/sci/ - Science & Math

[Anonymous No.5714268]

How do you learn Thermodynamics?

I just can't do it. Quantum I'm fine with. Complex mathematics? That's fine too. At least I know what I gotta learn.

Thermo is just a mess though. So many arbitrary definitions. So many "so using the result from 3 slides ago & this from lecture 7, we can clearly see this is = to that thing from lecture 3, and subtracting results a, b, and c from the lecture 9 result and combining with the above we can substitute into that thing on the last slide to obtain the following. Define this to be = to (some scientists name) and..."

I'm close to giving in and rote learning as much as possible to try and blag my way through the exam. It's what most students do anyway and seems to work for them.

[Anonymous No.5714281]

Not even like I can even ask my lecturer.

I'm not stuck on a single concept.

[Anonymous No.5714285]

Are you talking about thermodynamics with quasi-equilibrium states? Just read the book.

[Anonymous No.5714292]

>>5714285
What?

I'm British. Our universities set their own classes.

I think I will go get the book though since this is causing me so much grief. Usually the lecture notes suffice

[Anonymous No.5714296]

Thermo is just playing around with algebra until you get some sort of substitution and arrive at final answer. No application in real life at all.

[Anonymous No.5714298]

>>5714296

Yeah, but it's hard to keep track of 2,000 different formulae and where to substitute them in.

The notes even use the word "obviously" at one point: "This is obviously equation #1,031 rewritten in another form".

Yes, obviously

[Anonymous No.5714307]

Yes it's funny how thermodynamics is, imho, conceptually more difficult to grasp than QM and most of basic QFT.
Some people argue that it is hard to learn because, in modern terms, it would require contact geometry and people don't learn that in physics or engineering.
http://en.wikipedia.org/wiki/Contact_geometry
Related to this, I feel it's difficult because people tend to have a hard time understanding the potentials like Enthapy or Gibbs energy, and also why they are the. Secondly, the concept of System, Subsystem, bringing System in contact, these thing don't get enough emphasis.
And people underestimate or misinterpret the status of some quantities like temperature (which I, for the longest time, found to be more troublesome than e.g. spin)


However (!), your pic suggest that you really learn statistical physics, not pure thermodynamics, which is then a lot simpler, because of the propability intuition for entropy.
Do you learn for solid state physics?

In any case, do you have any specific questions or do you just want to whine a little.

Regarding you "question" I think one good way to learn is take the time and figure out the whole set of primitive assumptions, and then write down how all the things (down to heat coefficients) are defined - in terms of the initial quantities/functions.

[Anonymous No.5714320]

>>5714307
>However (!), your pic suggest that you really learn statistical physics,

That's taken from the 2nd half of the module. It's Thermal and Statistical Physics. I find the statistical half much easier.

I don't know what you mean by understand Gibbs Free energy, because it's simply defined for us in our notes. This is Gibb's Free energy. Now taking partial derivatives and comparing with this, then substituting into this...

[Anonymous No.5714376]

>>5714320
okay wait a sec, I'll write you a short physical motivation.

[Anonymous No.5714404]

(cont.)

Changing away from U(S,V) is desirable, but simply using <span class="math">U_{new}(T,V):=U(S(T,V),V)[/spoiler] does not work well, because the parameter T is defined using a derivative of the function U you want to express and since <span class="math">∂(U+c(V))/∂S=0[/spoiler] for all things c(V) constant w.r.t. S, you'd lose information. The Legendre transformations (or contact geometry) keeps track of that "gauge" and this is why you have all these things like Gibbs energy G, Enthalpy H, Free Energy F,...
For all combinations of re-expressions of your system defining quantity U in terms of "bad" parameters (being derivatives of the quantity itself) you want a new U-like function which is invertible (undlike an expression you took the derivative of).

The Gibbs free Energy
<span class="math">G(T,P):=-H^{*S \rightarrow T}=-(-U^{*V \rightarrow P})^{*S \rightarrow T}[/spoiler]
is connected to U by Legendre transformations with respect to both of the slopes defined above. That (a priori purely mathematical) involutive transformation is a relation between tangents and points of two functions f --> f^*.

The potential G is said to be "natural" in T and P. Similarly F has applications in which T and V is of specific interest.

Considering another potential than the internal energy U is particularly useful if the "slopes" are the parameters, which are held constant in the process you're interested in. Because then, of course, the new function doesn't change in them. For exmaple, if you know that P=const. in a process, then dP=0 and then the change of the enthalpy H(S,P) is given by

<span class="math">d H=\left( \frac{∂H}{∂S} \right)_P d S+0=Td S=\delta Q.[/spoiler]

I.e. the enthalpy change is just the heat flow.

(cont.)

[Anonymous No.5714403]

Okay, here some sort of motivation for the potentials:
(Gladly, I've written somethink like this before)

Arguable, the point of thermodynamics is to keep track of the energy distribution of your physical system.
In pure thermodynamics, you start out with the assumption that for the whole system of volume V you have a function U, the energy, which is additive w.r.t. breaking the system into subsystems. The volume is also additive V. You say these are extensive w.r.t. the systems. The expression U depends on V and possibly on other quantities X. In any case, U will depend on the mysterious internal value S. The "dynamical" behaviour of the theory is encoded in this quantity via the second law. (In statistical physics, the entropy has a simple interpretation. Okay now you have U(V,S).)

The variabes
<span class="math">T=∂U/∂S, -P=∂U/∂V[/spoiler]
give the amount of change of the internal energy U(S,V) w.r.t the extensive variables S and V. T and P themself are intensive/qunatity independend, because e.g. in ∂U/∂S you devide extensive things through extensive ones. (There are however other extensive variables like particle number, their slopes being the chemical potential). Okay so one energy change is heat and the other is work. The second law says the here introduced quantity T will equilibrate between systems. You can test the parameter T of one system by coupling it with a known one. And since you can measure T using gases (T=c·PV, and P and V is measurable) you want to work with T and not with S, which is something internal.

(cont)

[Anonymous No.5714406]

(cont.)

A typical such constant pressure situation where the enthalpy H(S,P) as well as the free enthalpy/Gibbs energy G(T,P) become relevant (the potentials, which depend on the intensive variable P is when you consider chemical reactions in a free area/a laboratory room, becuase then the preassure is just 1 atm whatever happens. Similarly, when a constant external termperature T is forced on your system, the change of the free energy F(T,V) is

<span class="math">d F=0+\left( \frac{∂F}{∂V} \right)_Td V=P d V=\delta W,[/spoiler]

and in this way the free energy relates to work the system can do (volume change in exchange for energy).

So again, the problem is that e.g. the intensive variable P is derived from U via a derivative <span class="math">P:=-(∂U/∂V)_S[/spoiler], and if you wanna use such a construct as a variable you might lose information. You can use U(S,P), or U(T,V) or other variant, but if you're just given these on the spot, you can't really reconstruct the true system.
The Potentials, i.e. U(S,V),S(U,V),H(S,V),F(T,V),G(T,P),... really contain the full information because they aren't just substitutes from U, but are derived via a Legendre transformation, which is designed to make a substitution, where one variable is a derivative.
An examples from mechanics is the Hamiltonian L(v,q) --> H(p,q), where q is space, v velocity, and p the momentum, which really just is proprtional to ∂L/∂v.

PS: In the case of G the result of the Legendre transformation (expressed in terms of the initial U) turns out to be the fundamental relation

U=G-PV+TS

(yo!)

[Anonymous No.5714418]

try the quantum approach to thermo, it is normally not covered as it is considered upper level. But I sometimes find the upper level stuff makes more sense then the introductory stuff because it is too simplified for me to get.

side question: What would happen if most of the random micro kinetic motion we call temperature aligned. Would the temperature drop and the normal macro kinetic energy rise making the object move?

[Anonymous No.5714455]

PPS: And yeah you could maybe read about the geometrical interpretation of the Legendre transformation.

And also, lastly, from practical POV it's clear that you are extremely interested in <span class="math">C(T)=\frac{\partial U}{\partial T}[/spoiler]. Because thermodynamics is about the energy distributions and T is the parameter you know best, experimentally. Since T is a derivative of U itself, the expressions for C get clumsy.
But in a way, much of material physics is just about finding the right microscopic theory, to math the C(T) you measure experimentally. I mean check out your pic. Statistical physics is kind of a machinery to get from a hamiltonian to some heat coefficients :D

The concept isn't unique to thermodynamical considerations. Qualitatively, heat capacities are somewhat like susceptibilies in electrodynamics, or ...I don't know linear response functions are related to their computation too.

>>5714418
if they would magically align (breaking momentumm conservation), then yes.

Let <span class="math">f(r,v,t)[/spoiler] be the distribution function, r, v being three dimensional vectors.
The average velocity V(r,t) is
<span class="math">V(r,t)=\frac{\int v·f(r,v,t)dv}{\int f(r,v,t)dv}[/spoiler]
Let <span class="math">w=V-v[/spoiler] be the relative velocity (of the random moving particles moving at v to the main flow V)
The kinetic energy is
<span class="math">\frac{1}{2}mv^2[/spoiler]
or
<span class="math">\frac{1}{2}mv^2=\frac{1}{2}mw^2+mw·V+\frac{1}{2}mV^2[/spoiler]
On average, the middle term doesn't contribute (because the signs turn around if you integrate over all of velocity space),
the last term clearly is the macroskopic kinetic energy, and hence the internal energy U(r,t) (proportional to the temperature) is coming from the small wild/random movements
<span class="math">U(r,t)=\frac{\int \frac{1}{2}w^2·f(r,v,t)dv}{\int f(r,v,t)dv}[/spoiler].
If particles allign so that most contributions form from <span class="math">v\sim V[/spoiler], then the contributions to <span class="math">w=v-V[/spoiler] drop and so does U and hence T.

[Anonymous No.5714497]

>>5714455

Woah thanks. I have heard the lecturer make reference to Legendre transforms, but we don't know what they are. He just mentions them by name alone. "Yeah you can also get this via a Legendre transform"

I think this module might be a victim of the "keep it simple for them to understand easier" mentality. He still wants to keep the difficult stuff for us to show us what the subject can really do, but is trying to introduce it using this strange convoluted framework. He needs to re-write lots of it and pluck stuff from thin air to fit the simplified version. It makes it much harder to follow.

I prefer stuff that's conceptually harder but actually involves lots less work. Fourier analysis in my other module is difficult to get your head round, but once you put in the effort there's not much to it and it's pretty easy.

This lecturer, on the other hand, has tried to keep it all to algebra and simple derivatives and it muddles it all up for us. He knows what he's doing as its a subset of the actual field for him, but for us it's our incomplete introduction to the entire subject and confusing as hell.

[Anonymous No.5714513]

Don't you feel guilty two timing on Emma Stone with Jane Levy?

[Anonymous No.5714803]

>>5714497
no, Emma is a frog and I haven't seen a relevant movie with her in ages.
And I like blondes, really.

[Anonymous No.5714808]

>>5714497
>>5714513
>>5714803
I misquoted here, obviously.

[Anonymous No.5716539]

[Anonymous No.5716550]

>>5714268
if you can't follow the lecture and understand what is happening just memorize the results.

[Anonymous No.5716618]

>>5716550

People like you are the reason bridges fall down.

[Anonymous No.5717414]

>>5714403
I'm not OP, but man, I must say thank you for this. I want to reach your level