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Old 1st March 2015, 09:21 AM   #1
TubbaBlubba
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Question about differential equations in thermodynamics

Yep, it's that time again. I have an introductory thermodynamics (Physic 103) exam in two weeks, and I have some general questions about differentail equations.

While there's a parallel course in basic multivariable calculus that is somewhat illuminating, and I'm reasonably comfortable with partial differential equations, gradients, Jacobians, multiple integrals and so on (I mean, it's mostly just extentions of introductory caluclus and linear algebra), the thermodynamics course material is, as I understand is usual, very handwavy in its treatment of (exact and inexact) differentials, differentiation and multivariable calculus (indeed banishing anything resembling a detailed treatment to a three-page appendix) - especially since the calculus material I'm used to uses standard epsilon-delta methodology.

There are more differential equations than I can count, but quite a few of them are on this form:

dU = (δU/δT)VdT + (δU/δV)TdV

This is, of course, not an unfamiliar form of equation. Indeed, since we have U = f(T, V), then the above equation would be equivalent to

dU = grad(U) dot (dT, dV)

Hum, so this seems analogous to Cauchy's definition dy = f'(x) dx. So, can I understand this equation in a sense more mathematical than "Some small change in the internal energy dU is given by the partial derivative with respect to temperature times some small change in temperature etc etc", as it is usually given? The equation U = f(T,V) apparently describes a three-dimensional surface (and f(T,V) = constant some two-dimensional level curve). How should I then understand the vectors (dT, dV, dU) and (dT,dV)? Can I think of (dT, 0, 0), (0, dV, 0), (0, 0, dU) as unit vectors in the dimension of their respective physical quantity? Or will I have to make peace with dT, dV, dU representing abstract, but real quantities of change?

And that's not even getting into the inexact differentials... but as I understand it, you get that when you don't have an actual equation Q = f(p,T,V) that you differentiate with respect to each quantity, but rather different equations for the various quantities.

I would be thankful for anyone who could help shed some light on this, or link to a good resource.
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Old 1st March 2015, 11:19 AM   #2
Darwin123
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Originally Posted by TubbaBlubba View Post
Can I think of (dT, 0, 0), (0, dV, 0), (0, 0, dU) as unit vectors in the dimension of their respective physical quantity?
The word 'dimension' is usually restricted to independent variables. It corresponds to the phrase 'minimal basis' in linear algebra. Dependent variables can't be dimensions.

The equations that you just wrote are a constraint on the three variables T,V and U. So the three unit vectors that you describe can't be linearly independent. Therefore, you really have a two dimensional space. You can choose any two of those variables as being the independent variables for this two dimensional space. You can choose two near combinations of the three variables to span the two dimensional space. However, the system is intrinsically two dimensional because of the physical constraint.

There are only two degrees of freedom in whatever problem you try to solve because the system is two dimensional.

If you choose any one of those variables as a dependent variable, then the other two span the two dimensional space. The choice of which two is a matter of convenience. However, there is a criterion for choosing the 'preferred' pair in terms of convenience.

Thermodynamic quantities can be classified as extrinsic or intrinsic depending on how they scale as you couple systems near equilibrium together.

If when you attach two identical systems together, and the quantity doubles, it is called an extrinsic system. Thus, volume is an extrinsic quantity. If you attach two systems of equal volume together, the volume doubles. Furthermore, U is an extrinsic quantity for the same reason.


If you attach two identical systems together , and the quantity remains the same, then the quantity is intrinsic. Here, T is an intrinsic property. If you attach two systems of identical temperature together, the combined system has the same temperature as the separate components.

It is convenient to lump the extrinsic quantities together and to lump the intrinsic properties together. Therefore, I would choose the U and the V as the independent degrees of freedom. The T would be the dependent system.

If you adapt U and V as independent variables, then the partial derivative of one versus the other is always zero. The partial derivative of T with respect to one of these independent variables will not always be zero.

I think the initial arbitrary choice confuses many students and even some experienced scientists who don't work with thermodynamics. The partial derivatives of U and V with respect to each other are zero only because of arbitrary choice. You have to choose them as independent variables. The physical constraint forced you choose two independent variables, but it didn't tell which of the two were independent.

This problem comes up in all the physics disciplines. Some of the mathematical difficulty in relativity comes from the use of partial derivatives, rather than the difficult physical concepts.


Originally Posted by TubbaBlubba View Post

Or will I have to make peace with dT, dV, dU representing abstract, but real quantities of change?
...
not even getting into the inexact differentials... but as I understand it, you get that when you don't have an actual equation Q = f(p,T,V) that you differentiate with respect to each quantity, but rather different equations for the various quantities.
A physical picture is useful only if the physical picture in some way takes the constraint in account. Consider,
Q=F(p,T,V)

If you use the chain law to find the perfect differential, dQ/dV. Then you get
dQ=c1 dp +c2 dT +c3 dV

where the c's represent partial derivatives. Sorry, my Google software doesn't have mathematical symbols.

There is usually a sentence that tells you that one of the differentials is zero. Suppose there is a sentence that tells you the system is isobaric. The pressure is constant. So dp=0. Then

dQ=c2 dT + c3 dV

Then, you can use T and V as independent variables. The partial derivative of one versus the other is zero. The variable q is a dependent variable. So dQ is a perfect derivative which can go in a total derivative.

Another way to say it is that Q 'explicitly' depends only on T and V. You don't have to know what p is to know what dQ is. Under isobaric conditions, you should write
Q=f(T,V)
rather than.
Q=f(p,T,V).

Since p is being held constant, the system is two dimensional. The second expression is correct.

In general, the perfect derivative is the only objective quantity that can be measured.

Boy, that didn't go well, did it?!
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Old 1st March 2015, 11:58 AM   #3
TubbaBlubba
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Right, I meant to say a two-dimensional surface embedded in three dimensions, as in, I think of the function U = f(T,V) for instance as describing some surface consisting of every collection of points (x,y,z) satisfying z = f(x,y), if that makes sense. The internal energy U would be the "height" of the surface above the xy-plane at any given point.

Irrespective of that, thanks a lot. I think I'm starting to understand this stuff, I guess I just really, really, don't like working with differentials... And I will remember your suggestion to lump extrinsic and intrinsic varibles together!

Maybe I should just start thinking of say dU, as whatever function, when integrated with respect to the dependent variables (as specified by the constraints) results in the function corresponding to U (again, according to the constraints), with dT, dV etc being the corresponding variables...
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Old 1st March 2015, 12:37 PM   #4
ctamblyn
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Originally Posted by TubbaBlubba View Post
Right, I meant to say a two-dimensional surface embedded in three dimensions, as in, I think of the function U = f(T,V) for instance as describing some surface consisting of every collection of points (x,y,z) satisfying z = f(x,y), if that makes sense. The internal energy U would be the "height" of the surface above the xy-plane at any given point.
That's a perfectly good mental picture. You have a surface in the 3-dimensional (T,V,U) space given by U = f(T,V). The partial derivatives (∂f/∂T)V and (∂f/∂V)T tell you how much your "altitude" (U-coordinate) changes when you move a unit amount in the T or V direction, respectively, while still remaining in the surface.

If you move an amount dT in the T direction as well as an amount dV in the V direction, and remain in the surface, your U coordinate must change by dU = (∂f/∂T)V.dT + (∂f/∂V)T.dV.
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Last edited by ctamblyn; 1st March 2015 at 12:39 PM.
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