Relativistic Collision of Particles

[Menisto]

Homework Statement

A particle of mass m whose total energy is twice its rest energy collides with an identical particle at rest. If they stick together, what is the mass of the resulting composite particle? What is its velocity?

Homework Equations

E = (gamma)mc^2
p = (gamma)mu

The Attempt at a Solution

The total energy of the first particle is twice its rest energy, or

E_{total} = (gamma)mc^2
= 2mc^2

or $$\gamma$$ = 2.

Using this, I find that the velocity of particle 1 is:
u = (sqrt{3})/{2}

Using this is both momentum and energy conservation equations yields the two comparable equations:

m_{final} = (gamma 1)m{1} / (gamma final)
and
m_{final} = sqrt{3}m{1}c / (gamma final) u{final}

Solving this, I get:

u_{final} = c

and

m_{final} = 0

The two answers in relation to each other seem alright, but what is happening here? Is this saying that the particles completely annihilated each other? What is special about the initial conditions that makes this happen?

 

[dextercioby]

Just write the conservation equations in this form$$E_{tot}_{1}+E_{tot}_{2}=E_{tot}_{\mbox{resulting particle}}$$

$$p_{1}+p_{2}=p_{tot}_{\mbox{resulting particle}}$$

Then you can solve them quite easily i guess.

 

[prasad_ind007]

Let v be the velocity of the particle before collision, and v1 be after collision. Similarly, 'm' be the mass before collision, and m1 after collision.

Now, as correctly stated, v = ($$\sqrt{3}$$/2)*c.
also, $$\gamma$$(v)=2.
So, from energy conservation, we get:
m1*$$\gamma$$(v1)=2m.

and, from momentum conservation, we get: v1=v.
this invariably leads to m1=m.

i think that this means: the first particle comes at rest, and the second particle travels with exactly the same velocity.

 

[Dick]

Let v be the velocity of the particle before collision, and v1 be after collision. Similarly, 'm' be the mass before collision, and m1 after collision.

Now, as correctly stated, v = ($$\sqrt{3}$$/2)*c.
also, $$\gamma$$(v)=2.
So, from energy conservation, we get:
m1*$$\gamma$$(v1)=2m.

and, from momentum conservation, we get: v1=v.
this invariably leads to m1=m.

i think that this means: the first particle comes at rest, and the second particle travels with exactly the same velocity.

There are TWO particles before the collision. Energy before the collision is the sum of BOTH their energies. And you don't need to explicitly compute v or gamma, just use conservation and E^2-p^2*c^2=m^2*c^4.

 

[prasad_ind007]

yup, i missed it. i will try it once again now. thank you for pointing out the mistake.

 

[prasad_ind007]

I am getting the final velocity to be $$\frac{c}{\sqrt{3}}$$
and combined rest mass as $$m\\*\\\sqrt{6}$$

please correct the answer if i have made a mistake once again.
thank you.

 

[Dick]

I am getting the final velocity to be $$\frac{c}{\sqrt{3}}$$
and combined rest mass as $$m\\*\\\sqrt{6}$$

please correct the answer if i have made a mistake once again.
thank you.

I think you have it right.