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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?

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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, 2014 at 17:47

4 Answers 4

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  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).
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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, 2017 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, 2014 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, 2014 at 15:31
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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. $\endgroup$ Apr 11, 2014 at 21:37
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    $\begingroup$ "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... $\endgroup$ Apr 11, 2014 at 23:57
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    $\begingroup$ @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. $\endgroup$ Apr 12, 2014 at 1:21
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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, 2014 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, 2014 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. $\endgroup$ Apr 11, 2014 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, 2014 at 18:22
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    $\begingroup$ You are missing the point of the question. See my comment to Flint72's answer. $\endgroup$ Apr 11, 2014 at 21:38

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