What is the notation you would use to express this theorem?

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In this video (the link should take you 23 minutes in, where) Professor Susskind writes a theorem. (He says E and S are functions of T and V.) I am just wondering if what he writes is an abuse of notation? And if so, how would you write it?

(Susskind gives his own proof from 26:30 to 36:00 if you want to see his way.)

I am tempted to prove it from the chain rule, like this:

Let E(V, T) = F(V, S(V, T))
(How can we be sure there is such an F(V, S)?)

Then partially differentiating both sides with respect to V (using the chain rule on the right side) yields:

∂E(V, T)/∂V = ∂F(V, S)/∂V + ∂F(V, S)/∂S • ∂S(V, T)/∂V

Then the final step is to subtract the last term to the other side.
The arguments of each function makes clear what is held fixed.
Both E and F represent energy, but we cannot put F = E because they are different functions.

Is this a valid way to articulate/prove it? Is Susskind’s notation valid? Any clarification is appreciated.
 

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  • #2
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what he writes is an abuse of notation?
I wouldn't qualify it so harshly: to me it's shorthand. I'm confortable because he says what it means and uses is consistently. I suspect your apparent dislike stems from the similarity with integration where the notation $${dF\over dx}=f \Rightarrow \int_a^b f \;dx = F\,\Big |_a^b$$ is standard. But from that you could -- with a bit of artistic freedom -- derive that ## F\,\Big |_a ## stands for: ##F## at fixed ##a##

However, there is an alternative: here (under 'Notation') you see a different way of writing it: $$\left ({\partial f\over \partial x}\right )_{y,z}$$which you indeed encounter more often in textbooks.


Let E(V, T) = F(V, S(V, T))
but we cannot put F = E
seems a bit self-contradicting...

:smile: Let's ask @Chestermiller:
Is this a valid way to articulate/prove it?
 
  • #3
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I'm having trouble understanding what you are trying to "prove." The thermodynamic equilibrium state of a pure single-phase substance is determined by specifying any two intensive properties. In this case, the two intensive properties selected are temperature and specific volume.
 
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I'm having trouble understanding what you are trying to "prove." The thermosynamic...
I should maybe not have used the word energy in the OP. The theorem is purely mathematical, (hence we’re in the math forum) it should be true for any functions (of course they must be differentiable).

## F\,\Big |_a ## stands for: ##F## at fixed ##a##
I have seen the following notation mentioned in passing in the past:
$$\frac{\partial F(x,y)}{\partial x}\Big |_{y=a}$$
My trouble is that now the notation is used like this:
$$\frac{\partial F(x,y)}{\partial x}\Big |_{z=a}$$
Where z is not an argument of F!! Is this notation mathematically meaningful? Is it common? I mean I don’t doubt Susskind’s mathematical consistency, but this sure is bothering me.
seems a bit self-contradicting...
Are you saying you would say E = F...? Surely they must (in general) be different functions?

For instance let’s say (just making stuff up) E(V, T) = V + T
and
S(V, T) = 2V + T
Then we would choose
F(V, S) = S - V
That way we get
F(V, S(V, T)) = 2V + T - V = V + T = E(V, T)
So you see that E and F are different functions. I said E ≠ F but maybe I should be extra clear and say E(x, y) ≠ F(x, y)

Kinda feel like I’m losing my sanity! Thanks for the replies.
 
  • #5
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Let E(V, T) = F(V, S(V, T))
Your writing, not mine. Does that answer
Are you saying you would say E = F...?
By way of 'and now for something completely different' (but hopefully more relevant than your example....:rolleyes:):

In the thermodynamic potentials a change of variables constitutes a different thermodynamic potential, e.g. (*) $$
H = U + pV \quad\quad H = H(S, p )\Rightarrow dH = dU + pdV + Vdp = TdS + V dp
$$change to ##T,p## by subtracting ##TS## : $$
G = H - TS = U + pV -TS \quad\quad G = H(T, p )\Rightarrow dG = dH - TdS - SdT = -SdT + Vdp
$$ all done with differentials, as you see.

Second derivatives lead to te Maxwell equations -- see in the link (further down) or here

(*) I left out the chemical potential -- not because they are irrelevant but to keep it simple in this context.

Kinda feel like I’m losing my sanity!
Nothing wrong with that, as long as it is temporary :wink:. Thermodynamics do that from time to time, and to many folks :confused:, including me.

And yes, it's just calculus so far, I know...:biggrin:
 
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  • #6
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I still want to say E ≠ F because I’m thinking of E as a function, which is to say it maps an ordered pair. The function F (despite also representing the same physical quantity) is a different mapping.

From what I gather, I should give up the conception that E represents a function and just view it as a physical quantity?

It seems so wrong to say E(V, T) = E(V, S) (seems to imply T = S) but I guess we just suppress the variables and get on with our lives?
 
  • #7
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I still want to say E ≠ F
And I agree, as should be clear from the above. So far the only one who wrote E=F was you. (either you wrote, or you were quoted)
 
  • #8
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And I agree, as should be clear from the above.
Sorry, I misunderstood what you were implying.
Can’t say I’m not still a bit bothered, but seeing the notation in your link is comforting.
Thanks for your time.
 
  • #9
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You're welcome. Interesting subject area !
 

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