Calculate relativistic com frame for two particles?

In summary, the final state is determined randomly by the matrix element for the process that we're dealing with.
  • #1
jason12345
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Does anyone know of a standard way of calculating the com frame velocity for two particles moving at arbitary velocities in the lab frame?

It's strange that this standard result isn't even in Goldstein's et al book
 
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  • #2
E2 - p2 = m2

βγ = p/m.
 
  • #3
The velocity of a single particle in terms of its energy and momentum is given by

$$\beta = \frac {pc}{E}$$

Given this, what would you expect the velocity of the "equivalent particle" representing the motion of a system of two particles to be? Or indeed, any number of particles?

You can get the result a bit more rigorously by using the Lorentz transformation for energy and momentum

$$p^{\prime} c = \gamma (pc - \beta E)$$

and requiring that the total momentum in the primed frame be zero, i.e. for a system of two particles ##p_1^{\prime} c + p_2^{\prime} c = 0##.
 
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  • #4
Calulate [tex]{\bf P=p_1+p_2)}[/tex], and [tex]E=E_1+E_2[/tex].
Then [tex]{\bf V=P}/E.[/tex].
 
  • #5
@itbell and Meir Achz, yes I see what you mean!

The key for me was visualising the disintegration of a particle into p1 and p2, while using conserving 4-momentum.

This really is a very elegant, beautiful result, which I can't find anywhere in my copy of Goldstein, 3rd edition, nor in any of the problems. Maybe it's mentioned in books devoted to the dynamics of particle collisions.
 
  • #6
Meir Achuz said:
Calulate [tex]{\bf P=p_1+p_2)}[/tex], and [tex]E=E_1+E_2[/tex].
Then [tex]{\bf V=P}/E.[/tex].

For c=1, but in general [tex]{\bf V=P}c^2/E.[/tex]

yes?
 
  • #7
jason12345 said:
The key for me was visualising the disintegration of a particle into p1 and p2, while using conserving 4-momentum.

This really is a very elegant, beautiful result, which I can't find anywhere in my copy of Goldstein, 3rd edition, nor in any of the problems. Maybe it's mentioned in books devoted to the dynamics of particle collisions.
Much of the formalism, including the concept of invariant mass during particle disintegration, is included here:
http://pdg.lbl.gov/2010/reviews/rpp2010-rev-kinematics.pdf
 
  • #8
jason12345 said:
For c=1, but in general [tex]{\bf V=P}c^2/E.[/tex]

yes?
I use light years.
 
  • #9
So in Fig 39.3 it shows the constraints of the final state.

What determines in what final state the system of equations will stabilize?
 
  • #10
The final state is located randomly somewhere inside the shaded area, with a probability distribution that is determined by the matrix element for the process that we're dealing with.

It's rather like asking "what determines exactly when a particular radioactive nucleus will decay?"
 
  • #11
So this is where deterministic measurement meets the road to philosophical physics.

This is where "squeezing" occurs?
 

FAQ: Calculate relativistic com frame for two particles?

1. What is a relativistic com frame?

A relativistic com frame is a coordinate system in which the total momentum of two particles is zero. This frame is used to simplify calculations involving the motion of two particles at high speeds, where relativistic effects cannot be ignored.

2. How do you calculate the relativistic com frame for two particles?

The relativistic com frame can be calculated using the following equation:
vcom = (m1v1 + m2v2) / (m1 + m2)
where m1 and m2 are the masses of the two particles, and v1 and v2 are their respective velocities.

3. Why is it important to calculate the relativistic com frame?

Calculating the relativistic com frame is important because it allows us to simplify complex calculations involving the motion of two particles at high speeds. It also helps us better understand the effects of relativity on the motion of particles.

4. Can the relativistic com frame be used for more than two particles?

Yes, the relativistic com frame can be used for any number of particles. The equation for calculating the com frame remains the same, but the masses and velocities of all particles must be taken into account.

5. How does the relativistic com frame differ from the center of mass frame?

The relativistic com frame and the center of mass frame are similar in that both involve a coordinate system where the total momentum of a system is zero. However, the relativistic com frame takes into account the effects of relativity, while the center of mass frame does not. The relativistic com frame is also used for particles moving at high speeds, while the center of mass frame is used for particles at low speeds.

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