Do massless particles have a meaning in Newtonian mechanics? Do they have a physical interpretation?

Because we don't have an equation of motion for massless particles, we don't have a gravitational potential. We could have electric potential but then we lack the equation of motion.

I don't see how they can be a physical object, but I would love to hear from you about this.


5 Answers 5


The other answers are right, but I'd like to give some practical advice as well.

You might encounter, for example, a block and tackle system where you are asked to treat the blocks as massless. What this means in practice is that you should enforce the constraint that the sum of forces on the blocks is $0$.

More generally, you might think of a massless Newtonian particle as a formal (not strictly physical) object which enforces the constraint $\sum F = 0$.


As you already pointed out, massless particles have no meaning in Newton mechanics. This comes back to the way in which mass is defined

\begin{equation} m = \frac{F}{a} \ . \end{equation}

Implying that a particle with $m=0$ would have no momentum and no kinetic energy and, hence, would be nothing at all.

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    $\begingroup$ "would have no momentum and no kinetic energy" — or it would have both, and experience infinite acceleration from any finite force. $\endgroup$
    – Ruslan
    Jun 22, 2020 at 18:03
  • $\begingroup$ @Ruslan Any non-zero force, although the idea of an infinite force is just funny... $\endgroup$
    – Michael
    Jun 23, 2020 at 6:49
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    $\begingroup$ This confuses algebra with physics. Mass is not defined by acceleration. Based on this idea, an object that is at rest or moving at a constant speed has an undefined mass. An object's mass is independent of its rate of acceleration and the forces acting upon it in Newtonian physics. $\endgroup$
    – JimmyJames
    Jun 23, 2020 at 14:56
  • $\begingroup$ Objects that are 'at rest' with respect to our reference frame are not experiencing 'no' acceleration. They are accelerating towards all of the other particles nearby with a force that varies inversely as the distance. The locations where there are larger numbers of particles, like the sun, generate a larger net acceleration. $\endgroup$ Jun 23, 2020 at 15:34
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    $\begingroup$ @MichaelTreanor That's interesting. When using F = ma in Newtonian mechanics, what do you plug in for a in the situation that an object is 'at rest' with respect to the frame? $\endgroup$
    – JimmyJames
    Jun 23, 2020 at 15:58

It's sometimes meaningful to describe a particle as having "negligible mass", i.e. to consider the limit where the particle's mass tends to zero. Sometimes, colloquially, such particles are described as "massless", although they should not be confused with massless particles like photons in special (or general) relativity, which have a zero rest mass but a non-zero momentum, and thus behave qualitatively differently from the "negligible mass" limit of Newtonian physics (where the particle's momentum also tends to zero).

One common example of such a situation occurs with orbital dynamics under Newtonian gravity. For example, consider a man-made satellite orbiting the Earth. Since the satellite's mass is vanishingly small compared to the mass of the Earth (or of the Sun or the Moon or any other planets or moons in the solar system) it has essentially no measurable effect on the movement of the Earth or any other bodies in the solar system. But it still follows a well defined trajectory that mathematically tends to a well defined limit as the satellite's mass decreases towards zero.

So, for calculations, it's often useful to treat the satellite's mass as if it were exactly zero, both so that we don't actually need to assign any specific mass to the satellite when performing the calculation and so that its influence on the trajectories of other bodies does not need to be calculated.

Of course, a zero mass means that any non-zero force applied to the satellite would result in an infinite acceleration, which is clearly unphysical. But as long as the satellite is only affected by gravity, this turns out to cause no real problems, since the gravitational forces exerted on the satellite by other bodies are also proportional to its mass, and these effects cancel each other out so that even a particle of negligible mass still experiences a gravitational acceleration that tends to a finite limit as the particle's mass tends to zero.

Fundamentally, this convenient cancellation reflects the fact that gravity is actually a fictitious force in general relativity — and thus also behaves like one under Newtonian physics, which is a low-mass, low-speed limit of general relativity. (See also this earlier answer I wrote to a related question.)


Newtonian mechanics by definition obey Newton's laws of motion.

The second law:

In an inertial frame of reference, the vector sum of the forces $F$ on an object is equal to the mass $m$ of that object multiplied by the acceleration $a$ of the object:

$F = ma$. (It is assumed here that the mass m is constant

If $m=0$ , $F=0$ by construction

so there is no kinematics to describe mathematically.


Massless particles have no meaning in Newtonian ("classical") mechanics.

The most common massless particle, the photon, has zero rest mass but since it always travels at the speed of light, it has momentum. There are no macroscopic objects in real life with these properties. It is real in the sense that it has measurable properties (energy or wavelength, velocity, and polarization) and its effects on other objects have measurable consequences (photoelectric effect, photoionization, etc.).

  • $\begingroup$ Possibly the point we are trying to come to is this: We can predict all of the motions of the planets and the stars very well using Newtonian physics. (ironically by using the interactions of light waves with those objects - I always smile when I think of that) How then can we explain this 'zero mass' hypothesis when it is plainly obvious that light is curved around large objects. We have observed many objects that are hidden behind other objects through this gravitational lensing effect. So ... how is it that a particle with no mass experiences the acceleration of gravity near those stars? $\endgroup$ Jun 23, 2020 at 15:42
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    $\begingroup$ Here is why. Einstein's model of general relativity contains a term which couples energy to gravity: anything possessing energy is therefore going to respond to a gravitational field. Because a photon carries energy, it responds to gravitational fields by getting its trajectory bent, as if it had mass. $\endgroup$ Jun 23, 2020 at 17:28
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    $\begingroup$ The photon is a quantum-mechanical object with no classical analogue This is just wrong. EM wave or point massless relativistic particle would be classical analogues (they are just not Newtonian mechanics objects), and btw there are massless representations of Galilei group. $\endgroup$
    – A.V.S.
    Jun 24, 2020 at 6:10
  • $\begingroup$ What would be the physical behavior of such massless representation, I can they have acceleration for example? Can a force act on them? Or are they just free particles by definition? $\endgroup$
    – TheoPhy
    Jun 28, 2020 at 7:34
  • $\begingroup$ @A.V.S.I have edited my answer. but tell me- do massless representations of the Galilei group exist in nature? $\endgroup$ Jun 28, 2020 at 7:42

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