta \omega}{2}\coth\big (\frac{\beta \omega}{2}\big )$$

yeah I'm getting the same as you now. As in, I have the book answer, but the book is missing a factor of $\beta$ on the first term

 

[Orodruin]

I have not done the explicit computation (just visualized how the terms would combine), but apart from the missing $\beta$ the overall identity looks quite reasonable to me.