4
$\begingroup$

Question: Why is (from an intuitive standpoint) the kinetic energy $T$ a fixpoint of the Legendre transformation, i.e. $\frac{\partial T}{\partial \dot q}\dot q-T = T$ for any general coordinate $q$? Can I derive this without Lagrangian or Hamiltonian mechanics?

Background: I noticed that I can derive $\frac{\partial T}{\partial \dot q}\dot q-T = T$ from Lagrangian / Hamiltonian mechanics via ($q$ is any general coordinate):

$$\begin{align} H & = \frac{\partial L}{\partial \dot q} \dot q - L \\ \implies\ T+V &= \frac{\partial (T-V)}{\partial \dot q} \dot q - (T-V) \\ \implies\ T+V &= \frac{\partial T}{\partial \dot q} \dot q - T + V \\ \implies\ T &= \frac{\partial T}{\partial \dot q} \dot q - T \end{align}$$

Now I asked myself why it makes intuitively sense that the kinetic energy is a fixpoint of the Legendre transformation. It doesn't seem to be a coincidence...

$\endgroup$
1
$\begingroup$

Let the Lagrangian have the traditional form

$$ \mathcal{L}(q,\dot{q};t) = \frac{\dot{q}^2}{2} - V(q) \, ,$$

where $V(q)$ represents the potential energy. We see that the condition

$$ \frac{\partial \mathcal{L}(q,\dot{q};t)}{\partial q} = - \frac{\partial V(q)}{\partial q} = 0 $$

describes extrema of the Hamiltonian

$$H(p,q) = p \dot{q} - \mathcal{L}(q,\dot{q};t) = \frac{p^2}{2} + V(q) \, , \quad \text{where} \qquad p = \frac{\partial \mathcal{L}(q,\dot{q};t)}{\partial \dot{q}} \, .$$

Since the Legendre transformation is defined as a sup, i.e for each evaluation point $p$ we have

$$ f^*(p) = \sup_{\dot{q}} \left( p \dot{q} - \mathcal{L}(q,\dot{q};t)\right) \, ,$$

the fact that the Legendre transform maps the kinetic energy to itself is equivalent to saying that the point where the potential energy vanishes, which is an extremum of the Lagrangian, will also be an extremum of the Hamiltonian. This is fundamental, since fixed points must have the same stability properties in either Lagrangian or Hamiltonian formulations: An unstable equilibrium in the Lagrangian description must be mapped to an unstable equilibrium in the Hamiltonian description, etc. The definition of the Legendre transform as a suppremum guarantees that the graph of $f^*(p)$ will be convex, being bounded above/below by the points where $V(q)=0$, i.e. by the kinetic energy.

All of the above can be generalized to non-convex functions using convex duals, but I don't know enough to elaborate.

Regarding your interest in Legendre transformations outside of coordinate-free formulations of mechanics (not talking about thermodynamics here), I would say they're simply not useful. Legendre transforms have no meaning in Newtonian mechanics apart from their abstract mapping of points into lines, which is quite useless unless your formulation takes place on a cotangent bundle.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.