I'm stumped. In this exercise, I've got a paramagnetic obeying a $m = \psi\left(\frac{H}{T}\right)$ law, where $m$ is the average magnetization (with the volume $V = 1$, so it's equivalent to the full magnetization), $H$ is the magnetic field intensity in $B = \mu_0 H$ and $T$ is of course the temperature. $\psi\left(\frac{H}{T}\right)$ is some known function.

I'm supposed to show that the internal energy $U(T, m)$ depends only on $T$ and not on $m$. Nnote the natural variables being "mixed" - as far as I know, the natural variables for $U$ are usually the entropy and the magnetization - extensive ones.

Looks simple enough, right? If $U$ does not depend on $m$, that must mean $\left(\frac{\partial U}{\partial m}\right)_T = 0$. For these natural variables, I would assume that the fundamental relation for the internal energy would be

$$dU = -S dT + \mu_0 H dm $$

But clearly, from this, $\left(\frac{\partial U}{\partial m}\right)_T = \mu_0 H$ - and I think I'm supposed to get a non-conditional zero, not just when there is no magnetic field. This is rather paradoxical to me and I think I'm getting something basic completely wrong. Help, please?

I also tried writing $dm$ as $\psi'\left(\frac{H}{T}\right) \left(\frac{dH}{T} - \frac{H dT}{T^2} \right)$, but then that would only let me get $\left(\frac{\partial U}{\partial H}\right)_T$, which seems unhelpful.

  • $\begingroup$ you have written the free energy not the internal energy, it should be $dU=TdS+\mu_0 H dm$ $\endgroup$ – hyportnex Feb 13 '19 at 19:15
  • $\begingroup$ 2nd hint: after correcting $dU$ divide both sides by $T$ $\endgroup$ – hyportnex Feb 14 '19 at 13:31

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