Thermodynamics basics.

DrDu

Andy, one counterexample may be sufficient. For an ideal gas, U is a linear function of T only and S can be expressed in terms of T and V (Sackur Tetrode equation). So e.g. T and V are sufficient to specify the state in all respects.

Urmi Roy

I think it's true...this is because the pressure at any point in a liquid is given by dgh (where d is the density,g is the acceleration due to gravity and h is the height of the column of the water above the point in concern.

So if you double the depth,the 'h' doubles,so P(pressure) =2(dgh).

Intuitively,we could say that the molecules at the depth 2h have to move around with twice the amount of kinetic energy than the molecules at 'h' in order to support the liquid above it.
(This is my personal idea...let's see what the experts say!)

P.S According to my intuition (that I put forward in the last paragraph),the temperature at the greater depths should be higher,as temperature depends upon the kinetic energy of the molecules....someone told me that this assumption is correct but due to rapid convection currents,we can hardly notice the temperature difference...is that right?

Andy Resnick

I don't know much about that expression, but it's only valid for an ideal gas. And we already know that an ideal gas can be discussed using extremely simplified models. But ideal gases don't display every behavior possible for real gases, let alone more complicated substances like liquids and solids.

Edit: and in any case, your 'counterexample' still has an odd number of state variables, which is the real point.

the_house

Can you elaborate a bit on this?

If I can shake out the cobwebs and try to remember correctly, in thermodynamics as it's usually taught in undergraduate and graduate courses, it is stated that all information about a thermodynamic system is contained in the relation for a thermodynamic potential--e.g., U(S,V,N1,N2,...) or the various Legendre transforms of this relation (e.g., enthalpy, Gibbs free energy, Helmhotz free energy, etc.). So then the state can be completely specified by 2+n quantities (where n is the number of conserved charges--in textbooks this is usually particle number for n species of particles). Any other quantities can be derived from the thermodynamic potential. Even your arXiv reference appears to say this is true in "classical mechanics", at least for n=0.

What has to be true for this not to be the case, and what exactly does it change?

Edit: OK, having skimmed the introduction to that arXiv reference, it says it's trying to answer the question. "Is there such a thing as ‘quantum thermodynamics’ where pressure or volume are represented as operators?" This work seems well outside the purview of standard thermodynamics and I believe it needlessly complicates the thread (titled "Thermodynamic basics"). In answering the original question, it's probably best just to state the standard thermodynamics answer: only 2+n quantities must be known to completely specify a thermodynamic system. If I've misunderstood something, let me know.

Andy Resnick

Unfortunately, not too much- I'm still learning it myself.

Other than the wiki page on contact geometry, there's some discussion in an appendix of Arnold's "Mathematical Methods of Classical Mechanics". There's a lot available on google scholar, but I'm not able to separate wheat from chaff

http://scholar.google.com/scholar?q=...=1&oi=scholart

the_house

Then I'll have to go with the assumption that all this business about quantizing contact geometries goes well beyond standard thermodynamics (if it's even related at all).

So I'll have to stick with my previous answer: in standard thermodynamics you can completely specify a thermodynamic state with 2+n variables, where n is the number of conserved quantities. As an explicit example, from earlier in the thread:

Assuming particle number is not conserved, you actually only need to know 2 of the variables. If you know V and T, then the entropy can be found by differentiating the relation for the Helmholtz free energy F(T,V): -S(T,V) = [dF/dT]_V (Sorry for the notation. I'm having trouble getting LaTeX to work). The pressure is P(T,V) = [dF/dV]_T. You can then plug in the values of S and P to determine the enthalpy H(S,P). (The enthalpy is related to F by a Legendre transform with respect to both T and V: H = F + TS + PV.)

Alternately, if you start with T and P, you can use the Gibbs free energy relation G(T,P) to calculate the entropy -S(T,P) = [dG/dT]_P, which you can then plug in for the enthalpy H(S,P). (Again, the enthalpy is related to G by a Legendre transform: H = G + TS).

SpY]

I was wondering about linear thermal expansion:If a rectangular plate with a hole in the middle is heated and expands, will the hole get bigger or smaller?
Does the whole thing just scale up (bigger hole) or swell like a donut (smaller)?

Studiot

Hello spy and welcome.

You should really start your own thread to ask separate questions, rather than adding to existing ones.

It is a good question which probably merits some discussion since there are several possible answers.

The short answer is that if the plate is free to expand around its outer edges the hole will also get bigger.

SpY]

Thanks for the tip, I'm just new new here. And thanks for a quick reply :)

Andy Resnick

I only know of one system with non-conservation of particles: photons. Otherwise, exempting chemical reactions, particle number is pretty well conserved.

I'm not really sure what your point is- when you write H = F + TS + PV or H = G +TS, how can you claim that thermodynamics does not have an odd number of state variables?

the_house

Any relativistic system will not conserve particle number. Even a non-relativistic system with fixed particle number will not have any associated chemical potentials. Mostly I just wanted to simplify things so there weren't a bunch of chemical potentials or conserved charges running around. Also, the only variables that were mentioned were T, P, V, S, and H, so I presumed there were no chemical potentials in this proposed system. In any case, the extension is straightforward. Instead of 2 variables to completely describe the system it takes 2+n.

This is a Legendre transform. It relates the various thermodynamic potentials. Incidentally, it is also the relationship between the Lagrangian and the Hamiltonian in classical mechanics.

I don't think I made that claim, although the relevance is unclear to me. It's certainly not something that was ever mentioned in any thermodynamics course I've taken. Everything I have actually stated is definitely standard thermodynamics.

Andy Resnick

I have never heard this- can you provide a reference?

Ah, ok. Oftentimes, thermodynamics is presented as some sort of 'average' statistical mechanics. That is, thermodynamics is based on a *mechanical* theory.

Since mechanical theories have a symplectic geometry while thermodynamics has a contact geometry, this cannot be the case.

George Jones

I have been doing some reading, and, while things are still fuzzy for me, I'm starting to see an overview. These variables are five possible coordinates for 5-dimensional thermodynamic phase space. The possible states of a system (in equilibrium) is a 2-dimensional (Legendre) submanifold of 5-dimensional thermodynamic phase space. In general, a thermodynamic system that has n degrees of freedom is an n-dimensional submanifold of the (2n + 1)-dimensional contact manifold that is thermodynamic phase space. n can be even or odd, but, trivially, 2n + 1 is always odd.

From the paper:
Other references:

http://sgrajeev.com/geometry-of-thermodynamics/

http://www.sci.sdsu.edu/~salamon/MathThermoStates.pdf.

The first link is to a blog posting by the author of the paper you cited. This posting works through some standard examples.

the_house

The way I stated this was way too strong, but for a system where there's enough energy for inelastic collisions (so particle types change and 2<->4, etc., processes also change particle number), it is unlikely for there to be any notion of conservation of particle number. It's certainly not a symmetry of our fundamental theories. There will still be associated chemical potentials for any other relevant conserved quantity though--electric charge, baryon number, etc.



Thanks, I think I'm starting to see how this fits together (especially after George's post, and looking at the blog post he linked to). If I understand correctly, the total number of state variables is (5+2n). I.e., there are always conjugate pairs plus one extra and this will always be odd. This is important for determining that the number of independent variables needed to specify the state is 2+n (note that my n--the number of conserved quantities--is the same as (n-2) in George's post). Note that this number of independent variables is not necessarily odd. I think there has been some confusion about this earlier in the thread.

This is made clear in the blog posting (and actually in your arXiv reference as well--see paragraph 4 on page 4). In the case of n=0, there are 5 variables (taken as U,T,S,P,V in these references), but only 2 of them are independent. Adding, e.g., a conserved particle number will add a conjugate pair N and mu (chemical potential) to the total number of variables, but only one of them is independent.


Thanks for that link. It was helpful.

DrDu

Let me count: T, one, V, two. Thats an even number.
Btw, similar equations for U and S are available for non-ideal gasses, e.g. the van der Waals gas:
http://en.wikipedia.org/wiki/Van_der_Waals_equation. The equations for U and S only depend on V and T, too, as in the case of the ideal gas.

Urmi Roy

Hi everyone!

I'm not sure if this is appropiate,especially with the debate that's going on about one of my questions regarding the number of state variables needed to define the system (I'm waiting for the debate to come to an end so that I get my final answer), however,my teacher taught something really interesting and I'm really curious to know about it.

Here's what it is:
Well,the Steady Flow Energy Equation (SFEE) basically says that when heat is supplied to to a thermodynamic sysytem or work is done on it,the energy so gained may be stored in the form of internal energy or pdv work( completely thermodynamic properties) or in the form of macroscopic enrgies(like kinetic energy of the fluid as a whole or its potential energy)....now whenever heat is provided to a system,it is supposed to supply kinetic energy to the molecules/atoms thereby increasing the temperature of the fluid.

However,according to the SFEE,supplying heat to the fluid in this way could also give kinetic energy to the fluid as a whole too! How is this possible?

Similarly,I can't see how the heat supplied to a system could provide potential energy to the entire fluid...I mean heat and flow work(pdV) are supposed to affect the molecules/atoms within the system,not the fluid as a whole!

This might have some relation with macroscopic and microscopic internal energies,but I'm not sure.

Also,another quick question: When we measure Cp(specific heat at constant pressure),is the process isothermal aswell as isobaric?

Studiot

When you heat something up it expands.

This expansion can do work and/or result in an increase in potential energy.

But you can't use all the input heat this way.

I'm sure you can think of lots of ways this might happen.