Angles in momentum four vectors

In summary: Would it be pcosa for x and psina for y?Yes, that's right! Now all you have to do is figure out how to transform that into the frame where the two photons have equal and opposite momenta.
  • #1
twinklestar28
21
0

Homework Statement



A particle decays to two photons. In the rest frame the two photons are emitted on the x-y plane, in opposite directions along a line that forms an angle alpha with the the x axis. Derive the momentum four vector of the two photons in the lab frame.


Homework Equations





The Attempt at a Solution



Photon 1 = (E1 sinθ1, E1 cos θ1 , 0 , E1)

Photon 2 = (-E2 sinθ2 , -E2 cos θ2 , 0 , E2)

I'm confused to what notations I should be using for the angles, can I use θ1 and θ2 for the angles of the two photons and then assume that alpha = (θ1+θ2) or is that the wrong way to go about it. Should alpha be in the vector equations?
Also I'm not sure if the signs are correct as the photons are moving in opposite direction in the x - y plane, could someone check my answer and point me in the right direction please :)
 
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  • #2
hi twinklestar28! :smile:
twinklestar28 said:
A particle decays to two photons. In the rest frame the two photons are emitted on the x-y plane, in opposite directions along a line that forms an angle alpha with the the x axis. Derive the momentum four vector of the two photons in the lab frame.

you're over-complicating it :redface:

in the rest frame, α is θ1 (and θ2 - 180°) :wink:

(but what is the relation between the rest frame and the lab frame? :confused:)
 
  • #3
ok so if i didn't sub in the values my vectors would be in the rest frame:

Photon 1 = (E1 sina, E1 cos a , 0 , E1)

Photon 2 = (-E2 sinθ , -E2 cos θ , 0 , E2)

a and θ is what they've asked for in the question, so to get the lab frame i use the lorentz transformation but i then have to show that the cosine of the angle between the two trajectories as a function of a is:

cos θ = β^2-1+^2sin^2a / 1-β^2cos^2a

how do i do this? i tried p'x^2 + p'y^2 + p'z^2 -(E/c)^2 = p^2 but i get complicated answers. Do i equate the momentum vector i found to the initial momentum?
 
  • #4
sorry the angle is meant to be

cos θ = β^2-1+β^2sin^2a / 1-β^2cos^2a

I forgot the β in the first part of the fraction..
 
  • #5
You missed tiny tim's point. In the rest frame, the photons have to have equal and opposite momenta because the total momentum has to be 0. So you should be able to write both four-momenta using one angle and one energy.
 
  • #6
oo ok can u check if my new vectors are correct please?

photon 1= (E1 sina, E1 cos a , 0 , E1)
Photon 2= (-E2 sinθ2 , -E2 cos θ2 , 0 , E2) = (0,E1,0,-E1)

using E1+E2=0 and since they are photons E=p

are these right, especially the signs?
 
  • #7
They don't look right to me. Don't just post your answers. Explain how you came up with them.
 
  • #8
they are my initial vectors that i came up with, but i changed theta 2 into 180 because it is in the rest frame and also expressed E2 in E1 since in the rest frame they are equal but opposite, am i completely off the track here? or am i getting close... :S
 
  • #9
The problem statement gives you a coordinate system to use. So let's say E is the energy of photon 1. As you noted, for a photon, E=p, so you know the magnitude of the photon's momentum. Now tell me, in terms of p and ##\alpha##, what are the x and y components of the photon's three-momentum?
 
  • #10
x component = (psina)^2
y component=(pcosa)^2
 
  • #11
No, that's not correct. Why are you squaring them for one thing? Also, the problem statement says, "the two photons are emitted on the x-y plane, in opposite directions along a line that forms an angle alpha with the the x axis." Try again.
 
  • #12
I was finding the magnitude, sorry got confused,
would it simply be

pcosa for x
-psina for y
 
Last edited:
  • #13
Why the negative sign in the y component?
 
  • #14
I put them both positive first and then i changed it, i think I am thinking in the one dimension so I am seeing the second photon in negative y quadrant, but looking in the x,y,z plane i think it would be positive, the signs r what i am most unsure of, is the rest correct?
 
  • #15
I'm not sure what you mean by the negative y quadrant. There are two quadrants where y<0.

I've attached a picture of what the problem describes. What are the components of the momentum of the upper right photon?
 

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Related to Angles in momentum four vectors

What are momentum four vectors?

Momentum four vectors are used in physics to describe the momentum and energy of a particle in four dimensions, taking into account the relativistic effects of time and space. They are represented as four-component vectors in space-time, with three components representing the momentum in the x, y, and z directions and the fourth component representing the energy.

How are momentum four vectors calculated?

Momentum four vectors are calculated by multiplying the energy (E) of a particle by the four-velocity (u). The four-velocity is a vector that describes the velocity of a particle in four dimensions, taking into account the effects of time dilation and length contraction.

What is the conservation of momentum in relation to four vectors?

The conservation of momentum states that the total momentum of a closed system remains constant over time. In terms of four vectors, this means that the sum of the four vectors before and after a reaction or interaction must be equal, ensuring the conservation of momentum in all four dimensions.

How are angles related to momentum four vectors?

Angles are used to describe the direction of the momentum in three-dimensional space. In the context of momentum four vectors, angles are used to determine the components of the vector in the x, y, and z directions, which can then be used to calculate the total momentum and energy of the particle.

What is the significance of momentum four vectors in particle physics?

Momentum four vectors are essential in the study of particle physics because they allow for the accurate description and analysis of high-energy particles moving at relativistic speeds. They also play a crucial role in understanding the conservation laws of energy and momentum in particle interactions, providing a deeper understanding of the fundamental forces of the universe.

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