# Internal energy of paramagnetic obeying m = function of H/T independent of magnetization?

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.

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