Four momentum of fused particles

AI Thread Summary
The discussion focuses on deriving an expression for the new mass of two fused particles after a collision, considering their initial speeds and the angle between their trajectories. The initial attempt to relate energy and momentum does not adequately incorporate the angle α, leading to confusion regarding the conservation laws in special relativity. Participants emphasize the importance of using Lorentz factors and correctly applying momentum conservation equations to solve for the unknowns. It is noted that the angle should only affect the momentum of one particle, and the number of equations should match the number of unknowns for a valid solution. Clarification on definitions and careful application of formulas is advised to resolve the issues presented.
LASmith
Messages
21
Reaction score
0

Homework Statement



two particles of masses m1 and m2 move at speeds u1, u2 respectively collide and fuse. If α is the angle between the two directions of motion before the collision, find an expression for the new mass m, in terms of m1 m2 u1 u2 and α


Homework Equations



Etot2 = p2c2+m2c4


The Attempt at a Solution



(m12 u12 c2 + m1 2 c4)1/2 + (m22 u22 c2 + m22 c4)1/2 = mc2

However this does not take into consideration the angle α, so I am unsure as to where this comes in.
 
Physics news on Phys.org
u_1,u_2 are speeds in the laboratory frame and, since there is an angle between the initial particles, there is net momentum of the system. So the final state particle cannot be at rest in the laboratory frame. You must also consider momentum conservation to solve the problem.
 
fzero said:
u_1,u_2 are speeds in the laboratory frame and, since there is an angle between the initial particles, there is net momentum of the system.

From this can I re-write the RHS of the equation to be = (m2u32c2 + m2c4)1/2

Squaring both sides I obtain

m12(u12+c2) + m22(u22+c2) + 2(m12m22(u12u22+u12c2+u22c2+c4))1/2 = m2(u32+c2)

Is this correct?

fzero said:
You must also consider momentum conservation to solve the problem.

So does momentum hold the same in special relativity as it does in general mechanics?
Therefore I would obtain

m1u1sin(α/2) - m2u2sin(α/2) = mu3sinβ

&

m1u1cos(α/2) + m2u2cos(α/2) = mu3cosβ

However this gives too many unknowns, I have attempted to find either u3 or β in terms of u1, u2 or α but have failed to find a connection, as the mass m is not just simply m1 + m2

Any helps would be much appreciated.
 
LASmith said:
From this can I re-write the RHS of the equation to be = (m2u32c2 + m2c4)1/2

Squaring both sides I obtain

m12(u12+c2) + m22(u22+c2) + 2(m12m22(u12u22+u12c2+u22c2+c4))1/2 = m2(u32+c2)

Is this correct?

You're missing Lorentz factors:

\vec{p} = \gamma m \vec{u}, ~~~\gamma= \frac{1}{\sqrt{1-(u/c)^2}}.

Also we know that E=\gamma mc^2, so we can simplify much of this.


So does momentum hold the same in special relativity as it does in general mechanics?
Therefore I would obtain

m1u1sin(α/2) - m2u2sin(α/2) = mu3sinβ

&

m1u1cos(α/2) + m2u2cos(α/2) = mu3cosβ

However this gives too many unknowns, I have attempted to find either u3 or β in terms of u1, u2 or α but have failed to find a connection, as the mass m is not just simply m1 + m2

Any helps would be much appreciated.

Having \alpha/2 in the formulas looks wrong. You can set the path of one of the incoming particles to be the x-axis, then the angle only appears in the momentum of the other two particles.

You have 3 unknowns (m,u_3,\beta) and 3 equations (E,p_x,p_y), so the unknowns are determined. Just double check the definitions of everything before plugging into formulas, since you do have a lot of mistakes.
 
Thread 'Help with Time-Independent Perturbation Theory "Good" States Proof'
(Disclaimer: this is not a HW question. I am self-studying, and this felt like the type of question I've seen in this forum. If there is somewhere better for me to share this doubt, please let me know and I'll transfer it right away.) I am currently reviewing Chapter 7 of Introduction to QM by Griffiths. I have been stuck for an hour or so trying to understand the last paragraph of this proof (pls check the attached file). It claims that we can express Ψ_{γ}(0) as a linear combination of...

Similar threads

Back
Top