2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ I believe the answer is $\mu \gamma_{rel} c^2$ $\endgroup$ Jun 14, 2023 at 4:58

1 Answer 1

0
$\begingroup$

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:

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

Thus,

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

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 $a=\mu$

$\endgroup$
3
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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