Differential forms

**omberlo** · August 22nd 2011, 01:24 PM

Hi everyone, I'm new here.

I'm an italian student who has an exam in applied thermodynamics soon. Through the whole course as well as the Physics one I have faced a lot of equations expressed with differentials, basically all of them. I have never been taught how to handle them though, all I was told is that a differential (for example dU) is a very small change in the value of U and even though I have studied derivatives of functions (multivariable functions too), integrals and double integrals, I don't feel like it's enough to fully understand the meaning of such expressions.

Let's take a concrete example: my notes say that the equation for the first law of thermodynamics is

$dU(S,V,N_i) = TdS - pdV + \sum_{i=1}^{n} \mu_i * dN_i$

and that T, p and MU_i are the partial derivatives of the internal energy U with respect to S,V and N_i respectively and as such they are also function of the same variables S,V and Ni. It makes sense because if I partial derivate that expression with respect to the variables S,V and N_i I obtain T p and MU_i respectively.
Now, the doubts I have are

1) How is the differential form of a function obtained? Is it simply the sum of the partial derivatives with respect to each variable multiplied by the differential of the variable?
I mean

$dU(S,V,N_i) = \frac{\delta U}{\delta S}dS+\frac{\delta U}{\delta V} dV+ \sum_{i=1}^{n} \frac{\delta U}{\delta N_i}dN_i$

If so, I can kind of grasp the meaning of it, since it basically means multiplying the change in the variable times the rate of change and sum the contributions of all the variables; but what are the advantages of using such an equation instead of the equation of the function itself?

2) If I wanted to calculate the actual change in the internal energy (dU) how would I proceed? Integrate that expression with respect to which variable? Am I supposed to integrate it like this:

$U_b - U_a = \int_a^b dU = \int_a^b TdS - \int_a^b pdV + \sum_{i=1}^{n} \int_a^b \mu_i dNi)$

But if so, what is the meaning of this operation? I mean, summing the integrals of different parts of a function with respect to different variables has no meaning in my book. Also, if I try this operation with, say, the function Z=x*y, I get

$dZ = \frac{\delta Z}{\delta x}dx+\frac{\delta Z}{\delta y} dy = ydx + xdy \\ \int dZ = \int ydx + \int xdy\ = 2xy$

which isn't correct.

I hope I've been clear enough, thanks in advance.

**vincisonfire** · August 22nd 2011, 05:14 PM

Hi,
In thermodynamics, you will find different "kinds" of energy. On of them is the one you stated above. There is also the Gibbs energy or Helmholtz energy for instance.
Depending on the system you are studying, you will use a different type of energy. T S P V $\mu$ N are all thermodynamic variables but you won't get anything if they are all allowed to vary. You need some control on your system. In the case were T P and $\mu$ are held fixed, you will use the energy as defined above.
1) How is the differential form of a function obtained? Is it simply the sum of the partial derivatives with respect to each variable multiplied by the differential of the variable?
Just like you said.
What are the advantages of using such an equation instead of the equation of the function itself? You can read off relevant quantities like T P and $\mu$ by varying S V and N respectively. It makes explicits the variables. It makes it easy to switch from one kind of energy to another using Legendre transforms. Probably later you will be ask to derive useful relations from there.

**vincisonfire** · August 22nd 2011, 05:19 PM

2) If I wanted to calculate the actual change in the internal energy (dU) how would I proceed? Integrate that expression with respect to which variable?
In thermodynamics, not only the endpoints matters. You actually needs to consider the path in between. This makes sense if you think of it. Consequently, you do not do a normal integral but rather a path integral.
You should thus write $\int dZ = \int ydx + \int xdy = \oint dt (y\dot{x}+x \dot{y})$ where $\dot{y} =\frac{dy}{dt}$ and dt is some parameter (could be time for instance).

**omberlo** · August 23rd 2011, 11:50 AM

Thanks for the quick answers vincisonfire.

However, I'm still unsure about the meaning of an equation expressed in differential form or better yet, how to "use" it.

Let's take again the equation for internal energy and semplify the chemical potential bit:

$dU = TdS - pdV$

is the above expression supposed to be used to calculate the change in internal energy ( $\Delta U$ ) of a thermodynamic process ONLY when T and P are constant throughout the whole process, as

$\Delta U = T * \Delta S -p * \Delta V$

or is it supposed to be used to calculate the $\Delta U$ of a thermodynamic process even if T and P aren't constant by somehow integrating it and obtaining some function of S and V, e.g.

$\Delta U = F( \Delta S ; \Delta V )$

or is it supposed to be used to calculate (or rather, estimate) $\Delta U$ even if T and P aren't constant, by dividing the whole process into very small processes where either T or p can be considered constant, and manually sum all the processes where T was constant and multiply them by the change in entropy ( $\Delta S$ ) of each of those processes, and adding them to the sum of all the processes where p was constant multiplied by the change in Volume ( $\Delta V$ ) of each process?

I mean something like:

$\sum_{i=1}^{n+m} dU_i = \sum_{j=1}^{n} T_j * \Delta S_j - \sum_{k=1}^{m} P_k * \Delta V_k$

or maybe none of the above? hah.

Thank you.

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