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Massless charged particle in electrostatic field

  1. Feb 9, 2014 #1

    vanhees71

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    Does anyone here know a source, where the equation of motion for a massless charged particle (which of course does not exist in Nature as far as we know) in a homogeneous electric field is actually solved? I googled, and there are some papers about this, but the equations are never actually solved, although they are pretty easy.

    My idea was to use the action principle in a covariant formulation with a Lagrange multiplyer to enforce the mass-shell condition for a massless particle, as follows. To that purpose I take the Lagrangian to read
    [tex]L=-\lambda \dot{x}_{\mu} \dot{x}^{\mu} +q x^3 E \dot{x}^0,[/tex]
    where the metric is [itex]\eta_{\mu \nu}=\text{diag}(1,-1,-1,-1)[/itex] and the speed of light is set to 1.

    In order to lead to a scalar action, the Lagrange multiplier must transform under reparametrizations of the "world-time parameter", [itex]\tau[/itex] (which is NOT the proper time of the particle, because we cannot define the proper time for a massless particle) as
    [tex]\tau \rightarrow \tau', \quad \lambda \rightarrow \lambda'=\lambda \frac{\mathrm{d} \tau'}{\mathrm{d} \tau}.[/tex]
    Thus, after deriving the equations of motion from the action principle, I can choose the "world-time parameter" such that [itex]\lambda=\text{const.}[/itex].

    Then the equations of motion read like for a massive particle
    [tex]\lambda \ddot{x}^0=q E \dot{x}^3, \quad \dot{\vec{x}}=q \dot{x}^0 E\vec{e}_3,[/tex]
    and the constraint
    [tex]\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau} \frac{\mathrm{d} x_{\mu}}{\mathrm{d} \tau} =0.[/tex]
    Solving this set of equations for the initial conditions that [itex]\dot{\vec{x}}(0)=u_0^1 \vec{e}_1[/itex]
    leads to the solution
    [tex]x^1=u_0^1 \tau, \quad x^2=0, \quad x^3=\frac{u_0^1}{\omega} \left [\cosh(\omega \tau)-1 \right ], \quad t=x^0=\frac{u_0^1}{\omega} \sinh(\omega \tau),[/tex]
    where
    [tex]\omega=\frac{q E}{\lambda}.[/tex]
    Now, I don't know, how to get rid of the arbitrary Lagrange parameter [itex]\lambda[/itex], which has the dimension of mass (or energy). The problem is that, even when I try to eliminate the arbitrary "world-time parameter" [itex]\tau[/itex] with the coordinate time, [itex]t[/itex], the dependence on the arbitrary Lagrange parameter doesn't vanish, and I don't see any other equation to fix it to some physically interpretable quantity.

    So my question is: Can I conclude that a classical massless charged particle in a electrstatic homogeneous field doesn't make sense or is there some way to fix [itex]\lambda[/itex] by physical quantities?
     
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  3. Feb 9, 2014 #2

    Jano L.

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    I do not understand your motivation for your Lagrangian (##x^3##?). One way to analyze your question is to consider ordinary eq. of motion for an electron in an external field and consider what happens as the mass is decreased to 0. I think the answer is obvious even without calculation: if the particle is very light, it accelerates violently under even weak external field and very quickly reaches speed close to ##c##. How quickly depends on the mass. If you set ##m=0## in the equation of motion, the equation breaks down. If you set ##m=0## in the solution, I think the particle will get the speed of light immediately while creating electromagnetic field that will be infinite even at infinity. Charged particle with ##m=0## seems like too absurd concept to me. But perhaps there can be more sensible solutions if you let both mass ##m## and charge ##q## let to zero. The results may depend on the order of taking limits though.
     
  4. Feb 9, 2014 #3

    bcrowell

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    I don't think a self-consistent solution exists. A charged particle has an electric field, and the electric field has a stress-energy tensor. That stress-energy has a nonvanishing time-time component in every frame of reference, so the particle has mass-energy.
     
  5. Feb 9, 2014 #4
    Electron, muon and tauon have very different rest masses. The true masses of up and down quarks are difficult to estimate due to quark confinement.

    Is it fundamentally possible to quote a mass that no particle of unit charge can be under?
     
  6. Feb 9, 2014 #5

    vanhees71

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    @Jano: The motivation is to study the motion of the particle in the most simple case, i.e., the motion in a homogeneous electric field in 3-direction. The potential energy in the Lagrangian is
    [tex]V=-q \vec{x}\cdot \vec{E}=-qE x^3.[/tex]
    Note that 3 is an inex here, not a power!

    @bctowell: mmaybe that's the solution to fix the Lagrange multiplier [itex]\lambda[/itex]! One can try to use the conserved total energy of the particle to eliminate [itex]\lambda[/itex] in favor of the total energy. I'll check this later.

    Another much more difficult question is whether the particle radiates (bremsstrahlung) and how to treat the radiative enrrgy loss, which is already difficult for massive particles.
     
  7. Feb 9, 2014 #6

    Vanadium 50

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    No such thing as a "tauon".

    A massless particle always moves at c. That means it instantaneously accelerates from moving at c in its initial direction to moving at c along the field lines. That, in turn means brehmsstrahlung - and rather a lot of it, since (once the particle is moving along the field lines) it goes up as 1/m8.

    [tex] P = \frac{E^2 q^4 p^6 c^3}{6\pi \epsilon_0 m^8} [/tex]

    where E is the applied electric field, q is the charge, and p is the momentum.

    You aren't going to get a consistent solution. Two many powers of m in the denominator, and they occur because one is assuming that the energy stored in the electric field of the massless particle is small compared to its mass - a contradiction.
     
  8. Feb 9, 2014 #7

    PeterDonis

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    There is if it means this:

    http://en.wikipedia.org/wiki/Tau_(particle [Broken])
     
    Last edited by a moderator: May 6, 2017
  9. Feb 9, 2014 #8

    vanhees71

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    Another solution of the eq. of motion, I've given in my original posting, of course is for the case the particle moves with c in 3-direction. Then it moves with c all the time as if there's no E-field at all. Then the Lagrange parameter trivially is eliminated when expressing the solution in terms of the coordinate time, [itex]t[/itex]. My solution doesn't suggest that the particle instantaneously goes into this solution when moving in another direction than the E-field, but it asymptotically reaches this state for [itex]t \rightarrow \infty[/itex]. The only question is, how to interpret the Lagrange parameter in terms of a physical quantity.

    Concerning the radiation problem, I don't think that you get the radiative energy loss by simply taking the limit [itex]m \rightarrow \infty[/itex] of the massive case.
     
  10. Feb 9, 2014 #9

    bcrowell

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    This is a problem that goes all the way back to the birth of relativity in 1905. In that era, people were working on all kinds of models of the electron, e.g., as a small sphere that suffered distortion when it moved through the ether. Once relativity came along, it became clear that there wasn't going to be a classical solution to the problem. A classical point charge's electric field has *infinite* mass-energy. This was one of the threads that led to QED.

    You can define a distance scale at which this kind of thing becomes a serious issue. The scale is called the classical electron radius. Roughly speaking, the field outside this radius has an energy that is enough to account for the electron's whole mass. You can define a classical muon radius, etc. Whatever that scale is, there is a corresponding time scale, which for an electron is 10^-23 s. When a force varies this rapidly, we have to consider the internal dynamics of the electron below the scale of the classical electron radius, or else we get all kinds of nasty problems. For instance, if you solve the equations of motion using the Lorentz-Dirac equation for the radiation reaction force, you get problems such as nonuniqueness of solutions and violations of causality. As an example of causality violation, the charge can accelerate before an external force is applied to it.

    So basically I think vanhees71 is trying to solve a problem that has been known since ca. 1905 not to have a consistent solution.
     
  11. Feb 9, 2014 #10

    Vanadium 50

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    I know that there is a tau lepton. There is no such thing as a "tauon".
     
    Last edited by a moderator: May 6, 2017
  12. Feb 9, 2014 #11

    PeterDonis

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    The Wiki page gives "tauon" as one of the names used to refer to the tau lepton. I know Wikipedia isn't always the best source, but I'm pretty sure I've seen "tauon" used in actual physics literature. ("Tauon" makes sense by analogy with "muon"; I've never seen anyone use the term "mu lepton".)
     
  13. Feb 9, 2014 #12

    Bill_K

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    If you look at the Wikipedia article that was cited, you'll see that tauon is listed as an alternate name. The article also goes on to mention the antitauon and the tauonium atom. Googling for "tauon" comes up with 462,000 hits. I'd have to say, therefore, that the name is acceptable usage.
     
  14. Feb 9, 2014 #13

    Vanadium 50

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    1903 and 1904, in fact. This was the Abraham-Lorentz model of the electron, and it was known even then to have inconsistencies: the most obvious of which is 4/3=1.

    Vanhees, the real problem with what I wrote is that there are eight powers of mass in the denominator and there are not eight powers of anything useful in the numerator to cancel them: q has to be non-zero, E is external and applied, c is a constant, and that leaves only six powers of p.

    The contradiction arises because a massless particle must have an instantaneous and complete reaction to any applied force, but changes to the electromagnetic field carried by this particle are limited to the speed of light.
     
  15. Feb 9, 2014 #14

    Vanadium 50

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    It is not acceptable usage. Googling gives about the same number of hits for "phlogiston", many of these references are for other things, and most of the particle physics links are pop sites put together by amateurs. Including the Wikipedia page. It's as wrong as "George Wershington" and should not be used.
     
  16. Feb 9, 2014 #15

    PeterDonis

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    Well, the term "phlogiston" is acceptable usage for describing that particular theory. The theory itself is not acceptable, but that doesn't mean the term isn't an acceptable way to refer to the theory.
     
  17. Feb 10, 2014 #16

    vanhees71

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    Ok, I guess you are right that there is no self-consistent solution, including radiation reaction, which is complicated even for massive particles. That wasn't the aim of my little calculation however. I just want to solve the much more simple problem of the motion of the particle without radiation-back-reaction.

    Do you know any papers/textbooks, where this problem is discussed really to the end; particularly, what's wrong with my "continuous" solution? In principle it gives a nice trajectory. The only issue is the physical meaning of the Lagrange parameter [itex]\lambda[/itex]. Is this somewhere discussed in the literature?
     
  18. Feb 10, 2014 #17

    vanhees71

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    Now I have an apparantly very simple solution. I hope, it's not somehow wrong. As I said, I don't worry about radiation-reaction issues here.

    The trick is to simply start in the non-covariant Hamiltonian formalism for a massive particle and then take the limit [itex]m \rightarrow 0[/itex]. This works (at least formally), because with the (canonical) momenta no problem with this limit occurs.

    So we start from the usual Lagrangian in the three-dimensional formalism
    [tex]L=-m\sqrt{1-\dot{\vec{x}}^2}-q \Phi, \quad \Phi=-E z.[/tex]
    Then we have
    [tex]\vec{p}=\frac{\partial L}{\partial \dot{\vec{x}}}=\frac{m \dot{\vec{x}}}{\sqrt{1-\dot{\vec{x}}^2}}.[/tex]
    The Hamiltonian is
    [tex]H=\vec{p} \cdot \dot{\vec{x}}-L,[/tex]
    which has to be expressed in terms of [itex]\vec{p}[/itex], leading to
    [tex]H=\sqrt{\vec{p}^2+m^2}+q \Phi.[/tex]
    In the massless limit we simply get
    [tex]H=|\vec{p}|+q \Phi.[/tex]
    The canonical equations of motion read
    [tex]\dot{\vec{p}}=-\frac{\partial H}{\partial \vec{x}}=q E \vec{e}_z, \quad \dot{x}= \frac{\partial H}{\partial \vec{p}} = \frac{\vec{p}}{|\vec{p}|}.[/tex]
    The solution for the momentum is
    [tex]\vec{p}(t)=\begin{pmatrix}
    p_{x0} \\ 0 \\ q E t
    \end{pmatrix}.[/tex]
    Integrating the equation for [itex]\vec{x}[/itex]
    finally gives
    [tex]\vec{x}=\frac{1}{q E} \begin{pmatrix}
    p_{10} \ln [(q E t + \sqrt{q^2 E^2 t^2+p_{x0}^2})/p_{x0}] \\ 0 \\
    \sqrt{q^2 E^2 t^2+p_{x0}^2}-p_{x0}
    \end{pmatrix}.[/tex]
    That seems to solve all my problems with the covariant formalism, expressing everything in physical quantities, and there is no singular behavior but a smooth trajectory.

    The speed is of course always [itex]\dot{\vec{x}}=|\vec{p}|/\vec{p}|=1[/itex] as it must be for a massless particle.
     
  19. Feb 10, 2014 #18

    bcrowell

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    Re #17,

    Is [itex]p_{10}[/itex] the same as [itex]p_{x0}[/itex]?

    If this solution is to be self-consistent, then there should be conditions under which the radiation reaction is negligible. What do you think those conditions are? Even in the case where px0=0, the particle's four-acceleration is nonzero.

    I have a physical objection to the solution as well, which is that you impute a momentum and energy to the particle that change with time, but these changes aren't connected in any physical way to any other observables. For example, an increase in the energy and momentum of an electromagnetic wave-packet would show up as an increase in the strength of the electric and magnetic fields.

    Another self-consistency condition is that this solution should make sense in other frames. In other frames, you'd have to include an [itex]A\cdot v[/itex] term in the Lagrangian.
     
  20. Feb 10, 2014 #19

    Jano L.

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    This last part does not look right. In case ##p_{x0} = 0## , if we differentiate by ##t##, we get velocity in the ##z## direction at ##t=0##
    $$
    \frac{dx}{dt} (t=0) = qEt,
    $$
    while there should be ##m## in the denominator (and thus increasing velocity for decreasing ##m##, while ##t >0## is held constant).
     
  21. Feb 11, 2014 #20

    vanhees71

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    You cannot have the initial condition [itex]\vec{p}=0[/itex], because this contradicts the on-shell constraint, i.e., that [itex]|\vec{v}|=1[/itex] for a massless particle! Of course [itex]p_{10}=p_{x0}[/itex] (sorry for the typo). So here's again the correct solution:
    [tex]
    \vec{x}=\frac{1}{q E} \begin{pmatrix}
    p_{x0} \ln \left [\frac{q E t + \sqrt{q^2 E^2 t^2+p_{x0}^2}}{p_{x0}} \right] \\ 0 \\
    \sqrt{q^2 E^2 t^2+p_{x0}^2}-p_{x0}
    \end{pmatrix}.
    [/tex]
    Concerning the radiation correction, I still have to evaluate the equivalent of the Larmor formula for the massless particle.
     
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