12
$\begingroup$

For relativistic massive particle, the action is $$\begin{align}S ~=~& -m_0 \int ds \cr ~=~& -m_0 \int d\lambda ~\sqrt{ g_{\mu\nu} \dot{x}^{\mu}\dot{x}^{\nu}} \cr ~=~& \int d\lambda \ L,\end{align}$$ where $ds$ is the proper time of the particle; $\lambda$ is the parameter of the trajectory; and we used Minkowski signature $(+,-,-,-)$. So what is the action for a massless particle?

$\endgroup$
1
  • 2
    $\begingroup$ A massless particle is necessarily relativistic; it travels at the speed of light. As such, it is a null ray and you cannot really say that $ds=d\tau$. Nor can you talk about the proper time of a massless particle. All you can do is re-work the Lagrangian of your classical theory until it has a kinetic term for a field and no mass term, then call that field a massless particle and that's its Lagrangian $\endgroup$ – Jim Apr 11 '14 at 17:47
12
$\begingroup$
  1. OP's square root action is not differentiable along null/light-like directions, which makes it ill-suited for a massless particle. So we have to come up with something else. One equation of motion for a scalar massless relativistic point particle on a Lorentzian manifold $(M,g)$ is that its tangent should be null/light-like $$ \dot{x}^2~:=~g_{\mu\nu}(x)~ \dot{x}^{\mu}\dot{x}^{\nu}~\approx ~0, \tag{A}$$ where dot denotes differentiation wrt. the world-line (WL) parameter $\tau$ (which is not proper time). [Here the $\approx$ symbol means equality modulo EOM.] This suggests that a possible action is $$ S[x,\lambda ]~=~\int\! d\tau ~L, \qquad L~=~\lambda ~\dot{x}^2, \tag{B} $$ where $\lambda(\tau)$ is a Lagrange multiplier. This answer (B) may seem like just a cheap trick. Note however that it is possible by similar methods to give a general action principle that works for both massless and massive point particles in a unified manner, cf. e.g. Ref. 1 and eq. (3) in my Phys.SE answer here.

  2. More importantly, the corresponding Euler-Lagrange (EL) equations for the action (B) are the null/light-like condition $$ 0~\approx ~\frac{\delta S}{\delta\lambda}~=~\dot{x}^2, \tag{C}$$ and the geodesic equations $$ 0~\approx ~-\frac{1}{2}g^{\sigma\mu}\frac{\delta S}{\delta x^{\mu}} ~=~\frac{d(\lambda\dot{x}^{\sigma})}{d\tau} +\lambda\Gamma^{\sigma}_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu},\tag{D} $$ as they should be.

  3. The action (B) is invariant under WL reparametrization $$ \begin{align} \tau^{\prime}~=~&f(\tau), \qquad d\tau^{\prime} ~=~ d\tau\frac{df}{d\tau},\cr \dot{x}^{\mu}~=~&\dot{x}^{\prime\mu}\frac{df}{d\tau},\qquad \lambda^{\prime}~=~\lambda\frac{df}{d\tau}.\end{align}\tag{E} $$ Therefore we can choose the gauge $\lambda=1$. Then eq. (D) reduces to the more familiar affinely parametrized geodesic equations.

References:

  1. J. Polchinski, String Theory Vol. 1, 1998; eq. (1.2.5).
$\endgroup$
4
  • $\begingroup$ Comment for later: Interestingly, the canonical 4-momentum $p_{\mu}:=\frac{\partial L}{\partial \dot{x}^{\mu}}= 2\lambda g_{\mu\nu}\dot{x}^{\nu}$ is proportional to the undetermined Lagrange multiplier $\lambda$. Momentum transfer with an exterior environment may be a way to fix $\lambda$. $\endgroup$ – Qmechanic Aug 6 '17 at 12:32
  • $\begingroup$ Are not the equations of motion supposed to tell everything about the trajectory? From (A), you cannot read out how light bends, only that it travels on a lightlike curve. So should (A) really be considered the equations of motion? $\endgroup$ – HelloGoodbye May 28 at 12:38
  • $\begingroup$ Also, how did you conclude from (A) that (B) is a possible action? $\endgroup$ – HelloGoodbye May 28 at 12:41
  • $\begingroup$ Hi @Hello Goodbye: Thanks for the feedback. I updated the answer. $\endgroup$ – Qmechanic May 28 at 13:28
1
$\begingroup$

It is conceptually possible to have a massless charged particle, although there are none that we know of. It is not true that the Lorentz force has to equal mass times acceleration. The momentum of a massless particle is a quantity indepenedent of its speed as all massless particles travel at the speed of light. The momentum $p$ is instead equal to $E/c$, the energy divided by the speed of light.

$\endgroup$
-2
$\begingroup$

For a massless particle we can not have a centre-of-mass frame.

Unfortunately I cannot yet add comments. Are you studying Classical Field Theory (CFT) or Quantum Field Theory (QFT)? My guess is CFT since this looks like a line out of a few lectures into a CFT course when you start to make find equations of motion.

In that case, for the (massless) photon, $A_{\mu}(x)$ say, we use the Maxwell Lagrangian, which is Lorentz invariant, and is given (in Heaviside-Lorentz units) by $$ \mathcal{L_{Max}} =-\frac{1}{4} \int d^4x \mathcal{F_{\mu \nu}} \mathcal{F^{\mu \nu}} $$ where $$\mathcal{F_{\mu \nu}} = \partial_{\mu}A_{\nu}(x) - \partial_{\nu}A_{\mu}(x) $$

$\endgroup$
11
  • 1
    $\begingroup$ I know this is the lagrangian of field which is quantized to particles. But I want to describe the lagrangian of a "classical" relativistic particle. $\endgroup$ – 346699 Apr 11 '14 at 15:19
  • 1
    $\begingroup$ What I have writen here is indeed the Lagrangian for a Classical relativistic field, $A_{\mu}(x)$. I don't quite understand what you mean by 'a field which is quantised to particles'? $\endgroup$ – Flint72 Apr 11 '14 at 15:31
  • 3
    $\begingroup$ Jim, Flint72, you are both missing the point of user34669's question. The question is not about writing down Lagrangian density for a field, but for a particle in the classical meaning of the word. The fact that people decided to use words photon and mass-less particle in connection with vector potential is irrelevant here. $\endgroup$ – Ján Lalinský Apr 11 '14 at 21:37
  • 2
    $\begingroup$ "All known massless relativistic particles are represented as fields in their respective Lagrangians." Which makes sense only in the quantum setting. The OP is asking in the classical setting... $\endgroup$ – Alex Nelson Apr 11 '14 at 23:57
  • 2
    $\begingroup$ @Flint72, the problem with such reasoning is that it requires QFT to underpin it. The OP was asking about a null geodesic. You produced a section of a bundle. Clearly these are inequivalent, both mathematically and physically (since null geodesics are spin 0 whereas your field is spin 1). If one were to try to write a particle as a field, it'd have its action $I=\int\delta(x-z(\lambda))\sqrt{g_{\mu\nu}\dot{z}^{\mu}\dot{z}^{\nu}}\,\mathrm{d}^{4}x\,\mathrm{d}\lambda$ where dots denote $\lambda$ derivatives, and we "fix" $z$ when considering variations. $\endgroup$ – Alex Nelson Apr 12 '14 at 1:21
-8
$\begingroup$

Particle with zero mass has to have also zero electric charge, otherwise the Lorentz formula for EM force acting on it cannot be used to find its acceleration according to $$ m\mathbf a = q\mathbf{E}_{ext} + q\frac{\mathbf v}{c} \times \mathbf B_{ext}. $$ However, particle with zero mass and zero charge has trivial equation of motion $$ 0=0 $$ and has zero effect on the EM forces on other particles. It seems like a vacuous concept.

$\endgroup$
6
  • 1
    $\begingroup$ are you saying that a photon, with 0 mass and 0 charge, has zero effect on the EM forces of other particles? $\endgroup$ – Jim Apr 11 '14 at 17:49
  • 1
    $\begingroup$ The description of a photon as a massless particle that mediates the EM force easily comes about in classical field theory. If a photon does not fit into the picture of what you refer to as "classical theory", then no other massless particle will either $\endgroup$ – Jim Apr 11 '14 at 17:59
  • 1
    $\begingroup$ In classical field theory EM force acts on bodies and this is described by Lorentzian formula for force density or by the Maxwell stress tensor. There are no photons in this theory. $\endgroup$ – Ján Lalinský Apr 11 '14 at 18:11
  • 1
    $\begingroup$ The field strength tensor, $F_{\mu\nu}$, as a term in the Lagrangian density describes a kinetic term for the $A_{\mu}$ field. In the classical theory, there is also no mass term for this field and there is a coupling to the EM forces. This is what describes the photon and that it mediates the EM force $\endgroup$ – Jim Apr 11 '14 at 18:22
  • 1
    $\begingroup$ You are missing the point of the question. See my comment to Flint72's answer. $\endgroup$ – Ján Lalinský Apr 11 '14 at 21:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.