consider a particle, A receiving energy from a second one,particle B in a one dimensional collision. E^2=p^2+m^2_0, EdE=pdp

Heergerneuu

Heergerneuu

Answered question

2022-09-25

consider a particle, A receiving energy from a second one,particle B in a one dimensional collision.
E 2 = p 2 + m 0 2
E d E = p d p
For particle A:
E A d E A = p A d p A ( 1 )
For Particle B:
E B d E B = p B d p B ( 2 )
Now
d E A =∣ d E B ( 3 )
From Conservation of linear momentum we have:
p A + p B = K
where k is a constant vector. Now,
d p A + d p B = 0
Or,
d p B = d p A
Or,
d p B =∣ d p A ( 4 )
Applying relations ( 3 ) and ( 4 ) to ( 1 ) and ( 2 ) we have:
E A E B = p A p B ( 5 )
A pair of particles cannot interact unless relation ( 5 ) is satisfied. Can we conclude that that relation ( 5 ) to be a restriction for 1 D collisions?
Now let's move to the general type of 3d collisions between a pair of particles A and B.
A frame is chosen where the particle B is initially at rest in it.
E B d E B = p B d p B ( 6 )
If the particle B is initially at rest the RHS of ( 6 ) is zero. But the LHS cannot be zero unless d E B = 0.
How does one get round this problem?

Answer & Explanation

unfideneigreewl

unfideneigreewl

Beginner2022-09-26Added 5 answers

You have assumed all the changes d E, d p etc. to be infinitesimal – infinitely small – because you have used the differential calculus. So you "theorem" doesn't apply to arbitrary collisions; it only applies to elastic collisions in which just an infinitesimal amount of energy and momentum is transmitted.
Indeed, for such collisions, your identity has to hold. One may also write it as
| p A | E A = | p B | E B
which is probably more insightful because it simply says that the two particles must have the same velocity. When it's so, they have the same rest frame. In that rest frame, one may transmit the momentum between the particles and the kinetic energy of both remains equal to the rest mass, up to negligible second-order terms.
However, you won't be able to find any finite collision of this sort because it would already violate the conservation laws: the second-order deviations would become important and couldn't be canceled.
If you lift the assumption that the changes of the energy and momentum are infinitesimal, you will be able to see that there are also nontrivial collisions in which the momentum and energy of both particles change by a particular finite amount. Those collisions are particularly comprehensible in the center-of-mass system in which the initial particles have the opposite values of the momentum and in this frame, the only change is that each particle reverts its sign of the momentum.

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Relativity

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?