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Suppose we have two point charges of rest masses $m_{1}, m_{2}$ and charges $q_{1}, q_{2}$. The electrostatic potential energy of the two objects is $U = k\cdot q_{1}q_{2}/r$ where $r$ is the distance separation. According to relativity, the mass of particle 1 is $m_{1}$, and the mass of particle 2 is $m_{2}$, however my understanding is that the mass of the two-particle system should be $m_{1}+m_{2}+U/c^{2}$.

Now I assume pushing with a force $F$ on particle 1 means it has an acceleration $a = F/m_{1}$. Pushing with a force $F$ on particle 2 means it has an acceleration $a = F/m_{2}$. However, it should be that somehow "pushing on the two-particle system" gives it an acceleration $a = F/(m_{1}+m_{2}+U/c^{2})$.

I am wondering how the last sentence can be made sense of. First of all, what would it even mean to push on the two-particle system (as opposed to pushing on each particle)? I can make sense of the "acceleration of the two-particle system" by taking the center of mass as the representative point of the two-particle system. However, we can't push on the center of mass since we can only push on the particles.

Second, if we could take care of the previous question, why does the two-particle system end up having more resistance to motion than each individual particle? How can we explain the change in inertia using forces and the EM field only?

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    $\begingroup$ It might be conceptually simpler to think about opposite charges forming a bound system… like a proton and an electron forming a hydrogen atom. The mass of a hydrogen atom is less than the sum of the masses of a proton and an electron. $\endgroup$
    – Ghoster
    Commented Aug 5, 2023 at 6:52
  • $\begingroup$ @Ghoster That's a good quantum example, although I'm trying to think about a classical electrostatics case. $\endgroup$
    – MaximusIdeal
    Commented Aug 5, 2023 at 9:48
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    $\begingroup$ In general, the concept of "mass" is not nearly as useful in relativity as in Newtonian mechanics, and for most purposes you should use "energy" instead. "Mass" is just the lower bound of the energy of the system. $\endgroup$
    – Eric Smith
    Commented Aug 5, 2023 at 16:46
  • $\begingroup$ The quantum mechanics is irrelevant to what you are trying to understand. $\endgroup$
    – Ghoster
    Commented Aug 5, 2023 at 17:21
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    $\begingroup$ @Dale What’s wrong with $U=k\,q_1q_2/r$? $\endgroup$
    – Ghoster
    Commented Aug 5, 2023 at 17:45

4 Answers 4

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First of all, what would it even mean to push on the two-particle system (as opposed to pushing on each particle)?

That the external force acts on at least one of the particles in a scenario where the two particles "move together", with similar averaged acceleration, so the system can be ascribed one common average acceleration.

If the particles come apart, average accelerations of both particles are no longer similar and then the system is not moving as a whole.

For example, two oppositely charged particles can remain close to each other due to attractive forces that keep the particles orbiting each other.

Or some other, non-electromagnetic force can keep two or many more particles close to each other. For example, particles forming the negative charge on curved conductive surface of a solid body (e.g. a metal sphere) may remain put there despite their repulsion, because there are also forces of constraint due to the solid body, keeping the charged particles from jumping out and spreading out into non-conducting air/vacuum.

Second, if we could take care of the previous question, why does the two-particle system end up having more resistance to motion than each individual particle? How can we explain the change in inertia using forces and the EM field only?

In classical mechanics, even if the force acted directly on single particle only, that force would "feel it has to accelerate the other particle as well". More correctly, the effect of the external force on the particle it acts on (its acceleration) is smaller when the other particle is attached or keeps close to the first particle. The acceleration is as if the external force acted on a hypothetical particle with greater mass, equal to sum of masses of the particles. In EM theory, this rule - the force effect being as if there was a particle with mass that is sum of masses of the particles - is no longer necessarily true.

When a charged particle moves with non-zero acceleration, it produces dynamic electric field, one part of which is proportional to acceleration of the particle (cf. the terms in electric field in https://www.feynmanlectures.caltech.edu/II_21.html - the last term is proportional to particle's acceleration).

This "acceleration electric field" may act on other nearby charged particles. In inductors, it is responsible for the induced electric field and thus induced EMF pointing against the acceleration, thus adding more inertia to mobile charge carriers forming the current, and thus greater effective mass. This is because all the accelerating charges have the same charge sign, so the induced field is proportional to acceleration, but points in the opposite direction.

If there is just a pair of positive and negative charge, the same kind of acceleration field is present due to both particles, and acts on the other particle in direction of its acceleration, thus decreasing effective mass of both particles. This is the simplest system manifesting "electromagnetic mass effect", an increase or decrease of effective mass due to EM interaction in the system.

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  • $\begingroup$ Does this mean that the mass of a system of charged particles is partly explained by Maxwell's equations? If so, I find that a pretty fascinating idea, because I've always thought of Maxwell's equations and mass as being completely independent and unrelated. $\endgroup$
    – Sophie Swett
    Commented Aug 5, 2023 at 16:57
  • $\begingroup$ Maxwell's equations are more of a detail here - the effect is due to part of interaction force being proportional to acceleration of the source particle, which I think is a general effect of special relativity. Also the mass defect effect has to be consistent with $\Delta E=\Delta mc^2$, which should hold in general, not just for EM field. Different force field not obeying the Maxwell equations could produce similar effect on effective mass. $\endgroup$
    – Ján Lalinský
    Commented Aug 5, 2023 at 17:19
  • $\begingroup$ But I think yes, part of observed mass can be ascribed to EM mass effect, described by Maxwell's equations and the Lorentz force formula. Feynman has a section in his lectures where he suggests proton and neutron have different mass, and part of the reason can be EM mass effect. There was a time when people thought maybe all mass is EM mass, but we don't believe that anymore. $\endgroup$
    – Ján Lalinský
    Commented Aug 5, 2023 at 17:23
  • $\begingroup$ Note however that usually (atoms, molecules) that EM mass effect is negative, i.e. bound systems have less mass than sum of masses of the components. To produce positive EM mass effect, one has to concentrate lots of charged particles of same sign into small region, and such systems require other stabilizing force. One can charge metal sphere somewhat, but the effect will be very hard to measure even at the point of critical electric field, where the charges start jumping out of the sphere. $\endgroup$
    – Ján Lalinský
    Commented Aug 5, 2023 at 17:31
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    $\begingroup$ An example where the positive EM mass effect may be present is atomic nucleus with high enough proton number. Lots of protons means lots of positive Coulomb energy. However, the system is stabilized by nuclear forces, which influence resulting mass as well. Here we have an example (nuclear forces) that the force field need not be Maxwellian in order to cause a change in effective mass. $\endgroup$
    – Ján Lalinský
    Commented Aug 5, 2023 at 17:32
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A classical analogy may be useful. Consider two masses $m_1$ and $m_2$ connected by a spring with mass $m_s$. Apply a force to $m_1$. Initially, only $m_1$ accelerates, as it takes time to accelerate the spring. As $m_1$ accelerates, its motion excites a wave on the spring. You may call the wave speed $c$. The force launching the wave reacts back on $m_1$, reducing its acceleration. Part of the spring accelerates while part remains in its original state of motion until the wave reaches $m_2$. Then, $m_2$ begins to accelerate.

So now, you have an accelerating, oscillating system. But, by momentum conservation, you can see that the center of mass of this mess follows $a = F/(m_{1}+m_{2}+m_3)$. Figuring out the individual trajectories and how much work the force does is left as an tedious exercise for the reader to avoid ツ.

The electromagnetic case is more complicated mathematically because the back reaction of the EM field on an accelerating point is an intractable problem, and the system will radiate.

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  • $\begingroup$ This is a very helpful analogy. Thank you! (You might find it nice to know that I actually asked a question about how to solve such a spring system here for the case where the spring is massless. And yes, I can see how if the spring were not massless, then it would contribute to the inertia.) $\endgroup$
    – MaximusIdeal
    Commented Aug 8, 2023 at 18:54
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The only answer I could think of is as follows

After the application of an impulse F (=F1+F2) on individual particles (separately) and get the same final speed v of the two particles (parallel to each other).

On the other hand the impulse F' is applied to the system of charged particles to get the same final speed v. Here

(F'-F)$\Delta$t = $\frac{U.v^2}{2c^2}$. (assuming v<<c)

Now if you want to bring the two separately moving particles together you have to supply the difference energy to the system.

I know energy and forces are a little bit mixed up in this explanation but I hope that might clear the confusion a little.

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Suppose we have two point charges of rest masses m1,m2 and charges q1,q2. The electrostatic potential energy of the two objects is U=k⋅q1q2/r where r is the distance separation. According to relativity, the mass of particle 1 is m1, and the mass of particle 2 is m2

The charges have extra weight and inertia. In the case of equal charges, it follows from symmetry that the extra mass must be divided equally among the particles.

(Where do you get that according to relativity, the mass of particle 1 is m1, and the mass of particle 2 is m2 )

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