Relativistic momentum calculation

In summary: P_1=-\frac{mv^2}{2}+\frac{mv^4}{4}$$$$P_2=-\frac{mv^2}{2}+\frac{mv^3}{3}$$in summary,In summary, the equation for conservation of momentum states that the momentum of a particle is equal to the sum of the momenta of its constituent particles.
  • #1
Saptarshi Sarkar
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Homework Statement
A particle X of mass M at rest disintegrates into a particle of mass m and another massless particle. Calculate the momentum of m.
Relevant Equations
What I figured:

From conservation of mass-energy
##Mc^2 = \frac {mc^2} {\sqrt {1 - \frac {v^2} {c^2}}}+ pc##

From conservation of momentum
##0 = \frac {mv} {\sqrt {1 - \frac {v^2} {c^2}}} + p##
The answer is required to be in terms of M,m and c only. But, I am not able to calculate the momentum of the m mass particle using the above two. Can anyone help me by telling me what I am missing?
 
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  • #2
Saptarshi Sarkar said:
Homework Statement:: A particle X of mass M at rest disintegrates into a particle of mass m and another massless particle. Calculate the momentum of m.
Homework Equations:: What I figured:

From conservation of mass-energy
##Mc^2 = \frac {mc^2} {\sqrt {1 - \frac {v^2} {c^2}}}+ pc##

From conservation of momentum
##0 = \frac {mv} {\sqrt {1 - \frac {v^2} {c^2}}} + p##

The answer is required to be in terms of M,m and c only. But, I am not able to calculate the momentum of the m mass particle using the above two. Can anyone help me by telling me what I am missing?

Try working with the quantities ##E## and ##p## for all particles involved.
 
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  • #3
Saptarshi Sarkar said:
The answer is required to be in terms of M,m and c only. But, I am not able to calculate the momentum of the m mass particle using the above two. Can anyone help me by telling me what I am missing?

You are missing that you *can* calculate the momentum of m using those equations. Two equations, two unknowns.
 
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  • #4
DEvens said:
You are missing that you *can* calculate the momentum of m using those equations. Two equations, two unknowns.

From the second equation I found the momentum to be equal to -p, but when I try to get the value to -p using the equation for conservation of mass-energy, I am not able to eliminate the v.
 
  • #5
Saptarshi Sarkar said:
From the second equation I found the momentum to be equal to -p, but when I try to get the value to -p using the equation for conservation of mass-energy, I am not able to eliminate the v.
Then don't have the ##v## in your equations in the first place. ##v## and ##\gamma## just get in the way.
 
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  • #6
Saptarshi Sarkar said:
Can anyone help me by telling me what I am missing?

Probably the most important equation in SR is $$E^2 = p^2c^2 + m^2c^4$$
Which holds for both massive and massless particles.
 
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  • #7
For a more systematic approach,try working with conservation of 4 momentum and using the fact that(c=1):

$$P.P=m^2 $$
$$P= P_1 + P_2$$
 
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1. What is relativistic momentum?

Relativistic momentum is the measure of the motion of an object in relation to the speed of light. It takes into account the effects of special relativity, such as time dilation and length contraction, on an object's momentum.

2. How is relativistic momentum calculated?

The formula for relativistic momentum is p = mv/√(1 - v^2/c^2), where p is momentum, m is mass, v is velocity, and c is the speed of light. This formula takes into account the relativistic effects on an object's mass and velocity.

3. What is the difference between classical and relativistic momentum?

Classical momentum is based on Newton's laws of motion and does not take into account the effects of special relativity. Relativistic momentum, on the other hand, considers the effects of high speeds and the speed of light on an object's momentum.

4. Why is relativistic momentum important?

Relativistic momentum is important because it allows us to accurately calculate the momentum of objects moving at high speeds, which is necessary for understanding and predicting the behavior of particles and systems in the universe.

5. Can relativistic momentum be greater than the speed of light?

No, according to the theory of relativity, nothing can travel faster than the speed of light. Therefore, relativistic momentum cannot exceed the speed of light.

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