If $x = e^t$, than what is the partial with respect to $\dot{x}$ of $\dot{x}^2-x$?

Question

The Motivation:

I am learning the basics of Lagrangian mechanics. The usual example of using Lagrangian mechanics is to find the equation of motion of a pendulum. Since this forum is about math and not physics, I will simply write down the expression which is confusing me.

The Lagrangian ($\mathcal{L}$) is

$$\mathcal{L} = \frac{m}{2}\left(l\frac{d{\theta}}{dt}\right)^2 - mgl(1-\cos(\theta))$$

(Where $m$ is the mass of the bob and $l$ is the length of the string.) The issue is when I try to find the quantity $\frac{\delta\mathcal{L}}{\delta \theta}$: the solution simply means treating $\dot{\theta}$ as a variable independent of $\theta$, even though since they are both functions of $t$ we can write down a relationship between the two.

The Problem:

So there's no point in making the algebra messy, so let's use an easier example. Let's pretend $\mathcal{L} = \dot{x}^2 - x$ and furthermore, let's pretend $x=e^t$.

This is a problem! Obviously, in this special case, $x=\dot{x}$, so wouldn't $\frac{\delta\mathcal{L}}{\delta\dot{x}}=2\dot{x}-1$? Apparently, $\frac{\delta\mathcal{L}}{\delta\dot{x}}$ is actually $2\dot{x}$.

We could also give a different example. Now, let's use the same $\mathcal{L}$, but let $x=e^{7t}$. In this slightly different case, $x=\frac{1}{7}\dot{x}$, so shouldn't $\frac{\delta\mathcal{L}}{\delta\dot{x}}=2\dot{x}-\frac{1}{7}$? Once again, apparently, $\frac{\delta\mathcal{L}}{\delta\dot{x}}$ is actually $2\dot{x}$.

The Question:

The reason I brought up the motivation for this problem is just in case the math is special for the case of Lagrangian mechanics (which I wouldn't expect).

Otherwise, why is it that when we take the partial of $\mathcal{L}$ with respect to $\dot{x}$, we can, with impunity, ignore the implicit relation between $\dot{x}$ and $x$ and just treat $\dot{x}$ as a variable completely independent of $x$?

Answer 1

The Lagrangian is a function that takes in functions $x(t)$ and $\dot{x}(t)$ and outputs a function $L(x,\dot{x})(t)$. The arguments in the Lagrangian, $x$ and $\dot{x}$, are treated as independent variables when working out derivatives. Thus we always have $\frac{\partial \dot{x}}{\partial x} = 0$ and $\frac{\partial x}{\partial \dot{x}} = 0$. Many authors use a different name for $\dot{x}$ like $p=\dot{x}$ to avoid this confusion (which is very common). In practice $L(x,\dot{x})$ can be treated as a normal function in two variables when working out the derivatives. Thus for your example $\frac{\partial L}{\partial x} = -1$ and $\frac{\partial L}{\partial \dot{x}}=2\dot{x}$.

It does not make sense to substitute in a function $x(t)$ in the Lagrangian and then asking what the derivatives $\frac{\partial L}{\partial x},\frac{\partial L}{\partial \dot{x}}$ are. If you substitute in a function like $x(t) = e^{t}$ you are left with a normal function and the functional information about how the Lagrangian looks like is lost.

Answer 2

It's been a while since I asked this, but I stumbled onto something that is conceptually helpful.

Imagine $x(t)$ takes a particle from point $a$ to point $b$. Pretend that the time it takes to go from $a$ to $b$ is 10 seconds and that it travels in a straight line. However, it might travel at a constant speed, or it might start out really fast and go slowly the rest of the way.

So now it is conceptually easy to accept that $x(t)$ and $\dot{x}(t)$ are like independent variables in the Lagrangian.