# Equations of Motion for Massless Particle in Potential

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• tomdodd4598
In summary, The Lagrangian for a massless particle in a potential, using the ##(-,+,+,+)## metric signature, is$$L = \frac{\dot{x}_\mu \dot{x}^\mu}{2e} - V,$$where ##\dot{x}^\mu := \frac{dx^\mu}{d\lambda}## is the velocity, ##\lambda## is some worldline parameter, ##e## is the auxiliary einbein and ##V## is the potential. The EL equations give us the EOMs$$\dot{x}_\mu \dot{x}^\mu = 0,$$$$\ddot{x}^\mu + \Gamma^\mu_{ tomdodd4598 TL;DR Summary How does one consistently solve the equations of motion for a massless particle in a non-constant potential? The Lagrangian for a massless particle in a potential, using the ##(-,+,+,+)## metric signature, is$$L = \frac{\dot{x}_\mu \dot{x}^\mu}{2e} - V,$$where ##\dot{x}^\mu := \frac{dx^\mu}{d\lambda}## is the velocity, ##\lambda## is some worldline parameter, ##e## is the auxiliary einbein and ##V## is the potential. The EL equations give us the EOMs$$\dot{x}_\mu \dot{x}^\mu = 0,\ddot{x}^\mu + \Gamma^\mu_{\sigma\rho} \dot{x}^\sigma \dot{x}^\rho - \frac{\dot{e}\dot{x}^\mu}{e} + e\partial^\mu V = 0,$$where ##\Gamma^\mu_{\sigma\rho}## are the Christoffel symbols of the metric ##\eta_{\mu\nu}## for some choice of coordinates. After this, I'm not sure how to proceed, for the following reasons. In the ##V=\text{constant}## case, the system is underdetermined, and we are free to choose some ##e##, such as setting ##e=1##. We then get the consistent EOMs$$\dot{x}_\mu \dot{x}^\mu = 0,\ddot{x}^\mu + \Gamma^\mu_{\sigma\rho} \dot{x}^\sigma \dot{x}^\rho = 0.$$In the general case, however, we seem to end up with the two EOMs being inconsistent. For example, choose Cartesian coordinates and an ##z##-direction potential such as ##V = z##, and choose initial conditions satisfying the null velocity condition such as ##\dot{x}^\mu \left(0\right) = \left(1,0,0,1\right)##. The EOM for the coordinates becomes$$\ddot{x}^\mu = \left(0,0,0,-1\right),$$yielding ##\dot{x}^\mu \left(\lambda\right) = \left(1,0,0,1-\lambda\right)##, which fails to satisfy the null velocity condition for ##\lambda \neq 0##. This section below was part of the original post, but Orodruin spotted an issue with the line of reasoning. We can solve for ##e## by multiplying the second EOM by ##\dot{x}_\mu## and plugging in the first EOM:$$\dot{x}_\mu \left( \ddot{x}^\mu + \Gamma^\mu_{\sigma\rho} \dot{x}^\sigma \dot{x}^\rho \right) + e \dot{x}_\mu \partial^\mu V = 0,e = -\frac{\dot{x}_\mu \left( \ddot{x}^\mu + \Gamma^\mu_{\sigma\rho} \dot{x}^\sigma \dot{x}^\rho \right)}{\dot{x}^\mu \partial_\mu V}.$$Unfortunately, ##e## is then undefined for ##\dot{x}^\mu \partial_\mu V = 0##, and the system of ODEs is horrendous regardless (Mathematica flat out refuses to solve them), and so I find it hard to believe that this is the correct approach. Last edited: Note that$$
0 = \nabla_{\dot x} \dot x_\mu \dot x^\mu = 2 \dot x_\mu \nabla_{\dot x} \dot x^\mu = 2 \dot x_\mu (\ddot x^\mu + \Gamma^\mu_{\nu\sigma}\dot x^\nu \dot x^\sigma)
$$from ##\dot x_\mu \dot x^\mu = 0##. So you cannot solve for ##e##, what you have is ##e\dot x^\mu \partial_\mu V = 0##. Orodruin said: So you cannot solve for ##e##, what you have is ##e\dot x^\mu \partial_\mu V = 0##. Ah, good spot! I have updated the original post to more clearly reflect my concern about inconsistency in light of this, but it does make me wonder... is this equation telling me that there is a constraint on what the potential can be, or at least what the initial conditions can be? tomdodd4598 said: Ah, good spot! I have updated the original post to more clearly reflect my concern about inconsistency in light of this, but it does make me wonder... is this equation telling me that there is a constraint on what the potential can be, or at least what the initial conditions can be? You should post your additional work on a new post and not as a "Spoiler" in the original. topsquark I don't think that this makes sense, because the Lagrangian used doesn't provide an invariant action. A possible action in the "square form" of the free-particle part for a massive particle would be$$L=\frac{m}{2} \dot{x}^{\mu} \dot{x}_{\mu} - g \sqrt{\dot{x}_{\mu} \dot{x}^{\mu}} \Phi(x),$$where ##\Phi## is a scalar field. In this form the world-line parameter is automatically an affine parameter for the solutions of the equation of motion, because$$p_{\mu} \dot{x}^{\mu}-L=\frac{m}{2} \dot{x}^{\mu} \dot{x}_{\nu} =\text{const}.$$For details see Sec. 2.4 in https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf I don't know, whether one can make this consistent for massless particles. topsquark PeroK said: You should post your additional work on a new post and not as a "Spoiler" in the original. Unfortunately I can't edit the OP now, so to clarify, the material in the spoiler was part of the original post which Orodruin pointed out was erroneous. The example of an inconsistency (using the ##V=z## potential) is the new content which I think explains my original doubt a bit more clearly. vanhees71 said: I don't think that this makes sense, because the Lagrangian used doesn't provide an invariant action. Could you explain what you mean by "invariant" in this context? I'm not sure I follow. vanhees71 said: A possible action in the "square form" of the free-particle part for a massive particle would be... In the massive case, I can use the Lagrangian$$L = \frac{\dot{x}_\mu \dot{x}^\mu}{2e} - \frac{em^2}{2} - V,$$and the EOM for ##e## is$$m^2 e^2 = - \dot{x}_\mu \dot{x}^\mu,$$which when substituted into the EOM for ##x^\mu## yields the exact same EOM as you get from the square root Lagrangian$$L = -m \sqrt{-\dot{x}_\mu \dot{x}^\mu} - V.$$Of course, in the massless case, none of this applies, but the rather interesting constraint ##e\dot x^\mu \partial_\mu V = 0## seems to emerge instead, and my example choice of ##V=z## would be inconsistent with it. Last edited: Have you read the section on the action principle my relativity manuscript, quoted above? The Lagrangians you write down are not leading to a Lorentz invariant action. That's all I'm saying. The correct Lagrangian for the motion of a massive particle in a scalar external field is$$L=-m \sqrt{-\dot{x}_{\mu} \dot{x}^{\mu}} -g \sqrt{-\dot{x}_{\mu} \dot{x}^{\mu}} \Phi(x).$$This is written in a form, where the action is not only Lorentz invariant but also parametrization invariant, because it is a homogeneous function of rank 1 wrt. ##\dot{x}^{\mu}##. Here your parameter is completely arbitrary. An alternative equivalent form is$$L=\frac{m}{2} \dot{x}_{\mu} \dot{x}^{\mu} -g \sqrt{-\dot{x}_{\mu} \dot{x}^{\mu}} \Phi(x).$$It has the advantage that here automatically the parameter becomes an affine parameter for the solutions of the equations of motion. Otherwise it's equivalent to the former choice of the Lagrangian. I don't know, whether one can work out the equations of motion for a massless particle in an external field. In any case there's no such thing in Nature. The only case, where such an idea has an application I know of is to use a naive notion of "photons" in general relativity to describe the propagation of light in a given background spacetime. These "photons" follow lightlike geodesics and define the ray-optical limit of classical electrodynamics (eikonal approximation). topsquark vanhees71 said: Have you read the section on the action principle my relativity manuscript, quoted above? I was confused by what you wrote/shared, because as you say, those Lagrangians don't really work for a massless particle, but I'm trying to use one that does :P Regardless, I think understand what your point is now: my choice of ##V=z## is not Lorentz invariant. It seems to me, then, that the interesting fact is that we can't just use any Lorentz invariant potential due to the constraint coming from the EOM. I'm not sure what you mean. In the 2nd form the constraint when using an affine parameter, that ##\dot{x}_{\mu} \dot{x}^{\mu}=\text{const}## is automatically fulfilled for the solutions of the equations of motion, if you make the interaction part a homogeneous function of rank 1 in ##\dot{x}^{\mu}##, and that's why this is the standard choice for the motion of a particle in an external scalar field. The other obvious such term,$$\dot{x}^{\mu} \partial_{\mu} \Phi=\frac{\mathrm{d} \Phi}{\mathrm{d} \lambda}
doesn't contribute to the equations of motion since it's the total derivative of a function containing only ##x^{\mu}##. You can add an arbitrary term of this kind without changing the equations of motion.

topsquark
vanhees71 said:
...that's why this is the standard choice for the motion of a particle in an external scalar field.
I understand that this is a legitimate Lagrangian, but I don't understand why we've introduced a scalar field, or indeed a Lagrangian that can't describe massless particles if, for example, ##\Phi(x)=0##. The reason I chose ##L = \dot{x}_\mu \dot{x}^\mu / \left(2e\right) - em^2 / 2 - V## is that I know it works for ##\left(m \neq 0, V \neq 0\right)##, ##\left(m \neq 0, V = 0\right)## and ##\left(m = 0, V = 0\right)##, but I have been struggling with the ##\left(m = 0, V \neq 0\right)## case.

The piece ##-V## is not homogeneous of first rank in ##\dot{x}^{\mu}##!

topsquark
vanhees71 said:
The piece ##-V## is not homogeneous of first rank in ##\dot{x}^{\mu}##!
I think I realize where I've been unclear: when I wrote "##V##", I didn't necessarily mean something that was just a function of position, i.e. it could be something like your scalar field term.

However, I don't see why we're necessarily restricted to such a scalar field term, or indeed the form of Lagrangian you propose. For example, could I not use my original Lagrangian with ##V=
\dot{x}^\mu A_\mu## (which you use with a massive square root Lagrangian in those notes)?

Sure, then you get electrodynamics with ##A_{\mu}## the four-vector potential of the em. field. I thought you were after the eom. for a (massless) particle in a scalar field.

Thanks for the help :)

I realize now that my EOMs were only valid for when the potential was a function of only ##x_\mu##, so that was another mistake on my part... regardless, I think things make sense now :)

## What is a massless particle?

A massless particle is a hypothetical particle that has no mass and therefore travels at the speed of light. It is often used in theoretical physics to simplify equations and models.

## What is the equation of motion for a massless particle in a potential?

The equation of motion for a massless particle in a potential is given by Newton's second law, F=ma, where F is the force acting on the particle and a is its acceleration. This equation can be used to determine the trajectory and behavior of the particle in the presence of a potential.

## How does the potential affect the motion of a massless particle?

The potential affects the motion of a massless particle by exerting a force on the particle. This force can either accelerate or decelerate the particle, causing it to change direction or speed. The specific effect of the potential depends on its shape and magnitude.

## What is the difference between a massless particle and a particle with mass in terms of equations of motion?

The main difference between a massless particle and a particle with mass in terms of equations of motion is that a massless particle does not experience inertia, meaning it does not resist changes in its motion. This results in simpler equations of motion for massless particles compared to particles with mass.

## Can a massless particle exist in reality?

Currently, there is no evidence of a massless particle existing in reality. However, massless particles are used in theoretical physics to simplify equations and models, and some particles, such as photons, have very low mass and can be approximated as massless in certain situations.

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