2
$\begingroup$

I am considering a case of collision in special relativity in which there is a change in the number of particles; for example, two particles colliding and sticking together to form a single particle. I was thinking about approaching a problem like this from the zero momentum frame, but then I realised that I did not know whether to consider the initial scenario (two particles approaching each othet) or the final scenario (a single particle travelling with speed v) to obtain the velocity of my zero momentum frame.

Something tells me that it should not matter which I choose, as if it did I could tell whether I am moving relative to some other frame, which would violate one of the principles of relativity. However mathematically I cannot see why it would be the case that these two considerations would lead to the same result for the velocity of the ZM frame.

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ Think about conservation of momentum, it will help you get to the answer :) $\endgroup$ – Ofek Gillon Apr 2 '17 at 17:46
  • $\begingroup$ @OfekGillon Ah I see! Silly i did not think of this! Thank you $\endgroup$ – Meep Apr 2 '17 at 20:24
1
$\begingroup$

Like @Ofek Gillon said, it is about conservation of momentum. In any frame it is conserved, so in the ZM frame before it is 0, after the collision in that same frame it is also zero. Both ZM frames are the same.

A lot of collision physics experiments are analyzed this way because it can be simpler, and after if one needs to one can Lorentz transform to any other frame.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Thank you for the credit :) , just one thing, it's spelled Gillon*. Have a wonderful day! $\endgroup$ – Ofek Gillon Apr 3 '17 at 17:42
  • $\begingroup$ @Ofek Gillon sorry for the misspell, I looked it twice but am pretty lousy with names, and used to know somebody with last name Guillen so it must have been a mental lapse. You have a great day and night as well! $\endgroup$ – Bob Bee Apr 4 '17 at 0:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.