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.

enter image description here

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

enter image description here

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

enter image description here

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

enter image description here

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?


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).

| cite | improve this answer | |
  • $\begingroup$ 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. $\endgroup$ – Timaeus Jan 17 '15 at 22:46
  • $\begingroup$ @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). $\endgroup$ – kristjan Jan 17 '15 at 23:10
  • $\begingroup$ Is $E_{interaction}$ the kinetic energy of the system in Ref 2? Is $E_{interaction} = c^2 (m_1( \gamma(v) -1) + m_2 (\gamma(v_2)-1))$? $\endgroup$ – philn Jan 20 '15 at 1:09
  • $\begingroup$ @user16098: Yes. $\endgroup$ – kristjan Jan 20 '15 at 6:07
  • $\begingroup$ @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. $\endgroup$ – philn Jan 20 '15 at 18:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.