In the realm of pre relativistic physics.


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

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.


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.

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    $\begingroup$ Related/possible duplicates: physics.stackexchange.com/q/7905/50583 and its linked questions $\endgroup$
    – ACuriousMind
    Jun 27, 2022 at 18:11
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    $\begingroup$ A massless object has to always travel at the speed of light, so using non-relativistic mechanics would be completely invalid. $\endgroup$ Jun 27, 2022 at 18:12
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    $\begingroup$ 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. $\endgroup$
    – hft
    Jun 27, 2022 at 18:48
  • $\begingroup$ 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 $\endgroup$ Jun 27, 2022 at 19:26
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    $\begingroup$ 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. $\endgroup$ Jun 28, 2022 at 20:40

5 Answers 5


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.

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    $\begingroup$ 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. $\endgroup$ Jun 27, 2022 at 23:35
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    $\begingroup$ @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. $\endgroup$ Jun 28, 2022 at 2:37
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    $\begingroup$ 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. $\endgroup$ Jun 28, 2022 at 9:34
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    $\begingroup$ @hft: Your summary is correct. $\endgroup$ Jun 30, 2022 at 11:47
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    $\begingroup$ @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. $\endgroup$ Jun 19, 2023 at 11:31

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.

  • $\begingroup$ 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. $\endgroup$ Jun 27, 2022 at 20:27
  • $\begingroup$ $$Y=mx$$ $$[ x=0]$$ 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 $\endgroup$ Jun 27, 2022 at 20:34
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    $\begingroup$ 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... $\endgroup$
    – user338734
    Jun 27, 2022 at 21:22
  • $\begingroup$ Then why constrained systems containing multiple massless pulleys in newtonian physics have finite accelerations? $\endgroup$
    – user326901
    Jun 28, 2022 at 4:26
  • $\begingroup$ @Shiva yes, I sense something that could be problematic there. Let me refresh my memory reading some problems and let me go back. $\endgroup$
    – user338734
    Jun 28, 2022 at 6:41

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.


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.


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.


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