Why don't we differentiate velocity wrt position in the Lagrangian?

Question

In Analytic Mechanics, the Lagrangian is taken to be a function of $x$ and $\dot{x}$, where $x$ stands for position and is a function of time and $\dot{x}$ is its derivative wrt time.

To set my question, lets consider motion of a particle along a line:

$$x: \mathbb{R} \to \mathbb{R} ~~as~~ t \mapsto x(t)$$

and take the Lagrangian to be:

$$L(x, \dot{x}) := \frac{1}{2}m\dot{x}^2 - V(x)$$

By applying the Euler-Lagrange equations:

$$ \frac{d}{dt}\left(\frac{\partial{L}}{\partial\dot{x}}\right) = \frac{\partial{L}}{\partial x}$$

we get back Newton's law of motion.

This follows formally where we consider $x$ and $\dot{x}$ as independent, but if we consider $\dot{x}$ as velocity, then it is indeed a function of position so when we partially differentiate $L$ wrt $x$ the $\dot{x}$-terms shouldn't vanish and this messes up the derivation. What am I misunderstanding?

edit

I am beginning to think that this is a non-question as some people seem to have suggested and I am being confused by the hand-wavy math in basic physics textbooks.

In the above example, let just consider $C$ to be the configuration space of the particle. Locally $C$ is given by the co-ordinate function $x: U \to \mathbb{R}$, the tangent bundle $\pi: TC \to C$ being locally trivial has as co-ordinate functions above $U$: $(x\circ \pi) \oplus dx: TU \to \mathbb{R^2}$.

We take the Lagrangian to be simply a functional: $TC \to \mathbb{R}$, which when written locally is in terms of $x$ and $\frac{\partial}{\partial x}$.

But, what about the dot in $\dot{x}$?

Say the particle traces out $\gamma: I \to C$, where interval $I$ is time. Let $\gamma(0) = p \in C$. Locally in terms of $x$, since we have $\gamma_{\ast}(\frac{d}{dt}\rvert_0)$ in $T_{p}C$, we get $dx(\gamma_{\ast}(\frac{d}{dt}\rvert_0)) = \dot{x \circ \gamma}(0)$, which we can abuse notation and write as $\dot{x}(p)$, so the corresponding point on the bundle looks like $(x(p), \dot{x}(p))$.

Does this make sense or have is there something I am still missing?

Answer 1

This is an excellent question! And the answer has its roots back in the origin of the Euler-Lagrange equations as solutions to Hamilton's variational principle.

Recall that the Euler-Lagrange equations are the result of extremising the action $$S(q[t]) = \int_0^T L(q, \dot q)\,dt.$$ It is conventional to write $L$ as function of $q$ and $\dot q$ (and this will lead to the cleanest formulation of the Euler-Lagrange equations) but it is by no means rigidly required: you could add $q^2$ as a third argument to the functional $L$, if you really wanted to, and of course $q$ and $q^2$ are not independent.

Now the usual derivation is to look at variations $\delta q(t), \delta q(0) = \delta q(T) = 0$, which yields

$$\frac{d}{ds}S(q[t]+s\delta q[t])\Bigg\vert_{s\to 0} = \int_0^T \left[(D_1 L)(q, \dot q)\delta q + (D_2 L)(q,\dot q) \delta \dot q\right]\,dt,$$ where $D_iL$ means differentiating $L$ with respect to the $i$th parameter. Notice that the above in no way requires that the values that are plugged in to $L$ (namely $q$ and $\dot q$) are independent! $D_1L$ is often written as $\frac{\partial L}{\partial q}$, but this is an abuse of notation: we are merely differentiating the two-parameter function $L$ with respect to its first argument, without knowing or caring what we will eventually plug into $L$.

Now we apply the usual integration by parts,

$$\int_0^T \left[(D_1L)(q,\dot q) - \frac{d}{dt}\left[(D_2L)(q,\dot q)\right]\right]\delta q = 0,$$ where the time derivative, unlike the two derivatives of $L$, happens after plugging in $q$ and $\dot q$, and we recover the usual equations of motion by taking $\delta q$ to be bump functions at all times in $(0,T)$.

Answer 2

This is an excellent question (and I'm biased because I also have it) no one seems able to answer it well. Here are some threads about it I have found which discuss it:
(1) https://physics.stackexchange.com/questions/168551/independence-of-position-and-velocity-in-lagrangian-from-the-point-of-view-of-ph
(2) https://physics.stackexchange.com/questions/885/why-does-calculus-of-variations-work

The best I have been able to discern is the following: the Euler-Lagrange equation essentially defines the velocity to be the derivative of the position with respect to time. Without assuming the Euler-Lagrange equation, velocity is NOT the time derivative of position.

When we consider phase space, $q$ and $\dot{q}$ are just variables, so I will denote $\dot{q}=r$. I.e. in general NO relationship holds between $q$ and $\dot{q}$.

So the Lagrangian is just a regular function of two variables.

(In the simplest case, phase space is just $\mathbb{R}^3$ with axes $t,q,r$).

The Euler-Lagrange equation then gives us a particular curve in phase space (NOT all of phase space).

The curve is in three dimensions and we can project it onto the $tq$ plane.

On this plane, the curve happens to define $q$ implicitly as a function of $t$ (I don't know how to prove this or exactly why it's true, but it does seem to be the case).

So we have a function, denote it $x : \mathbb{R} \to \mathbb{R}$ such that for all $t$: $x(t)=q$, i.e. $x: t \mapsto q$.

For some reason (which I also don't know and can't prove) $x(t)$ is a differentiable function of $t$. So $x'(t)$ is well-defined for all $t$.

Therefore, for every point $t \in \mathbb{R}$, we have a point in $\mathbb{R}^2$=tq-plane which is $(t, x(t))=(t,q)$ defined implicitly by the projection of the curve given by the solution to the Euler-Lagrange equation.

Therefore, given any point in the projection of the curve in the tq-plane, we consider the original point in $\mathbb{R}^3$ it was projected from --> $(t,q,r)=(t,x(t),r)$.

Now the Euler-Lagrange equations are such that it turns out that $x'(t)=r$ (I don't know how to prove this either) -- hence our point on this curve in 3-space can be written as $(t,x(t),x'(t))=(t,x(t),r)=(t,q,r)$.

Since physicists already know in advance that, for the curve in 3-space defined by the Euler-Lagrange equation which they will be considering (note that they don't consider any remaining part of phase space where this is NOT true), it will be true that $r=x'(t)$ and $q=x(t)$, they just call the variables $x$ and $\dot{x}$ i.e.

The notation that physicists use is a sloppy abuse of notation that assumes in advance that only the solution curve of the Euler-Lagrange equation (which is equivalent to the principle of least action) will hold (since they always assume that it holds).

So in some sense (if you consider the entire $r$ axis in phase space, and not just those points which are part of the solution curve of the Euler-Lagrange equation) $\dot{x}$ isn't the velocity of $x$, so they really are just independent variables.

I would love to see to a rigorous proof of all of this myself, but so far this is the only answer that has made sense to me.

A picture that sort of explains the idea of what I am trying to say:

Answer 3

I think your are getting confused by the notations of the partial derivative. If $L$ is a function of two variables, $L:(x,y)\mapsto L(x,y)$, $\dfrac{\partial L}{\partial x}$ is the derivative with respect to the first variable and $\dfrac{\partial L}{\partial y}$ w.r.t to the second (if you write these derivatives $\partial_1 L$ and $\partial_2 L$, there is no possible confusion). Then $x$ and $y=\dot x$ are two independent variables of the function $L$ and $\partial_2 L$ is the derivative of $L$ w.r.t. the second variable. Here, you need to evaluate this second derivative in a particular value (i.e. $\dot x$).

In short, $x$ and $\dot x$ are independent variables of $L$.

Answer 4

[Second answer, separately because it is different]

Maybe this will help: This independency of position and velocity can be seen in Newton's equation: it is a second order differential equation. To solve it (uniquely) for a 1-dof system, you need two initial conditions (initial position and initial velocity).

Answer 5

That the Lagrangian of the particle moving at small speeds can be expressed in the stated form is an experimental observation. The principle of least action limits the manner in which the particle can explore its phase space. For the application of the least action principle to determine the equation of motion, the coordinate, and the velocity would have to be treated independently then. It is possible to think of a situation where the Lagrangian would depend also on $\ddot{x}$, in which case $x$, $\dot{x}$ and $\ddot{x}$ would be treated independently. Our world does not seem to behave that way though.