# Relativistic dynamics problem

1. Jul 10, 2010

### PhMichael

1. The problem statement, all variables and given/known data
A particle having a $$3m$$ rest mass is moving with velocity $$v$$ towards another particle, which is at rest and having a rest mass of $$m$$. After the collision, two new particle are formed, each having a rest mass of $$2.5m$$

I'm required to find the minimum velocity for which such a process occurs.

2. The attempt at a solution

(1) conservation of energy:

$$\gamma \cdot 3mc^{2} + mc^{2} = 2 \sqrt{p^{2}c^{2}+(2.5m)^{2}c^{4}}$$

(2) conservation of momentum:

$$\gamma \cdot 3mv = 2p$$

solving these two equations with $$\gamma = \frac{1}{\sqrt{1-v^{2}/c^{2}}}$$, we get:

$$v = \frac {\sqrt{21}}{5} c$$

This answer is correct, but I have several questions regarding my own solution (LOL):

1) why is the energy of both particles equal?

2) why is equation no.(2) correct? in other words, why is the momentum of the new particles equal? and why is it assumed that they both will move along the same direction that the 3m particle has moved?

Last edited: Jul 10, 2010
2. Jul 10, 2010

### kuruman

The answer to this question (and the next) lies in the statement of the problem "find the minimum velocity for which such a process occurs". What must be true if the velocity is minimum? Hint: Think center of mass.

See above.

3. Jul 12, 2010

### PhMichael

Well, actually, I didn't even consider the center of mass frame in my solution and still, with great luck, I got it right ... Can you please expand your explanation!? ... to be more specific, I don't really underatand the momntum equation.

4. Jul 12, 2010

### kuruman

Minimum velocity of the projectile particle means that there is just barely enough energy to produce the two particles. What does that say about the energy of the daughter particles in their CM frame? What should that energy be?

5. Jul 27, 2010

### PhMichael

Let me see if I got that right:

I should look for the final situation in which the two particles exist with the minimum possible energies; in this case, they will have no velocity relative to each other, that is, the two product particles will move together as a glob with respect to the lab frame.

right?

6. Jul 28, 2010

### kuruman

Right, and in the center of mass frame instead of the lab frame that velocity would be .....?

7. Jul 28, 2010

### PhMichael

well this glob's velocity is with respect to the lab frame so if i'm going to take it with respect to the glob's frame, it's supposed to be zero. right?

8. Jul 28, 2010

### kuruman

You missed the point of what I am trying to say to answer your initial question. Imagine being an observer moving at the velocity of the center of mass. That's a velocity that is constant as long as momentum is conserved. When the glob is formed, it has zero velocity with respect to its center of mass. Therefore, it is also zero with respect to the observer. So if you transform back to the lab frame, since the particles making up the glob have the same mass and are at rest with respect to the CM frame, they should have the same energy in the lab frame.

Last edited: Jul 28, 2010
9. Jul 28, 2010

### PhMichael

actually, I've used the conserved invariant $$E^{2} - (pc)^{2 } = (mc^{2})^{2}$$ in the following manner:

Before collision - lab frame:

Total energy: $$E+mc^{2}$$
Total momentum: $$p$$

where E and p stand for the energy and momentum of the 3m particle, respectively.

After collision - glob's frame:

Total energy: $$2(2.5m)c^{2} = 5mc^{2}$$
Total momentum: $$0$$

Now, i'll used the mentioned invariant:

$$(E+mc^{2})^{2}-(pc)^{2} = (5mc^{2})^{2}$$

from this I get the energy of the 3m particle: $$E=\frac{15}{2}mc^{2}$$, and using the fact that $$E=\frac{3mc^{2}}{\sqrt{1-(v/c)^{2}}}$$, i can find the velocity to be:
$$v=\frac{\sqrt{21}}{5}c$$
just like before, but now with a bit of understanding =)