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While reading David Tong's Statistical Physics lecture notes (https://www.damtp.cam.ac.uk/user/tong/statphys.html) I came across this weird identity in page 26 (at the end of the 1.3.4 free energy subsection)

$$ \frac{\partial}{\partial \beta} = -k_B T^2 \frac{\partial}{\partial T} $$

whereas $ \beta $ is as usual $ \frac{1}{k_B T} $. I can't wrap my head around this identity, so I would appreciate any help in understanding this.

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    $\begingroup$ It is just the chain rule of differentiation. $\endgroup$
    – hft
    Commented Sep 16, 2023 at 23:31
  • $\begingroup$ Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. $\endgroup$
    – Community Bot
    Commented Sep 16, 2023 at 23:52

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...I came across this weird identity...

$$ \frac{\partial}{\partial \beta} = -k_B T^2 \frac{\partial}{\partial T} $$

This is just the chain rule of differentiation.

whereas $ \beta $ is as usual $ \frac{1}{k_B T} $...

Yes, so that means that $$ \frac{dT}{d\beta} = -k_B T^2 $$

And therefore: $$ \frac{df}{d\beta} = \frac{dT}{d\beta}\frac{df}{dT} = -k_B T^2 \frac{df}{dT} $$

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    $\begingroup$ Thanks a lot! This is very helpful. $\endgroup$
    – duodenum
    Commented Sep 17, 2023 at 12:14

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