# Minimum Potential energy required to behave like a turning point in relativistic case?

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.

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

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 $$a=\mu$$

• 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.. Jun 14, 2023 at 8:10
• @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. Jun 14, 2023 at 8:52
• 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. Jun 14, 2023 at 11:10