The n-Category Café

I think if you take the definition of Laplace transform and take the temperature $\to 0$ limit, you’ll see it becomes the Legendre transform.

More precisely: according to Gibbs and Boltzmann, for any temperature $T > 0$ there’s a map from energies to relative probabilities given by

$$E \mapsto exp(-E/T)$$

This lets us take the rig of relative probabilities

$$([0,\infty), + , 0, \times, 1)$$

and pull back the entire rig structure to the set of energies

$$\mathbb{R} \cup \{+\infty\}$$

We get a rig $\mathbb{R}^T$ which is isomorphic to

$$([0,\infty), + , 0, \times, 1)$$

When we take the $T \to 0$ limit, this rig $\mathbb{R}^T$ reduces to the rig

$$(\mathbb{R} \cup \{+\infty\}, min, +\infty, +, 0)$$

Now: take the definition of Laplace transform and write in terms of the rig $\mathbb{R}^T$, and then take the $T \to 0$ limit. You’ll get one of the standard definitions of the Legendre transform!

Some hints:

1. I think the exponential in the Laplace transform disappears when we describe it using the rig $\mathbb{R}^T$ instead of the isomorphic rig $$([0,\infty), + , 0, \times, 1)$$ The exponential is in the isomorphism, as mentioned above.
2. The integral in the definition of Laplace transform becomes the minimum in the definition of Legendre transform - or maximum, depending on your conventions. I like minimizing energy and action, but these guys like maximizing. It’s not a big deal.
3. I don’t mention the definition of Legendre transform using minimization in my course notes, but you can find it on Arnold’s book on classical mechanics, and you’ll see it’s basically equivalent. I’m secretly minimizing something by setting its derivative to zero.
4. Innocent bystanders who don’t know about any of this stuff should read this, and then look at some books on “tropical mathematics” or “idempotent analysis”.

Christine writes:

Section 4.3, page 15: “From now on, assume that L is strongly regular” - You are also assuming that L is not an explicit function of time, right?

Be careful! In this situation, ‘time’ can mean two drastically different things — the parameter describing a path $$\gamma: [t_0,t_1] \to M,$$ or a coordinate on the manifold $M$!

In this class, the Lagrangian is always a function

$$L: T M \to \mathbb{R}$$

where $M$ is some manifold called the ‘configuration space’. A point in $T M$ is a pair $(q, \dot q)$, so the Lagrangian can depend on $q$ and $\dot q$, but not on the parameter $t$ parametrizing a path in $M$.

But note: in applications to the physics of a single particle, $M$ can be space or spacetime! When we’re doing general relativity and studying a particle moving around on a curved spacetime, we should always take $M$ to be spacetime. In fact, it’s even good to do this in special relativity!

In this situation we get a Lagrangian of the form $L(q,\dot q)$. There’s no explicit dependence on the parameter $t$. So, Noether’s theorem applies and we get a conserved quantity from time translation invariance. But, this quantity is zero!

This is the simplest place to start understanding ‘the problem of time’.

To see how this works in special relativity, see section 3.4.3 in my classical mechanics book. To see how it works in general relativity, see section 3.8.

But if $L$ is $L(q,\dot{q},t)$ (for instance, with a time varying potential)? I think it would be interesting to qualitatively mention in brief the difficulties involved.

If the Lagrangian depends on the parameter $t$ that parametrizes the path in $M$, everything becomes difficult and mathematically ugly. So, this sort of time-dependent Lagrangian is something I’d rather not discuss here.

It might be good to talk about parameter-dependent Lagrangians in my general-purpose classical mechanics course — and I do intend to keep expanding the book for that course. But, it’s not something I want to discuss when I’m trying to find the most elegant way to do everything… and then categorify it all.