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Point particle in scalar potential

  1. Jan 17, 2014 #1
    What happens to a particle in a scalar potential [itex]U(t,x)[/itex]? I have been living under a belief that the equation of motion would be

    [tex]
    \frac{d}{dt}\frac{m\dot{x}(t)}{\sqrt{1- \frac{\|\dot{x}(t)\|^2}{c^2}}} = -\nabla_x U(t,x(t))
    [/tex]

    but I just proved that this isn't Lorentz invariant, and therefore cannot be the answer.

    Or to be more precise, I have had some doubts about the previous equation, and in some contexts I have been living under a belief that the equation

    [tex]
    \frac{d}{dt}\frac{\big(m + \frac{U(t,x(t))}{c^2}\big)\dot{x}(t)}{\sqrt{1- \frac{\|\dot{x}(t)\|^2}{c^2}}} = -\nabla_x U(t,x(t))
    [/tex]

    would have a better chance of being Lorentz invariant. Now when I looked this more carefully, I was unable to to prove the invariance claim for this to be true or false.
     
    Last edited: Jan 17, 2014
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  3. Jan 17, 2014 #2

    vanhees71

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    That's an interesting point in the dynamics of relativistic point particles.

    An elegant way to define covariant equations of motion is to use covariant action functionals. The idea is to introduce a world parameter [itex]\lambda[/itex] and write the action functional as
    [tex]A[x]=\int \mathrm{d} \lambda L(x,\dot{x}),[/tex]
    where [itex]x[/itex] is the space-time four-vector of the particle in Minkowski space and [itex]\dot{x}=\mathrm{d} x/\mathrm{d} \lambda[/itex].

    Then one can demand that the action functional is invariant under reparametrizations of [itex]\lambda[/itex], because this is rather a parameter to describe the world line of the particle rather than a physical parameter. As can be shown, one way to establish this is to demand that the Lagrangian is a homogeneous function of [itex]\dot{x}[/itex] of degree 1, i.e., that for any number [itex]\sigma[/itex]
    [tex]L(x,\sigma \dot{x})=\sigma L(x,\dot{x}) \; \Leftrightarrow \; \dot{x}^{\mu} \frac{\partial L}{\partial \dot{x}^{\mu}}=L. \qquad (*)[/tex]
    If then [itex]L[/itex] is a scalar you get covariant equations.

    For the kinetic term this leads to
    [tex]L_0=-m \sqrt{\dot{x}_{\mu} \dot{x}^{\mu}}.[/tex]
    A possible coupling to a scalar field is
    [tex]L=-g \sqrt{\dot{x}_{\mu} \dot{x}^{\mu}} \Phi(x).[/tex]
    The equations of motion then read
    [tex][m+g \Phi(x)] \ddot{x}^{\mu}=(\eta^{\mu \nu}-\dot{x}^{\mu} \dot{x}^{\nu})\partial_{\nu} \Phi.[/tex]
    It's easy to see that this equation fulfills the constraint
    [tex]\dot{x}_{\mu} \ddot{x}^{\mu}=0,[/tex]
    which should hold in order to be compatible with the on-shell condition for a classical particle, i.e.,
    [tex]p_{\mu} p^{\mu}=m^2.[/tex]

    For a more detailed discussion along these lines, see

    A. O. Barut, Electrodynamics and classical theory of fields and particles, Dover Publications (1980)
     
  4. Jan 17, 2014 #3

    Meir Achuz

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    What do you mean by 'scalar potential'?
    If U is a Lorentz scalar, then its effect is included by setting the RHS of your second equation to zero.
     
  5. Jan 17, 2014 #4
    Such equation would fail to reduce to the

    [tex]
    m\ddot{x}(t) \approx -\nabla_x U(t,x(t))
    [/tex]

    in the non-relativistic limit.
     
  6. Jan 17, 2014 #5
    The first equation of the opening post comes from a Lagrange function

    [tex]
    L(t,x,\dot{x}) = -mc^2\sqrt{1 - \frac{\|\dot{x}\|^2}{c^2}} - U(t,x)
    [/tex]

    and the second one from

    [tex]
    L(t,x,\dot{x}) = -\Big(mc^2 + U(t,x)\Big)\sqrt{1 - \frac{\|\dot{x}\|^2}{c^2}}
    [/tex]

    What these mean is not clear to me. The notation with mu- and nu-indices doesn't clear things for me, because the derivatives with respect to time or some other parameter still appear confusing.
     
  7. Jan 17, 2014 #6

    vanhees71

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    This Lagrangian most probably won't lead to covariant equations of motion, because the action functional is not a scalar!

    Which book are you using to learn this stuff? It's not clear to me how you do this without the usual vector-component notation with upper and lower indices for four-vectors.
     
  8. Jan 17, 2014 #7
    I'm not following any book. I've collected knowledge from different sources and gone through lot of thinking over the years. For example, did you know that the Lorentz transformation can be written as

    [tex]
    t' = \frac{t - \frac{x\cdot u}{c^2}}{\sqrt{1 - \frac{\|u\|^2}{c^2}}}
    [/tex]

    [tex]
    x' = x - \frac{x\cdot u}{\|u\|^2} u + \frac{\frac{x\cdot u}{\|u\|^2} - t}{\sqrt{1 - \frac{\|u\|^2}{c^2}}} u
    [/tex]

    for [itex]t,t'\in\mathbb{R}[/itex] and [itex]x,x'\in\mathbb{R}^3[/itex]. I know that many people don't know, because the books usually mention the special case [itex]u=(u^1,0,0)[/itex] and then proceed to the notation [itex]\overline{x}^{\mu}=\Lambda^{\mu}{}_{\nu}x^{\nu}[/itex]. Even in general cases it is possible to work out transformations with more primitive touch too. Lately I've been working on how all these quantities in the formulas transform, and been trying to find out if they keep the same form.
     
  9. Jan 17, 2014 #8
    Isn't this the same thing as my second Lagrangian? With [itex]\lambda=t[/itex], to be more precise, but this was supposed to work with reparametrizations.
     
  10. Jan 17, 2014 #9

    vanhees71

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    Ok, so you use the non-covariant (1+3)-dimensional notation. Then I understand, why you have a hard time to get correct equations of motion! It's way harder to find these equations using non-covariant approaches!

    The great thing with the approach I tried to explain in my previous posting is that you can immediately use the coordinate time, [itex]t[/itex], as the world parameter [itex]\lambda[/itex] (I'm setting [itex]c=1[/itex] all the time by the way). Since the action is independent on the world parameter and written in terms of a scalar world parameter the action is Lorentz invariant, also then you get equations of motion that are correct relativistic equations.

    In this case the Lagrangian for a particle moving in a scalar field reads
    [tex]L=-\sqrt{1-\dot{\vec{x}}^2} [m+g \Phi(t,\vec{x})].[/tex]
    Now you get directly equations for the three-vectors [itex]\vec{x}[/itex]. Although now the notation is not covariant any longer, because you don't use a scalar world parameter anymore but the coordinate time, [itex]t[/itex] in a fixed Galilei frame in Minkowski space, the physics content of the equation of motion, derived from the variational principle of least action, is the same as before, and thus it's a valid relativistic equation of motion. To prove that it is covariant under Lorentz transformations directly is, however, a bit more cumbersome than in the covariant notation, because in the latter case it's simply obvious.
     
  11. Jan 17, 2014 #10
    So you are sure that my second equation is Lorentz invariant in the end?

    If it can be proven elegantly, then it should also be possible to prove it not elegantly, and I would like the not elegant way, because it appears more primitive and certain. And it is always good to prove stuff several different ways.

    The quantity [itex]L[/itex] itself is not scalar for fixed spacetime point on the trajectory, so I don't see how the invariance of the extrema paths is supposed to be seen.

    (Lot of editing... whoops... I hope nobody has been quoting this lately...)

    The answer must be that the integral

    [tex]
    \int L(t,x(t),\dot{x}(t))dt
    [/tex]

    is invariant.
     
    Last edited: Jan 17, 2014
  12. Jan 17, 2014 #11
    Ok. I managed to prove that the action integral is Lorentz invariant, by using the ordinary change of variables in integrals and some stuff with three component vectors. Key intermediate results were the formulas

    [tex]
    dt = \frac{\sqrt{1 - \frac{\|u\|^2}{c^2}}}{1 - \frac{u\cdot v}{c^2}} dt'
    [/tex]

    and

    [tex]
    1- \frac{\|v'\|^2}{c^2} = \frac{\big(1 - \frac{\|v\|^2}{c^2}\big)\big(1 - \frac{\|u\|^2}{c^2}\big)}{\big(1 - \frac{u\cdot v}{c^2}\big)^2}
    [/tex]

    So I think I have shown rather primitively that the second equation of motion is Lorentz invariant. Although I don't like the fact that I had to resort to the invariance of the action integral. Surely the transformation properties of momentum, energy and gradients should allow a more direct proof too.
     
    Last edited: Jan 17, 2014
  13. Jan 17, 2014 #12

    Meir Achuz

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    I'm sorry. I did my first post without much thought.
    Using the relativistic Hamiltonian for a Lorentz scalar potential, I get almost your first equation, with the right hand side being [itex]-\sqrt{1-v^2}\nabla U[/itex].
    Of course, I still might have made a mistake.
     
  14. Jan 22, 2014 #13
    There was a mistake in my formulas. Neither of the two equations of motion in the first post is Lorentz invariant. The second Lagrangian mentioned in post #5 implies the equation of motion

    [tex]
    D_t \frac{(m + \frac{U(t,x(t))}{c^2})\dot{x}(t)}{\sqrt{1 - \frac{\|\dot{x}\|^2}{c^2}}} = -\sqrt{1 - \frac{\|\dot{x}\|^2}{c^2}} \nabla_x U(t,x(t))
    [/tex]

    No wonder I couldn't get my invariance proof working...

    I can tell where the mistake came from. In the Lagrangian the interaction comes by replacing [itex]m[/itex] with [itex]m+\frac{U(t,x(t))}{c^2}[/itex], so it might look as if the potential besides the mass does the trick also in the equation of motion.

    You were right here. Unfortunately it took me couple of days to get on the track...
     
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