Action integral: Just a constant?

[Wminus]

Hi!

While testing my knowledge of analytical mechanics I stumbled across a fallacy that I am unable to resolve. Could you assist me?

When you finish integrating a Lagrangian over the time domain $[t_1,t_2]$, shouldn't its position $q(t)$ and position dot $\dot{q(t)}$ variables take the values of $q(t_2),q(t_1),\dot{q(t_1)},\dot{q(t_2)}$? Hence the action integral is just a function of the endpoints, and since the variation at the endpoints is zero the action integral should be zero for ALL parametrization of $q$ and $\dot{q}$.

So have I found a glaring flaw in the theory of physics? :P Or is there something I am missing?

thanks for all input :)

 

[ShayanJ]

The action integral is a definite integral and so surely is a constant for a given path!
In fact the action integral is a functional which means it takes a function ($q(t)$) as an input and gives a number as an output. What we do in the calculus of variation to get the EL equations, is trying to find out which $q(t)$ minimizes the action. There is no flaw here!

 

[Wminus]

The action integral is a definite integral and so surely is a constant for a given path!
In fact the action integral is a functional which means it takes a function ($q(t)$) as an input and gives a number as an output. What we do in the calculus of variation to get the EL equations, is trying to find out which $q(t)$ minimizes the action. There is no flaw here!

But what about the variation being zero at $t_1$ and $t_2$? After you integrate the Lagrangian over the time domain, you insert $t_1$ and $t_2$ into the phase-space coordinates, and then the action becomes a "function" of the beginning and starting coordinates.

Also, say have a Lagrangian $L(q, \dot{q},t)$. Since you always can add a function $F(q,\dot{q},t)$ to a Lagrangian without changing the physics, can't you in principle set $F(q,\dot{q},t) = -L(q,\dot{q},t)$, and so get a new lagrangian $L' = L + F = 0$ ?

 

[ShayanJ]

But what about the variation being zero at $t_1$ and $t_2$? After you integrate the Lagrangian over the time domain, you insert $t_1$ and $t_2$ into the phase-space coordinates, and then the action becomes a "function" of the beginning and starting coordinates.

Its not like that. You can't say the action is both a function of the path and the end points. Only one at a time. Also even when you say the action is a function of the end points, it can only be a function of one of the end points, not both because it would be meaningless.
If you want to consider the action to be a function of the path, then you should fix the end points and so the action becomes a function of the path in the space of paths that cross those end points. If you change the end points, you're changing the domain and so you're changing the function. This case is the familiar Lagrangian mechanics.

Also, say have a Lagrangian $L(q, \dot{q},t)$. Since you always can add a function $F(q,\dot{q},t)$ to a Lagrangian without changing the physics, can't you in principle set $F(q,\dot{q},t) = -L(q,\dot{q},t)$, and so get a new lagrangian $L' = L + F = 0$ ?

That's wrong. You can only add to the Lagrangian a function of the form $\frac{d}{dt} F(q,t)$, not just any arbitrary function!

 

[Wminus]

Thank you for taking the time to help me. I still am confused about this stuff though:

Its not like that. You can't say the action is both a function of the path and the end points. Only one at a time. Also even when you say the action is a function of the end points, it can only be a function of one of the end points, not both because it would be meaningless.
If you want to consider the action to be a function of the path, then you should fix the end points and so the action becomes a function of the path in the space of paths that cross those end points. If you change the end points, you're changing the domain and so you're changing the function. This case is the familiar Lagrangian mechanics.

But why isn't it a function of both? Changing the path of the integration and/or changing its endpoints will both alter the action.

And is it not correct that after the integration is performed, $q(t)$ and $\dot{q(t)}$ cease to be free variables as $t_1$ and $t_2$ are inserted to produce a definitive integral? If so, shouldn't the variance of this be zero since the endpoints are considered to be fixed for the variation??

 

[ShayanJ]

But why isn't it a function of both? Changing the path of the integration and/or changing its endpoints will both alter the action.

Its true that changing either of them changes the action but it doesn't mean we have to let both be variables. The point is, we want to know if we consider two points in space, what path the particle takes to go from one to the other. Note that we don't specify any two special points, so there is no restriction.

And is it not correct that after the integration is performed, q(t)q(t) and q(t)˙\dot{q(t)} cease to be free variables as t1t_1 and t2t_2 are inserted to produce a definitive integral? If so, shouldn't the variance of this be zero since the endpoints are considered to be fixed for the variation??

It seems to me that you don't understand Lagrangian mechanics yet. You should spend more time studying it and also using it. But for now, I can only say that what we do to find out the EL equations, is like when we find the maximums or minimums of a function. The function gives a constant at a particular point but this is irrelevant because we want to know in which points the function has minimums or maximums.

 

[PeroK]

Thank you for taking the time to help me. I still am confused about this stuff though:

But why isn't it a function of both? Changing the path of the integration and/or changing its endpoints will both alter the action.

You've got good answers from Shyan above. I'd like to add that most mechanics texts are not fully specific about how the Lagrangian is defined. So, to resolve your questions fully, you may have to go back to your text and note carefully how the Lagrangian is defined and what it is (and is not) dependent on.

There's another thread here that might help:

https://www.physicsforums.com/threads/lagrangian-invariance-short-question.807289/#post-5070417

 

[Wminus]

So basically we "choose" to vary either the end points or the paths when using the $\delta S$ notation, where $S$ is the action?

This is why here http://en.wikibooks.org/wiki/Classical_Mechanics/Lagrange_Theory#Is_the_Lagrangian_unique.3F the variation in $F$ disappears while the variation in $S$ survives; they define that $\int_{t_1}^{t_2} \frac{dF}{dt} dt$ is a function of end point coordinates, while $S$ is a function of paths?

 

[lightarrow]

So basically we "choose" to vary either the end points or the paths when using the $\delta S$ notation, where $S$ is the action?
This is why here http://en.wikibooks.org/wiki/Classical_Mechanics/Lagrange_Theory#Is_the_Lagrangian_unique.3F the variation in $F$ disappears while the variation in $S$ survives; they define that $\int_{t_1}^{t_2} \frac{dF}{dt} dt$ is a function of end point coordinates, while $S$ is a function of paths?

Don't know what you have understood about action and lagrangian, but the lagrangian *cannot*, in general, be written as the time derivative of a function, otherwise the action would only depend on the end points and not on the entire path between them. In that document they show how you can add to your lagrangian the time derivative of a function and so get the same variation of the action, for the reason I've written up, and not that *every* lagrangian can be written in that way.
Don't know if that helps.

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