Momentum in spec relativity

In summary, to solve the problem of a particle at rest decaying into two smaller particles with masses m1 and m2, you need to find their energies and 4-momenta while conserving relativistic momentum. This can be done by using the equation for Mc and substituting m2=0, then solving for E2. The energy equation reduces to E=pc for a massless particle, and p is the same for both particles in this scenario.
  • #1
thenewbosco
187
0
here is my question and my solution, id like to see where i have gone wrong and how to rectify it:

A particle of mass M, at rest, decays into two smaller particles, masses m1 and m2. What are their energies and 4-momenta (given: [tex] \sqrt{p^2 + m_1^2c^2}+\sqrt{p^2 + m_2^2c^2}=Mc[/tex] [tex]E_1 = \sqrt {m_1^1c^4 + p^2c^2} [/tex] [tex] E_2=\sqrt{m_2^2c^4+p^2c^2}[/tex]? then the part i am working on: solve the problem again for m2=0. solve the equations for p and e1, and take the limit m1 -> 0.

i find it kind of ambiguous, but to clarify, is the correct answer obtained by using the equation for Mc and subbing m2=0 in there, then using the two equations you then have to solve for E2?
 
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  • #2
Relativistic momentum has to be conserved. The decaying particle has no momentum, so the two decay products must have equal and opposite momenta. Since the masses are assumed known, you just need to find their respective momenta subject to conservation of energy (your first equation).

For a massless particle, the energy equation reduces to E = pc
 
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  • #3
i can just put into the first equation that m2=0 correct?
 
  • #4
thenewbosco said:
i can just put into the first equation that m2=0 correct?
Yes. I can't post just yes, so p is the same for both, which I think you already know.
 

1. What is momentum in special relativity?

Momentum in special relativity is a measure of the quantity of motion of an object, taking into account its mass, velocity, and direction. It is a fundamental concept in physics and is closely related to the concept of energy.

2. How is momentum different in special relativity compared to classical mechanics?

In special relativity, momentum is described by the relativistic momentum equation, which takes into account the effects of time dilation and length contraction at high speeds. This differs from classical mechanics, where momentum is described by the classical momentum equation, which does not account for these effects.

3. How does momentum relate to the theory of relativity?

The theory of relativity, specifically special relativity, is based on the principle that the laws of physics are the same in all inertial reference frames. This includes the law of conservation of momentum, which states that the total momentum of a closed system remains constant over time.

4. Can momentum be transferred between particles in special relativity?

Yes, momentum can be transferred between particles in special relativity through interactions such as collisions or radiation. However, the relativistic momentum equation must be used to accurately calculate the transfer of momentum at high speeds.

5. How is momentum used in practical applications of special relativity?

Momentum plays a crucial role in many practical applications of special relativity, such as in particle accelerators, nuclear reactors, and spacecraft propulsion systems. It is also used in the study of astrophysics and cosmology, where the high speeds and energies of celestial objects require the use of special relativity.

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