4 momentum mass non-conservation, special relativity

In summary: This could be the case (if you have done some experiment or some theory) but it needs to be stated.In summary, the conversation discusses the collision of a particle with mass M and speed v with a stationary mass m, resulting in two particles with mass µ emerging at angles with respect to the x-axis. The equations for conservation of 4-momenta are given, and it is argued that the angles of exit of the particles must be symmetric about the x-axis. Three independent equations for the velocity u, mass µ, and angle θ of the new particles are obtained, but the third equation is not explicitly stated. In the non-relativistic limit, the expressions can be expanded in the small parameter v/c, with the
  • #1
Konhbri
2
1

Homework Statement


A particle with mass M and speed v along the positive x-axis hits a stationary mass m. Two particles, each with mass µ, emerge from the collision, at angles with respect to the x-axis.

(a) Write the equation for conservation of the 4-momenta, for arbitrary angles θ_1, θ_2 of exit.

(b) Argue that the angles of exit of the ‘after’ particles must be symmetric about the x-axis;
call the angle θ.

(c) Obtain three independent equations for the velocity u, mass µ, and angle θ of the new particles, in terms of the initial quantities.

(d) Check the non-relativistic limit; for example, expand your expressions in the small parameter v/c. Note the constraint that in this case there is very little energy available for the sum of masses after to be greater than the sum of masses before, while in general it is possible for mass to be created.

Homework Equations


(a)
(Gamma for velocity v is denoted as 'G', gamma for velocity u is denoted as 'g')
GMc(1,v/c,0,0)+mc(1,0,0,0)=
gµc(1,(u/c)*cos(θ_1),(u/c)*sin(θ_1),0)+gµc(1,(u/c)cos(θ_2),(u/c)sin(θ_2),0)

therefore

(GMc+mc, GMc*(v/c),0,0)=2(gµc, gµc*(u/c)*cos(θ),0,0)$ (This is shown in part B)

The Attempt at a Solution



(b) The universe is isotropic and the particles are identical therefore if the particles switch places the situation should be identical, therefore their mass and velocity are the same and the angles are opposite in direction

(c) mass: GMc+mc=2gµc
velocity: GMc*(v/c)= 2gµc*cos(θ)
angle:?? I can't figure this third independent equation out

I haven't attempted part (d) as I have yet to complete (c)
 
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  • #2
Konhbri said:
(b) The universe is isotropic and the particles are identical therefore if the particles switch places the situation should be identical, therefore their mass and velocity are the same and the angles are opposite in direction
Opposite in direction relative to what?
Your argument doesn't specify a reference frame but the statement is not true in every reference frame.

For (c) you can consider the momentum in two separate dimensions.
Konhbri said:
mass: GMc+mc=2gµc
That is the energy in the system, not the mass.
 
  • #3
Konhbri said:
(a)
(Gamma for velocity v is denoted as 'G', gamma for velocity u is denoted as 'g')
GMc(1,v/c,0,0)+mc(1,0,0,0)=
gµc(1,(u/c)*cos(θ_1),(u/c)*sin(θ_1),0)+gµc(1,(u/c)cos(θ_2),(u/c)sin(θ_2),0)

therefore

(GMc+mc, GMc*(v/c),0,0)=2(gµc, gµc*(u/c)*cos(θ),0,0)$ (This is shown in part B)

Just to add that you have assumed in a) that the two particles have the same speed. This is equivalent to assuming they have the same angle.(Simply consider the y-momentum.)

I suggest you need some justification that the two particles have the same speed.
 

Related to 4 momentum mass non-conservation, special relativity

1. What is 4-momentum and how does it relate to mass non-conservation?

4-momentum is a concept in special relativity that combines the energy and momentum of a particle into a single four-dimensional vector. In special relativity, mass is not conserved, but rather is a component of the energy-momentum vector. This means that mass can be converted into energy and vice versa, leading to the concept of mass-energy equivalence.

2. How does special relativity explain the non-conservation of mass?

Special relativity explains the non-conservation of mass through the concept of mass-energy equivalence. According to the famous equation E=mc^2, mass and energy are interchangeable and can be converted into one another. This means that in certain situations, mass can be "lost" or converted into other forms of energy, such as kinetic energy or radiation.

3. Can you provide an example of mass non-conservation in special relativity?

One example of mass non-conservation in special relativity is the process of pair production, where a high-energy photon can spontaneously create a particle-antiparticle pair. In this process, the energy of the photon is converted into the mass of the particles, violating the conservation of mass but not the conservation of energy.

4. How does 4-momentum account for the conservation of energy and momentum?

4-momentum accounts for the conservation of energy and momentum by treating them as components of a single vector. In a closed system, the total 4-momentum must remain constant, meaning that any changes in energy or momentum must be balanced by corresponding changes in the other component. This allows for the conservation of energy and momentum to be maintained even in situations where mass is not conserved.

5. What implications does mass non-conservation have for our understanding of the universe?

The concept of mass non-conservation in special relativity has significant implications for our understanding of the universe. It allows for the conversion of mass into energy, which plays a crucial role in many natural phenomena, such as nuclear reactions and the behavior of particles in high-energy environments. It also helps to explain the relationship between matter and energy, and how they are fundamentally connected in our universe.

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