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I should determine the spontaneous magnetization of an ferromagnet below its critical temperature $T_{c}$ by only knowing the derivatives:

$$ \left ( \frac{\partial M}{\partial H } \right )_{T}=\frac{a}{1-T/T_c}+3bH^{2} $$ where a and b are some real constants

$$ \left ( \frac{\partial M}{\partial T } \right )_{H}= \frac{1}{T_{c}}\frac{f(H)}{(1-T/T_c)^{2}}- \frac{1}{2}\frac{M_{0}}{T_{c}}\frac{1}{(1-T/T_c)^{1/2}} $$

with $M_{0}$,$T_{c}$,a and b are constants and f(H) is some function with the property f(H=0)=0

a)determine f(H) b)determine M(T,H)

Now I would start by writing down

$$ dM = \left ( \frac{\partial M}{\partial H } \right )_{T}dH + \left ( \frac{\partial M}{\partial T } \right )_{H} dT $$

This tells me how I would obtain the M in terms of the derivatives but I don't know how I would obtain the function f(H). Can anyone give me a hint how I would start the determination of f(H).

$\mathbf{Edit}$: I tried now to do the integration what gives: $$ M(T,H)=\frac{aH}{1-T/T_{c}}+bH^{3}+\frac{f(H)}{1-T/T_{c}}+M_{0}(1-T/T_{c})^{1/2} $$

If I compute now again the derivative with respect to H I obtain: $$ \left ( \frac{\partial M}{\partial H } \right )_{T}=\frac{a}{1-T/T_c}+3bH^{2}+\frac{1}{1-T/T_{c}}\frac{\partial f(H)}{\partial H} $$

This is now not the same as in the exercise sheet. From here I don't know how to go on. Is this already wrong or is it somehow possible from this equation to determine f(H). My first intuition was to say f(H) is just zero, because then the derivative with respect to H would be satisfied and so would be the f(H=0)=0 criterion

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  • $\begingroup$ You don't need to list an edit history. An edit history is available for those who want to see it. Just edit the question if you need to. $\endgroup$ Oct 2 '18 at 19:34
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I was asking the same question in the mathematics forum and they figured out an answer for me: https://math.stackexchange.com/questions/2940391/obtain-function-from-its-partial-derivatives

For completeness I will also sketch this answer here. Starting by doing the integration of the partial derivative with respect to H

$$ M(H,T)=\frac{aH}{1-T/T_{c}}+bH^{3}+\phi(T) $$

Now making the derivative with respect to T:

$$ \frac{\partial M(T,H)}{\partial T}=\frac{1}{T_{c}}\frac{aH}{(1-T/T_{c})^{2}}+\phi'(T) $$

By comparing the coefficients with the supplied derivative. $f(H)=aH$ and $\phi'(T)=-\frac{1}{2}\frac{M_{0}}{T_{c}}\frac{1}{(1-T/T_{c})^{1/2}}$.

The rest is then obtained by just inserting for f(H).

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