Four momentum proton-proton scattering question

Click For Summary
SUMMARY

The discussion focuses on deriving the kinetic energy T' of a particle scattered elastically at an angle θ in a fixed target proton-proton scattering experiment. The relationship is established as T' = T cos² θ / (1 + ((T/2M)sin² θ)). Key equations involve the four-momentum vectors P1, P2, P3, and P4, with the Mandelstam variables s, t, and u defined in terms of these vectors. Participants emphasize the importance of conservation of energy and momentum in solving the problem, with specific references to algebraic manipulation and the use of the cosine rule.

PREREQUISITES
  • Understanding of four-momentum in particle physics
  • Familiarity with Mandelstam variables (s, t, u)
  • Knowledge of conservation laws in elastic collisions
  • Proficiency in algebraic manipulation of equations
NEXT STEPS
  • Study the derivation of Mandelstam variables in particle collisions
  • Learn about four-momentum conservation in relativistic collisions
  • Explore the cosine rule in the context of scattering angles
  • Practice problems involving elastic scattering and kinetic energy calculations
USEFUL FOR

Physics students, particle physicists, and anyone interested in understanding elastic scattering processes and the application of four-momentum in collision theory.

Jdraper
Messages
51
Reaction score
0

Homework Statement


In a fixed target experiment a particle of mass M and kinetic energy T strikes a stationary particle of mass M. By evaluating s, t and u in the laboratory frame and using the above relation, or otherwise, show that the kinetic energy T' of the particle scattered elastically at an angle θ is given by

T' = T cos^2 θ / (1 + ((T/2M)*sin^2 θ))

Homework Equations


We know that:
We define P1 as the four momentum of the incoming proton:
P1 = (T+M,0,0,T)
P2 as the stationary proton:
P2 = (M,0,0,0)
P3 as the initially incoming proton after the scattering event:
P3 = (T'+M, 0, T'sinθ, T'cosθ)
P4 as the initially stationary proton after the scattering event:
P4 = (E4+M, P4) where P4 in the bracket is a 3 vector not 4.

We also know:

s = (P1 + P2)2 , t = (P1 − P3)2 and u = (P1 − P4)2

and

s + t + u = ∑ mi2 = 4m2

The Attempt at a Solution


I have attempted this multiple times over a few days, I realize that P4 and therefore u need to be eliminated as they aren't in the solution but I am unsure how.

Any help/ suggestions would be appreciated. Thanks, John.
 
Physics news on Phys.org
Jdraper said:

Homework Statement


In a fixed target experiment a particle of mass M and kinetic energy T strikes a stationary particle of mass M. By evaluating s, t and u in the laboratory frame and using the above relation, or otherwise, show that the kinetic energy T' of the particle scattered elastically at an angle θ is given by

T' = T cos^2 θ / (1 + ((T/2M)*sin^2 θ))

Homework Equations


We know that:
We define P1 as the four momentum of the incoming proton:
P1 = (T+M,0,0,T)
.

You seem to be using the KE instead of the momentum in your four-vector.
 
PeroK said:
You seem to be using the KE instead of the momentum in your four-vector.
Hey, forgot to mention that I'm using the convention that c=1.
 
Jdraper said:
Hey, forgot to mention that I'm using the convention that c=1.

Okay, but you still don't have ##T = p##:

##T = E - M## and ##p^2 = E^2 - m^2##
 
PS The problem comes out from conservation of energy and momentum, although I haven't looked at how to do it using your equations for ##s, t, u, m##.
 
PeroK said:
Okay, but you still don't have ##T = p##:

##T = E - M## and ##p^2 = E^2 - m^2##

You've opened up a can of worms in my understanding here. My professor uses this convention and I have never really questioned it.

So you are saying:

##P=(E^2 -M^2)^{0.5}##
therefore as ##E=T+M## we can rewrite this as:

##P=((T+M)^2 + M^2)^{0.5} = (T^2 +2TM)^{0.5}##
 
Jdraper said:
You've opened up a can of worms in my understanding here. My professor uses this convention and I have never really questioned it.

So you are saying:

##P=(E^2 -M^2)^{0.5}##
therefore as ##E=T+M## we can rewrite this as:

##P=((T+M)^2 + M^2)^{0.5} = (T^2 +2TM)^{0.5}##

Yes. I used that equation to get the result.
 
PeroK said:
Yes. I used that equation to get the result.
So, just to check, the first four momentum vector would be:

##P_1 = (T+M, 0,0, (T^2 + 2TM)^{0.5})##
 
Jdraper said:
So, just to check, the first four momentum vector would be:

##P_1 = (T+M, 0,0, (T^2 + 2TM)^{0.5})##

Yes.
 
  • #10
Jdraper said:
We also know:

s = (P1 + P2)2 , t = (P1 − P3)2 and u = (P1 − P4)2

and

s + t + u = ∑ mi2 = 4m2

I've had a quick look at using this equation. I must be missing something but I don't see the trick. It looks harder to me than starting from the equation for conservation of momentum using the cosine rule. Perhaps I'm missing a trick but the term in ##P_1P_4## is a nuisance.
 
  • #11
PeroK said:
I've had a quick look at using this equation. I must be missing something but I don't see the trick. It looks harder to me than starting from the equation for conservation of momentum using the cosine rule. Perhaps I'm missing a trick but the term in ##P_1P_4## is a nuisance.

Currently trying to do it your way by conservation of momentum:

## (P_1 + P_2)^2 = (P_3 + P_4)^2 ##

If you want to read further into them, s,u and t are called the mandelstam variables. Yes, it's very difficult to cancel.The algebra becomes very messy.
 
  • #12
Here is what my lecturer said when I emailed him a few days ago regarding this question:

For q1b in 2014 you first need to write down explicitly the expressions for s, t, and u in terms of m, T, T’ and theta. Then there are several options. One way is to take the t variable and, using s + t + u = 4m^2, express it via s and u in terms of their explicit expressions obtained in the first step. You then equate this expression to the explicit expression of t.

After that it’s just a tedious and longish algebra and you come up with the answer.
 
  • #13
Jdraper said:
Here is what my lecturer said when I emailed him a few days ago regarding this question:

For q1b in 2014 you first need to write down explicitly the expressions for s, t, and u in terms of m, T, T’ and theta. Then there are several options. One way is to take the t variable and, using s + t + u = 4m^2, express it via s and u in terms of their explicit expressions obtained in the first step. You then equate this expression to the explicit expression of t.

After that it’s just a tedious and longish algebra and you come up with the answer.

Maybe my way is better! Here's a summary of the first steps. I kept it in terms of ##E_i## and ##p_i## to begin with and then used ##E = T + M## half way through.

1) Get an equation for ##E_4^2## using conservation of energy.

2) Get an equation for ##p_4^2## using conservation of momentum and the cosine rule. Substitute ##p_i = E_i - M##, which gives you another equation for ##E_4^2##.

3) Use 1) and 2) to eliminate ##E_4^2##.

(This is a fairly standard approach for collisions, so it's worth being familiar with). In this case, you should get:

##E_1E_3 - M(E_1 - E_3) - M^2 = p_1p_3 cos\theta##
 
  • Like
Likes   Reactions: Jdraper
  • #14
PeroK said:
Maybe my way is better! Here's a summary of the first steps. I kept it in terms of ##E_i## and ##p_i## to begin with and then used ##E = T + M## half way through.

1) Get an equation for ##E_4^2## using conservation of energy.

2) Get an equation for ##p_4^2## using conservation of momentum and the cosine rule. Substitute ##p_i = E_i - M##, which gives you another equation for ##E_4^2##.

3) Use 1) and 2) to eliminate ##E_4^2##.

(This is a fairly standard approach for collisions, so it's worth being familiar with). In this case, you should get:

##E_1E_3 - M(E_1 - E_3) - M^2 = p_1p_3 cos\theta##

Thank you, think I've followed your method by eliminating ##E_4 and P_4 ## but I've become bogged down in the algebra. Will have another attempt and let you know
 
  • #15
Jdraper said:
Thank you, think I've followed your method by eliminating ##E_4 and P_4 ## but I've become bogged down in the algebra. Will have another attempt and let you know

I substituted ##E = T + M## in the left-hand side and simplified.
 
  • #16
Just got it solved. I used conservation of 4 momenta to derive P4. After that work out s, u, t and use s+t+u= 4m^2
 
Last edited:

Similar threads

  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 21 ·
Replies
21
Views
5K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 4 ·
Replies
4
Views
4K
Replies
3
Views
2K