Help with Relativistic Collision

Click For Summary
SUMMARY

This discussion focuses on deriving relativistic momentum without using the concept of relativistic mass, as preferred by the user. The scenario involves two identical billiard balls in an elastic collision, analyzed from the perspective of an observer moving horizontally with one of the balls. The user proposes that the change in vertical momentum for both balls should be equal, leading to a contradiction in their calculations. The discussion highlights the complexities of applying the gamma function in this context and questions the validity of the inverse gamma function.

PREREQUISITES
  • Understanding of relativistic momentum and its derivation
  • Familiarity with the gamma function in special relativity
  • Knowledge of elastic collisions in physics
  • Basic principles of momentum conservation
NEXT STEPS
  • Study the derivation of relativistic momentum without using relativistic mass
  • Explore the implications of the gamma function in relativistic physics
  • Investigate the concept of momentum conservation in non-inertial frames
  • Learn about the mathematical treatment of elastic collisions in special relativity
USEFUL FOR

Physics students, educators, and anyone interested in advanced concepts of relativistic mechanics and momentum conservation.

SamRoss
Gold Member
Messages
256
Reaction score
36
I'm stuck trying to complete this derivation of relativistic momentum without reverting to relativistic mass (a concept I don't like). Those who have read Richard Feynman's "Six Not So Easy Pieces" will realize that I'm really just taking his setup but instead of introducing relativistic mass to maintain conservation of momentum I'm trying to redo the end without it. Maybe someone can help.

I start with two identical billiard balls moving diagonally towards each other and then bouncing off in an elastic collision. The motion can be thought of as forming an "X" shape where the top "v" is ball #1 moving, let's say, down left and then up left. The bottom upside down "v" is ball #2 moving up right and then down right.

If the collision is seen by an observer moving horizontally to the right with the same horizontal velocity as ball #2 then to him the collision would look like a "Y". In other words, ball #2 simply moves up and down with no horizontal movement while ball #1 whizzes by even faster than before. For this observer, it is clear from the picture that horizontal momentum is conserved before and after the collision - ball #2 has no horizontal movement while the horizontal velocity of ball #1 is unchanged. This leaves us to check the conservation of vertical momentum.

Let w be the vertical velocity of ball #2. The vertical velocity of ball #1 can be determined by doing a composition of velocities. This turns out to be w\gamma-1(u) where u is the horizontal velocity of ball #1. Note the inverse gamma is a function of u as opposed to w. For simplicity's sake let's let w\gamma-1(u)=s.

Now we get to the last step and this is where I'm having trouble. My method is actually not so much to derive the relativistic momentum as it is to propose it and then show that it results in the conservation of momentum. If we propose \gammamv (m being invariant mass) then the change in vertical momentum of ball #2 before and after the collision should be 2\gamma(w)mw while the change in vertical momentum of ball #1 is 2\gamma(s)ms.

I would think that these two things should be equal but it's not turning out that way on my paper. Am I just making a careless error and these things really are equal or is there something wrong with my reasoning?
 
Last edited:
Physics news on Phys.org
You can always replace relativistic mass with the energy or use ##\gamma m##.

The inverse gamma function of a velocity doesn't make sense. The gamma function assigns a real number to a velocity, the inverse would assign a velocity to a real number. Plugging in a velocity can't be meaningful.
 

Similar threads

  • · Replies 67 ·
3
Replies
67
Views
7K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 15 ·
Replies
15
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 36 ·
2
Replies
36
Views
5K
Replies
10
Views
1K