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What is the meaning of relativity rapidity? I know that the rapidity of a particle in general is defined as $\text{tanh}^{-1} \left( \frac{v}{c} \right) = \phi$.

I encountered a problem that talks about the relative rapidity. In particular two particles have a relative rapidity and collide to form a third particle. I don't understand how that's sufficient to compute the mass of the third particle since we don't know the direction of the velocity or the original velocities beforehand.

Is it always possible to do this?

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  • $\begingroup$ If you knew the relative velocity between the two particles, would that be sufficient? $\endgroup$
    – Sten
    Commented Sep 2, 2023 at 23:39
  • $\begingroup$ That's what confuses me. Is it sufficient to know the relative velocity or do we have to know the actual velocities? $\endgroup$ Commented Sep 2, 2023 at 23:41
  • $\begingroup$ More specifically, if one of the particles is traveling at $v_1$ and the other at $v_2$ versus one at $0$ and the other at $v$. Doesn't that affect the mass of the particle created post collision? Why is it sufficient to know just the relativity velocity $\endgroup$ Commented Sep 2, 2023 at 23:49
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    $\begingroup$ You could shift both particles' velocities just by changing the velocity of yourself, the observer. Does that change the mass of the particle produced in the collision? $\endgroup$
    – Sten
    Commented Sep 2, 2023 at 23:55

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We might need some specifics of what's given in the problem.

We start with conservation of total 4-momentum for a single particle after the collision: $$ \tilde m_1 + \tilde m_2 = \tilde M,$$ where $\tilde m_1$ is the 4-momentum of a particle with invariant mass $m_1$, etc.

The above 4-momentum equation can be visualized in energy-momentum space as a "triangle with three timelike legs" (the sum of two 4-momentum vectors tip-to-tail equals a single 4-momentum vector). So (since the sum of two vectors lies on the plane spanned by those vectors), this problem reduces to a problem in (1+1)-dimensions.

  • The relative-rapidity is ${\rm arccosh}(\hat m_1 \cdot \hat m_2)$ [involving the Minkowski-dot product of their 4-velocities] assuming the $(+,-,-,-)$ signature convention.
    [Think "angle between two unit-vectors": $\gamma_{rel}=\cosh\phi_{rel}=\hat m_1 \cdot \hat m_2$.]
  • (Note: $v_{rel}=\tanh(\phi_{rel})= \tanh(\phi_2-\phi_1)=\frac{\tanh\phi_2 - \tanh\phi_1}{1-\tanh\phi_2\tanh\phi_1}$.
    That is, the relative-rapidity $\phi_{rel}={\rm arctanh}(v_{rel})$, the inverse-hyperbolic-tangent of the relative-velocity [the velocity of $m_2$ in the frame of $m_1$].

The mass $M$ of the output particle (the magnitude of $\tilde M$) can be gotten by the Minkowski version of the law of cosines. (I'll leave the details to you.) Do you know the input masses?

(Possibly useful: my answers to Momentum diagram for two colliding Particles and How to determine particle energies in center of momentum frame? )

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