Relativistic energy and momentum 2

[roman15]

Homework Statement

proton hits a stationary proton which then produces a pion with the following reaction
p + p = p + p +pion
if the initial proton has just enough energy to produce the pion what are the speeds of the final protons and pion in the laboratory frame?
m(pion)=0.144m(proton)

Homework Equations

E=E(rest)+K

The Attempt at a Solution

since the incident proton had just enough energy to make the pion, that means its final kinetic energy was zero, so final speed of zero right?

using relativistic energy equations i got that
K(initial of incident proton)=K(final of proton 2) + K(pion) + E(rest of pion)
but I am not really getting anywhere with this equation cause i don't know the initial speed of the incident proton

 

[tiny-tim]

hi roman15!

if the initial proton has just enough energy to produce the pion what are the speeds of the final protons and pion in the laboratory frame?

try it in the centre of mass frame first

 

[roman15]

what do you mean the centre of mass frame? of which particle?
and I am not sure if this will help cause even in the other frames i won't have anymore information that what i have now right?

 

[tiny-tim]

the centre of mass of the whole system …

no, you won't have any more information, but it should make the maths simpler

 

[roman15]

K I asked my TA an they said to look at the zero momentum frame so that the final speeds of both protons is zero

 

[tiny-tim]

the centre of mass frame is the zero momentum frame!

 

[roman15]

oh I am sorry, I've never heard of it being referred to as that
so what should i do?
should i find the speed of the frames relative to each other using the lorentz transformations by setting:
u'1=u'2

 

[roman15]

oops i mean set
u'1=-u'2
because in this zero momentum frame they must have the same but opposite speeds

 

[tiny-tim]

oops i mean set
u'1=-u'2

yup!

 

[roman15]

ok so i got a quadratic equation for v
(u1/c^2)v^2 - 2v + u1 = 0
i got this by setting u2=0 because its at rest in the initial frame of the question
so i guess i have to use the quadratic formula to find v?

im not sure what to do know even with the relative speed of the frames, all i know is that after the collision, both protons will be at rest, because they have just enough energy to make the collision occur

 

[tiny-tim]

… i got this by setting u2=0 because its at rest in the initial frame of the question

now I'm completely confused

i thought you were intending to use the zero momentum frame (the centre of mass frame), with the initial velocities equal and opposite?

 

[roman15]

well because in the question, one proton is at rest and the other collides with it
so I am trying to find the frame that sees the two protons having equal but opposite speeds, this other frame is S'
but in the frame that the questions speaks of, one proton has speed u1 and the other has a speed u2=0

is this wrong? what did you think of doing?

 

[vela]

Your approach is valid, but it makes the problem harder than it needs to be.

Tim is suggesting you solve for the energy and momentum of the protons in the center-of-mass frame first. Once you know those values, you can calculate the velocity of the protons in that frame. Then it's straightforward to figure out the boost required to get back to the lab frame.

 

[roman15]

ok well in the centre of mass frame I am getting that for energy:
2K(proton) = m(pion)c^2 + K(pion)
because the final kinetic energy of the protons is zero and the rest mass energy of the protons just cancel out on each side

for momentum:
gamma*m(proton)*u1 + gamma*m(proton)*(-u1) = gamma(pion)*m(pion)*u3
0=gamma(pion)*m(pion)*u3

are those correct?

 

[vela]

Yes, those are correct. The kinetic energy of the protons turn into the pion, and at the lowest energy, the pion is created at rest in the COM frame.

 

[roman15]

ok you a lot easier
so in the centre of mass frame all the particles have a final velocity of zero then?
but how do i find the speed between the lab frame and the centre of mass frame?

 

[roman15]

would i just use the energy equation because if the pion is also at rest in the COM frame, then the kinetic energy of the two protons is just equal to the rest mass of the pions so i have to solve for v in the gamma of the kinetic energy of the protons?

2(gamma-1)*m(proton)*c^2 = m(pion)*c^2

 

[roman15]

oh wait the gamma in that equation is the gamma of the protons so first i have to solve for u'
ok so using this i got that the initial speeds of the protons in the COM frame was 0.36c
then knowing that the initial speed of the one proton in the lab frame of reference was zero i found that the speed between the frames was 0.36c
since all the particles have a final speed of zero in the COM frame after the collision, then using the lorentz transformations I am getting that they are all moving with speed 0.36c in the lab frame after the collision
this doesn't seem right to me because i thought that in the lab frame the incident proton has a final speed of zero, but i checked with conservation of momentum (because that the next part of this question) and momentum was conserved if all the particles had a final speed of 0.36c

 

[vela]

That's correct. At the lowest energy, the particles are at rest relative to each other, so in the lab frame, they all move together with the same velocity.