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Relativistic Lagrangian and Hamiltonian for a free particle

  1. Nov 24, 2015 #1

    dyn

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    Hi.
    I am working through a QFT book and it gives the relativistic Lagrangian for a free particle as L = -mc2/γ. This doesn't seem consistent with the classical equation L = T - V as it gives a negative kinetic energy ? If L = T - V doesn't apply relativistically then why does the Hamiltonian H = T + V as the total energy still apply ? Thanks
     
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  3. Nov 25, 2015 #2

    pervect

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    Momentum is the partial of the Lagrangian with respect to ##\dot{x}##, which in this case would be velocity. So if you want positive momentum for a positive mass and a positive velocity, you want the negative sign in the Lagrangian - as v increases, gamma increases, and the Lagrangian becomes less negative as written, meaning the momentum is positive.
     
  4. Nov 25, 2015 #3

    Orodruin

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    It does not give a negative kinetic energy. In the non-relarivistic limit you have ##-1/\gamma = -\sqrt{1-v^2}\simeq -1+v^2/2## which is just the regular classical Lagrangian up to an additive constant. L = T-V does not hold relativistically, the kinetic term is not quadratic in ##\dot x##.
     
  5. Nov 25, 2015 #4

    pervect

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    The energy function (which is the total energy, the Hamiltonian in different v) is ##v\frac{\partial L}{\partial v} - L##)
    How are you calculating the energy? Given the Lagrangian, ##L(x,\dot{x},t)## one usually uses the legendre transformation

    $$H = v\,\frac{\partial L}{\partial \dot{x}} - L$$

    which comes out positive and equal to $$\frac{mc^2}{\sqrt{1-(v/c)^2}}$$
     
    Last edited by a moderator: Nov 25, 2015
  6. Nov 25, 2015 #5

    dyn

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    Thanks for your replies. So L = T - V does not apply relativistically. Does H = T + V apply ? If not , does H still give the total energy ?
     
  7. Nov 26, 2015 #6

    stevendaryl

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    The general relationships for Lagranian mechanics are these:

    [itex]p_\mu = \frac{\partial L}{\partial U^\mu}[/itex]
    [itex]H = p_\mu U^\mu - L[/itex]

    where [itex]U^\mu = \frac{dx^\mu}{ds}[/itex], for [itex]s[/itex] the "evolution parameter" (not necessarily time--relativistically, it is often proper time). Only in the special case where [itex]L[/itex] is quadratic in [itex]U^\mu[/itex] (that is, [itex]L[/itex] involves only zeroth, first and second powers of [itex]U^\mu[/itex]) does [itex]L = T - V[/itex] or [itex]H = T+V[/itex].
     
    Last edited: Nov 26, 2015
  8. Nov 26, 2015 #7

    pervect

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    No, H in the relativistic case is given by the Legendre transformation as in my post #4. (Which works for both relativistic and non-relativistic physics).

    Basically, the least-action principle when applied relativistically results in the principle of maximal aging, see for instance http://www.eftaylor.com/leastaction.html.

    [add]In other words, the expression for L is proportioanl to ##d\tau##, because ##d\tau = dt / \gamma## and ##L = -mc^2 / \gamma## and ##mc^2## is constant for a free particle.
     
    Last edited: Nov 26, 2015
  9. Nov 26, 2015 #8

    dyn

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    Thanks. So neither L = T-V or H = T+V apply relativistically. But does the Hamiltonian always give the total energy ?
     
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