4 momentum is particle physics

fengqiu
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Okay so i have a question for you guys
if I have a positron striking an electron at rest to create 2 pions( + and -) and I want to calculate the minimum kinnetic energy that the electrons can possesses to create these pions... then the created pions will be at rest correct?
so this gives me two four vectors
[Ee++Ee-,Pe+C,0,0)
the other one being
[Epi++Epi-,0,0,0]
now what really confuses me is.
1) momentum isn't conserved? we're still in the same frame of refernce ie lab frame, shouldn't momentum be conserved?
2) is energy conserved? if so when i equate the energies and equate the four vector length (invariant mass) i get different answers?

Thanks for your help!

Adam
 
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fengqiu said:
then the created pions will be at rest correct?
Incorrect, this would - as you have noticed - violate conservation of momentum.

The pions will be created at relative rest, i.e., they will be moving with the same velocity.
 
Orodruin said:
Incorrect, this would - as you have noticed - violate conservation of momentum.

The pions will be created at relative rest, i.e., they will be moving with the same velocity.
oh okay... so if they have the same velocity... that makes a bit more sense...
OKAY I'm still confused about the energies. could i not just solve the equation by equating the energies?

Cheers\

Adam
 
Yes. You can do either, equating the invariant masses is simpler.

In order for us to figure out where you go wrong, you must provide us with your actual computations, i.e., you must show us what you do, not only tell us what you do.
 
i'm just crunching through the calculations now, I'll get back to you if i get the right answer, sorry one more question is that how do you know the pions are going at equal velocities?
 
fengqiu said:
i'm just crunching through the calculations now, I'll get back to you if i get the right answer, sorry one more question is that how do you know the pions are going at equal velocities?

They do not need to do so if there is more energy available. The invariant mass of the two pions is the smallest when they do and therefore it represents the threshold value, i.e., when the electrons need the least invariant mass - translating to the lowest energy possible.
 
Orodruin said:
The invariant mass of the two pions is the smallest when they do

To specify, call the pion 4-momenta ##p_1## and ##p_2##, respectively. This leads to
$$
s = (p_1+p_2)^2 = 2 m_\pi^2 + 2 p_1\cdot p_2.
$$
You can compute the inner product ##p_1 \cdot p_2## in any frame since it is Lorentz invariant. Computing it in the rest frame of the first pion gives ##p_1 = (m_\pi,0)## and ##p_2 = (m_\pi + T_\pi,p_\pi)##, where ##T_\pi## and ##p_\pi## are the kinetic energy and momentum of the second pion in the rest frame of the first. Hence
$$
s = 4 m_\pi^2 + 2m_\pi T_\pi \geq 4 m_\pi^2,
$$
with equality when ##T_\pi = 0##, i.e., when the pions are at relative rest.
 
you, Orodruin are the man!
thanks for your help
it turns out that we get the same equation for energy equivalence and invariant mass
that makes ALOT more sense
 
fengqiu said:
you, Orodruin are the man!

Not really, I have just been teaching this stuff for six years and make a living as a particle physicist ...
 
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