# Conservation of relativistic momentum

I'm trying to prove that momentum, $\rho = m \, v \, \gamma(v)$, is conserved in all frames of reference. I'm having problems with the following situation that I made; momentum is not conserved to large speeds!

A mass $m_1$ (red) and $m_2$ (green) are traveling together.

At some point in time, an internal explosion happens. So that $m_1$ stops moving and $m_2$ gets some velocity $v_2'$.

In the reference frame of $v$ before the explosion, nothing is happening:

And after the explosion, $m_1$ moves left with a speed $v$.

Using Relativistic Velocity Transformation, I find $v_2' = \displaystyle\frac{v+v2}{1+v \cdot v_2 /c^2}$

The momentum of the system shouldn't change in both frames of reference. As a result of conservation of momentum, equation (1) and (2) (shown below) should always be true.

Just to make sure everyone is on the same page, I'm using $\gamma(v) = \displaystyle\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

$\textbf{Ref 1}$

$(m_1 + m_2) \, v \, \gamma(v) = m_2 \, v_2' \, \gamma(v_2')$ (1)

$\textbf{Ref 2}$

$m_1 \, v \, \gamma(v) = m_2 \, v_2 \, \gamma(v_2)$ (2)

Subtracting (2) from (1) and simplifying gives:

$v \gamma(v) + v_2 \gamma(v_2) = v_2' \gamma(v_2')$ (3)

Unfortunately, (3) is false; I tested it with random numbers. For values $v,v_2 << c$, (3) is correct. $\textbf{Question}$

What assumptions/approach is incorrect? Did I simply make a mistake somewhere? Is the explosion itself the problem?

## 1 Answer

You have to take into account the energy initially stored as the interaction energy of the particles. As mass and energy are just related by a factor of $c^2$, the initial rest mass of the "composite" particle is $m_1 + m_2 + E_{interaction}/c^2$. Indeed, note that in your calculations, the relativistic mass (or total energy) is not conserved (this can easily be seen in the reference frame two setup).

• The expression for momentum, $\vec{p}= \frac{m\vec{v}}{\sqrt{1-v^2/c^2}}$, uses the rest mass, $m$, for the $m$ term. Commented Jan 17, 2015 at 22:46
• @Timaeus: Indeed. However I don't see what is this to do with my answer. If in the rest frame of the composite particle, there is some interaction energy $E$, so the total 4-momentum of this system is $(m_{eff}, 0, 0, 0)$, where $m_{eff} = m_1 + m_2 + E{interaction}/c^2$. The 4-vector of this system transforms exactly the same way as would a 4-vector of a particle of mass $m_{eff}$. Thus $p = \gamma m_{eff} v$ for this system, whether we look at it any system or as a composite particle (as could be implied from my answer). Commented Jan 17, 2015 at 23:10
• Is $E_{interaction}$ the kinetic energy of the system in Ref 2? Is $E_{interaction} = c^2 (m_1( \gamma(v) -1) + m_2 (\gamma(v_2)-1))$? Commented Jan 20, 2015 at 1:09
• @user16098: Yes. Commented Jan 20, 2015 at 6:07
• @kristjan This is my attempt at accounting for $E_{interaction}$ (I'm calling it $E$) $y(v) v (m_1+m_2 + \frac{E}{c^2}) = y(v_2') v_2' m_2$ (1) \* Accounting for $\frac{E}{c^2}$: $y(v) v (m_1+m_2 + (m_1(y(v)-1)+m_2 (y(v_2)-1)) = y(v_2' v_2' m_2$ (2) (2) doesn't seem correct when I plug in random numbers. Commented Jan 20, 2015 at 18:06