Inspired by this question a normal extension would be to ask:

What is the minimum potential energy required (to behave as a turning point) for an elastic collision between $2$ point particles $A$ and $B$ with velocity $v_A^{\mu}$ and $v_B^{\mu}$ in a relativistic classical mechanics? (Assume number of particles are conserved)

I suspect the action should be of the form:

$$ S = -\underbrace{m_A \int d^4 x \delta^3(q^i_A(\tau_A) - x^i)}_{\text{Action of particle A}} -\underbrace{m_B \int d^4 x \delta^3(q_B^i(\tau_B) - x^i)}_{\text{Action of particle B}} - \underbrace{V_0 \int \delta^3 (x^i- q^i_c)d^4 x}_{\text{Potential behaving as turning point}} $$

where $m_k$ is the mass of the $k$'th particle, $q_k^i$ is $i$'th component of the $k$'th particle, $V_0$ is a constant, $\tau$ is the proper time and $q_c$ is the intersections of the geodesics $A$ and $B$.

P.S: Do check the question on why a naive application of Total energy $\geq$ Potential Energy will not suffice.

  • $\begingroup$ I believe the answer is $\mu \gamma_{rel} c^2$ $\endgroup$ Jun 14, 2023 at 4:58

1 Answer 1


Disclaimer: this answer is heuristic. We know,

$$p_1^\mu + p_2^\mu = C^\mu$$

Where $C^\mu$ is a constant $4$ vector. Taking the inner product and differential:

$$ d (p_1^\mu p_{2\mu}) = 0 $$


$$ m_1 m_2 \gamma_{rel} c^2 = \lambda$$

where $\gamma_{rel}$ is the gamma factor in the where the velocity is the relative velocity and $\lambda$ is a constant. We associate the relative energy $\propto \gamma_{rel} c^2$. We know irrespective of which frame we are in the energy should be positive if we subtract the relative energy. As an anology think of:

$$ \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_1v_2^2 - \frac{1}{2}\mu v_{rel}^2 \geq 0$$

where $\mu$ is the reduced mass. Now similarly we have to find $a$:

$$ \gamma_1 m_1 c^2 + \gamma_2 m_2 c^2 - a \gamma_{rel}c^2 \geq 0 $$

We also know he arithmetic mean $\geq $ than the harmonic mean. Hence,

$$ \gamma_1m_1c^2 + \gamma_2 m_2 c^2 \geq (\frac{1}{ \gamma_1 m_1 c^2} + \frac{1}{ \gamma_2 m_2 c^2})^{-1} \geq \frac{\gamma_1 \gamma_2 m_1 m_2 c^4 }{\gamma_2 m_2 c^2 + \gamma_1 m_1 c^2} \geq \mu \gamma_{rel}c^2$$

we have morphed the equation so that we have something invariant in all frames. Thus $a=\mu$

  • 1
    $\begingroup$ This derivation is not clear to me, and I suspect not correct. The final inequality that you arrive at is, for low velocities, completely dominated by the rest mass term. I should suspect that if you want a kinematic equation, you should have to subtract this term. However I would say that a more rigorous derivation is probably needed.. $\endgroup$
    – Jakob KS
    Jun 14, 2023 at 8:10
  • $\begingroup$ @JakobKS about the last inequality I use: $m_2/\gamma_1 + m_1/\gamma_2 \leq m_2 + m_1$. But yes I would prefer a better derivation. $\endgroup$ Jun 14, 2023 at 8:52
  • $\begingroup$ While that inequality is true, I'm not convinced that it can be used to solve the problem. In particular, I can't follow the reasoning for saying that $a = \mu$, nor that the inequality $\gamma_1 m_1 c^2 + \gamma_2 m_2 c^2 - a \gamma_{rel} c^2 \geq 0$ is sufficient to solve the problem in the first place. $\endgroup$
    – Jakob KS
    Jun 14, 2023 at 11:10

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