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The relativistic Lagrangian

  1. Nov 26, 2009 #1
    In J.D. Jackson's Classical Electrodynamics, an argument is made in support of the assertion that the relativistic Lagrangian [itex]\mathcal L[/itex] for a free particle has to be proportional to [itex]1/\gamma[/itex]. The argument goes something like this:
    1. [itex]\mathcal L[/itex] must be independent of position and can therefore only be a function of velocity and mass.
    2. [itex]\gamma \mathcal L[/itex] must be a Lorentz scalar.
    3. The only available Lorentz invariant function of the 4-velocity is [itex]c^2 = v_\mu v^\mu[/itex].
    From this, it is (according to Jackson) "obvious" that the relativistic Lagrangian for the free particle has to be
    [tex]
    \mathcal L = -mc^2 / \gamma.
    [/tex]
    I guess I can see why this should be the case, given that the Euler-Lagrange equations need to be satisfied and that the Lagrangian needs to have the appropriate units. What I *don't* get is where (2) and (3) come from. Can someone please explain?
     
    Last edited: Nov 26, 2009
  2. jcsd
  3. Nov 26, 2009 #2
    (When I initially made this post, my question was incomplete. I've updated it. Sorry!)
     
  4. Nov 26, 2009 #3

    bcrowell

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    Can you give a page number? I can't seem to find what you're referring to.

    #3 seems obvious to me, but not #2. Re #3, what other scalar could you make out of the 4-velocity besides c? (Of course, you could make c5, etc.) Re #2, I don't know, but I'm guessing this is because the action has to be a scalar...?
     
  5. Nov 26, 2009 #4
    Sure. It's pgs. 580-81.

    A question about (3)...am I to infer that the only Lorentz invariant function of any 4-vector is its scalar product?
     
  6. Nov 26, 2009 #5

    bcrowell

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    Hmm...on pp. 580-581, I have the tail end of section 12.2, "On the question of obtaining the magnetic field..." This is in the 2nd ed. of Jackson. I don't see the material you're referring to. Do you have a later edition?

    I think so, except that you can obviously take the norm of a 4-vector and push it through any function you like that accepts a scalar input and is itself Lorentz-invariant. E.g., you can multiply the norm of the v 4-vector by the mass and get the norm of the momentum, or take the norm of the v 4-vector to the 5th power.
     
  7. Nov 26, 2009 #6
    Yes, I think we've got different editions; I'm pretty sure I'm using the 3rd Ed. The discussion I'm talking about shows up at the start of Chapter 12, in any event.
     
  8. Nov 26, 2009 #7

    bcrowell

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    Okay, on p. 573, the second page of section 12.1 in my 2nd edition of the book, I have an argument that [itex]A=\int_{\tau_1}^{\tau_2} \gamma L d \tau [/itex], and since A and [itex]\tau[/itex] are supposed to be invariant, [itex]\gamma L [/itex] has to be invariant as well. This seems reasonably sensible to me, although I wouldn't have had the confidence to state the same argument in the same somewhat breezy form without having thought long and hard about all the details that are not explicitly given. E.g., it seems plausible to me that we need to require Lorentz invariance for A, but it's not obvious that this is really true. If someone told me that A could be non-invariant, but all the predictions of the theory about experimental observables would still be 100% invariant, I wouldn't have had a snappy comeback to prove they were wrong.
     
  9. Nov 27, 2009 #8
    Ok. But I can understand an attempt to make the Lagrangian Lorentz invariant. (Is "covariant" another word for "Lorentz invariant"?) What I can't quite understand is this statement: "since A and [itex]\tau[/itex] are supposed to be invariant, [itex]\gamma L [/itex] has to be invariant as well." WHY is this true? I have a pretty rigorous training in higher mathematics, and it's trained me to need (and demand) a REASON for this! Maybe the integral can do something strange to [itex]\mathcal L[/itex] that makes it unnecessary for [itex]\gamma \mathcal L[/itex] to be Lorentz invariant...I certainly can't think of any reason why this can't happen!
     
  10. Nov 27, 2009 #9

    bcrowell

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    More or less. The term is used kind of loosely by physicists, with several different, but related, definitions. "General covariance" refers to the property of GR that its predictions are invariant under any smooth change of coordinates. "Covariant" can also be used as the opposite of "contravariant," to describe tensors with upper versus low indices. In the present context, however, it basically means the same thing as "Lorentz invariant."

    Yeah, I agree that Jackson is leaving a lot to the reader's imagination. I think it might be more transparent if you rewrite the equation for the action in differential form, as [itex]dA/d\tau=\gamma L[/itex]. The left-hand side is the quotient of two infinitesimally small numbers. (I hope you're not allergic to infinitesimals. They can be just as rigorous as limits, as shown by Robinson in the 60's.) Now [itex]d\tau[/itex] is Lorentz-invariant, and say we want dA to be Lorentz-invariant as well. Then dividing these two Lorentz-invariant quantities gives another Lorentz scalar. Therefore [itex]\gamma L[/itex] has to be a scalar as well. This is a very common mode of reasoning in relativity. You start with things that you know behave as well-defined tensors, and combine them to get new tensors. E.g., [itex]m dv/d\tau[/itex] is constructed out of a scalar, a 4-vector, and a scalar, so it produces a valid 4-vector (the momentum 4-vector).

    The main thing that sticks out to me as maybe needing more justification in my own reasoning above is that it's not necessarily obvious that dA has to be Lorentz-invariant, or even that A does. The only experimental observables are essentially incidence relations between world-lines, i.e., do they cross or not. It's conceivable that A and/or dA could be non-Lorentz-invariant, and yet you'd get observables that would be Lorentz-invariant. But I think the general philosophy is that all the machinery of tensors and least-action are designed so that you never, ever write anything down on the paper that isn't *manifestly* a valid relativistic equation.
     
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