Why doesn't the relativistic formula for momentum appear consistent with collisions?

The relativistic formula for momentum is $$p = \frac{mv}{\sqrt{1 - \dfrac{v^2}{c^2}}} \,.$$

In the following example, I apply the formula in the most basic way possible to the addition of velocities. I calculate the momentum before the collision, $p_{0}$, and the momentum after the collision, $p_{1}$. Blindly using these formulas, I come to the conclusion that $p_{0} \neq p_{1}$. I suspect most beginning students of special relativity would not be able to find fault with this argument; it is therefore worth answering and of broad interest to the Stack Exchange community.

Consider two objects of mass $m$ initially at rest that, after some combustion, move apart from one another with velocity $v$ in the positive $x$ direction and negative $x$ direction. Move to the frame of reference that moves at velocity $v$ in the positive $x$ direction with respect to the initial position of the blocks at rest.

Before the collision, apparent velocity of the two objects joined at rest is $-v$. Therefore, \ $$p_{0} = -2\frac{mv}{\sqrt{1 - \dfrac{v^2}{c^2}}} \,.$$

After the collision, the object moving in the positive $x$ direction appears stationary with respect to the frame of reference. Applying the formula for the addition of velocities, the velocity of the object moving in the negative $x$ direction is, $$v_{-} = \frac{-2v}{1 + \dfrac{v^2}{c^2}} \,.$$

Therefore, the total momentum of the system is $$p_{1} = \frac{mv_{-}}{\sqrt{1 - \dfrac{v_{-}^2}{c^2}}} = -2\dfrac{mv}{1-\dfrac{v^2}{c^2}} \,.$$

Thus, clearly, $p_{0} \neq p_{1}$.

• The resolution is that the rest mass $m$ of the reactants can change in relativistic collisions. For example, your example could have been describing the process $\mu^+ \mu^- \to e^+ e^-$ during which the parameter $m$ in your formulas changes from $m_\mu$ to $m_e$. – knzhou Jul 2 '16 at 19:33
• This is also a much better written question than your previous one! Thanks for improving it. – knzhou Jul 2 '16 at 19:34
• This story must obey energy conservation, so the final energy must have come from somewhere -- in this case, from the internal energy of the objects before the collision. The new feature of relativity is that this energy, which was in the objects all along, contributes to their mass before collision via $E = mc^2$. – knzhou Jul 2 '16 at 19:39
• Mass is lost in conventional chemical explosives. It's just such a small fraction that you don't need to worry about it under most conditions. But if you are proposing to impart relativistic speeds to the two masses then you will need to keep track of it. – dmckee Jul 2 '16 at 20:07
• Yup, and the compressed spring will have more mass than when it's uncompressed! – knzhou Jul 2 '16 at 20:15

Let's consider the frame in which initially both the masses are at rest to be the frame $O$. In frame $O,$ momentum conservation is trivially followed because of the symmetry of the problem. For the energy conservation, we require that $M = m \sqrt{1-v^2}$, where $m$ is the initial rest mass of each of the particles and $M$ is the final rest mass of each of the particles.

Now, let's observe the situation from the point of view of an observer $O'$ moving with a velocity $v$ in the positive $x$ direction. In this frame, the initial momentum is

$$p_i = -\dfrac{2mv}{\sqrt{1-v^2}}$$

and the final momentum is

$$p_f = \dfrac{M\bigg(\dfrac{-2v}{1+v^2}\bigg)}{\sqrt{1-\bigg(\dfrac{-2v}{1+v^2}\bigg)^2}} = -\dfrac{2Mv}{1-v^2} = -\dfrac{2mv}{\sqrt{1-v^2}}$$ if $M=m\sqrt{1-v^2}$, which is consistent with what we derived from energy conservation in $O$.

For $O'$, the initial energy is

$$E_i = \dfrac{2m}{\sqrt{1-v^2}}$$

and the final energy is

$$E_f = \dfrac{M}{\sqrt{1-\bigg(\dfrac{-2v}{1+v^2}\bigg)^2}} + M = \dfrac{2M}{1-v^2} = \dfrac{2m}{\sqrt{1-v^2}}$$ for $M=m\sqrt{1-v^2}$, which is again consistent with all of the previous considerations.

So, taking into account the change in rest mass of the particles because of the change in their structure during combustion (or whatever process that accelerates them) we can see that in both the frames, both the energy and momentum conservation laws can be consistently maintained.