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Conceptual questions on Lagrangian formalism

  1. Aug 11, 2014 #1
    Hi all,

    I have a (hopefully) quick conceptual question that I'd like to clear up with your help.

    Is the following argument for why we treat position and velocity as independent variables in the Lagrangian correct? (referring to classical mechanics) :

    The Lagrangian, [itex] \mathcal{L}[/itex] of a given physical system contains a full description of the dynamics of the system. As space and time are considered to be homogeneous (and in addition, space to be isotropic), if one can fully specify the configuration of a system at a given instant in time, then one can (in principle) determine its configuration at any later time. This is done by finding a specific form for the Lagrangian of the system from which one can determine the evolution of the system.

    In order to fully describe the state of the system at a given instant in time [itex] t[/itex], and hence, be able to determine its state at some later time [itex]t+\delta t[/itex], it is sufficient to specify the coordinates and velocities of all of the particles constituting the system at the initial time [itex]t[/itex]. Specifying the coordinates of each particle alone is insufficient, as one cannot determine the dynamics of the system (and hence, how it will evolve) without knowing the corresponding velocities of the constituent particles.

    At a given instant in time [itex]t [/itex], the [itex]i^{th} [/itex] particle within a given system, has a given coordinate [itex]q_{i}\left(t\right) [/itex] of fixed value (for that instant in time. Where [itex]q_{i} [/itex] is a generalised coordinate, working in 1-dimension for simplicity). In principle, for a coordinate value [itex]q_{i} [/itex] at time [itex]t[/itex], a particle can have any number of different velocities [itex]\dot{q}_{i}\left(t\right) [/itex] (in essence, the coordinates and velocities that specify the initial conditions of the system are independent). As such, one treats both coordinates [itex] q_{i}(t)[/itex] and velocity [itex] \dot{q}_{i}(t)[/itex] as independent variables in order to fully describe the state of the system at a given instant in time [itex] t[/itex] and hence determine its dynamics.

    Sorry, I know this argument is a bit convoluted, but hopefully you get the gist of it.
     
    Last edited: Aug 11, 2014
  2. jcsd
  3. Aug 12, 2014 #2

    jedishrfu

    Staff: Mentor

    Your description seems reasonable and rather legalistic so perhaps theres a simpler way to describe it and bring out the beauty of Lagrangian Mechanics.

    The way I look at it is:

    For any given closed physical system, if we know the initial positions of particles then we know the potential energy of the system but nothing of its kinetic energy.

    If we know only the initial velocities of the particles then we know the kinetic energy of the system.

    If we know both then we can use the principle of least action (or stationary action as Susskind likes to say) to
    to find the path where T - V = 0 and that is the path the system will take over time.

    Until we know the path where T - V = 0 we can't say how p and v will vary with time so we can treat them as independent variables.
     
  4. Aug 12, 2014 #3
    Thanks, yes that is the kind of explanation I was going for - that in order to fully specify the state of a mechanical system and hence predict how it will evolve, one must know the coordinates and velocities of all the constituent components of the system. Knowing one or the other of these two variables is not enough. It is only after varying the path of the system that the "functional relationship" between coordinates and velocity come to light, through the relationship [itex]\delta\dot{q}_{i} \left(t\right) = \frac{d}{dt}\left(\delta q_{i}\right) [/itex]. Like you say, as initially we do not know how the coordinates vary in time, we must treat the velocity as an independent variable.

    One slight point (sorry to be nit-picking), but shouldn't it be the path where [itex]\delta S = \int^{t}_{t_{0}} \delta\mathcal{L} \;dt = \int^{t}_{t_{0}} \delta\left( T-V\right)\;dt =0 [/itex]? (although I'm guessing you're giving a rougher argument for brevity, so apologies in advance if this is the case).
     
    Last edited: Aug 12, 2014
  5. Aug 12, 2014 #4

    jedishrfu

    Staff: Mentor

    Yes you are more correct.
     
  6. Aug 13, 2014 #5
    As a follow-up, is it correct to say that the reason coordinates [itex]q_{i}\left(t\right)[/itex] and corresponding velocities [itex]\dot{q}_{i}\left(t\right)[/itex] are treated as independent in the Lagrangian is because the positions and velocities of the particles constituting the mechanical system can be chosen independently in the initial conditions. We can not obtain [itex]\dot{q}_{i}\left(t\right)[/itex] from [itex]q_{i}\left(t\right)[/itex] as we do not know explicitly how [itex]q_{i}\left(t\right)[/itex] depends on time, for a given mechanical system, until the appropriate variations of the path of the system between two instants in time are carried out. Hence we must treat the two variables as independent (at least initially). The "functional" relationship between [itex]q_{i}\left(t\right)[/itex] and [itex]\dot{q}_{i}\left(t\right)[/itex] [itex]\left(\text{i.e.}\;\; \delta\dot{q}_{i} = \frac{d}{dt}\left(\delta q_{i} \right)\;\right) [/itex] only arises upon varying the Lagrangian to obtain the equations of motion. Would this be an appropriate summary of the situation?
     
    Last edited: Aug 13, 2014
  7. Aug 16, 2014 #6
    There is a much simpler explanation. When the variation of the action integral is computed, we use the chain rule, which works regardless of any dependencies among the variables. Then we get rid of ##\delta \dot q## via the integration by parts; that certainly does assume a lot of dependency between ##q## and ##\dot q##, following which we get rid of both the integral and ##\delta q##, and the resultant equations, which seemingly suggest independence between ##\dot q## and ##q##, remain an eternal source of confusion for students :).
     
  8. Aug 17, 2014 #7
    Thanks, yes that's true. My confusion was more about making the Lagrangian a function of velocity as well as position in the first place, but I think I've cleared it up in my mind somewhat:

    Although the function [itex]\dot{q}\left(t\right)[/itex] is the derivative of [itex]q\left(t\right)[/itex] with respect to time, [itex]t[/itex], at a given point in time the value of [itex]\dot{q}\left(t\right)[/itex] is not related to the value [itex]q\left(t\right)[/itex] at that point (since a value is just a number, not a function). Hence, at each given point in an interval, say [itex]t\in\left[t_{0},t_{1}\right][/itex], the values of [itex]q\left(t\right)[/itex] and [itex]\dot{q}\left(t\right)[/itex] can assume values independently of one another. Therefore, it makes sense to define a function [itex]\mathcal{L}[/itex] that is a function of both [itex]q\left(t\right)[/itex] and [itex]\dot{q}\left(t\right)[/itex] as each of the possible paths that could be taken from one configuration of the system at [itex]t=t_{0}[/itex] to another, at some later time [itex]t=t_{1}[/itex] will depend on the positions and the velocities of the constituent particles. Since at any point [itex]t\in\left[t_{0},t_{1}\right][/itex] the values [itex]q\left(t\right)[/itex] and [itex]\dot{q}\left(t\right)[/itex] are independent, they can be varied independently.

    Is this a fair assessment?
     
  9. Aug 17, 2014 #8
    It is true that if you just view the Lagrangian as a function of its arguments, then you have to specify coordinates and velocities independently.

    But as soon as you start dealing with real or virtual trajectories and variations, that is not a correct assumption. A function and its derivative are intrinsically related, so if you vary one, that puts at least a constraint on the variation of another. But, lo and behold, when we vary the action, we can use the chain rule that treats the Lagrangian as just a function of independent variables, and then we use some trickery that rids us of variations entirely, leaving only the "independent" derivatives of the Lagrangian.
     
  10. Aug 17, 2014 #9

    vanhees71

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    Science Advisor
    2016 Award

    This also bothered me a lot when I learnt about the Lagrange formalism the first time. The reason is that physicists get used to a short-hand notation, knowing what they mean by convention.

    Once one should look at it in the right way. In the Lagrange formulation of the Hamilton principle of least action one starts with generalized coordinates, [itex]q^k[/itex] ([itex]k[/itex] is a superscript index similar to writing upper and lower indices in careful treatments of the Ricci calculus in vector spaces, where one clearly distinguishes vectors and covectors=linear forms).

    The Lagrange function, [itex]L(q,v)[/itex] is a function of the [itex]q^k[/itex] and the generalized velocities [itex]v^k[/itex] (sometimes also explicitly of time, but that's a minor complication).

    The action is a functional of curves in configuration space, parametrized as functions of time, [itex]q^k(t)[/itex]. It's defined as
    [tex]S[q]=\int_{t_1}^{t_2} \mathrm{d} t L[q(t),\dot{q}(t)].[/tex]
    The mathmatically strict notation for the functional derivative, would be
    [tex]\frac{\delta S}{\delta q^k}=\left [\frac{\partial L(q,v)}{\partial q^k} \right ]_{q=q(t),v=\dot{q}(t)}- \frac{\mathrm{d}}{\mathrm{d} t} \left \{\left [\frac{\partial L}{\partial v^k} \right]_{q=q(t),v=\dot{q}}\right \}.[/tex]
    Now this notation is pretty clumsy for everyday use, and the physicists simply write
    [tex]\frac{\delta S}{\delta q^k}=\frac{\partial L}{\partial q^k}-\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{q}^k}.[/tex]
    In this short-hand notation, when taking a partial derivative with respect to [itex]\dot{q}^k[/itex] one consideres these [itex]\dot{q}^k[/itex] as independent variables from the [itex]q^k[/itex], but when taking the total time derivative of an expression, you have to read [itex]\dot{q}^k[/itex] as the time derivative of [itex]q^k[/itex].
     
  11. Aug 17, 2014 #10
    Thanks voko and vanhees71, appreciate your help on the matter. I understand what you're describing, but I guess my main issue was trying to justify why we treat the Lagrangian as dependent on velocities as well as positions in the first place?! Is it simply that we know that the state of a given system can be completely specified if one knows the positions and velocities of all of the particles constituting the system, and as such, as we don't know what the positions of the particles vary with time (and hence how their velocities vary) we must consider both parameters as (in some sense) independent parameters that the Lagrangian depends on at each point along its path (at least initially, before performing any variations)? It is only when we vary the path taken between two fixed configurations that we gain a dependence between [itex]\dot{q}[/itex] and [itex]q[/itex], i.e. the variation is velocity is equal to the time derivative of the variation in the position.
     
  12. Aug 17, 2014 #11
    Yes, your understanding is correct. A given Lagrangian describes the "design" of a given physical system. The system, according to the design, can be in various states, and the full specification of any given state requires every generalized coordinate and every generalized velocity.

    This is consistent with the fundamental fact that a system of second-order differential equations (which is what Euler-Lagrange equations are) needs positions and velocities to specify a unique solution.
     
  13. Aug 17, 2014 #12
    Cheers voko, appreciate the time you've spent helping me clear up my conceptual understanding.
     
  14. Aug 18, 2014 #13
  15. Aug 18, 2014 #14
    Thanks for the link :) Looks like it'll be a good read.
     
  16. Aug 18, 2014 #15
    Interestingly, I found this argument http://physics.stackexchange.com/questions/885/why-does-calculus-of-variations-work given by "Qmechanic" (4th (main) answer down in the thread, starts with "Here is my answer, which is basically an expanded version of Greg Graviton's answer.

    The question of why one can treat position and velocity as independent variables...") on physics stackexchange which was kind of what I was trying to explain originally, but didn't do a very good job. It sounds convincing to me, but is it correct?
     
  17. Aug 18, 2014 #16
    I do not really see any new content in that discussion. The Lagrangian is defined as a function of both coordinates and velocities (the book I referenced explains why that is so); the variations that seem to treat coordinates and velocities as independent functions only seem so. All of that has been said here and is repeated on stackexchnage.
     
  18. Aug 18, 2014 #17
    Yeah, you're right, just thought I'd add it on for completeness and to verify it (I tend to trust physics forums more than I do stackexchange). Thanks anyway.
     
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