1
$\begingroup$

In a paramagnetic system, we have:

$$N = N_\uparrow + N_\downarrow$$.

If we have a large system, with $N >> 1$, is it generally okay to assume $N_\uparrow \approx \frac{N}{2}$ and $N_\downarrow \approx \frac{N}{2}$? I would like to make this approximation, but I'm having a hard time justifying it and determining why it is correct.

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ Well, if you assume that there is no interaction between the spins and that there is no external magnetic field, this follows immediately from the law of large numbers (for a collection of N independent Bernoulli random variables of parameter $1/2$). $\endgroup$
    – Yvan Velenik
    Commented Nov 10, 2013 at 9:37

2 Answers 2

Reset to default
3
$\begingroup$

That can depend on the temperature, among other things, so it is generally not OK:-) .

EDIT: I mean it can depend on the temperature if, e.g., the external magnetic field is not zero, and it can depend on the external magnetic field.

Improve this answer
$\endgroup$
2
  • $\begingroup$ So I thought my approximation was reasonable because in a large two-state paramagnetic system, the multiplicity is very sharply peaked about $N_\uparrow = \frac{N}{2}$. This means the most likely macrostate for the system will be $N_\uparrow = N_\downarrow = \frac{N}{2}$. Is this not true? Does this not justify the original statement? I'd like to show this rigorously if possible. $\endgroup$
    – vectorize7891
    Commented Nov 10, 2013 at 1:13
  • $\begingroup$ @vectorize7891: If, e.g, the external magnetic field is not zero and the temperature is close to zero, pretty much all spins will be directed along the external field. So I don't think you will be able to justify your statement at all without some caveats, let alone rigorously. $\endgroup$
    – akhmeteli
    Commented Nov 10, 2013 at 1:37
0
$\begingroup$

So a "pure" paramagnetic system has a positive magnetic susceptibility and for which the dipoles don't interact. Usually the effect of the "thermal bath" outweighs an applied magnetic field so the net magnetization is 0. I would say that in general the answer is yes seeing as it takes a Squid magnetometer to detect a paramagnetic system, but it is certainly not always true:

You can just look at the boltsmann distribution: $$\frac{N_{\uparrow}}{N_{\downarrow}} =exp{\frac{-E_{\uparrow}+E_{\downarrow}}{kT}}$$ For no external magnetic field: $E_{\uparrow}=E_{\downarrow}$, $$\frac{N_{\uparrow}}{N_{\downarrow}} = 1$$ Therefore, in the absence of a magnetic field (and at finite T) the two states are equally populated (randomized) and $$N_{\uparrow}\approx N_{\downarrow}\approx\frac{N}{2}$$ Even if we apply a magnetic field, at say room temperature, the denominator in the Boltzmann distribution will overwhelmingly outweigh the difference between $E_{\uparrow}$ and $E_{\downarrow}$ ($kT>>\mu_B B$). If you want to see when this is valid just plot the distribution with realistic numbers.

If you want to get fancy about it: $$ E=-B\sum_{i}S_i $$ When you solve for the partition function of this system and take the limit N-->$\infty$the expectation value of the magnetization is: $$ \frac{\langle M \rangle}{N}=-\frac{\phantom{X} \partial F}{N\partial B}=\mu_B tanh\frac{\beta\mu_B B} {kT}$$ Take a look at the behavior of this equation and check this text (~page 30) if you want to see how it is derived . I think the behavior of this equation is what akhmeteli was referring to.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.