I am reading on the Principle of Least Action from a historical perspective. I am also trying to make sense of it from a contemporary point of view -- though my training in contemporary physics is modest.

Maupertuis, who first introduced the PLA, was able to derive laws of collision from this. Maupertuis, in the case of inelastic collision, tries to minimize the quantity $$m_{1} \left(v_{1}-v_{f} \right)^{2} + m_{2} \left( v_{f}-v_{2} \right)^{2}$$

How could I arrive at a similar expression? Or was his proposal for this quantity just a convenient guess.

For me, the Lagrangian $L$ would be defined as $$\frac{1}{2}mv_1(t)^2 + \frac{1}{2}mv_2(t)^2$$

I don't see where I can write the initial and final velocities. I think part of my confusion is that Maupertuis speaks of the "change of velocities" whereas (according to my understanding) such a change has no place in the contemporary formalism.

I also have some conceptual problems, because I want to see the "path" is going from pre-collision to post-collision. I don't see how this fits with the textbook definition of $L=K-U$ (and $U=0$). Does one need to take the integral of the Lagrangian from $t-\epsilon$ to $t+\epsilon$ and do something then? Attempt:

$$S= \int_{t-\epsilon}^{t+\epsilon} \frac{1}{2}mv_1(t)^2 + \frac{1}{2}mv_2(t)^2 \ dt$$

and this needs to be minimized. Not sure this gives the same result as Maupertuis.


You might be making more a big deal about this than it needs. First review what an inelastic collision does. It decreases the total kinetic energy of the system of parts by as much as is possible while conserving momentum. That's the definition. So in the center of momentum frame they just come together and stick together so that all the kinetic energy goes away.

In the other frames there is some net momentum so that has to be there but otherwise you minimize the kinetic energy as much as possible. You want as much energy to be lost (it is the opposite extreme of an elastic collision). In an elastic collision no energy is lost. In an inelastic collision you do the opposite and lose as much as you can lose. It's just the definition.

Maupertuis, [...] tries to minimize the quantity $$m_{1} \left(v_{1}-v_{f} \right)^{2} + m_{2} \left( v_{f}-v_{2} \right)^{2}$$

How could I arrive at a similar expression?

In the frame with zero total momentum there is nothing stopping you from having zero momentum for each part and hence zero kinetic energy for each part. So the total kinetic lost is $$p^2/m_1+(-p)^2/m_2=2(m_1+m_2)p^2/(m_1m_2).$$

But that is exactly the expression you have above. And in fact you could write it as $(\Delta p_1)^2/m_1+(\Delta p_2)^2/m_2$ where there are two possible meanings to the delta. In the center of momentum frame it is the change in momentum before and after. And in another frame it is the relative momentum so you can think of it as a frame invariant measure of how the momentum differs from the average. But I this is really a pretty post hoc analysis. So if that was my answer it would seem ...

Or was his proposal for this quantity just a convenient guess.

But there is more. Firstly, while energy is frame dependent, changes in kinetic energy at a point are not frame dependent. So if we computed the change of kinetic energy in the center of momentum frame then that really is the change in kinetic energy in all frames.

Secondly, we know we are trying to minimize the kinetic energy as much as possible subject to a constraint. The constraint is conservation of momentum. So $m_1v_1+m_2v_2=(m_1+m_2)v_f.$

Minimizing $m_1v_f^2+m_2v_f^2$ subject to $m_1v_1+m_2v_2=(m_1+m_2)v_f$ is a real problem with a real solution.

And so you might do it one way. Get an answer. Then when checking your work you can do a post hoc analysis and convince yourself it couldn't have been anything else even though that wasn't how you came up with it. Note I am not saying that any of what I did above was at all historically accurate.

For me, the Lagrangian $L$ would be defined as $$\frac{1}{2}mv_1(t)^2 + \frac{1}{2}mv_2(t)^2$$

If you don't have a potential energy that is going to take all that energy then you aren't going to get an inelastic collision. If your final conditions are the particles together at a location and your initial conditions are them located at their positions then that free particle Lagrangian is going to have them cruise in a straight line to the final position at just the speed where they collide at that final last moment. Which doesn't tell you anything about what happens next.


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