Lagrangian - How can we differentiate with respect to time if $v$ not a function of time?

Question

In the Lagrangian itself, we know that $v$ and $q$ don't depend on $t$ (i.e - they are not functions of $t$ - i.e., $L(q,v,t)$ is a state function.)

Imagine $L = \frac{1}{2}mv^2 - mgq$

Euler-Lagrange is given by $\frac{d}{dt}\frac{\partial L}{\partial v} = \frac{\partial L}{\partial q}$

This yields: $\frac{d}{dt}(mv) = -mg$

Now we take time derivative of $mv$ which seems would only be possible if $v$ was a function of time in the beginning, but we know in $L$, $v$ is not a function of $t$, so how can we take derivative of it with respect to to $t$?

Answer 1

In the Lagrangian itself, we know that $v$ and $q$ don't depend on $t$ (i.e - they are not functions of $t$ - i.e $L(q,v,t)$ is a state function.)

Imagine $L = \frac{1}{2}mv^2 - mgq$

OK.

Euler-Lagrange is given by $\frac{d}{dt}\frac{\partial L}{\partial v} = \frac{\partial L}{\partial q}$

Not really--Or at least this is vague, since you do not indicate that the arguments are evaluated at the classical path. This vagueness seems to be the cause of your confusion.

The Euler-Lagrange equation you wrote above is only true for the correct classical path, which I will call $\ell(t)$ (to avoid the above-mentioned symbolic confusion). We have to evaluate at $q = \ell(t)$ and $v = \dot \ell(t)$.

When you derive the Euler Lagrange equation you determine the classical path $\ell(t)$ by finding the path $\ell(t)$ that minimizes $$ S[q] = \int_{t_1}^{t_2} L(q(t),\dot q(t), t)dt\;. $$

In other words: $$ S_{min}=S[\ell(t)] = \int_{t_1}^{t_2} L(\ell(t),\dot\ell(t),t)dt $$

It is only this correct classical path that satisfies the equation of motion. So, strictly, you should be writing: $$ \frac{d}{dt}\left.\frac{\partial L}{\partial v}\right|_{q=\ell(t), v=\dot \ell(t)} = \left.\frac{\partial L}{\partial q}\right|_{q=\ell(t), v=\dot \ell(t)}\;. $$

The function $L(a,b,c)$ has three inputs, and we need to know which input we are taking the partial derivative with respect to. In physics we usually do this by writing the derivative with respect to a symbol that we have agreed represents the correct position such as $\frac{\partial L}{\partial a}$ to indicate partial derivative with respect to the first argument.

Suppose you wanted to use some different notation such as $L^{(i)}$ to mean the derivative with respect to the ith argument. Then the equation of motion would read: $$ \frac{d}{dt}L^{(2)}(\ell(t), \dot \ell(t), t) = L^{(1)}(\ell(t), \dot\ell(t), t)\;, $$ which could further be "simpified" to: $$ (L^{(2)})^{(1)}\dot\ell +(L^{(2)})^{(2)}\ddot\ell +(L^{(2)})^{(3)}=L^{(1)} $$

This yields: $\frac{d}{dt}(mv) = -mg$

No, strictly speaking, it yields: $$ \frac{d}{dt} \left(m \dot \ell(t)\right) = -mg\;, $$ since you need to have evaluated $v$ at $v=\dot \ell(t)$.

Answer 2

The main point is that

1. On one hand, the arguments $(q,v,t)$ of the Lagrangian $(q,v,t)\mapsto L(q,v,t)$ are independent (so that it e.g. makes sense to talk about partial derivatives of $L$).

2. On the other hand, for the solutions$^1$ $t\mapsto q(t)$ to the Lagrange equations, it is assumed that the position $q(t)$ and the velocity $v(t)=\frac{dq(t)}{dt}\equiv\dot{q}(t)$ do depend on time $t$.

For further details, see e.g. my related Phys.SE answer here.

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$^1$ Notabene: Be aware that in the physics literature the same notation is often used for a function, a value of a function, and the codomain variable.

Answer 3

This problem is solved by writing the Euler Lagrange equation more explicitly like this :

$$\frac{d}{dt} \Bigr (\frac{\partial L(q,v)}{\partial v}\Bigr |_{(q,v)=(q(t), \dot {q}(t))}\Bigr )=\frac{\partial L(q,v)}{\partial q}\Bigr |_{(q,v)=(q(t), \dot{q} (t))}$$

So, after differentiating L, we are evaluating the derivative at $(q(t), \dot {q} (t))$. This is where both the LHS and the RHS become a function of time.

Answer 4

Indeed, the Euler-Lagrange equations are two, exactly as the Hamilton equations. Unfortunately, almost always one of the two is not written: $$\dot{q}(t)=\frac{dq}{dt}$$ $$\frac{d}{dt}\frac{\partial L(t,q(t),\dot{q}(t))}{\partial \dot{q}}=\frac{\partial L(t, q(t), \dot{q}(t))}{\partial q}$$ where one is looking for a curve $$t\mapsto (q(t),\dot{q}(t)).$$

The Legendre transformation to prove the equivalence of Lagrangian and Hamiltonian formulations strongly relies upon this viewpoint.

A reference.

Answer 5

Now we take time derivative of $mv$ which seems would only be possible if $v$ was a function of time in the beginning, but we know in $L$, $v$ is not a function of $t$...

This isn't actually true. The Lagrangian you provided certainly is independent of time (meaning it has no explicit dependence on $t$), but this does not mean that $v$ and $x$ are independent of time. They are, in fact, both implicitly time-dependent, which also makes the Lagrangian implicitly time-dependent. Hence taking the time derivative of an implicit function of time is quite sensible.

See also the Euler-Lagrange equation on Wikipedia where the whole derivation relies on the implicit time-dependence of $v$ and $r$. It is also discussed in the Wikipedia article on Lagrangian mechanics. I suspect most textbooks cover this as well, but it's been a while since I've looked at one.

Answer 6

Consider a map of the Earth. Latitude and longitude are independent. But then consider a path between, say, Denver and Tokyo. Along that path, latitude is a function of longitude. The map is a tool for finding paths.

Similarly, the Lagrangian is a tool for finding paths through the state space. The velocity and position don't depend on time in the space, but they do along any particular trajectory.