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Legendre transform, better explained (2017)

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Re: Legendre transform, better explained (2017)

#11

Post by LolWolf »

It’s neat! To be fair, as a physicist, I did not understand the Legrendre transform essentially until taking convex optimization (where it is known as the Fenchel conjugate).

Many sources, but all of them are reasonable and give a constructive definition that actually explains what it does: we can characterize a function either by its graph, or its supporting hyperplanes (when it is a closed, convex function).

While the observation is almost silly, it has very deep consequences for different characterizations of problems and other constructions!

Re: Legendre transform, better explained (2017)

#13

Post by shiandow »

crdrost wrote: post #6

I, too, spent a long time staring at expressions like “half-invert p(x, v) to get v(x, p) s.t. p(x, v(x, q)) = q then the Legendre transform is H(x, p) = p v(x, p) – L(x, v(x, p))” And I did come to one of the same conclusions as this article, which is that if we're talking pure mathematics, these “thermodynamic” expressions like (∂L/∂v)_x, (∂L/∂p)_x are deeply easy to get confused about and in fact you should just s…

I know programmers like to blame mathematicians for writing functions with lots of one letter variable names, but it's the physicists who insist on doing so without defining any of them.

You want to know what V is? It's clearly the potential, we've defined it six papers ago! Oh you were wondering what it's type was, well it's usually a scalar field. No, don't write the parameter as t that changes the whole meaning!

Re: Legendre transform, better explained (2017)

#15

Post by archgoon »

crdrost wrote: post #6

I, too, spent a long time staring at expressions like “half-invert p(x, v) to get v(x, p) s.t. p(x, v(x, q)) = q then the Legendre transform is H(x, p) = p v(x, p) – L(x, v(x, p))” And I did come to one of the same conclusions as this article, which is that if we're talking pure mathematics, these “thermodynamic” expressions like (∂L/∂v)_x, (∂L/∂p)_x are deeply easy to get confused about and in fact you should just s…

> I believe I could walk up to any college senior in physics and they wouldn't know that “the determinant is the product of the eigenvalues,"

Unless things have gotten significantly worse in physics education in the past decade, I'd be happy to take the other side of that bet.

I will also be willing to bet they could prove it.

The problem you'd have with physicists is convincing them that there are matrices that aren't diagonalizable.

Re: Legendre transform, better explained (2017)

#16

Post by shiandow »

prof-dr-ir wrote: post #9
crdrost wrote: post #6

I, too, spent a long time staring at expressions like “half-invert p(x, v) to get v(x, p) s.t. p(x, v(x, q)) = q then the Legendre transform is H(x, p) = p v(x, p) – L(x, v(x, p))” And I did come to one of the same conclusions as this article, which is that if we're talking pure mathematics, these “thermodynamic” expressions like (∂L/∂v)_x, (∂L/∂p)_x are deeply easy to get confused about and in fact you should just s…

Agreed, the way thermodynamics is often taught is such a mess. My personal and controversial [0] take is that the free energy should really be seen as the Legendre transform of the entropy, not of the energy. I know it is ultimately semantics, but this viewpoint makes the passage from the micro-canonical to the canonical ensemble so much nicer. In particular, the saddle point approximation for the canonical partition…

I'm not even sure if it makes sense to view it as a Legendre transform. Or well, it is one, I'm just not sure if it's a good definition.

You get the free energy for 'free' if you use a Lagrange multiplier to maximize entropy while keeping the energy fixed (temperature is the inverse of that Lagrange parameter). In one fell swoop this shows why temperature is a thing and why minimizing the free energy is important.

The Legendre transform just returns the value of the constraint from the minimized function, but at that point why bother?

I do agree that it makes more sense to see the fee energy as a Legendre transform of the entropy, that's kind of what you end up doing if you minimize entropy in this way.

Re: Legendre transform, better explained (2017)

#18

Post by mydogcanpurr »

crdrost wrote: post #6

I, too, spent a long time staring at expressions like “half-invert p(x, v) to get v(x, p) s.t. p(x, v(x, q)) = q then the Legendre transform is H(x, p) = p v(x, p) – L(x, v(x, p))” And I did come to one of the same conclusions as this article, which is that if we're talking pure mathematics, these “thermodynamic” expressions like (∂L/∂v)_x, (∂L/∂p)_x are deeply easy to get confused about and in fact you should just s…

> I think this is because we introduce a complicated way to calculate determinants and then we use determinants to calculate the eigenvalues?

Yes, the determinant should be taught and defined as the volume of the parallelepiped in n-dimensions defined by the columns of the given square matrix. This perspective makes it immediately obvious that the eigenvalues scale the parallelepiped in each of its dimensions (a basis of eigenvectors makes it even simpler). Of course the volume (determinant) must be the product of these scaling factors (eigenvalues)! Since algebra is too convenient for solving problems, this geometric intuition is often an afterthought if it's even taught at all.

Re: Legendre transform, better explained (2017)

#20

Post by lupire »

mydogcanpurr wrote: post #18
crdrost wrote: post #6

I, too, spent a long time staring at expressions like “half-invert p(x, v) to get v(x, p) s.t. p(x, v(x, q)) = q then the Legendre transform is H(x, p) = p v(x, p) – L(x, v(x, p))” And I did come to one of the same conclusions as this article, which is that if we're talking pure mathematics, these “thermodynamic” expressions like (∂L/∂v)_x, (∂L/∂p)_x are deeply easy to get confused about and in fact you should just s…

> I think this is because we introduce a complicated way to calculate determinants and then we use determinants to calculate the eigenvalues? Yes, the determinant should be taught and defined as the volume of the parallelepiped in n-dimensions defined by the columns of the given square matrix. This perspective makes it immediately obvious that the eigenvalues scale the parallelepiped in each of its dimensions (a basi…

What trash math classes were you all in that didn't teach all of this?
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