# Physical Interpretation of the Graph of the Legendre Transform?

Look at the following diagram from the first link:

I was trying to think of the simplest example to interpret this physically: consider the graph of kinetic energy versus magnitude of velocity: $T(v) = \frac{1}{2}mv^2$. (i.e. $x:=v$)

Then $s= \frac{d}{dv} T(v) = mv = p$ the magnitude of the momentum obviously.

Furthermore, $sv= pv = mv^2$. $F=T(v)=\frac{1}{2}mv^2$, and since $sv = F+G$, it follows that $G(s)=mv^2 - \frac{1}{2}mv^2 = \frac{1}{2}mv^2 = \frac{s^2}{2m}=\frac{p^2}{2m}$.

However, for a general Legendre transform, $G$ is just the negative of the intercept of the tangent line -- so what is the physical interpretation of the fact that the negative of the intercept of the tangent line to the graph of $T(v)=\frac{1}{2}mv^2$ is always equal to $T(v)$ itself? Is this mathematical fact a result of isotropy of space or some other symmetry we assume (presumably a similar but not identical fact would hold for relativistic mechanics, for example)?

What is the physical significance of the fact that $T(v)=F(x)=G(s)=T(p)$?

Also in the second link, it describes the Legendre transform as transforming between "conjugate variables". What does it mean physically for two quantities to be conjugate variables? What is the physical significance of the fact that velocity and momentum are conjugate variables? Why are they conjugate variables? Does it relate somehow to $F(v)=G(p)$?

These questions are almost certainly very silly, so I very much appreciate your patience with me and any help which you might be able to provide.

Note: This question is NOT a duplicate of this question; I am asking about a specific aspect of the Legendre transform.

EDIT: apparently this is supposed to mean something about velocity and momentum being "convex conjugates" -- but I still don't see what the physical interpretation of "convex conjugate" is nor of the graphical relationship it is supposed to imply.

• I think the answer probably has to do with the isotropy of space since the free particle Lagrangian must be a function of the velocity squared. Since mv^2 describes a parabola, that means the slope of the tangent lane uniquely traces out a parabola. The slopes of G are fixed by the slopes of the original function. One can then parameterize the transformed function G in terms of the slopes and G will also be a parabola by integration of these same very slopes. Convex conjugation is the generalization of the Legendre transform to vector valued functions, which is the case for velocity/momentum. – Daniel Kerr Jul 18 '16 at 18:35

This is a partial answer which I will hopefully come back to and expand.

The property of being its own Legendre transform is unique to the pure quadratic kinetic energy $T(v)=\frac12 mv^2$.

• As a simple example, consider $T(v)=\frac14Av^4$. Here the Legendre momentum is $$p=\frac{\partial L}{\partial v}=\frac{\partial T}{\partial v}=Av^3,$$ so the velocity is $v=v(p)=\sqrt[3]{p/A}$, and the Legendre-transformed kinetic energy is $$\tilde T(p)=\frac14Av(p)^4 =\frac14\frac{p^{4/3}}{A^{1/3}}.$$ This is a valid kinetic energy, but it does not coincide with our original quartic.

• As a more physical example, consider the relativistic kinetic energy, $$T(v)=-\frac{m c^2}{\gamma}=-mc^2\sqrt{1-v^2/c^2},$$ for which the Legendre conjugate variable is $$p=\frac{\partial L}{\partial v}=\frac{\partial T}{\partial v}=\frac{mv}{\sqrt{1-v^2/c^2}}=\gamma mv,$$ so the velocity is $$v=v(p)=\frac{cp}{\sqrt{p^2+m^2c^2}}$$ and the Legendre-transformed kinetic energy is $$\tilde T(p)=-\frac{m^2c^3}{\sqrt{p^2+m^2c^2}}$$ which, again, does not coincide with our original function. In particular, your hopes of a similar result in relativistic kinematics are not quite realized.

Neither of these results says much about what what the Legendre transform means, but they do say a fair bit about what it takes for the Legendre-transformed kinetic energy to coincide with the original form; in particular, this cannot have anything to do with isotropy or a naive symmetry argument.

That said, the quadratic Lagrangian can in fact be derived from very basic arguments, which as I recall was done in a terse but clear fashion in the beginning of Landau's mechanics book.

So, again, a partial answer - but hopefully enough to help steer this in the right direction.

• Yes in Landau's book a derivation of the Lagrangian for a free particle is given based on symmetries. – Chill2Macht Jul 19 '16 at 19:00
• I would love to see if you have anything more to add this weekend -- the grace period ends this Sunday. – Chill2Macht Jul 21 '16 at 21:29
• @William I'll see if I have enough time. I looked in Landau, and there is indeed a derivation of the quadratic Lagrangian, but I don't think it's particularly deep. It relies (on top of isotropy and homogeneity of space, which reduce you to functions of $v^2$) on imposing the requirement that all Galilean boosts reduce to gauge transformations, and that gives you $L\propto v^2$. Honestly, to me that looks conceptually orthogonal to the self-Legendre-transformation thing. – Emilio Pisanty Jul 21 '16 at 21:47