Lagrangian total time derivative for the velocity

[gionole]

I have a couple of questions and I'd appreciate if you could answer them one by one.

1. Here , It says the following:

The variable $\dot{q}$ is **independent** of the variables $t$ and $q$ otherwise the derivative
$\frac{\partial L(t,q, \dot{q})}{\partial \dot{q}}$ would be of dubious interpretation. The relation between $q$, $t$, and $\dot{q}$ is given by the second equation $\frac{dq}{dt}= \dot{q}$ and **it is valid only along the solution of the equations we are looking for.** The variables appearing in $L(t,q,\dot{q})$ are **independent of each other** before imposing the EL equations.

I wonder what is meant by independent ? When we vary the path $q(t) -> q(t) + \epsilon k(t)$ in the variational calculus, we also say that when that happens, $\dot q(t)$ changes and becomes $\dot q(t) + \dot k(t)$. Could you provide an example of the counter argument that how it would look if they were not independent and what wouldn't work ?

2. My question is now why 𝑓 can be dependent of $\dot q$ since only functions of the type ##\frac{d}{dt} f(q,t) leave the EL equations invariant ? Basically this is the same question as on the link, but I couldn't understand the answer.

 

[vanhees71]

It would be less confusing for students starting to learn about the action principle if one would define the lagrangian as a function $L(q,v,t)$, where $q$, $v$, and $t$ are $(2n+1)$ independent variables and then the action as a functional (!!!) of a curve in configuration space, parametrized by time, $q(t)$. Then
$$A[q]=\int_{t_1}^{t_2} \mathrm{d} t L[q(t),\dot{q}(t),t].$$
Now all quibbles should be clearly resolved.

Lazy as we physicists are we use the same symbol $\dot{q}$ on the one hand for the independent "generalized velocity" variables, $v$, as arguments of the Lagrangian and the time derivatives for the (arbitrary!) trajectories $\dot{q}(t)$ in the integral defining the action-function functional.

Now the variational principle used in Hamilton's action principle is that one looks for stationary points of the functional. Taking $q(t)$ being the stationary point, one thus has to evaluate
$$\left .\frac{\mathrm{d}}{\mathrm{d} \epsilon} A[q+\epsilon k] \right |_{\epsilon=0},$$
The $k(t)$ must obey the boundary conditions $k(t_1)=k(t_2)=0$. Then you get, doing the usual integration by parts the Euler-Lagrange equations
$$\frac{\partial}{\partial q} L[q(t),\dot{q}(t),t]-\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial}{\partial v} L[q(t),\dot{q}(t),t)]=0.$$

 

[gionole]

I get all that you explained, but I was asking a different question.

1. I don't get why $\dot q$ are independent of $t$ and $q$(especially $t$). The way we derive Euler lagrange is we vary $q(t)$ to $q(t) + \epsilon k(t)$ and $\dot {q(t)}$ to $\dot {q(t)} + \dot{k(t)}$. You see $\dot q(t)$ was changed as well when we changed $q(t)$. So how are they "independent" ?

2. I understand that adding total time derivative of $\frac{d}{dt} f(q,t)$ doesn't change equation of motion, but they say that adding $\frac{d}{dt} f(q,\dot q, t)$ doesn't change either but only if we have some more info. I'm asking what that would be.

 

[gionole]

About Question 1, I found so many questions asked by others. So I will first spend time to figure it out and update this thread later.

 

[vanhees71]

I tried to explain right this. In the ususal physicists' sloppy notation the expression $L(q,\dot{q},t)$ is meant as a function of $q$, $\dot{q}$, and $t$, being independent variables as far as the partial derivatives are concerned. A total time derivative of such a function means, again by definition, to plug in a trajectory $q(t)$ with $\dot{q}=\dot{q}(t)$ being the time derivatives, and then take the time derivative, i.e., with the chain rule for multi-variable derivatives,
$$\frac{\mathrm{d}}{\mathrm{d} t} f(q,\dot{q},t)=\dot{q} \cdot \frac{\partial f}{\partial q} + \ddot{q} \cdot \frac{\partial f}{\partial \dot{q}} + \frac{\partial f}{\partial t}.$$

 

[gionole]

In the Lagrangian formalism, $L$ itself is NOT a functional and $q$ and $\dot q$ are not functions of $t$, but as soon as we write the action, it becomes a functional.

Before: $L(q, \dot q, t)$

After: $S = \int_{t_1}^{t_2} L(q(t), \dot q(t), t)$

What is it that allows us to decide that: "ah, now, we can now start saying $q$ is a function of $t$ and $\dot q$ then is automatically the derivative of it ? Why can we do that ?" Is there some solid explanation about this ? I mean before we wrote an action, $L$ was thought to be the function of $q, \dot q$ which didn't depend on t, but in action, $q$ and $\dot q$ both depend on $t$

 

[vanhees71]

That's what I'm saying; $L$ is a function of the $(2f+1)$ independent variables, $q$, $v$, and $t$, and $S$ is a functional of the trajectories $q(t)$ and you have to set $v=\dot{q}(t)$, as you wrote yourself, defining $S$. That's a definition. The problem of understanding arises from the physicists's somewhat sloppy notation to write $\dot{q}$ instead of $v$.

 

[jbergman]

In the Lagrangian formalism, $L$ itself is NOT a functional and $q$ and $\dot q$ are not functions of $t$, but as soon as we write the action, it becomes a functional.

Before: $L(q, \dot q, t)$

After: $S = \int_{t_1}^{t_2} L(q(t), \dot q(t), t)$

What is it that allows us to decide that: "ah, now, we can now start saying $q$ is a function of $t$ and $\dot q$ then is automatically the derivative of it ? Why can we do that ?" Is there some solid explanation about this ? I mean before we wrote an action, $L$ was thought to be the function of $q, \dot q$ which didn't depend on t, but in action, $q$ and $\dot q$ both depend on $t$

@vanhees71 is correct. A clearer way to see it as follows.

Imagine starting with $L(a, b, c)$, a function of three variables a, b, and c.

Now we turn that into a functional by as you say plugging it into an Action with $q$, $\dot q$, and $t$ to get,

$$ A = \int L(q, \dot q, t) dt $$

Now we calculate $\delta A$ by adding to q another function of t, $h(t)$ where $h$ is 0 at our starting and ending times we are integrating over.

Plugging in we get

$$\delta A = \int L(q + h, \dot q + \dot h, t) dt - L(q, \dot q, t) dt $$

Now do a taylor expansion of L for the first term in the integral.

As a function, the Taylor expansion of L where we are only varying the first two parameters is

$$ L(a + \delta a, b + \delta b, c) = L(a, b , c) + \frac{\partial L}{\partial a} \delta a + \frac{\partial L}{\partial b} \delta b$$

Now, in our functional we just setting $a = q$ and $\delta a = h$ and similarly $b = \dot q$ and $\delta b = \dot h$.

Rewriting our variation of the action in terms of a,b, and c we get.

$$\delta A = \int \left[L(a, b , c) + \frac{\partial L}{\partial a} \delta a + \frac{\partial L}{\partial b} \delta b - L(a, b, c) \right]dc $$

Now we can perform a change of variables under what we said above, along with $t=c$ and canceling terms.

$$\delta A = \int \left[ \frac{\partial L}{\partial q} h + \frac{\partial L}{\partial \dot q} \dot h \right]dt $$

I hope this shows that the partial derivatives of $L$ really have nothing to do with the fact that q and $\dot q$ are related.

 

[gionole]

I think my final confusion comes to this point and I believe this is why I had hard time understanding all this.
Quite a big text, but have no choice.

After reading tremendous amount, I still got confusions.

Ok, we say that $q$ and $v$ are independent variables. The best explanation I found about this so far is the following: Imagine we got the following Action. So we're looking for a path such as inserting it in the $S$ and inserting velocity profile on that path gives the minimum value for $S$

$S = \int_{t_1}^{t_2} \frac{1}{2}mv(t)^2 - mgx(t)$

Now, read this: https://physics.stackexchange.com/a/168555/366606 (It's important for my question)

So in the integral above, we don't yet know the path, so what we do in the software programming sense is brute force it - pick a $t_1 + dt$, then for that $t_1 + dt$, pick random $x$ and for that $t$, pick a random $v$. Then, pick $t_1 + dt + dt$ and come up with random $x$ and random $v$ there and continue doing it for all next instances of $t$ until reaching $t_2$. We now came up with some constructed path $x(t)$ and $v(t)$ on that path. This path doesn't mean that it's true yet. Now we do the same process again and pick $t_1 + dt$, but now choose some other random numbers for $x$ and $v$ at that instant. Now, we came up with another constructed path and velocity. Since we do it randomly, $v(t)$ is not derivative of $x(t)$ for those constructed paths. We observe for whichever $x(t), v(t)$ S is minimum and thats what our true path is.
So in the nutshell, there are infinite combinations, but at some point, we would have come up with such: $x(t), v_1(t)$ and $x(t), v_2(t)$. Here, $x(t)$ is not true path, but some constructed path. The reason I have $v_1(t)$ and $v_2(t)$ is due to lots of combinations, at some point, I'd exactly come up with random numbers for $v$ for $x(t)$ such that it would match the derivative of $x(t)$. So let's say $v_2(t)$ is like this that matches $\dot x(t)$.

Question 1: Now, check - https://physics.stackexchange.com/a/891/366606 (only first reply) on which it's said $v(t)$ and $\dot x(t)$ only match on the real, true path, but With infinite combinations, we could have come up with non-true path on which I would have chosen random numbers for $v(t)$ in such order that it would match the derivative of that non-true constructed path $x(t)$. So not quite clear why it says $\dot x(t)$ matches $v(t)$ only on true path.

Question 2: When we do this brute force, why do we need to be choosing random numbers for $v$ at each instance during interval ? We could just choose random $x$ and use the velocity on that point by $\frac{x(t) - x(t-dt)}{dt}$. Why do we consider/brute force such paths where velocity not matches derivative of the path ? I don't find the "initial condition"(which says that at initial condition, velocity and position are unrelated), well it is there, but we could be only consructing/considering paths that match the derivative of $x(t)$ to $v(t)$.

 

[vanhees71]

That's nonsense!

Once more: the action is a functional (!!!), i.e., a mapping between trajectories in configuration space $q(t)$ and the real numbers. It's defined with help of a Lagrange function, which is a function of the independent variables $q$, $v$, and $t$, $L(q,v,t)$. The action functional (!!!) is then given by
$$A[q]=\int_{t_1}^{t_2} \mathrm{d} t L[q(t),\dot{q}(t),t].$$
Here $q$ can be any trajectory, for which this integral is defined.

Then there's Hamilton's principle, which says that the trajectory really taken by the particle is a stationary point of the action function among all trajectories with fixed initial an final points $q(t_1)$ and $q(t_2)$, which leads to the Euler-Lagrange equations as the equations of motion.

 

[gionole]

That's nonsense!

Once more: the action is a functional (!!!), i.e., a mapping between trajectories in configuration space $q(t)$ and the real numbers. It's defined with help of a Lagrange function, which is a function of the independent variables $q$, $v$, and $t$, $L(q,v,t)$. The action functional (!!!) is then given by
$$A[q]=\int_{t_1}^{t_2} \mathrm{d} t L[q(t),\dot{q}(t),t].$$
Here $q$ can be any trajectory, for which this integral is defined.

Then there's Hamilton's principle, which says that the trajectory really taken by the particle is a stationary point of the action function among all trajectories with fixed initial an final points $q(t_1)$ and $q(t_2)$, which leads to the Euler-Lagrange equations as the equations of motion.

I get that, but have you checked the links I posted ? they explain why the variables are independent. Especially this one - https://physics.stackexchange.com/q...ty-in-lagrangian-from-the-point-of-view-of-ph

 

[vanhees71]

Hm, I don't understand what the poster there means. I'd rather rely on textbooks than random postings on the internet.

 

[gionole]

Hm, I don't understand what the poster there means. I'd rather rely on textbooks than random postings on the internet.

I meant the answer. you see here https://physics.stackexchange.com/a/168555/366606 how he explains: "pic a time t, then pick a random r, then random v". My huge analysis and questions in #9 were based on this. If you don't understand this either, let me know and I will forget.

 

[vanhees71]

This is utter nonsense. We don't deal with any stochastic equations here. We have a function $L:\mathbb{R}^{2f+1}$, $(q,v,t) \mapsto L(q,v,t)$, from which a functional of trajectories $q:\mathbb{R} \rightarrow \mathbb{R}^f$ is defined by
$$A[q]=\int_{t_1}^{t_2} \mathrm{d} t L[q(t),\dot{q}(t),t]$$
with $\dot{q}(t)=\mathrm{d} q(t)/\mathrm{d}t$ being the time derivative is defined. Nothing here is random in any way. It's all just standard multi-variable calculus.

 

[gionole]

Question 1: So as soon as you write the following integral $A[q] = \int_{t_0}^{t_1} dt L[q(t), \dot q(t), t]$ and try to find minimum, you assume that $q(t)$ is the true path and if so, $v(t)$ would definitely be the derivative of $\dot q(t)$. Correct ? i think I understand this.

Question 2: my whole confusion started here: https://physics.stackexchange.com/a/985/366606 - you see how he in the beginning when he has integral, doesn't still say that $\dot q$ is not derivative of $q(t)$ ? Whats the reason of this ?

 

[vanhees71]

Question 1: Yes. In the standard way you assume $q(t)$ being a stationary point of the functional, which implies that its first functional derivative at this point must vanish. Now
$$\frac{\delta A[q]}{\delta q(x)}=\frac{\partial L}{\partial q}-\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{q}},$$
where also the fixed boundary conditions $\delta q(t_1)=\delta q(t_2)=0$ have been taken into account.

Question 2: but in the integral $\dot{q}$ IS the time derivative of $q(t)$. As I said, rather read a good textbook on analytical mechanics than random forum postings on the internet. Some good books are

the 2nd edition (avoid at all costs later editions!) of Goldstein, Classical Mechanics
A. Sommerfeld, Lectures on Theoretical Physics vol. 1
Landau&Lifshitz vol. 1
F. Scheck, Mechanics

For a more mathematical treatment:

V. I. Arnold, Mathematical Method of Classical Mechanics

 

[jbergman]

I think my final confusion comes to this point and I believe this is why I had hard time understanding all this.
Quite a big text, but have no choice.

After reading tremendous amount, I still got confusions.

Ok, we say that $q$ and $v$ are independent variables. The best explanation I found about this so far is the following: Imagine we got the following Action. So we're looking for a path such as inserting it in the $S$ and inserting velocity profile on that path gives the minimum value for $S$

$S = \int_{t_1}^{t_2} \frac{1}{2}mv(t)^2 - mgx(t)$

Now, read this: https://physics.stackexchange.com/a/168555/366606 (It's important for my question)

So in the integral above, we don't yet know the path, so what we do in the software programming sense is brute force it - pick a $t_1 + dt$, then for that $t_1 + dt$, pick random $x$ and for that $t$, pick a random $v$. Then, pick $t_1 + dt + dt$ and come up with random $x$ and random $v$ there and continue doing it for all next instances of $t$ until reaching $t_2$. We now came up with some constructed path $x(t)$ and $v(t)$ on that path. This path doesn't mean that it's true yet. Now we do the same process again and pick $t_1 + dt$, but now choose some other random numbers for $x$ and $v$ at that instant. Now, we came up with another constructed path and velocity. Since we do it randomly, $v(t)$ is not derivative of $x(t)$ for those constructed paths. We observe for whichever $x(t), v(t)$ S is minimum and thats what our true path is.
So in the nutshell, there are infinite combinations, but at some point, we would have come up with such: $x(t), v_1(t)$ and $x(t), v_2(t)$. Here, $x(t)$ is not true path, but some constructed path. The reason I have $v_1(t)$ and $v_2(t)$ is due to lots of combinations, at some point, I'd exactly come up with random numbers for $v$ for $x(t)$ such that it would match the derivative of $x(t)$. So let's say $v_2(t)$ is like this that matches $\dot x(t)$.

The above scenario is an ok mental model but not exactly true. The problem with above of picking a random x at each point is that it also determines the v's. So you can't pick both as you say. My suggestion would be to read the first chapter of Gelfand's calculus of variations to see how we discretize things. What you do instead of what you said is to add a random increment function h like I did in the derivation above. Did you look at that?

The problem you seem to be hung up on is how to do a Taylor expansion of the Lagrangian and some aspects of partial derivatives.

Question 1: Now, check - https://physics.stackexchange.com/a/891/366606 (only first reply) on which it's said $v(t)$ and $\dot x(t)$ only match on the real, true path, but With infinite combinations, we could have come up with non-true path on which I would have chosen random numbers for $v(t)$ in such order that it would match the derivative of that non-true constructed path $x(t)$. So not quite clear why it says $\dot x(t)$ matches $v(t)$ only on true path.

That is nonsense. Read Greg Graviton's answer there.

 

[jbergman]

The above scenario is an ok mental model but not exactly true. The problem with above of picking a random x at each point is that it also determines the v's. So you can't pick both as you say. My suggestion would be to read the first chapter of Gelfand's calculus of variations to see how we discretize things. What you do instead of what you said is to add a random increment function h like I did in the derivation above. Did you look at that?

The problem you seem to be hung up on is how to do a Taylor expansion of the Lagrangian and some aspects of partial derivatives.

That is nonsense. Read Greg Graviton's answer there.

Another way to explain this might be the following, imagine we start with a functional like,

$$ A = \int L(a(t), b(t), t) dt $$ and derive the Euler Lagrange Equations by setting the variation to 0.

Now, after you've derived the conditions for a minimum you just say for you specific problem $b(t) = \dot a(t)$. That might be closer to what people are trying to suggest with respect to independent variation of both position and velocity.

 

[gionole]

I actually understand the whole derivation of EL. It seems like internet confuses more when they mean that $q$ and $\dot q$ are independent. In the $L$(without action), we have $L = \frac{1}{2}mv^2 - mgx$. You can plug in totally unrelated values into it. Ok.

Let's upgrade $L$ to $L = \frac{1}{2}mv(t)^2 - mgx(t)$. If the particle didn't yet travel the path, ofc, $v(t)$ can't be derivative of $x(t)$, because $x(t)$ is not a path, but look at it as a function of $t$(just gives you some values when you plug in $t$, but doesn't mean particle yet travelled this path)..

Well, I get that, but who really cares that they're independent variables ? Thats what I don't get. why do we care to even ask this question whether they're dependent or independent ?

---

@jbergman

That is nonsense.

did you mean what I said or whats said on (https://physics.stackexchange.com/a/891/366606 (only first reply) ?
`
---

If we really care about that they're independent at first, how did we come up with $L = K - U$ ? Clearly, K and U are never independent, so again, what usefulness does it provide that K and U(velocity and position) are independent ? what do we use this information for ?

 

[jbergman]

I actually understand the whole derivation of EL. It seems like internet confuses more when they mean that $q$ and $\dot q$ are independent. In the $L$(without action), we have $L = \frac{1}{2}mv^2 - mgx$. You can plug in totally unrelated values into it. Ok.

Let's upgrade $L$ to $L = \frac{1}{2}mv(t)^2 - mgx(t)$. If the particle didn't yet travel the path, ofc, $v(t)$ can't be derivative of $x(t)$, because $x(t)$ is not a path, but look at it as a function of $t$(just gives you some values when you plug in $t$, but doesn't mean particle yet travelled this path)..

Well, I get that, but who really cares that they're independent variables ? Thats what I don't get. why do we care to even ask this question whether they're dependent or independent ?

So there are two perspective's here, from the point of view of taking the total derivative of $L$ we treat $q$ and $\dot q$ as independent in the Lagrangian. This is a standard thing we do often in calculus. However, they aren't actually idependent and we only vary our path $q(t)$ when computing the variation of the action.

did you mean what I said or whats said on (https://physics.stackexchange.com/a/891/366606 (only first reply) ?

What was said on that question was mostly nonsense. Greg Graviton's and QMechanic's answers were the only accurate ones I saw there.

If we really care about that they're independent at first, how did we come up with $L = K - U$ ? Clearly, K and U are never independent, so again, what usefulness does it provide that K and U(velocity and position) are independent ? what do we use this information for ?

We don't as you have rightfully stated. Again treating them as independent is just a calculation device in the derivation of the Euler Lagrange equations. Things like this are commonly done, though in physics and calculus. For instance we can turn a second order ODE into a system of first order ODEs like,

$$ F = \ddot x $$ which we can transform into a system of first order ODEs as follows.

$$ \begin{align}
v = \dot x \\
F = \dot v
\end{align}$$

 

[vanhees71]

You are really overcomplicating things. In short: In the expresion $L(q,\dot{q},t)$ the variables $q$, $\dot{q}$ and $t$ are meant to be independent variables as long as you consider partial derivatives with respect to these variables. Total time derivatives mean that you consider a trajectory, $q(t)$, in configuration space and $\dot{q}(t)$ as the time derivative, i.e., for a function of these variables you have the total time derivative,
$$\frac{\mathrm{d}}{\mathrm{d} t} f(q,\dot{q},t)=\dot{q} \cdot \frac{\partial f}{\partial q} + \ddot{q} \cdot \frac{\partial f}{\partial \dot{q}} + \frac{\partial}{\partial t} f,$$
which simply uses the chain rule of multi-variable calculus.

 

[gionole]

Thanks to both of you so much. I was overcomplicating it for sure. Now it makes sense.