Four momentum of fused particles

In summary, the problem involves two particles of masses m1 and m2 moving at speeds u1 and u2 respectively, colliding at an angle α and fusing into a new mass m. The formula for the new mass is given by (m1u1^2c^2 + m1^2c^4)^1/2 + (m2u2^2c^2 + m2^2c^4)^1/2 = mc^2, taking into consideration the momentum conservation of the system. However, Lorentz factors need to be included in the equations.
  • #1
LASmith
21
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.
 
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  • #2
[itex]u_1,u_2[/itex] 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.
 
  • #3
fzero said:
[itex]u_1,u_2[/itex] 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.
 
  • #4
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:

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

Also we know that [itex]E=\gamma mc^2[/itex], 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 [itex]\alpha/2[/itex] 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 ([itex]m,u_3,\beta[/itex]) and 3 equations [itex](E,p_x,p_y)[/itex], so the unknowns are determined. Just double check the definitions of everything before plugging into formulas, since you do have a lot of mistakes.
 
  • #5


I would first clarify the context of this problem. Is it related to nuclear fusion, particle physics, or some other field? This information would be important in determining the appropriate equations and assumptions to use in solving the problem.

Assuming this is related to nuclear fusion, I would approach the problem by first considering conservation of momentum and energy. The total momentum before the collision is equal to the total momentum after the collision, and the total energy before the collision is equal to the total energy after the collision. This can be expressed as:

m1u1 + m2u2 = (m1 + m2)v

and

m1c2 + m2c2 = (m1 + m2)c2 + KE

where v is the final velocity of the fused particle and KE is the kinetic energy of the fused particle.

Next, I would use the Law of Cosines to relate the angle α to the final velocity v:

v2 = u12 + u22 - 2u1u2cosα

Substituting this into the equations for conservation of momentum and energy, we can solve for the final mass m:

m = (m1m2u1u2cosα + m1m2c2) / (c2 - u1u2cosα)

This expression takes into consideration the angle α and the masses and velocities of the two particles before the collision. However, it is important to note that this is a simplified model and may not accurately reflect the complex processes involved in nuclear fusion. Further research and experimentation would be needed to fully understand the behavior of fused particles.
 

FAQ: Four momentum of fused particles

1. What is the four momentum of fused particles?

The four momentum of fused particles is a physical quantity that describes the total energy and momentum of a system of fused particles. It takes into account both the spatial momentum and the energy of the particles.

2. How is the four momentum of fused particles calculated?

The four momentum of fused particles is calculated by adding together the four-momenta of each individual particle in the system. The four-momentum of a particle is given by its energy multiplied by a four-vector with components (1, px, py, pz), where px, py, and pz are the components of the particle's momentum in the x, y, and z directions, respectively.

3. Why is the four momentum of fused particles important?

The four momentum of fused particles is important because it is a conserved quantity in all interactions between particles. This means that the total four momentum before and after a fusion event must be equal. It is also a useful tool for understanding the dynamics of particle interactions and making predictions about their behavior.

4. How does the four momentum of fused particles relate to special relativity?

The four momentum of fused particles is a concept that arises from special relativity. In this theory, the energy and momentum of a particle are not conserved separately, but rather as a combined quantity known as four momentum. This is because energy and momentum are related by the speed of light in special relativity, and the four momentum takes this into account by combining them into a single four-vector.

5. Can the four momentum of fused particles be observed experimentally?

Yes, the four momentum of fused particles can be observed experimentally in high energy physics experiments. By measuring the energies and momenta of the particles involved in a fusion event, scientists can calculate the total four momentum of the system. This allows for the validation of theoretical predictions and the study of fundamental particles and their interactions.

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