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For relativistic massive particle, the action is $$S ~=~ -m_0 \int ds ~=~ -m_0 \int d\lambda ~\sqrt{ g_{\mu\nu} \dot{x}^{\mu}\dot{x}^{\nu}} ~=~ \int d\lambda \ L,$$ 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?

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    $\begingroup$ 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
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I) The equation of motion for a scalar massless relativistic point particle on a Lorentzian manifold $(M,g)$ is

$$ \tag{A} \dot{x}^2~:=~g_{\mu\nu}(x)~ \dot{x}^{\mu}\dot{x}^{\nu}~\approx ~0, $$

where dot denotes differentiation wrt. the world-line parameter $\tau$ (which is not proper time). [Here the $\approx$ symbol means equality modulo eom.] Thus a possible action is

$$ \tag{B} S[x,\lambda ]~=~\int\! d\tau ~L, \qquad L~=~\lambda ~\dot{x}^2, $$

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 here.

II) The corresponding Euler-Lagrange (EL) equations for the action (B) reads

$$ \tag{C} 0~\approx ~\frac{\delta S}{\delta\lambda}~=~\dot{x}^2, $$

$$ \tag{D} 0~\approx ~-\frac{1}{2}g^{\sigma\mu}\frac{\delta S}{\delta x^{\mu}} ~=~\frac{d(\lambda\dot{x}^{\sigma})}{d\tau} +\lambda\Gamma^{\sigma}_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu}. $$

III) The action (B) is invariant under world-line reparametrization $$ \tag{E} \tau^{\prime}~=~f(\tau), \qquad d\tau^{\prime} ~=~ d\tau\frac{df}{d\tau},\qquad \dot{x}^{\mu}~=~\dot{x}^{\prime\mu}\frac{df}{d\tau},\qquad \lambda^{\prime}~=~\lambda\frac{df}{d\tau}.\qquad$$ Therefore we can choose the gauge $\lambda=1$. Then eq. (D) reduces to the familiar geodesic equation.

References:

  1. J. Polchinski, String Theory Vol. 1, 1998; eq. (1.2.5).
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  • $\begingroup$ 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
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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.

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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 $$\mathcal{F_{\mu \nu}} = \partial_{\mu}A_{\nu}(x) - \partial_{\nu}A_{\mu}(x) $$

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ I agree, this is an example of a Lagrangian for a massless particle $\endgroup$ – Jim Apr 11 '14 at 17:48
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    $\begingroup$ 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
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    $\begingroup$ @JánLalinský The question explicitly asks about massless relativistic particles. All known massless relativistic particles are represented as fields in their respective Lagrangians. If we have missed the point, please tell us what the classical meaning of "particle" is. From what I know, there is no major distinction between a Lagrangian density involving kinetic, interaction, and/or massive terms of a field and the classical idea of a particle. The Lagrangian might have to describe multiple interacting fields and states, but all particles can be broken down to that on some level. $\endgroup$ – Jim Apr 11 '14 at 21:47
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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.

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ You are missing the point of the question. See my comment to Flint72's answer. $\endgroup$ – Ján Lalinský Apr 11 '14 at 21:38

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