Dimensional analysis on an Ising model solution

[JD_PM]

Homework Statement

The mean field solution for the Ising model is:

$$m = tanh[\beta (mJz + H)]$$

I wanted to carry out a dimensional analysis in order to verify the equation.

Homework Equations

$$m = tanh[\beta (mJz + H)]$$

The Attempt at a Solution

Knowing that:

$$[m] = \frac{A}{L}$$

$$[\beta] = \frac{T^2}{ML^2}$$

$$[J] = \frac{ML^2}{T^2}$$

$$[H] = \frac{A}{L}$$

As dimensions of $\beta$ and $J$ cancel out and $z$ is dimensionless you get the desired dimensions, so no problem with the $\beta mJz$ component.

However $\beta H$ does not yield the desired dimensions, so I guess I made a mistake coping on my lecture notes and the equation should be:

$$m = tanh[\beta (mJz) + H)]$$

Do you agree?

 

[TSny]

Check your notes to see how each symbol is defined. I suspect that $m$ is defined as a dimensionless quantity and that both $H$ and $J$ are energies.

 

[scottdave]

I'm not familiar with this particular problem, though I did go look and read a little about Ising.

But I know this. Every formula or equation that I have encountered has e or any number raised to a power, the power must be dimensionless. tanh() is essentially raising e to something and its negative and subtracting and dividing those quantities. That means the argument of tanh() should be dimensionless. And tanh itaelf should also be dimensionless. For those reasons, I agree with @TSny that m should be dimensionless.

 

[mastrofoffi]

Of course I do find dimensional analysis to be of great use, but I think your approach here is kind of cumbersome: dimensional analysis is useful if you can pick what the "interesting" dimensions in your problem are. Pulling out ampere(not totally sure what that $A$ means, but I guess it must be ampere), mass, time in the context of Ising model seems useless(at best) if not counterproductive. What I would care about here is just energy.

Since you have the solution, it should be clear to you that $m$ is a dimensionless quantity due to the fact that $tanh$(and likewise all the transcendental functions) needs a dimensionless argument and returns a dimensionless value, as @scottdave noted.
In case you didn't know the solution yet, you can tell that your field $H$ has the dimensions of an energy simply by looking at how the energy of the system is defined; you have
$$ E = -\sum_{i,k}J_{ik}\sigma_i\sigma_k + \sum_i\sigma_iH_i $$
where $\sigma_i$ is the "spin" of site $i$, $H_i$ is the field at site $i$ and $J_{ik}$ is the coupling energy of sites $i,k$.
Clearly you want the two addends to have the same dimensions; in principle you have an infinite number of possibilities of assigning the units(these are just conventions in the end) but given the units of spin, you can tell what the units of $J$ and $H$ are. Since usually the spins are introduced as dimensionless quantities, $[\sigma]=1$, you can easily see that you must have $[J] = [H] = [E]$ i.e. $J$ and $H$ both have dimensions of energy, making the first formula (dimensionally) correct. This, combined with the fact that the magnetization is defined as the expectation value of the spins, is consistent with $[m]=1$.

In the end I guess that the point of my post is that no one likes to work with weird combinations of units, so just keep it as simple as possible.

 

[JD_PM]

Thank you all.

$H$ is the magnetic field, which has dimensions of amperes per meter. I do not see how these dimensions are equivalent to the energy ones...

 

[TSny]

Can you write the Hamiltonian for the system as it was given to you in your notes?

 

[JD_PM]

Can you write the Hamiltonian for the system as it was given to you in your notes?

It was provided by @mastrofoffi:

$$E = -\sum_{i,k}J_{ik}\sigma_i\sigma_k + \sum_i\sigma_iH_i$$

 

[TSny]

It was provided by @mastrofoffi:

$$E = -\sum_{i,k}J_{ik}\sigma_i\sigma_k + \sum_i\sigma_iH_i$$

OK. (Is this identical to what is in your notes?) What do you find if you do a dimensional analysis of this equation?

 

[JD_PM]

OK. (Is this identical to what is in your notes?) What do you find if you do a dimensional analysis of this equation?

Yes, it is the same equation that I have.

You get energy dimensions. However I still do not know why amperes per meter are equivalent to energy dimensions...

 

[TSny]

When the ising model Hamiltonian is written this way, the quantity $H$ represents the magnetic field multipliled by the magnetic moment $\mu$ of a spin. So, $H$ in the Hamiltonian has the dimensions of energy. People still refer to $H$ as the "magnetic field" even though the actual field would be $H/\mu$. Thus, if $H$ is the quantity as given in the Hamiltonian, it would have energy units.
$H/\mu$ would have SI units of A/m.

Also, the magnetization $m$ is usually normalized by $\mu$ so that $m$ is dimensionless.