# Thermodyamics, total differentials and integration.

1. Sep 19, 2008

### stephen_E

Hi everyone! Im a UK grad student working in acoustics. My own background is in EE, so i am largely self taught in thermodynamics. Consequently id really appreciate any insight any of you real physicists can give me with my problem!

Following derivations in books i can derive the enthalpy, H_{A}, of an acoustic medium to be as follows (S is strain, T is stress, D and E are electric filed and electric displacement,B and H are magnetic field, $$\theta$$ is temperature and $$\sigma$$ is entropy. Furthermore, the problem is assumed to be adiabatic.

$$H_{A} &= U -(ST)-(ED)-(BH)$$
(where $$dU = dW + dQ = T_{i}dS_{i} + E_{i}dD_{i} + H_{i}dB_{i} + \theta d\sigma$$), therefore:
$$dH_{A} &= - S_{i}dT_{i} - D_{m}dE_{m} - B_{m}dH_{m} + \theta d\sigma$$

so:
$$S_{i}=\frac{-\partial{H_{A}}}{\partial{T_{i}}}$$
etc etc for the other non infinitesimal quantities etc.

My problem comes when I expand Si and the other non-state variables from dH_A individually into total differentials ie

$$dS_{i} &=(\frac{\partial{S_{i}}}{\partial{T_{i}}})_{E,H,\sigma}dT_{i} + (\frac{\partial{S_{i}}}{\partial{E_{m}}})_{H,\sigma}dE_{m} + (\frac{\partial{S_{i}}}{\partial{H_{m}}})_{H,E}dH_{m} + (\frac{\partial{S_{i}}}{\partial{\sigma}})_{H,E}d\sigma$$

I need to convert this expression into a non-infinitesimal form, ie so the LHS is S and not ds. I know the correct answer from my textbook is:

$$S_{i} &=(\frac{\partial{S_{i}}}{\partial{T_{i}}})_{E,H,\sigma}T_{i} + (\frac{\partial{S_{i}}}{\partial{E_{m}}})_{H,\sigma}E_{m} + (\frac{\partial{S_{i}}}{\partial{H_{m}}})_{H,E}H_{m} + (\frac{\partial{S_{i}}}{\partial{\sigma}})_{H,E}d\sigma$$

However I don't see how to obtain this and the book glosses over the details of how to do this! Obviously I need to perform some kind of integration here, but how do I do this?! The thing that is throwing me is that last term involving entropy and temperature. I can see how to integrate this for terms T,E and H ie a term of the form
$$(\frac{\partial{S_{i}}}{\partial{T_{i}}})_{E,H,\sigma}dT_{i}$$
converts to $$(\frac{\partial{S_{i}}}{\partial{T_{i}}})_{E,H,\sigma}T_{i}$$.

Can someone explain to me the significance of $$(\frac{\partial{S_{i}}}{\partial{\sigma}})_{H,E}d\sigma$$ since this term doesn't seem to transform in the same fashion?

Also: Should i ditch the infinitesimal notation for S? At least one book ive seen assumes S is a function of the form $$S(T,E,H,\sigma)$$ and expands this function using a Maclaurin series: this seems to avoid the tricky infinitesimal quantities. Is this a better approach? Generally, using infinitesimals in calculus seems to be a bit of a minefield, as they are not very rigorous.

Thanks!
-=+Stephen Ellwood+=-
Ultrasound Research Group, Leeds University, UK

2. Sep 19, 2008

### stephen_E

Just thought i should add one more detail: the subscript indices here are related to the fact that the material this derivation refers to is anisotropic - ie properties vary in different directions.

However this shouldn't drastically effect the outcome here.

3. Sep 19, 2008

### Mapes

I'm with you up to here, but this can't be correct; you can't integrate and be left with $d\sigma$. Shouldn't the last term be

$$\int^{\sigma_2}_{\sigma_1} \left(\frac{\partial{S_{i}}}{\partial{\sigma}}\right)_{H,E,T}d\sigma\,\mathrm{?}$$

It seems like you could move forward by expanding as

$$\left(\frac{\partial{S_{i}}}{\partial{\sigma}}\right)=\left(\frac{\partial{S_{i}}}{\partial{\theta}}\right)\left(\frac{\partial \theta}{\partial{\sigma}}\right)$$

where the first quantity is related to thermal expansion and the second to heat capacity. Does this help?

4. Sep 19, 2008

### atyy

I haven't the foggiest idea really - my feeling is that an equation of state is being used - like PV=nRT for an ideal gas. Perhaps some context may help? What are the titles of the book and chapter? What is the final goal of the derivation?

5. Sep 19, 2008

### Count Iblis

Euler's theorem for homogeneous functions is used here, see here for a similar problem:

Note that you should call H the magnetization (commonly denoted by M), E the electric polarization (commonly denoted by P).

6. Sep 19, 2008

### stephen_E

Hi guys I can confirm that the formulae I have given are correct; there is another derivation in the same book where the guy uses a very similar approach but this time on an isothermal not adiabatic problem.

I'm assuming that he wouldn't make the same mistake twice; furthermore if you can't preform the step ive outlined here you cannot prove the fundamental equations of piezoelectricity which would be a major showstopper!

The book in question is "Principles of Acoustics" Volume IA, edited by Warren P Mason, published by Academic Press - its very old though (c. 1950's)!

Many thanks for your reply Count Iblis, ill look into this Euler's theorem business. It sounds like it could be just the key to solving this.