Special Relativity and Particle Decay

In summary, we are given the rest masses of a particle and its decay products and asked to calculate the total energy, momentum, and speed of the particles produced in the decay. Using the equations E^2=p^2c^2 + E_0^2 and E=\gamma m_0c^2, we find that each particle has a total energy of 300 MeV, a momentum of 259.8 MeV/c, and a speed of √3/2 times the speed of light.
  • #1
TheTallOne
1
0

Homework Statement


A particle with rest mass [tex]M_0[/tex] can decay at rest into a pair of particles each with rest
mass [tex]m_0[/tex]. Calculate the following in the rest inertial frame of the original particle
using [tex]M_0c^2=600[/tex]MeV and [tex]m_0c^2=150[/tex] MeV .
(i) The total energy of each particle produced in the decay.
(ii) The magnitude of the relativistic linear momentum of each particle produced in
the decay.
(iii) The speed of the particles produced in the decay.

Homework Equations


[tex]E^2=p^2c^2 + E_0^2[/tex] [1]
[tex]E=\gamma m_0c^2[/tex] [2]

The Attempt at a Solution


i)Ok, first I'm assuming that the new particles will have equal energy, so each E=300MeV
ii) I think I'm meant to use equation [1], where E=300 and E0=150. This gives [tex]p=\frac{259.8}{c}=0.866[/tex]. This is the part I'm unsure on (and possibly the next one). If the method is correct, what are the units?
iii) I use the fact that [tex]E=\gamma m_0c^2[/tex], giving [tex]\gamma=2 \Rightarrow v=\frac{\sqrt{3}}{2}c[/tex]

Thanks very much for helping!
 
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  • #2
TheTallOne said:
ii) I think I'm meant to use equation [1], where E=300 and E0=150. This gives [tex]p=\frac{259.8}{c}=0.866[/tex]. This is the part I'm unsure on (and possibly the next one). If the method is correct, what are the units?

That's correct. You can work out the units: E^2-E0^2 has units of MeV^2, so p has the unit MeV*s/m. To get the result in the familiar kg*m/s, just convert MeV to J before doing the calculation.
 
  • #3
TheTallOne said:

Homework Statement


A particle with rest mass [tex]M_0[/tex] can decay at rest into a pair of particles each with rest
mass [tex]m_0[/tex]. Calculate the following in the rest inertial frame of the original particle
using [tex]M_0c^2=600[/tex]MeV and [tex]m_0c^2=150[/tex] MeV .
(i) The total energy of each particle produced in the decay.
(ii) The magnitude of the relativistic linear momentum of each particle produced in
the decay.
(iii) The speed of the particles produced in the decay.

Homework Equations


[tex]E^2=p^2c^2 + E_0^2[/tex] [1]
[tex]E=\gamma m_0c^2[/tex] [2]

The Attempt at a Solution


i)Ok, first I'm assuming that the new particles will have equal energy, so each E=300MeV.
Good. You're not really assuming the answer though; you're deducing it from symmetry.
ii) I think I'm meant to use equation [1], where E=300 and E0=150. This gives [tex]p=\frac{259.8}{c}=0.866[/tex]. This is the part I'm unsure on (and possibly the next one). If the method is correct, what are the units?
The method is correct. Just write the units in explicitly, and you'll see how they turn out.

[tex](pc)^2 = E^2 - E_0^2 \rightarrow pc = \sqrt{(300~\textrm{MeV})^2 - (150~\textrm{MeV})^2} = 259.8~\textrm{MeV}[/tex]

So your answer is in units of MeV/c.
iii) I use the fact that [tex]E=\gamma m_0c^2[/tex], giving [tex]\gamma=2 \Rightarrow v=\frac{\sqrt{3}}{2}c[/tex]

Thanks very much for helping!
Since you have the momentum, you can also find the velocity using β=v/c=pc/E. Either method is fine, though.
 

FAQ: Special Relativity and Particle Decay

What is special relativity?

Special relativity is a theory developed by Albert Einstein that describes the behavior of objects moving at high speeds. It is based on the principle that the laws of physics are the same for all observers, regardless of their relative motion.

How does special relativity affect particle decay?

Special relativity predicts that particles moving at high speeds will experience time dilation, which means that time will pass more slowly for them compared to an observer at rest. This can affect the rate at which particles decay, as measured by an observer, and can also lead to the creation of new particles from the energy released during decay.

What is the difference between special relativity and general relativity?

Special relativity deals with the behavior of objects in inertial frames of reference, while general relativity extends this to non-inertial frames and includes the effects of gravity. Both theories are based on the principle of relativity and have been extensively tested and confirmed by experiments.

Can special relativity explain the behavior of particles at the quantum level?

Special relativity is a classical theory and does not fully account for the behavior of particles at the quantum level. However, it has been successfully incorporated into quantum field theory, which is a more comprehensive framework for understanding the behavior of particles at the subatomic level.

How does special relativity impact our understanding of the universe?

Special relativity has revolutionized our understanding of the universe by providing a more accurate and comprehensive understanding of the behavior of objects at high speeds. It has also led to the development of other important theories, such as general relativity and quantum mechanics, which have further expanded our understanding of the universe.

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