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Lagrangian intuition

  1. Jul 4, 2010 #1

    nomadreid

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    Starting on the topic of the Lagrangian, I have been told not to try to make intuitive sense, but just accept the nice differential equations which it goes into. Fine, but it should at least make basic sense. That L= T-V should be stationary means then d(T-V) = dT-dV = 0, i.e., dT=dV, which translates into that if the kinetic energy increases, then the potential energy increases by the same amount. In what sort of context is this supposed to make sense?
    Thanks in advance.
     
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  3. Jul 4, 2010 #2

    radou

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    You can think of it as "nature being lazy" in a sense. If the potential energy increases, the Lagrangian decreases, since a system "wants" chooses for its trajectory the path of least action, which is the one that minimizes the Lagrangian.
     
  4. Jul 4, 2010 #3

    nomadreid

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    Thank you, radou; this is the kind of explanation I am looking for. I am still a bit puzzled. Let us take the idea of the potential energy increasing to bring T-V down. Nature being lazy would seem to me to favor decreasing the potential energy, down to as low as possible, which is why things are attracted to a gravitational center. To increase the potential energy of an object in a gravitational field would mean that it would be taken further away from the center, which is not a path that nature takes without added energy.
    I'm missing something in the necessary conditions, I am sure.
     
  5. Jul 5, 2010 #4

    Born2bwire

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    A gravitational potential is conservative. As such, any change in the potential from point A to point B will have a corresponding change in the energy of the object. The Lagrangian path is the stationary path, which may not necessarily be the path that maximizes or minimizes the action. This is an important point because it allows us to find multiple paths for a given system. For example, a beam of light incident upon a multilayered medium will have multiple paths of reflection and transmission. Each of these paths must be a valid solution for the associated Lagrangian (via Fermat's principle or QFT Path Integral formulations). Back to the gravitational potential, don't forget that it is T-V, the kinetic energy of the object plays an important role and it is the kinetic energy that allows for the paths that differ from just plunging into the potential well (a stable orbit for example).
     
  6. Jul 8, 2010 #5

    Dale

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    This is not correct. Consider for example a mass m dropped from the origin in a uniform gravity field g pointing in the -x direction. We know from Newtonian mechanics that the solution is x = -1/2 gt². The kinetic energy is T = 1/2 mx'² and the potential energy is U = mgx. Substituting in the solution for x gives us T = 1/2 m(-gt)² and U = mg(-1/2 gt²) = -T. You were not clear in the math what you were differentiating wrt, but from what you said you are assuming it is time so dT/dt = mg²t and dU/dt = -mg²t = -dT/dt. This is just a counter example to show that your premise is wrong. The Lagrangian is NOT conserved.
     
  7. Jul 9, 2010 #6

    Dale

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    I think I understand the confusion here. What is stationary is not the Lagrangian, but rather the action (the integral of the Lagrangian). And it is stationary not wrt small changes in time, but wrt small variations in the trajectory of the system in state space. Remember, this is a variational approach so we are varying functions and calculating functionals.
     
    Last edited: Jul 9, 2010
  8. Jul 9, 2010 #7

    Cleonis

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    Try reading the article "www.eftaylor.com/pub/ActionFromConsEnergy.pdf"[/URL][/u] written by Jozef Hanc and Edwin F. Taylor.

    The're putting in a serious effort to build up to an intuitive understanding of the principle.

    There's more where that came from, the authors have collaborated on several other articles. Links to those articles are avaible on the [u][PLAIN]http://eftaylor.com/leastaction.html" [Broken]
    page on Taylor's website.
     
    Last edited by a moderator: May 4, 2017
  9. Jul 10, 2010 #8

    Cleonis

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    As Dalespam points out: the condition is that the time integral of the Lagrangian, varied over a range of trajectories, is an extremum.

    I take an example from the discussion by Hanc and Taylor, linked to in post #7

    A particle travels a distance in space, and during that trajectory there is at one point a jump in potential energy. (For instance, a particle travels a long pipe, enters a linear accelerator, and then exits again. That is, the section where the acceleration occurs is very short compared to the total journey. For simplicity let's say the exit pipe is just as long as the entry pipe.)

    In a diagram the worldline of the particle will consist of two straight sections, at some angle to each other.

    Along each individual section there is constant kinetic energy and constant potential energy. At the point where the two straight sections meet there is an exchange of energy: the potential energy decreases with a particular amount, and the kinetic energy increases with that amount.

    Now vary the worldline away from the actual worldline, by varying the angles of the two sections. (What that does is that it shifts the duration of the two sections, keeping the total duration of the journey the same.)
    Along any worldline that isn't the actual worldline the change in kinetic energy will not match the change in potential energy. Another way of saying this: if one section of the worldline is too steep the other section will be too shallow.

    Following Hanc and Taylor the following action can be defined for this particular case:

    [tex] S = (T - V)_1 t_1 + (T - V)_2 t_2 [/tex]

    This action has a minimum for the case of the actual worldline.

    This can be generalized to the case of summing over a sequence of infinitisimally short sections, with a different potential between each pair of adjacent sections.
    The actual worldline is the one where at each passage from section to section the exchange of kinetic energy and potential energy matches.

    As we know, summing over a sequence of infinitisimally short sections is integration. You evaluate the time integral of the Lagrangian, over a range of variations of the wordline. This evaluation reaches an extremum in the case of the actual worldline.


    Hanc and Taylor would be uncomfortable with this presentation, because they want to derive the actual worldline from the action principle, rather than derive the action principle from a precalculated wordline. All along their aim is to show that the action principle is more profound than the newtonian calculation.
     
  10. Jul 10, 2010 #9

    Cleonis

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    I'm going to do something dodgy now, I'm going to copy a section from the original article by Hanc and Taylor, "www.eftaylor.com/pub/ActionFromConsEnergy.pdf"[/URL][/u], except that I'm going through the steps [i]in reverse[/i].

    Recapitulating the setup of post #8 :
    A particle travels a distance in space, and during that trajectory there is at one point a jump in potential energy. (For instance, a particle travels a long pipe, enters a linear accelerator, and then exits again. That is, the section where the acceleration occurs is very short compared to the total journey. For simplicity let's say the exit pipe is just as long as the entry pipe.)

    Length of entry pipe : d
    length of exit pipe : d
    Kinetic energy [itex] E_k [/itex]
    Potential energy [itex] E_p [/itex]

    The goal is to try a range of worldlines, each one traversing the jump point at a different point in time. Just one wordline among those satisfies the condition of energy being conserved:

    [tex] - \left (\tfrac{1}{2}m{v_1}^2 + E_{p,1} \right ) + \left (\tfrac{1}{2}m{v_2}^2 + E_{p,2} \right ) = 0 [/tex]

    Rewriting to make all terms explicitly a function of a single variable 't'.

    [tex] - \frac{md^2}{2t^2} - E_{p,1} + \frac{md^2}{2(t_{total}-t)^2} + E_{p,2} = 0 [/tex]

    Now I'm going to regard the above equation as a [i]derivative with respect to time[/i] of another function. Then - dropping constants - the primitive is as follows:

    [tex] \frac{md^2}{2t} - E_{p,1}t + \frac{md^2}{2(t_{total}-t)} - E_{p,2}(t_{total}-t) [/tex]

    Once again a rewrite, changing the notation of the time intervals:

    [tex] \tfrac{1}{2}m{v_1}^2t_1 - E_{p,1}t_1 + \tfrac{1}{2}m{v_2}^2t_2 - E_{p,2}t_2 [/tex]


    Which is the action as defined in post #8

    [tex] S = (T - V)_1 t_1 + (T - V)_2 t_2 [/tex]

    Of course this is not a rigorous demonstration. But it will help, I think, in moving towards an intuitive understanding that the demand that the trajectory for which the action is an extremum is mathematically equivalent to the demand that energy is to be conserved.
     
    Last edited by a moderator: Apr 25, 2017
  11. Jul 11, 2010 #10

    Dale

    Staff: Mentor

     
    Last edited by a moderator: Apr 25, 2017
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