E-P invariant on relativistic mecahnics problem

In summary, to find the speed of the new particle after the collision, you need to use the conservation of energy and momentum equations. This will give you the correct answer of v=\frac{1}{3}c.
  • #1
Froskoy
27
0

Homework Statement


A particle of mass [itex]m[/itex] moving at speed [itex]\frac{3}{5}c[/itex] collides with an identical particle at rest, and forms a new particle of mass [itex]M[/itex] which moves off at speed [itex]v[/itex]. Find v.

Homework Equations


E-P invariant: [itex]E_1^2-p_1c^2=E_2^2-p_2^2c^2=\mathrm{const.}[/itex]
Momentum: [itex]p^2=m^2c^2\left({\gamma_v^2-1}\right)[/itex]
Energy: [itex]E=\sum_i\gamma_im_ic^2[/itex]

The Attempt at a Solution


There is a single particle after the collision, so [itex]E_2^2-p_2^2c^2=M^2c^4[/itex]
where [itex]E_2=\gamma_vMc^2[/itex]
and [itex]p_2^2=M^2c^2\left({\gamma_v^2-1}\right)[/itex]

so

[tex]
\gamma_v^2M^2c^4-M^2c^4\left({\gamma_v^2-1}\right)^2=M^2c^4\\
\gamma_v^2-\left({\gamma_v^4-2\gamma_v^2+1}\right)=1\\
\left({\gamma_v^2-1}\right)\left({\gamma_v^2-2}\right)=0
[/tex]

Reject the [itex]\gamma_v^2=1[/itex] solution, since this would mean [itex]v=0[/itex].

The [itex]\gamma_v^2=2[/itex] solution gives [itex]v=\frac{\sqrt{2}}{2}c[/itex], which is incorrect - the answer is [itex]v=\frac{1}{3}c[/itex].

I expect there's something fundamentally wrong with my method, but not sure what? Thank you!
 
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  • #2
Froskoy said:

Homework Statement


A particle of mass [itex]m[/itex] moving at speed [itex]\frac{3}{5}c[/itex] collides with an identical particle at rest, and forms a new particle of mass [itex]M[/itex] which moves off at speed [itex]v[/itex]. Find v.


Homework Equations


E-P invariant: [itex]E_1^2-p_1c^2=E_2^2-p_2^2c^2=\mathrm{const.}[/itex]
Momentum: [itex]p^2=m^2c^2\left({\gamma_v^2-1}\right)[/itex]
Energy: [itex]E=\sum_i\gamma_im_ic^2[/itex]


The Attempt at a Solution


There is a single particle after the collision, so [itex]E_2^2-p_2^2c^2=M^2c^4[/itex]
where [itex]E_2=\gamma_vMc^2[/itex]
and [itex]p_2^2=M^2c^2\left({\gamma_v^2-1}\right)[/itex]

so

[tex]
\gamma_v^2M^2c^4-M^2c^4\left({\gamma_v^2-1}\right)^2=M^2c^4\\
\gamma_v^2-\left({\gamma_v^4-2\gamma_v^2+1}\right)=1\\
\left({\gamma_v^2-1}\right)\left({\gamma_v^2-2}\right)=0
[/tex]
You shouldn't have squared the factor ##(\gamma_v^2-1)## in the first line. With this approach, all you'll end up with is M=M, so it's not helpful.
Reject the [itex]\gamma_v^2=1[/itex] solution, since this would mean [itex]v=0[/itex].

The [itex]\gamma_v^2=2[/itex] solution gives [itex]v=\frac{\sqrt{2}}{2}c[/itex], which is incorrect - the answer is [itex]v=\frac{1}{3}c[/itex].

I expect there's something fundamentally wrong with my method, but not sure what? Thank you!
You haven't used any information from before the collision. Use conservation of energy and momentum.
 

Related to E-P invariant on relativistic mecahnics problem

1. What is the E-P invariant in relativistic mechanics?

The E-P invariant, also known as the energy-momentum invariant, is a fundamental concept in relativistic mechanics that states that the total energy and momentum of a system remains constant regardless of the reference frame in which it is observed. It is derived from Einstein's famous equation, E=mc^2, and plays a crucial role in understanding the behavior of particles at high speeds.

2. How is the E-P invariant related to the conservation of energy and momentum?

The E-P invariant is closely related to the conservation of energy and momentum, as it states that the total energy and momentum of a system remain constant in all reference frames. This means that even though energy and momentum may appear to change in different reference frames, the total amount remains the same. This is a key principle in understanding the behavior of particles in relativistic mechanics.

3. Can the E-P invariant be violated?

No, the E-P invariant is a fundamental law of nature and cannot be violated. It is a consequence of the laws of conservation of energy and momentum and has been experimentally verified in numerous experiments. Any violation of the E-P invariant would require a fundamental change in our understanding of the laws of physics.

4. How is the E-P invariant used in practical applications?

The E-P invariant is used in various practical applications, such as in particle accelerators and nuclear reactors. It helps scientists and engineers predict the behavior of particles and systems at high speeds and energies, and is crucial in designing and operating these complex systems. It is also used in the field of astrophysics to understand the behavior of particles in extreme environments such as black holes and supernovae.

5. Are there any limitations to the E-P invariant?

The E-P invariant has been extensively tested and has been shown to hold true in all known physical phenomena. However, it does have limitations when applied to extreme conditions such as near the speed of light or in highly curved spacetimes. In these cases, more advanced theories, such as general relativity, are needed to fully understand the behavior of particles and systems.

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