Help with Relativistic Collision

In summary, the conversation revolves around trying to complete a derivation of relativistic momentum without using the concept of relativistic mass. The speaker is using a setup from Richard Feynman's "Six Not So Easy Pieces" and is trying to redo the end without using relativistic mass. The speaker asks for help in determining if their reasoning is correct and whether there is an error in their calculations. They also mention the possibility of using energy or the equation ##\gamma m## instead of relativistic mass.
  • #1
SamRoss
Gold Member
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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[itex]\gamma[/itex]-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[itex]\gamma[/itex]-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 [itex]\gamma[/itex]mv (m being invariant mass) then the change in vertical momentum of ball #2 before and after the collision should be 2[itex]\gamma[/itex](w)mw while the change in vertical momentum of ball #1 is 2[itex]\gamma[/itex](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?
 
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  • #2
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.
 

1. What is a relativistic collision?

A relativistic collision is a collision between particles moving at speeds close to the speed of light. This type of collision takes into account the principles of special relativity, which states that the laws of physics should be the same for all observers, regardless of their relative motion.

2. How is momentum conserved in a relativistic collision?

In a relativistic collision, momentum is conserved by taking into account the relativistic mass and velocity of the particles. This means that the total momentum before the collision is equal to the total momentum after the collision.

3. What is the equation for calculating relativistic momentum?

The equation for calculating relativistic momentum is p = mγv, where p is the momentum, m is the rest mass of the particle, γ is the Lorentz factor, and v is the velocity of the particle. This equation takes into account the increase in mass at high velocities.

4. How does energy change in a relativistic collision?

In a relativistic collision, energy is not conserved in the traditional sense. Instead, it is converted from kinetic energy to mass and back again. This can be described by the famous equation E=mc², which states that energy is equal to mass times the speed of light squared.

5. What are some real-world examples of relativistic collisions?

Some examples of relativistic collisions include particle accelerators, such as the Large Hadron Collider, where particles are accelerated to near-light speeds and collide with each other. Cosmic ray collisions in space also involve relativistic speeds and can be studied to understand high-energy processes in the universe.

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