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In the realm of pre relativistic physics.

$$\vec{p}=m\vec{v}$$

$$F= \frac{dp}{dt}= m\vec{a}$$

If there exists an electric field in space, the force experienced by it would be $$F= q\vec{E}$$

Applying newtons laws:

$$q\vec{E} = m \vec{a}$$

This an equation stating a relation between the force, and the acceleration of an object.

Given $$m=0$$

It follows that $$q\vec{E} = 0$$

Since we assumed in the beginning that there exists and electric field.

$$q=0$$

Thus a massless object must have no charge. [Or newtons laws don't work for massless objects even in pre relativistic physics]

My question is: is my reasoning correct? And is there a generalisation of this to the relativistic equations.

Qmechanic
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jensen paull
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    Related/possible duplicates: https://physics.stackexchange.com/q/7905/50583 and its linked questions – ACuriousMind Jun 27 '22 at 18:11
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    A massless object has to always travel at the speed of light, so using non-relativistic mechanics would be completely invalid. – Sahand Tabatabaei Jun 27 '22 at 18:12
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    It's a little unfair, since you start off by assuming you can use the kinematic equations that govern the motion of a classical particle with mass. You then do some manipulations and find what seems to be a contradiction when you set the mass to zero. Because your starting place seems inappropriate it calls your conclusions into question. – hft Jun 27 '22 at 18:48
  • I wouldn't call it a contradiction. yes a is undefined, but q=0 is an inevitability, not a contradiction. But yes, m=0 does seemingly violate newtons law when finding acceleration atleast. But yes, I already knew newtonian mechanics wouldn't be correct – jensen paull Jun 27 '22 at 19:26
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    A "proof" requires axioms. Here your axioms apparently include "Newton's laws hold when interpreted according to the rules of algebra for finite real numbers". If so then it is valid, because $m=0$ does then imply $ma=0$. But physics is often done under what are in effect other axioms of algebra, e.g. distribution theory, so your choice is not necessarily the only reasonable one. – Nate Eldredge Jun 28 '22 at 20:40

5 Answers5

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As was pointed out in the comments above, one has to use relativistic mechanics to talk meaningfully about massless particles; you can't just write $F = ma$ and expect it to work. And, indeed, it turns out that we can come up with a reasonable question of motion for a charged massless particle if we use the machinery of relativistic dynamics.

The equation of motion of a massive charged relativistic particle is $$ \frac{d p^\mu}{d\tau} = \frac{q}{m} F^{\mu}{}_\nu p^\nu, \tag{1} $$ where $m$ is the particle's rest mass, $q$ is its charge, $p^\mu$ is its four-momentum, $F^{\mu \nu}$ is the field strength tensor, and $\tau$ is the proper time along the (massive) particle's world-line. In particular, $\tau$ serves mainly as a parameter that traces out the particle's world-line through spacetime.

If we multiply both sides by $m$ and introduce a new parameter $\lambda = \tau/m$, it turns out that the above equation is equivalent to $$ \frac{d p^\mu}{d \lambda} = q F^\mu {}_\nu p^\nu. \tag{2} $$ What's more, under this parameterization we have $$ \frac{d x^\mu}{d \lambda} = m \frac{d x^\mu}{d \tau} = p^\mu. \tag{3} $$

The new parametrization (2) has a perfectly well-behaved limit as $m \to 0$; in fact, this is a conventional parametrization for the worldlines of massless particles. Defining our $x$-direction to be the direction of the field, we have $F^{01} = - F^{10} = E_x$, and so the equations of motion are \begin{align*} \dot{p}^t &= q E_x p^x & \dot{p}^x &= q E_x p^t & \dot{p}^y = \dot{p}^z = 0 \tag{4} \end{align*} where dots denote differentiation with respect to $\lambda$. This can then be solved (in principle) for the four-momentum $p^\mu$ as a function of $\lambda$. If desired, one can then find the trajectories $t(\lambda)$, $x(\lambda)$, etc. in parametric form, and (in principle) invert the first of these to obtain $x(t)$, $y(t)$ and $z(t)$.

This means that the energy ($p^t$) and $x$-momentum ($p^x$) of a massless charged particle will increase as it travels. But you can also show from the equations of motion (4) (try it!) that the quantity $p_\mu p^\mu = -(p^t)^2 + (p^x)^2 + (p^y)^2 + (p^z)^2$ is constant with respect to $\lambda$. So these equations ensure that a massless charged particle (with $p_\mu p^\mu = - m^2 = 0$) stays massless as it travels; and that means that it always travels at the speed of light, even as its energy and momentum change.

Michael Seifert
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  • Massless charged particles have very interesting properties then! I haven't learned tensors yet, however atleast with what I know, I am confused how conservation of energy\momentum applies here. How can the momentum of the particle increase if the speed doesn't? I only know of field momentums contribution to the total momentum, outside of the formula $p= \gamma m_{0}\vec{v}$, which I presume wouldn't apply here. What form does this momentum take? Obviously you state p in the eq of motion but intuatively what "stores" this momentum. [If that is a better phrasing] – jensen paull Jun 27 '22 at 21:31
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    I think a massless charged sphere is impossible because the field energy contributes some non-zero mass and the material of the sphere cannot have negative mass to compensate this. Also, in the limit of a pointlike entity the field energy blows up unless the charge goes to zero. – Andrew Steane Jun 27 '22 at 23:35
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    @AndrewSteane: Your last objection would also rule out massive charged particles, not just massless ones; it's a well-known problem with point charges in classical electrodynamics. Also, note that the fields have momentum, not just mass-energy, and so it's entirely possible that when we take all of that into account we still have $m^2 = E^2 - p^2 = 0$ for the system. I'll admit I haven't done those calculations, though. – Michael Seifert Jun 28 '22 at 02:37
  • @jensenpaull: It's best to think about this in analogy with a photon. All photons travel at the speed of light, and they all have $E = pc$. But some have more energy and some have less energy. Asking what "form" the momentum takes for a photon isn't really the right question to ask; it just has a certain amount of momentum. The only difference between a photon and your hypothetical charged massless particle is that it's possible for an electric field to do work on this particle, changing its energy (and, correspondingly, its momentum.) – Michael Seifert Jun 28 '22 at 02:43
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    I think there is a certain similarity to gravitational forces because while a photon doesn't have "rest mass", all existing photons do have mass, and experience "forces" in "gravitational fields" potentially increasing their energy and momentum -- much like a charge would in an electric field -- even though they don't get any faster. – Peter - Reinstate Monica Jun 28 '22 at 09:34
  • Photons have energy and momentum,stored in the fields[atleast EM context], which is the reason they experience gravity. But why say they have mass? Yes it is true they experience gravitational effects but they don't actually have "mass" do they, its due to the momentum and energy not mass? Rest mass is zero like you said. And E=mc^2 doesn't apply for photons. [Although you could say an object with mass E/c^2 acts as though it has the same pull as light, that's not the same thing as saying light actually has mass though)?]. – jensen paull Jun 28 '22 at 14:03
  • @MichaelSeifert Just for the sake of completeness, it seems like the implied answer to OP's actual question is: No. Can you confirm that your answer to "[i]s this proof that massless objects cannot be charged?" is "No, this is not proof that massless objects cannot be charged." – hft Jun 30 '22 at 02:08
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    @hft: Your summary is correct. – Michael Seifert Jun 30 '22 at 11:47
  • Is the resulting force Lorentz invariant? It seems like the parametrization $\lambda$ of the massless particle worldline would end up depending on the choice of inertial frame you initially did this in. – Maximal Ideal Jun 18 '23 at 14:17
  • @MaximalIdeal: Can you elaborate on why you think that would be? $\tau$ along a particular particle's worldline is invariant between reference frames, and so is its mass $m$, so $\lambda = \tau/m$ would be independent of reference frame as well. – Michael Seifert Jun 18 '23 at 18:21
  • @MichaelSeifert I think I agree now. The limit $m\rightarrow 0$ seemed deceiving, because as $m$ changes, the trajectory of the particle also changes, so we might need to consider some kind of limit of the particle's trajectory itself as well. I would love to see a more rigorous version of this process. [...] – Maximal Ideal Jun 19 '23 at 03:23
  • Anyways, I agree the resulting parametrization is Lorentz invariant, because $\tau$, $m$ are invariant (as you said), so the paths $x_{m}(\lambda)$ are the same regardless of what inertia frame you use (their coordinate representations $x_{m}^{\mu}(\lambda)$ respect Lorentz transformations across changes in inertial frames), and so the resulting limiting path $x_{m=0}(\lambda)$ is independent of what inertial frame was used (its coordinate representation $x_{m=0}^{\mu}(\lambda)$ respects Lorentz transformations across changes in inertial frames). – Maximal Ideal Jun 19 '23 at 03:24
  • If I wanted to avoid the limiting $m\rightarrow 0$ process, would it be fine to use the equation (assume inertial reference frame) $$ \frac{d^{2}x^{\mu}}{d\lambda^{2}} = qF^{\mu}{}{\nu}\frac{dx^{\mu}}{d\lambda} $$ directly in place of $(2)$ and define $p^{\mu} = \frac{dx^{\mu}}{d\lambda}$? We'd have to use an appropriate initial condition $p^{\mu}(0) = p{0}^{\mu}$ with $g_{\mu\nu}p_{0}^{\mu}p_{0}^{\nu} = 0$ to make sure momentum is light-like, but I think the result gives exactly the same as in your post without having to worrying about taking any limits. – Maximal Ideal Jun 19 '23 at 03:36
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    @MaximalIdeal: Yes, it's fine to just start with that definition — in fact, it's a conventional parametrization for the worldlines of massless particles, as mentioned in the other question I linked to. – Michael Seifert Jun 19 '23 at 11:31
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I disagree with your conclusion that $qE=0$ if $m=0$. My interpretation is that in such a case the acceleration is infinite, so the product is well defined. It is only natural to think that in Newtonian mechanics a massless object will experience infinite acceleration if it experiences any finite force. For a given force, the smaller the mass you use, the larger the acceleration you get.

Note about massless pulleys and similar situations: Given what I said above, how can a pulley of zero mass have a finite angular acceleration? Should it not be infinite? Well, the acceleration needs to be infinite only if the force is finite. For zero force, the acceleration is arbitrary. So there is no conflict: massless pulleys can have any acceleration (determined by the assumptions/constraints of the problem, like rope does not slip), as far as they feel or make no finite force. But all this simplifications or lack of rigour works because they mathematically work in the limit. So real physical problems in which the pulley has very low mass behave as if the mass of the pulley were zero and the acceleration given by the problem constraints. When you take the limit of mass goes to zero, you can chose that the force is proportional to the mass, so the less mass the less force. In the limit of zero mass, the force will be zero but the acceleration will remain finite.

  • Mathematically the conclusion that qE=0 is correct. Why do you dissagree? $$qE=ma$$ $$\frac{qE}{m}= a$$ $$ \frac{qE}{0} = a$$ therefore acceleration is undefined not infinite we aren't dealing with limits, the product of m and a is defined, it is 0. Y=mx when x=0, y=0. – jensen paull Jun 27 '22 at 20:27
  • $$Y=mx$$ $$[ x=0]$$ then $$y = m*0=0$$ you can't then say "$\frac{y}{x}= m$", thus m is infinite. m can be anything. There are infinitely many m's that satisfy those conditions, hence its undefined. But what you can say for certain, is that when x=0 -> y=0. Which the the same for m=0 -> qE=0 – jensen paull Jun 27 '22 at 20:34
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    The most you can say is that the acceleration is undefined, not that a charged massless particle cannot exist. In such a case you would need to complete your theory. One way is to propose that there are no charged massless particles. Another is to propose that such particles have infinite acceleration. Only experiment can decide in such a case. The experiment says that Newtonian mechanics is wrong in many other ways, so... –  Jun 27 '22 at 21:22
  • Then why constrained systems containing multiple massless pulleys in newtonian physics have finite accelerations? –  Jun 28 '22 at 04:26
  • @Shiva yes, I sense something that could be problematic there. Let me refresh my memory reading some problems and let me go back. –  Jun 28 '22 at 06:41
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The correct answer is your parenthetical possibility that "Newton's laws don't work for massless objects in nonrelativistic physics." This has nothing to do with electricity or electric charge: you can't have a meaningful massless particle in Newtonian physics at all, because in that case $F = ma$ would imply that any net force would induce an infinite acceleration, which is difficult to make mathematically rigorous.

We sometimes talk about "massless pulleys" in the context of Newtonian physics, but these aren't actually dynamical objects - they're just an idealization representing the fact that we semi-magically require that the tensions in both ropes coming out of the pulley must be equal. (In fact, when we talk about a "massless pulley", we usually actually mean that the pulley has a negligible moment of inertia, not mass.)

Whether a charged massless particle could exist in relativistic physics is a more interesting and complicated question, but that is out of the scope of your question.

tparker
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I think a massless electron traveling at $c$ under classical Maxwell's Eq. would be problematic, but:

The Standard Model says electrons were massless and charged before Spontaneous Symmetry Breaking, but there wasn't enough time to cause problems.

Also: gluons are massless and carry color/anti-color charge. Of course, they're confined so they never get out to cause problems.

JEB
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As described in Michael Seifert's answer, the question requires relativity, not Newtonian mechanics. As explained in the comments by Andrew Steane and Michael Seifert, the question also requires quantum mechanics. So we're in the realm of quantum field theory.

Massless charged particles are problematic in QFT, because then we would get pair production at arbitrarily low energies. This would manifest itself in ordinary blackbody radiation, for example, and that's not what we observe. There are various other issues, but this is the most obvious. It's one of the reasons that Penrose's CCC theory, at least in its original version, was totally non-viable.

Bleh Bleh
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