Four momentum proton-proton scattering question

AI Thread Summary
In a fixed target experiment, a particle with mass M and kinetic energy T collides with a stationary particle of the same mass, leading to an elastic scattering at an angle θ. The kinetic energy T' of the scattered particle is derived using the four-momentum vectors and the Mandelstam variables s, t, and u. The discussion highlights the importance of conservation of energy and momentum in deriving the relationship T' = T cos^2 θ / (1 + ((T/2M)*sin^2 θ)). Participants emphasize the need to eliminate certain variables and navigate through complex algebra to arrive at the solution. Ultimately, the method involves substituting expressions for energy and momentum to simplify calculations and derive the final result.
Jdraper
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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.
 
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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##
 
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  • #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
 
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