Relativistic collision and invariant s

In summary, the invariant s = (P1+P2)2 can be written as a function of masses and energies in both the center of momentum frame and the lab frame where b is at rest. In the lab frame, the equation becomes 2E'1E'2+m012+m022 = 2E1E2+E12+E22. If s is very large compared to the masses, then the energy E'1 will mainly depend on E'2 and not the masses or momentums. When particle 2 is at rest, the energy E'1 needed to get a certain value of ##\sqrt{s}## will not depend on anything. When the particles collide head-on with equal energy, the energy
  • #1
carllacan
274
3

Homework Statement


Write the invariant s = (P1+P2)2 as a function of masses amd energies of the process

1+2 → 3+4

in the center of momentum frame and on the lab frame, in which b is at rest. Interpret the result.

Homework Equations

The Attempt at a Solution


For the CoM frame I have:
s = (P1+P2)2 = (E1+E2, p1 + p2)2 = (E1+E2)2
(and similar for particles 3 and 4)

In the lab frame (primed symbols are used for variables in this frame)
s' = (P'1+P'2)2 = (E'1+E'2, p'1)2 = (E'1+E'2)2 + |p1|2 = (E'1+E'2)2 + m012-E'12 = 2E'1E'2+m012+m022
(and similar for particles 3 and 4)

Is this right? I still don't feel secure with this problems .Is if is right, can we get any interesting conclusions from the result? We must have s = s', so
2E'1E'2+m012+m022 = 2E1E2+E12+E22

And I get to E'1E'2 = E1E2 + |p1|2. If the energies were being summed I would interpret it as the fact that the energies in the lab frame is the sum of the energies in the CoM frame plus the momentum of the particles, or something like that, but this means nothing to me.
 
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  • #2
carllacan said:
In the lab frame (primed symbols are used for variables in this frame)
s' = (P'1+P'2)2 = (E'1+E'2, p'1)2 = (E'1+E'2)2 + |p1|2 = (E'1+E'2)2 + m012-E'12 = 2E'1E'2+m012+m022
I highlighted a typo (by the way: this forum supports LaTeX).

Before you interpret the result: you know E'2. What happens if you are interested in very large invariant ##\sqrt{s}##? 100 or 1000 times the masses, for example.
Compare this to a collider where b is not at rest.
 
  • #3
mfb said:
I highlighted a typo (by the way: this forum supports LaTeX).

Before you interpret the result: you know E'2. What happens if you are interested in very large invariant ##\sqrt{s}##? 100 or 1000 times the masses, for example.
If s is very large compared to the masses then I guess the energy E'1 will depend mostly on E'2, but not on the masses or the mometums.
Compare this to a collider where b is not at rest.[/QUOTE]
If I repeat the calculus with particle 2 having momentum I get an expression where the momentums and the angle between them matter. Is that what you mean?
 
  • #4
What is E'2 for a particle at rest? This is the first question, everything below relies on that answer.

Can anything depend on that? You can assume that particle masses are fixed.
carllacan said:
If I repeat the calculus with particle 2 having momentum I get an expression where the momentums and the angle between them matter. Is that what you mean?
No. For two particles of equal mass m, what do you need as E'1 to get ##\sqrt{s}=100m## if particle 2 is at rest?
What do you need if they collide head-on with equal energy?
 

1. What is a relativistic collision?

A relativistic collision is a type of collision between particles that involves high speeds close to the speed of light. This type of collision takes into account the effects of special relativity, such as time dilation and length contraction.

2. What is the invariant s in relativistic collisions?

The invariant s in relativistic collisions is a quantity that remains constant in all frames of reference. It is a combination of the energies and momenta of the particles involved in the collision, and is used to calculate the outcomes of the collision.

3. How is the invariant s calculated?

The invariant s is calculated using the formula s^2 = (E1 + E2)^2 - (p1 + p2)^2c^2, where E1 and E2 are the energies of the particles and p1 and p2 are their momenta. The speed of light, c, is also included in the equation.

4. What is the significance of the invariant s in relativistic collisions?

The invariant s is significant because it remains constant in all frames of reference, allowing scientists to make accurate predictions about the outcomes of relativistic collisions. It also helps to understand the relationship between energy and momentum in these high-speed collisions.

5. How does special relativity affect relativistic collisions?

Special relativity affects relativistic collisions by taking into account the effects of high speeds on the measurement of time and distance. It also introduces the concept of mass-energy equivalence, which plays a crucial role in calculating the invariant s in these collisions.

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