Statistical mechanics: Sums of exponentials with sums.

[Beer-monster]

Homework Statement

I'm working through an example from class and the textbook, but I'm confused about how the steps progress mathematically.

The example involves the Gibb's partition for a paramagnet.

[tex] \sum_{s} exp(\beta \mu B \sum_{i}^{N} s) [/tex]

Where s = -a,-a+1...a for each spin.

Am I right in thinking the sum in the exponent is a sum over this particles in the sample (N). Since a sum in an exponent is a product of the exponentials we get:

[tex] [\sum_{s} exp(\beta \mu B s)]^{N} [/tex]

However, I'm not sure what do with the second sum. I can't see a neat way proceed. Can anyone point me in the right direction?

 

[G01]

The second sum is a sum over spin states. If you're dealing with quantum systems with a discrete number of spin states, then it's probably best to write the sum explicitly and re-express the result in a more simple form. HINT: use some hyperbolic trig identities.

If you are treating the spin classically and the value of s can take on any value [itex]s_o cos( \theta )[/itex] (where theta is the angle from the z axis), then your best best will be to convert the sum to an integral in theta.

 

[Beer-monster]

Thanks for the response. I'm treating the spins as quantum. I managed to solve the taking out the factor of a and treating the rest as a geometric series i.e.

[tex][e^{-\beta\mu B a} \sum_{s=1}^{(2a+1)} exp(\beta \mu B s)]^{N} [/tex]

Which, using the formula for the sum of a finite geometric series gives:

[tex][e^{-\beta\mu B a} \frac{1-e^{(2a+1)\beta\mu B}}{1-e^{\beta \mu B}}]^{N}[/tex]

Problem, is I can't seem to resolve that down to a hyperbolic function and I'm sure it should. Any ideas?

 

[G01]

Oh ok. So, your dealing with a quantum spin system of arbitrary integer or half integer spin, not just spin 1/2. That makes it slightly more complicated but, not by too much.

The sum of the finite geometric series is the way to go.

I don't think the partition function can be simplified directly to a single hyperbolic trig function as it stands, but for large values of a, your expectation values for the Energy, Magnetization, Specific heat, etc. definitely can be.

Try this: Keep the partition function in exponential form, but simplify and multiply through the exponential,[itex]e^{-\beta \mu B a}[/itex]. Take [itex]\frac{\partial Z}{\partial \beta}[/itex] with everything in exponential form. Then you can use that, along with Z to find all the state variable expectation values. Then for large a, these quantities will be expressible in terms of only hyperbolic trig functions.

 

[paranormal]

I understand your confusion and can provide some clarification on the steps involved in this example.

First, let's break down the formula and understand what each part represents. The Gibbs partition function is a fundamental concept in statistical mechanics and is used to calculate the thermodynamic properties of a system. In this case, we are looking at a paramagnet, which is a system of particles with magnetic moments that can align with an external magnetic field (represented by B).

The first sum, \sum_{s}, is a sum over all possible spin states for a single particle. In this case, s can take on values from -a to a, representing the possible spin orientations of the particle. The second sum, \sum_{i}^{N}, is a sum over all N particles in the sample. This means that we are considering the total spin of the sample, taking into account all the individual spins of each particle.

Now, let's look at the exponent term, which is the key to understanding this formula. The exponent contains two terms - \beta \mu B and \sum_{i}^{N} s. The first term, \beta \mu B, is a constant that represents the interaction between the magnetic moment of a particle and the external magnetic field. The second term, \sum_{i}^{N} s, is the total spin of the sample, which is a sum of all the individual spins of the particles.

By combining these two terms, we can see that the exponent represents the energy of the system, which is dependent on both the external magnetic field and the total spin of the sample. This is why we have a sum of exponentials in the first place - it allows us to consider all possible combinations of spin orientations and their corresponding energies.

Now, going back to your question, you are correct in thinking that the sum in the exponent is a sum over all particles in the sample. However, it is important to remember that the second sum is also a part of the exponent. So, we cannot simply take it out of the exponent and treat it as a product of exponentials.

In order to simplify the expression, we can use some mathematical techniques such as the binomial theorem or Taylor series expansion. These techniques allow us to rewrite the exponent as a polynomial, which can then be manipulated further to obtain the desired result.

I hope this explanation helps you understand the steps involved in this example and how to proceed with it. Remember