# Why is the Legendre transformation the correct way to change variables from $(q,\dot{q},t)\to (q,p,t)$?

I always found Legendre transformation kind of mysterious. Given a Lagrangian $$L(q,\dot{q},t)$$, we can define a new function, the Hamiltonian, $$H(q,p,t)=p\dot{q}(p)-L(q,\dot{q}(q,p,t),t)$$ where $$p=\frac{\partial L}{\partial \dot{q}}$$. Here, we are also expressing $$\dot{q}$$ as a function of $$(q,p,t)$$ by inverting $$p=\frac{\partial L}{\partial \dot{q}}$$. This way of defining the new function of $$(q,p,t)$$ from a function of $$(q,\dot{q},t)$$ is called the Legendre transformation; $$H$$ is called the Legendre transdform of $$L$$.

But I might have defined a function of $$(q,p,t)$$ by a simpler route. Take $$L(q,\dot{q},t)$$ and simply re-express it as a function of $$\tilde{L}(q,p,t)$$ without doing any Legendre transformation. If we are interested in changing variables from $$\dot{q}\to p$$, this is as good.

• My question is, why not work with the function $$\tilde{L}(q,p,t)$$? An inelegant thing about $$\tilde{L}(q,p,t)$$ (as opposed to $$H(q,p,t)$$ obtained by making a Legendre transformation) is that we cannot find an equation of motion for $$\tilde{L}(q,p,t)$$. Also, it has no energy interpretation. Is there something more to it (mathematically and physically)? Why is the Legendre transformation always the correct way to go from $$(q,\dot{q},t)\to (q,p,t)$$?

Legendre transformation is necessary to switch to new independent variables: $$q, \dot{q}, t\rightarrow q,p,t$$. The differential of $$H$$ is: $$dH = \frac{\partial H}{\partial q}dp + \frac{\partial H}{\partial q}dq + \frac{\partial H}{\partial t}dt = \dot{q}(p)dp - \frac{\partial L}{\partial q}dq - \frac{\partial L}{\partial t}dt,$$ i.e. $$H$$ is genuinely a function of $$q,p,t$$, whereas the differential of $$\bar{L}$$ still requires knowing $$\partial\dot{q}/\partial p$$, even it is parametrized by $$p$$.

It is the same Legendre transformation (although with a different sign) that is used in thermodynamics to switch between different thermodynamic potentials, i.e. between different sets of independent variables and responses.

When you switch from Lagrangian mechanics to Hamiltonian mechanics, you are not just making a change of variables, but you are moving from a problem set on the tangent bundle $$TM$$ to a problem set on the cotangent bundle $$T^*M$$. Furthermore, you are gaining a whole new symplectic structure.

Keep in mind that your goal is to solve the equations of motion. As already observed in a comment above, if you just make a change of variables, what equation do you get? To derive the Lagrange equations, you minimize the action functional $$S=\int \mathcal{L}(t,q,\dot{q})dt$$ by varying $$q$$ and $$\dot{q}$$. If you try to apply a similar treatment without introducing the Hamiltonian function, you will encounter some difficulties.

1. Explaining why OP's proposal won't work is usually more difficult than just showing how the standard construction works, but let us try: Besides practicality, a problem with OP's proposal $$\tilde{L}(q,p,t)$$ is that it is not self-contained. To deduce the EOMs$$^1$$, we need more information than what the function $$\tilde{L}(q,p,t)$$ itself provides, e.g. a relation between $$\dot{q}$$ and $$p$$.

2. Example: A non-relativistic charge in an EM background: The Lagrangian is $$L~=~\frac{m}{2}{\bf v}^2+q({\bf v}\cdot{\bf A}-\phi).$$ Then $${\bf p}~=~\frac{\partial L}{\partial {\bf v}}~=~m{\bf v}+q{\bf A}.$$ So $$\tilde{L}~=~\frac{{\bf p}^2}{2m} - q\underbrace{\left(q\frac{{\bf A}^2}{2m}+ \phi\right)}_{=\phi_{\rm eff}},$$ i.e. from $$\tilde{L}$$ we don't know how much of $$\phi_{\rm eff}$$ is due to electric and/or magnetic potentials, respectively, even though they lead to different physics.

3. In contrast, an important virtue of both the Lagrangian and the Hamiltonian formulations is that they each are self-contained formulations. Knowing the Lagrangian xor the Hamiltonian gives us the EOMs.

4. Concerning the standard Legendre transformation, see also e.g. this & this related Phys.SE posts.

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$$^1$$ The equation for $$\tilde{L}$$ can be transcribed from the equation for $$L$$, but it will contain other functions as well.

• Are you saying that knowing Lagrange's equation (in terms of $L(q,\dot{q},t)$) is not enough to determine the differential equation for $\tilde{L}(q,p,t)$? Jul 16, 2020 at 13:52
• $\uparrow$ No. I updated the answer. Jul 16, 2020 at 13:56