11

Consider two particles $A$ with mass $m$ and $B$ with mass $M$ ($M>> m$), and each particle has charge $Q$. Suppose they are at a distance $r$ ($r$ is of the order $10^{10}m$) apart at a time $t$. Due to electric force of attraction both particle comes closer to each other.

The force exerted on particle $B$ due to particle $A$ at a time $t$ will be due to the field exerted by $A$ at a time $t-\Delta t$, when particle $A$ was a distance of $r+\Delta r$ from the position of particle $B$ at time $t$ (this is because electric field travels at the speed of light and not instantaneously). Therefore the force experienced by particle $B$ will be $\dfrac{Q^2}{4\pi\epsilon_0(r+\Delta r)^2}$.

Similarly the force exerted on particle $A$ due to particle $B$ at a time $t$ will be due to the field exerted by $B$ at a time $t-\Delta t'$, when particle $B$ was a distance of $r+\Delta r'$ from the position of particle $A$ at time $t$. Hence the force experienced by particle $A$ will be $\dfrac{Q^2}{4\pi\epsilon_0(r+\Delta r')^2}$.

Force on the two particles is approximately equal (because $\Delta r$ ,$\Delta r'<<r$) therefore $\Delta r'<<\Delta r~$ because the mass of particle $B$ is much larger than the mass of particle $A$. Then this would lead to the conclusion that force exerted on particle $A$ due to particle $B$ is not equal to the force exerted on particle $B$ due to particle $A$ at time $t$. So does this situation violate Newtons Third Law?

Asher2211
  • 437
  • 2
    Newtons third law is not correct when considering electrodynamics. Which leads to the need for electromagnetic momentum. – jensen paull Apr 08 '22 at 19:30
  • $\Delta r$ is the change in separation between the charges; it doesn't matter which particle moved most, does it? – Philip Wood Apr 08 '22 at 19:33
  • 2
    Much of the relevant physics has been discussed in the current answers, I will add that the physical mechanism through which the missing momentum (that is required to maintain conservation of momentum) is carried away by the electromagnetic field is the emission of electromagnetic radiation, which is predicted to be emitted by Maxwell equations whenever a charged particle accelerates. –  Apr 09 '22 at 07:58
  • @PhilipWood It does if you take into account retarded potentials like the OP is doing. –  Apr 09 '22 at 08:08
  • You are a deep thinker! – tryst with freedom Apr 09 '22 at 10:14

3 Answers3

12

Yes! Electromagnetic forces violate Newton's Third Law. The reason is that the Third Law is essentially a statement about momentum conservation. Notice that ensuring that $$\sum_i \mathbf{F}_i = \mathbf{0}$$ means ensuring $$\frac{\mathrm{d}}{\mathrm{d} t} \mathbf{P}_{\text{tot}} = \mathbf{0}.$$

However, the electromagnetic field itself carries momentum as well. Hence, electromagnetic momentum also enters the equation and ends up spoiling the Third Law: there can be different variations of momenta in the different particles because the momentum in the electromagnetic field accounts for the difference and ensures momentum is overall conserved.

Níckolas Alves
  • 17,567
  • So if we include electromagnetic momentum then newtons third law is conserved, am I correct? – Asher2211 Apr 08 '22 at 19:35
  • @Asher2211 Not really. Newton's Third Law is a statement about forces which can be reinterpreted as a statement about momentum. The statement about momentum still holds once we consider the contributions of the electromagnetic field, but the forces on the particle's won't add up to zero and there is no way (at least that I know of) of defining a force acting on the field itself. As mentioned in another answer, once you treat the speed of light as being finite, Newtonian mechanics no longer applies, so it is not an issue to lose the third law – Níckolas Alves Apr 08 '22 at 19:48
  • 1
    In short: we no longer have the third law, it fails when you consider forces acting at a distance and with a finite speed of propagation (as per your argument). However, we still have momentum conservation, which is more general than the third law and essentially reduces to it in the relevant cases of Newtonian mechanics – Níckolas Alves Apr 08 '22 at 19:49
  • 1
    Though the original statement by Newton was in terms of impulse (change of momentum) , not directly about forces. – aschepler Apr 09 '22 at 13:34
4

If you want to take into account (non-infinite) speed of light you are definitely beyond the scope of Newton laws. So it definitely does not apply here.

vtm11
  • 184
  • 1
  • 1
    True, but Newton's second law can be carried over to special relativistic mechanics. On the other hand, Newton's third law cannot. So, what the OP is pointing out is a non-trivial tension between special relativity and Newton's third law. –  Apr 09 '22 at 08:01
1

Your, excellent, question as such is inconsistent as Newton mechanics can only be applied at speeds $v\ll c$. However, it is really about momentum conservation, which holds at any speed.

Indeed the Coulomb force is the retarded force. However, the two retarded forces obey Newton's first law, as the retarded time is the same for both particles. See the illuminating description in Griffith's Electrodynamics, Ch.8. What is not included in your question is the magnetic force. The momentum given by the Poynting vector does obey Newton's third law, but only if the field contributions are included. Besides non-radiative contributions, which in my opinion are paradoxical, Poynting's vector distribution also contain radiative contributions, which describe the momentum carried off by radiation. The latter of course violate the simple two particle picture of the original question.

my2cts
  • 24,097