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Which Lagrangian is correct for SR?

  1. Mar 30, 2013 #1
    Is: [tex]\mathcal L=-\frac{m}{2} u^\alpha u_\alpha [/tex]

    a correct Lagrangian for SR (assuming the parameter is proper time rather than world time)?

    It leads to the correct EOM when plugged into the Euler-Lagrange equation, [tex]m\frac{du^\alpha}{ds}=0 [/tex]

    Or is this the correct Lagrangian:

    [tex]\mathcal L=-m \sqrt{u^\alpha u_\alpha} [/tex]

    which also leads to the correct EOM, [tex]m\frac{du^\alpha}{ds}=0 [/tex]?
     
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  3. Mar 30, 2013 #2

    WannabeNewton

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    For a free particle the lagrangian is ##\mathcal{L} = -mc^{2}\sqrt{1 - \frac{v^{2}}{c^{2}}}## (Landau and Lifgarbagez - Classical Theory of Fields, page 26).
     
  4. Mar 30, 2013 #3
    I am not sure if the first one can give the geodesic equation. It also fails to have reparametrization invariance.
     
    Last edited: Mar 30, 2013
  5. Mar 30, 2013 #4

    Bill_K

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    The Lagrangian is not unique. Any of these three may be used, depending on what you have in mind. The first two have the virtue of being relativistically invariant, and they're valid if you're writing the action in terms of the particle's proper time: I = ∫L dτ.

    Their disadvantage comes if you have more than one particle, and hence more than one proper time. In that case it's necessary to use the third form with the action I = ∫L dt.
     
  6. Mar 30, 2013 #5
    I have a question concerning the constraint that the 4-velocity squared is 1. Naively, the first Lagrangian gives -m/2 if you enter the constraint in, while the second Lagrangian gives -m. So it seems the second Lagrangian gives the correct action, since if the Lagrangian is -m, then the action is:

    ∫(-m) ds=-m∫ds

    The first Lagrangian is smaller by 1/2, which seems like it should matter when adding an interaction such as the E&M one, but it doesn't!

    Also the equation of motion for the second Lagrangian I think is derived like this:

    [tex]\frac{d}{ds} \frac{\delta (-m\sqrt{u^\alpha u_\alpha})}{\delta u^\beta}=F [/tex]
    [tex]\frac{d}{ds}\left(-m\frac{u_\beta}{\sqrt{u^\alpha u_\alpha}}\right)=F [/tex]

    Now if you set the 4-velocity squared equal to 1 in the denominator, you recover the EOM. So is the rule this: you can only insert a constraint after you take derivatives with respect to the coordinate fields to get the Euler-Lagrange equation?

    That Lagrangian makes more sense to me, but Landau and Lifgarbagez on page 64 of the same book derive it with 2nd Lagrangian (set A=0 in his example). The problem is he varied the action rather than using the Euler-Lagrange equations, and I got lost on when he was allowed to use the constraint that the 4-velocity squared equals 1.
     
  7. Mar 30, 2013 #6

    Bill_K

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    The value of the Lagrangian doesn't have any significance. The Lagrangian is a functional form to be varied, that's all it is.

    That's correct. You only get to use the constraint after the equations of motion have been derived.

    In addition to giving the Lagrangian, you also have to specify what's being varied. In the first case, uμ is not an independent variable, it's understood to be a shorthand for dxμ/dτ, and it's really xμ(τ) that's being varied.
     
  8. Mar 30, 2013 #7
    I thought for constraints, you can plug in the constraint immediately into the Lagrangian. As an example take the Lagrangian of a free particle in polar coordinates in 2 dimensions.

    For the constraint r=1, you can just set r=1 into the Lagrangian! Then the kinetic term involving the time derivative of the radial field is zero (since dr/dt=0), and the kinetic term for the angular field has a pure number multiplying it, r2(dθ/dt)2 becomes (dθ/dt)2.
     
  9. Mar 30, 2013 #8

    Bill_K

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    Quote from relativity book by Ohanian:
    I believe what he's saying is that the condition (dxμ/dτ)(dxμ/dτ) = 1 is not an a priori constraint, rather it's a dynamical consequence derived from the equations of motion after you have them.

    Similarly, in your 2-dimensional particle example, from the equations of motion one can derive the conservation of angular momentum, L = mr(dθ/dt)2, but you can't just go back and make this replacement in the Lagrangian.
     
  10. Mar 30, 2013 #9
    Thanks. That makes sense.

    If you're interested, here is something interesting I found:

    http://arxiv.org/pdf/hep-th/0009128v2.pdf

    Only look at equations 5 and 6. If you solve the E-L equations for η (η has no dynamics so it's a constraint field), and plug this η back into equation 5, you recover √(uμuμ)! So what the author seems to be claiming is that you can get rid of the square root, but you have to add the constraint field.

    But it seems you can directly use (1/2)[uμuμ) without an η field, so maybe the author has it wrong?
     
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