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I Lagrangian paradox

  1. Feb 24, 2017 #1
    Let me begin by saying I know I'm doing something wrong here, but I'm having trouble seeing what it is. This is a reformulation of https://www.physicsforums.com/threads/plugging-eom-into-lagrangian.905099/, where I've reduced the issue to a much simpler problem. Moderators, feel free to close my original post

    Start with a 1-dimension particle at x(t) undergoing some force F(t), with the action,
    [itex]S[x] \equiv \int{\left(\dfrac{1}{2}m\dot{x}^2 + Fx\right)dt}[/itex].
    Under arbitrary variations [itex]x(t)\rightarrow x(t)+\delta{x}(t)[/itex], the variation in the action is given by,
    [itex]\delta{S} = \int{\left(-m\ddot{x}+F\right)\delta{x}dt}[/itex] + 'surface' terms.
    The surface terms can be eliminated by choosing [itex]\delta{x}[/itex] to vanish on the boundary, and by requiring [itex]\delta{S}=0[/itex] we get the Euler-Lagrange equation [itex]F = m\ddot{x}[/itex].

    Now, if in addition to our initial assumption about the action we add in [itex]F = m\ddot{x}[/itex] as a postulate, we arrive at a paradox. The action can be reduced to,
    [itex]S'[x] = \int{\left(\dfrac{1}{2}m\dot{x}^2 + m\ddot{x}x\right)dt} = \int{\left(-\dfrac{1}{2}m\dot{x}^2 + m\dfrac{d}{dt}\left(\dot{x}x\right)\right)dt}[/itex].
    The right-hand term can be dropped and we're left with just,
    [itex]S'[x] = \int{\left(-\dfrac{1}{2}m\dot{x}^2 \right)dt}[/itex].
    This results in the Euler-Lagrange equation [itex]\ddot{x}=0[/itex], meaning that [itex]F=0[/itex]. Clearly plugging the EOM back into the Lagrangian is a tricky operation, but if done carefully it should still produce valid results, which it doesn't seem to be doing here

    *note* Interestingly, repeating the same procedure for a spring-type force with action,
    [itex]S[x] \equiv \int \left(\dfrac{1}{2}m\dot{x}^2 - \dfrac{1}{2} kx^2\right)dt[/itex],
    results in [itex]S'[x] = 0[/itex]. This is exactly what I would expect. If there is only a single EOM, and you make it one of your postulates, the resulting action is trivial. Arriving at a conflicting action just has me confused
     
  2. jcsd
  3. Feb 24, 2017 #2

    Vanadium 50

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    It shoudln't work at all.

    1. The Lagrangian's value doesn't matter (it's often zero). What matters is its functional form. Plugging in anything canges the functional form.
    2. The point of the Lagrangian formalism is to obtain the EOMs. Plugging them in before you've obtained them is logically impossible.
     
  4. Feb 24, 2017 #3
    Well in this example I agree it seems pretty dumb, but if you had a Lagrangian with multiple DOF you could conceivably want to remove one. For example, integrating out a field in a QFT, which basically follows this same procedure if I remember it correctly. You use the EOM to replace one field with an infinite series that falls off rapidly in the energy range of interest, allowing you to drop all but the leading order term and work with an EFT.

    What bothers me is that the new functional form is inconsistent with the original one, not that it has a different value. I would expect any remaining DOF to retain equivalent EOM, and if there are none I would expect a trivial L~0
     
  5. Feb 28, 2017 #4
    ^bump, anyone else?

    Also, I'd just like to point out there is no logical flaw here, since I added the EOM as an additional assumption in the second example. The "paradox" is that adding the EOM as an assumption results in an inconsistent "EOM" with additional constraints (F=0)
     
  6. Mar 1, 2017 #5

    Vanadium 50

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    There is no paradox. The problem is not with the Lagrangian formalism. It's because you insist on performing an invalid operation.
     
  7. Mar 1, 2017 #6
    I know there is no paradox or problem with the Lagrangian formalism, that's why I put quotes around it. Im trying to understand why this is an invalid operation. How is this any different fundamentally than integrating a heavy field out of a field theory? You're removing a degree of freedom and looking at the resulting theory, which should be equivalent to the original if you include the extra assumption you've made.
     
  8. Mar 1, 2017 #7

    Vanadium 50

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    As I said in post #2, you are changing the functional form of the Lagrangian. All that matters is the functional form. That's why it is invalid.
     
  9. Mar 1, 2017 #8
    I don't think you're understanding the point I'm trying to make. Of course changing the functional form of the Lagrangian changes the EOM.

    Let me abstract this a litte:
    - Imagine you have some theory T described by Lagrangian L
    - Then take some theory T' which also includes L but adds some additional assumption A.
    - The EOM describing T and T' will clearly be different, but if you apply A to T you should arrive at a theory equivalent to T' no matter when you apply it.
    - Applying A before solving for the EOM should be exactly the same as applying it afterwards.

    In my example, A happens to be an EOM of T, which definitely makes things weirder, but I don't see why this would change anything in my logic above.

    If you're claiming I'm doing something invalid, could you please:
    a) Point out my logical flaw in the above paragraph
    b) Explain how integrating a field in a field theory is fundamentally different from what I'm doing

    I can accept that what I'm doing is wrong, but the whole point of posting this is to understand why. Until you can address the two issues above you're basically just telling me "You're wrong, trust me", which doesn't help me at all
     
  10. Mar 1, 2017 #9
    For the same reason that dividing by zero is an invalid operation.
    It is outside of the assumptions that need to be made so that the theory applies.
     
  11. Mar 1, 2017 #10

    Vanadium 50

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    You're right. That's exactly what I am doing. I can show you how to do it right, but if that's not enough for you, you're going to need to find someone else to help you.
     
  12. Mar 1, 2017 #11
    If somebody asked me why dividing by zero was an invalid operation I would be able to give more of a detailed answer than just "because it is". That's all I'm asking for here. And as I've mentioned 3 times now, this "invalid" operation is nearly identical to how you integrate out fields! Surely that isn't invalid right?
     
  13. Mar 1, 2017 #12
    Vanadium, unless I'm mistaking what you mean, I already know how to do it "right". You simply ignore my second case and use the first one to derive the EOM.

    If I'm wrong, there must be a logical flaw in those 4 lines I placed above. I'm simply asking you to show me what that is, so I can understand where I went wrong.

    As I've mentioned, there definitely are situations in which plugging additional equations into a Lagrangian is not only valid, but it produces useful results. So to just dismiss this operation as invalid isn't helpful, especially when you haven't even acknowledged those situations.
     
  14. Mar 1, 2017 #13
    Please describe your question as an idea instead of just throwing up formulas which might mean something or not.\\\\
     
  15. Mar 2, 2017 #14

    PeroK

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    The
    notation hides what is really going on with the maths. For example, ##S## is this case is a function from the set of particle trajectories to ##\mathbb{R}##. If you choose a specific trajectory you get a specific real number. Not something that you can optimise.

    A simpler example again is a free particle, where you have:

    ##S(\gamma) = \int_{\gamma} T = \int_{t_1}^{t_2} \frac12 mv^2 dt##

    Where the expression on the right represents a paramaterisation of the trajectory ##\gamma##, although notationally this detail is not shown.

    The normal approach when proving the Lagrangian result is to recast ##\gamma## using a real parameter ##\alpha## and a general variation on your trajectory, ##\eta##, then differentiate with respect to that real parameter, and finally, use the arbitrariness of the variation to get the result.

    If you plug in any specific trajectory, ##\gamma_0##, e.g. the solution where the particle has constant velocity, ##v_0##, you just get a number:

    ##S(\gamma_0) = \frac12 mv_0^2 (t_2 - t_1)##

    It's not a function, or a derivative or anything to be minimised. It's just a number.

    PS although, of course, as we have used the actual path it is the smallest such number.
     
    Last edited: Mar 2, 2017
  16. Mar 3, 2017 #15
    Thank you PeroK, I came to a similar conclusion the other night while I was falling asleep, and forgot about it until now. The answer I had arrived at was:
    in the first case, the Lagrangian is a functional L[x], as I had written. However, once you plug in [itex]F=m\ddot{x}[/itex] x can no longer be varied and is fixed by F. So you just end up with L = constant, which as you said can't be minimized in any meaningful way.

    Basically, I had continued to treat x as a dynamical variable after I had already assumed it to be fixed. There is nothing invalid about plugging EOM into the Lagrangian, you just need to be careful
     
  17. Mar 3, 2017 #16

    PeroK

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    Although many physics text books adopt this style of using ##x## rather ambiguously as a spatial variable and a curve/trajectory, I think it pays to distinguish between these things. Note also that if you were differentiating wrt ##x## (or ##\gamma##), the curve, you would need a metric on your multi-dimensional space of particle trajectories. Whereas, the trick is to characterise all variations on a trajectory by a fixed variation and a real parameter; thereby reducing the problem to one of single-variable calculus.

    If you get caught up in an apparent paradox, as you did, the quickest and best solution is usually to fall back on an element of mathematical rigour, which usually exposes the problem.
     
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