Relativistic dynamics does not reduce to Newtonian dynamics?

Question

Background

So I realised the following:

1. What enables us to talk about kinematics and dynamics separately is Newton's first law:

A body remains at rest, or in motion at a constant speed in a straight line, unless acted upon by a force.

2. Hence, one can ask questions of kinematics and dynamics independently. For example, consider an ideal elastic collision between $2$ point like objects. Let the potential be of the form: $$ V=\begin{cases} 0 & r_1 \neq r_2 \\ V_0 & r_1 = r_2 \end{cases} $$ Where $r_i$ is the position of the $i$'th particle and $V_0$ is the minimum potential energy to behave as a turning point. Note, $V_0 \to 2 V_0$ will result in the same equations of motion (and thus kinematics). What should be the value of $V_0$ is a question of dynamics.
3. When one takes the low energy/velocity limit of the action in relativistic mechanics and reaches Newtonian mechanics. This is actually a statement of the kinematics of special relativity reducing to the kinematics of non-relativistic classical mechanics.
4. Thus, if one were to pose the question what is the value of $V_0$ in the theory of Newtonian mechanics (where the answer is $V_0 = \frac{1}{2} \mu v_{rel}^2$) versus relativistic mechanics (I believe the answer is $V_0 = \mu\gamma_{rel}c^2$) where $v_{rel}$ is the relative velocity and $\gamma$ is it's corresponding gamma factor, in the limit $v_{rel}/c \to 0$, it is a possibility that they may have different answers. In fact, these answers do not match.

Question

I haven't heard of this before, I suspect there should be a flaw? If so, can someone point it out?

Answer 1

I hope that I've understood your question correctly. Let me rephrase the question to make clear what I will try to answer.

(1) We want to describe the elastic collision of two particles.

(2) We choose to model the particles as "hard spheres", i.e. they interact via a potential energy $$V(x_1,x_2) = \begin{cases} V_0 \ &\text{if } \ |x_1 - x_2| < d \\ 0 \ &\text{otherwise} \end{cases},$$

where $d$ is the diameter of the sphere which I've allowed to be finite to avoid mathematical issues (cf. the discussion in the comments). We could also make the potential differentiable by introducing "ramping" function that ramps up the potential from 0 to $V_0$ in a differentiable manner. Note that this potential is not relativistically invariant because the quantity $|x_1 - x_2|$ is not invariant, but as only the height of the potential barrier ends up being important for this problem, I don't think this is a major issue.

(3) The question you are asking (as I understand) is what should be the minimum value $V_0$ of the potential wall be in order for two given particles to bounce back, rather than pass right through each other.

To answer the question, we make use of the fact that for an elastic collision both momentum and energy are conserved. Let us go to a frame where the total momentum is zero, i.e. $$m_1 \gamma_1 \dot{x}_1 + m_2 \gamma_2 \dot{x}_2 = 0.$$ The total energy of the system is $$E = m_1 \gamma_1 c^2 + m_2\gamma_2 c^2 + V = \text{const.}$$

Because we chose a frame in which the total momentum is zero, we can say that in order for the particles to bounce off each other, $V_0$ must be sufficiently large such that $\dot{x}_1 = \dot{x}_2 = 0$ is a solution to the energy equation. Inserting this, we find that we require $$V_0 > E - m_1 c^2 - m_2 c^2 = (\gamma_1 - 1) m_1 c^2 + (\gamma_2 - 1)m_2 c^2.$$

In the classical limit, $m_i (\gamma_i - 1) c^2 \approx \frac{1}{2} m_i v_i^2$, and we find that in this case the requirement becomes $$V_0 > \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 = \frac{1}{2} \mu v_\mathrm{rel}^2,$$

which matches the classical result derived in the other answer.