Solution of basic special relativity problem

  • #1
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Homework Statement


This is more of a solution check than an actual problem, but I didn't see any other fitting thread to post this on. I'm starting to learn special relativity and have received 2 starting questions on the subjects, I solved them to the best of my abilities but have no actual way of checking my answer and I am positive I have made some mistakes. I also need some clarification on the last parts of both questions. Help on this regard would be very much appreciated (keep in mind I am a total self taught beginner and these are my first questions...).

1) 2 identical particles, each particle having a rest mass of ##m_0##, are moving towards each other with equal speeds 0.7c. As a result of the collision a single particle is created.
i) What is the rest mass ##M_0## of the created particle?
ii) How would you explain the difference between ##M_0## and the total mass of the particles that created it ##2m_0##?

2) A photon with an energy of 4 MeV goes by a gold nucleus and turns into an electron-positron pair, who are moving in opposite speeds. Assume that the electron and positron receive equal amounts of energy, and that the energy given to the gold nucleus is negligible.
i) What is the mass of each particle?
ii) What is the speed of each particle?
iii) What is the speed of the positron relative to on observer moving with the electron?
iv) What is the role of the gold nucleus in the question?

Homework Equations


(It is possible I've misinterpreted the equations)
Kinetic energy of moving particle : ##K_E=\frac {mc^2} {\sqrt{1- \frac {v^2} {c^2}}}-mc^2## where ##m## is the rest mass.
Total energy of moving particle: ##E = K_E+mc^2=\frac {mc^2} {\sqrt{1- \frac {v^2} {c^2}}}## where ##m## is the rest mass.
Velocity addition: ##u=\frac {v+u'} {1 + \frac {vu'} {c^2}}## where ##u## is the velocity of C relative to A, ##v## is the velocity of B relative to A and ##u##' is the velocity of B relative to A.
Energy of electron-positron pair production: ##E = hv-2m_0c^2## where ##v## is the frequency of the photon, ##h## is plank's constant and ##m_0## is the rest mass of the electron/positron.
These are the main relevant equations I can think of.

3. The Attempt at a Solution

1) i) The energy of each particle prior to the collision is ##=\frac {m_0c^2} {\sqrt{1- \frac {(0.7c)^2} {c^2}}}=\frac {m_0c^2} {\sqrt{0.51}}##.
Since the resulting particle is at rest, its energy is ##M_0c^2##. Using conservation of energy:
##M_0c^2=\frac {2m_0c^2} {\sqrt{0.51}}##
##M_0=\frac {2m_0} {\sqrt{0.51}}=2.8m_0##
ii) My guess would be that some of the kinetic energy was converted into rest mass energy, but I don't really have any solid explanation to base that on.

2) i) ##4 MeV=4⋅1.6⋅10^{-13} J=6.4⋅10^{-13}J##
##6.4⋅10^{-13}=2mc^2##
##m=3.56⋅10^{-30} kg##
ii) The kinetic energy of the system is the rest mass energy subtracted from the total energy. The rest mass of an electron/positron is 9.1⋅10^{-31}, thus: ##2K_E=6.4⋅10^{-13}-2⋅9.1⋅10^{-31}c^2=4.762⋅10^{-13}J##
The kinetic energy of either particle is thus ##K_E=2.38⋅10^{-13}J##
Using the formula for kinetic energy ##K_E=\frac {mc^2} {\sqrt{1- \frac {v^2} {c^2}}}-mc^2##:
##2.38⋅10^{-13}J=\frac {9.1⋅10^{-31}c^2} {\sqrt{1- \frac {v^2} {c^2}}}-9.1⋅10^{-31}c^2##
##2.38⋅10^{-13}J=\frac {8.19⋅10^{-14}} {\sqrt{1- \frac {v^2} {c^2}}}-8.19⋅10^{-14}##
##3.2⋅10^{-13}=\frac {8.19⋅10^{-14}} {\sqrt{1- \frac {v^2} {c^2}}}##
##\sqrt{1- \frac {v^2} {c^2}}=\frac {8.19⋅10^{-14}} {3.2⋅10^{-13}}=0.256##
##1- \frac {v^2} {c^2} = 0.0655##
##v^2=8.41⋅10^{16}##
##v=2.9⋅10^8 \frac m s##
iii) According to relativistic velocity addition: ##v'=\frac {2⋅2.9⋅10^8} {1+\frac {(2.9⋅10^8)^2} {c^2}}=2.99⋅10^8##
iv) I don't really know, I know it has to do with conservation of momentum, since the electron-positron pair has 0 net momentum but the photon does not, an extra factor must be applied for the process to occur. But I don't know the details.

If anyone would actually take the time to check this tedious post and provide some insight on the last parts of both questions I would be very grateful.
 

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
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Your work looks very good to me. I think you have answered part (iv) pretty well. The gold nucleus must recoil with a small velocity in order to conserve linear momentum. But the kinetic energy of this recoil is very small. Your answer for (iii) does not include a sufficient number of significant figures. The result is closer to 3.00 x 108 m/s than to 2.99 x 108 m/s. Instead of expressing your velocities in terms of m/s, it would be nicer to express them as fractions of c.
 
  • #3
127
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Yeah I see the result in (iii) is ##2.9983⋅10^8##, glad to hear that the actual operations that I did were correct. Is my guess for part (ii) of question 1 also correct? and If so is my explanation sufficient?
 
  • #4
TSny
Homework Helper
Gold Member
13,096
3,416
Yeah I see the result in (iii) is ##2.9983⋅10^8##, glad to hear that the actual operations that I did were correct. Is my guess for part (ii) of question 1 also correct? and If so is my explanation sufficient?
Yes, your work for part (ii) looks correct to me.
 
  • #5
127
12
Ok then, thanks for the help!
 

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