I'm assuming questions about the third law being invalid in relativity have already been asked on this site, but I'm specifically asking about how momentum is still conserved inspite of it.

Consider only two particles in the universe. In some reference frame, they're moving with velocities $v_1(t)$ and $v_2(t)$ at time $t$.

The relativistic momentum is:

$$\gamma (v_1(t))m_1v_1(t)+\gamma (v_2(t))m_2v_2(t)$$

The relativistic momentum after a small time $dt$ is:

$$\gamma (v_1(t+dt))m_1v_1(t+dt)+\gamma (v_2(t+dt))m_2v_2(t+dt)$$

If both these quantities are same, we can set them equal. After setting them equal, we get:

$$\gamma (v_1(t+dt))m_1v_1(t+dt)-\gamma (v_1(t))m_1v_1(t)=-(\gamma (v_2(t+dt))m_2v_2(t+dt)-\gamma (v_2(t))m_2v_2(t))$$

Dividing both sides by $dt$ and letting $dt\rightarrow 0$, we get:



So what's wrong with the above?

Edit- That other question is about General Relativity and the answers do not address the computations I've provided here. I want to know what's wrong with this specifically

Edit- After reading the other links, what I got is that the momentum conservation, as I've stated stated it in this post, is incorrect. This is because I did not account for the momentum of the field. So that means that Newton's third law is also correct if we extend the notion of force from particle-particle interactions to particle-field interactions. Is this correct? And in what sense do fields carry momentum? What is the mass and velocity of fields?

  • $\begingroup$ I apologise, I believe I misinterpreted your question in my answer that I have now deleted, and now I believe it is closed as it has already been answered. Again, my apologies. $\endgroup$ Sep 6, 2020 at 14:32
  • $\begingroup$ @ACuriousMind perhaps chose the wrong question as a duplicate. A better choice would have been Griffiths' argument that Newton's third law is invalid in special relativity. $\endgroup$ Sep 6, 2020 at 14:45
  • $\begingroup$ Thanks, I've added that one and its duplicate target, too. $\endgroup$
    – ACuriousMind
    Sep 6, 2020 at 14:50
  • $\begingroup$ @Ryder, what you have done in your question is to ignore that except for point masses that are touching one another, it takes time for a signal to propagate. You have implicitly assumed that signals transfer instantaneously. Momentum is conserved in special relativity, but that requires accounting for the momentum held by the electromagnetic field. $\endgroup$ Sep 6, 2020 at 14:50
  • $\begingroup$ There is no reason to reopen this question. The top answer to the question I posed as a duplicate to this question explicitly addresses the issue raised by the author of this question, and does so in the context of special relativity rather than general relativity. $\endgroup$ Sep 6, 2020 at 14:58

1 Answer 1


Good question, and the answer is that Newton's third law remains valid in Special Relativity as long as one applies it the right way, and that means it has to be applied locally at each event where forces are acting, not non-locally by comparing a force at some location $A$ with another force at some other location $B$. The third law applies to forces acting in a pair at any one location, say $A$.

When a force acts at the boundary between solid objects, this is straightforward. Each object pushes on the other.

When a force acts throughout a solid, one can analyse it the same way; for the details you need to invoke the concept of pressure and/or tension and stress. This is done in full via the stress-energy tensor.

The case where people say the third law breaks down is for example when charged objects attract or repel one another at a distance. It is true that in such cases the force on one object is not necessarily of equal size and opposite direction to the force on the other object. But one should ask: how is the force arising? It arises by an interaction between the charge on any given body and the electromagnetic field right there at the body. If the force causes the body to accelerate, for example, then, by conservation of momentum, one must find that momentum is moving out of the field and into the accelerating body. Force is, by definition, rate of change of momentum. One concludes that there is a pair of forces: one acting on the charged body, and the other, equal and opposite, acting on the electromagnetic field. These forces cause momentum to go into the charged body, and an equal and opposite momentum to go into the electromagnetic field. They are both present at the same location. They are equal and opposite.

It might seem odd to think of a force acting on a field, but the stress energy tensor makes no distinction between matter and field. Anything that can carry momentum can in principle have a force act on it.

P.S. The calculation offered in the original question is ok if the two particles are colliding at a single event, but if they are at different places and the momenta are changing with time (owing to the force from a field, for example) then you can't assume each particle's momentum will change by the same amount during some small time. In this case the sum of the two particles' momenta is not constant, because they are interacting with a third party, namely the field.

  • $\begingroup$ If fields carry momentum, do fields have mass? Also, in solids, the fundamental particles are still at a small distance. So is the stress-energy tensor just an approximate way of analysing it? $\endgroup$
    – Ryder Rude
    Sep 9, 2020 at 4:51
  • 1
    $\begingroup$ Yes, fields have mass in the following senses. They have at each point an energy-density. Integrated over volume and divided by $c^2$, that is mass. Also, when a charged particle accelerates, the field it produces has to be put in motion too, and this requires that momentum be delivered to the field. What we choose to call the inertial mass of the particle includes a contribution from the fields around it. The stress-energy tensor is exact in a classical model, and quantum physics has a quantum version of it. $\endgroup$ Sep 9, 2020 at 9:05
  • 2
    $\begingroup$ I wish I could upvote this more than once. The fact that the third law is often misapplied, and so is mistakenly thought to be "invalid" in certain situations, is a very important point that's missed by a lot of answers on this site, and you hit the nail on the head here. $\endgroup$ Sep 9, 2020 at 11:03

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