- #1

gionole

- 281

- 24

Imagine we have ##L = \frac{1}{2}m\dot x^2##.

If we do: ##x(t) -> x(t) + \epsilon(t)##, we transform the whole trajectory and We will arrive at ##\frac{dP}{dt} = 0## (momentum conserved).

Even though particle won't move on the transformed trajectory, we still do the transformation. The way we arrived to momentum conservation is by the same way as Euler lagrangian derivation as such ##x(t)## is a true path, so its action must be minimum, so action of ##x(t) + \epsilon(t)## must be less. The only way this is proved is by "action of true path" must be minimum. Is there other type of proof ? The reason I'm asking is with this proof, it's like we're saying, "yeah, we know particle moves on ##x(t)## path, but what if it used a different path(##x(t) + \epsilon(t)##), and then we show it can't and arrive at momentum conservation. So it's NOT the kind of proof where particle was at ##A## and then moved to ##B## and momentum didn't change and then moved to C and momentum still didn't change, because since the whole trajectory was transformed, particle wouldn't appear on ##C## as ##C## is on the old path.

So to me, this doesn't seem like understandable for some reason. The better way would be that we know particle moves on the true path ##x(t)## and on that path, there we got ##A, B, C## points and we show when particle moves between them, momentum is conserved. That would be the real, more understandable way, but I guess, we can't do that.

So why are we saying: "when we do transformation, momentum is conserved ?" - particle doesn't actually start using new ##x(t) + \epsilon(t)## trajectory at all, so what's the point of saying the "transformation" at all ? I understand the derivation but not the intuitive idea of considering transformation. I think this "transformation" is the same thing as using the "action must be minimum" over the true path and nothing else - so it seems like the mathematical trick and not the intuitive idea of "coordinate transformation".

Hope I explained well enough, if not, let me know.