Positron-electron collision can someone check my answer?

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The discussion revolves around calculating the energy in the center of mass (CM) frame for a collision between a positron and an electron, each with 500 GeV of energy. The initial calculation for the CM energy is confirmed as correct at 1000 GeV. A subsequent question addresses how to achieve the same CM energy by colliding a beam of positrons with stationary electrons, leading to confusion regarding the application of relativistic mechanics versus Newtonian mechanics. Participants emphasize the importance of using the invariant mass and the correct relativistic formulas to relate energies and momenta in different frames. The conversation highlights the necessity of understanding the fundamental principles of relativistic physics to solve the problem accurately.
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Homework Statement



a positron and electron travel in opposite directions each with energy 5ooGeV and collide head-on.
what is the energy in the centre of mass (cm) of the collision?



The Attempt at a Solution



E(cm) = E(e+) + E(e-) = 500GeV + 500GeV = 1000GeV

is this right?

thanks
 
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Yes.
 
okay then thanks for that...
also could you just look at this as well? i wanted to try it before i put it up

if the same energy in the cm were to be achieved by colliding a beam of positrins with a target of stationary electrons, what energy of the positron would be required?

in cm frame

v = positron vel. in lab frame
v+' = positron vel. in cm
v-' = electron vel. in cm
vcm = cm vel.

v+' = v-vcm
v-' = vcm

[STRIKE]m[/STRIKE][v+'] = [STRIKE]m[/STRIKE][v-']
so Ecm = 1/2m[v+']2 + 1/2m[v-']2 = m[v+']2

v+' = v-vcm
v-' = vcm
so v - vcm = vcm therefore vcm= v/2

so Ecm = m[v-vcm]2 = m[v/2]2
and then v2 = (4Ecm)/m

Kinetic energy of beam = 1/2mv2 = 1/2m[(4Ecm)/m] = 2Ecm = 2000GeV

is this right?
thanks
 
No, you're mixing up Newtonian mechanics with relativistic mechanics. You need to be a bit more careful.
 
where is it that I've gone wrong?
what would you suggest to try and get the right answer?
 
For one thing, the kinetic energy isn't equal to 1/2 mv2 in relativity. Also, you can't just add and subtract velocities like you did. You have to use the relativistic velocity-addition formula.

The quantity E2-(pc)2 is an invariant. Use that to relate the quantities in the two frames.
 
ok so i tried this instead

mass of e+ and e- =>
m2c4 = (E+mc2)2 - p2c2
= E2 - p2c2 + 2Emc2 + m2c4
= 2Emc2 + 2m2c4

and as mc2<<E
m2c4 = 2Emc2

now this is where I'm abit unsure:
can i say: m2 = (1000GeV)2?
cause this would give
(1000GeV)2 = 2E(0.51MeV)
so E = 1.96x1018eV
 
indie452 said:
ok so i tried this instead

mass of e+ and e- =>
m2c4 = (E+mc2)2 - p2c2
How did you get this?
 
(E+mc2)
is the sum of the energies of the 2 particles. the e- is stationary so i thought that its energy is just mc2 in the lab frame
p2c2 i didnt put a p(e+) +p(e-) because the momentum of e- is zero so its just the mom. of the e+
 
  • #10
What about the lefthand side of the equation?
 
  • #11
as in that i should have put (2m)2c4 on the left hand side?
 
  • #12
No, I'm trying to ascertain why you set it to m2c4 in the first place and why later you ask if it should be (1000 GeV)2. The variable m stands for the mass of the electron, right?

The RHS of your equation is fine. I just want to know what you think that quantity, E2total-(ptotalc)2 should equal and why it should equal that. This is a key point in solving this problem.
 
  • #13
well i set it to m2c4 cause this is the invariant mass and so i can find this in the lab frame and it will be the same in the cm frame.
when i found the relation for this i went to the cm frame to use its Ecm to find the energy i need for the beam.

now while writing this i think i should have done:
m2c4 = 2Emc2 = 2(1000GeV)(0.52MeV)c2
m2 = 1x1018eV
m=1GeV
and so E=mc2 = (1GeV)c2

is this right? or am i still failing in my understanding of it?
 
  • #14
indie452 said:
well i set it to m2c4 cause this is the invariant mass and so i can find this in the lab frame and it will be the same in the cm frame.
It's the invariant mass of the two-particle system, not of the electron, so you shouldn't use m, the mass of the electron, on the LHS.
when i found the relation for this i went to the cm frame to use its Ecm to find the energy i need for the beam.

now while writing this i think i should have done:
m2c4 = 2Emc2 = 2(1000GeV)(0.52MeV)c2
m2 = 1x1018eV
m=1GeV
and so E=mc2 = (1GeV)c2

is this right? or am i still failing in my understanding of it?
No, you're mixing up quantities between the two frames. Look at it this way:

E2cm - p2cm = E2lab - p2lab

where the lab frame is where the target is stationery and where RHS of the equation is the same as the RHS of your equation. What should be on the left, I think, should be clear to you now.

Edit: Out of habit, I left out the factors of c, but you should be able to figure out where they should be.
 
  • #15
ok so on the left the Ecm is 1000GeV,
is pcm2c2 = Ecm2 - m2c4

...wait i don't think that helps...
my lecturer didnt really go over how to get the momentum. the only thing he metioned ws the 4momentum but i don't think that is what i use here
 
  • #16
What is the total momentum in the center-of-mass frame?
 
  • #17
wait is it just zero?...thats what i got from doing some rearrangement (too much to write out)
 
  • #18
Yes, the center-of-mass frame is the frame where the total momentum is 0.
 
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