Question

A proton and antiproton collide with equal and opposite momentum at the SPS collider in Geneva to form a single massive particle which is 8 times more massive than a proton. What is the velocity (as a fraction of c) of the proton beam?

Answer 1

I think this is what to do. it's is a two-parter - first use conservation of energy, followed by a simple calculation using the momentum.

We know that the total energy of each particle is given by
\[E^2=p^2c^2+(mc^2)^2\]
So
\[E_p = \left(p_p^2c^2+(m_pc^2)^2\right)^{1/2} \quad ; \quad E_a = \left(p_a^2c^2+(m_ac^2)^2\right)^{1/2} \quad ; \quad E_n=m_nc^2\]
Where Ep is the energy of the proton, and Ea is the energy of the antiproton, and En is the energy of the new particle (which has no momentum so only has energy equal to its rest mass)
Conservation of energy yields
\[E_p+E_a = E_n\]
\[\left(p_p^2c^2+(m_pc^2)^2\right)^{1/2} +\left(p_a^2c^2+(m_ac^2)^2\right)^{1/2}=m_nc^2\]
We also know that the mass of the proton and antiproton are the same, and the square of the their momenta would also be the same. The mass of the new particle is 8 times the mass of either the proton or antiproton, so we can rewrite that as
\[2\left(p^2c^2+(mc^2)^2\right)^{1/2}=8mc^2\]
Squaring both sides gives
\[4\left(p^2c^2+(mc^2)^2\right)=64(mc^2)^2\]
From there you can easily solve for the square of the momentum of one of the particles.

Next, we look back at the energy of either particle
\[E^2=p^2c^2+(mc^2)^2\]

Rearrange for the momentum
\[p^2= \frac{E^2}{c^2}-m^2c^2\]
then substitute in for E
\[E=\gamma mc^2\]
where
\[\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\]
which gives
\[p^2= m^2c^2(\gamma - 1)\]
\[p^2=m^2c^2\left(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1\right)\]

You can substitute in the value for the momentum that you found, and solve for v - you should never have to plug in the mass of the proton/antiproton because they cancel out. The velocity is incredibly high - over .99c - but this seems the agree with the speeds seen in the SPS

http://www.quantumdiaries.org/2011/04/24/the-cern-accelerator-complex/

Hope this helps! ^_^

Answer 2

Also, keep c in letter form, not numerical. It helps immensely.

Answer 3

So
\[p^2=15m^2c^2\]
and
\[15m^2c^2=m^2c^2\left(\frac{1}{\sqrt{1-v^2/c^2}}-1\right)\]
\[16=\frac{1}{\sqrt{1-v^2/c^2}}\]

Answer 4

hey. sorry on not responding back. I was logged out and forgot my login passwords. I saw the response but yea....couldnt Best Response it.
Hey dude. Your awesome.

Answer 5

Thanks! ^_^ You had a fun one!