# Relativistic dynamics does not reduce to Newtonian dynamics? [closed]

## 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.

1. 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.
2. 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.
3. 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?

• You haven't really demonstrated a problem here, and it looks like there's a lot more information than needed to set up what you're asking. You might want to try rephrasing your question Jun 14, 2023 at 5:33
• @SeñorO I've linked the information required now. Hopefully, that suffices? Jun 14, 2023 at 5:58
• Reason for downvote? Jun 14, 2023 at 6:26
• (Not the downvoter) Your construction is flawed, your potential does not model elastic collision, if interpreted in the standard way in terms of Sobolev spaces it is just the same as $V = 0$. Also note that your construction is also flawed in the sense, that you have to fit your potential to the initial conditions: Why should your particles have a different potential in dependence of the initial velocities? Jun 14, 2023 at 7:25
• @SebastianRiese different point particles can have different potential. These potentials are independent of particle velocity. All this calculation implies is one will just see a potential $V_0$ above which stability is no longer there for elastic collisions. How much energy should I collide my particles with to get $V_0$? I use the condition $E \geq V$. Usually people do this with finite radius and set the potential as infinity: en.m.wikipedia.org/wiki/Hard_spheres Jun 14, 2023 at 7:46

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.