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Spiral.. Physics
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Modern Physics
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The Special Theory of Relativity:
Dynamics
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Copyright © 2003
Paul D’Alessandris
Spiral Physics
Rochester, NY 14623
All rights reserved. No part of this book may be reproduced or transmitted in any
form or by any means, electronic or mechanical, including photocopying, recording,
or any information storage and retrieval system without permission in writing from
the author.
This project was supported, in part, by the National Science Foundation. Opinions expressed are those of
the author and not necessarily those of the Foundation.
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The Special Theory of Relativity:
Dynamics
Relativistic Momentum, Force and Energy
Once Einstein revolutionized our understanding of space and time, physicists were faced with
a monumental task. All of physics, before Einstein, was based on the idea of absolute space
and time. Once these concepts were found to be erroneous, all of classical physics had to be
re-examined in this light. In this section, we will “re-examine” our understanding of
momentum, force, and energy.
Momentum
In classical physics, momentum is defined as
p  mv
However, using this definition of momentum results in a quantity that is not conserved in all
frames of reference during collisions. However, if momentum is re-defined as
p  mv
it is conserved during particle collisions. Therefore, experimentally, momentum is defined by
the above equation.
Force
Once nature tells us the proper formula to use for calculating momentum, mathematics tells us
how to measure force and energy. Force is defined as the time derivative of momentum
F
dp
dt
(In classical physics, where p = mv, this reduces to F = ma.) Substituting the correct
relationship for momentum yields
d (mv)
dt
d
dv
F  ( )mv  m( )
dt
dt
2
d
v
F  mv [(1  2 ) 1 / 2 ]  ma
dt
c
F
3
Use the chain rule to evaluate the derivative of ,
1
v2
2v dv
F  mv[ (1  2 ) 3 / 2 ( 2 ) ]  ma
2
c
c dt
2
v
v
F  mv[ 2 (1  2 ) 3 / 2 a]  ma
c
c
Factor out the common factor ma,
v2
2
F  ma[ c 2  1]
v
1 2
c
Find a common denominator and simplify,
v2
v2

1

2
c2 ]
F  ma[ c
v2
1 2
c
1
F  ma[
]
v2
1 2
c
3
F   ma
This is the relativistically correct form of Newton’s Second Law.
Energy
The kinetic energy of an object is defined to be the work done on the object in accelerating it
from rest to speed v.
v
KE   Fdx
0
Using our result for force derived above yields
v
KE    3 madx
0
4
The variable of integration is x, yet the integrand is expressed in terms of a and v (v is hidden
inside ). To solve this problem,
v
KE    3 m(
dv
)dx
dt
v
dx
)dv
dt
0
KE    3 m(
0
v
KE    3 mvdv
0
v
mvdv
v2
0
(1  2 ) 3 / 2
c
KE  
This integral can be done by a simple u-substitution,
v2
c2
2v
du   2 dv
c
u  1
v
KE  
KE
KE
KE
KE
mc 2 du
2 0 u 3 / 2
mc 2

[2u 1 / 2 ]
2
1 v
 mc 2 [
]0
v2
1 2
c
1
 mc 2 [
 1]
v2
1 2
c
2
 mc  mc 2
Rearranging this yields,
mc 2  KE  mc 2
ETotal  KE  E Rest
Einstein identified the term mc2 as the total energy of the particle. Thus, the total energy is
the sum of the kinetic energy and a completely new form of energy, the rest energy. Particles
have rest energy just by virtue of having mass. In fact, mass is simply a form of energy.
5
Momentum and Energy
An electron is accelerated through a potential difference of 80 kilovolts. Find
the kinetic energy, total energy, momentum and velocity of the electron.
The following collection of equations express the relationships between momentum, energy,
and velocity in special relativity. (Momentum is often easier expressed as “pc” rather than “p”
as you will see once you begin working problems.)
p  mv
v
pc  mc2 ( )
c
Etotal  mc2
Etotal  KE  mc2
KE  (  1)mc2
Etotal  ( pc) 2  (mc2 ) 2
2
The last equation is particularly useful in that it allows a direct relationship between energy
and momentum without the need to calculate the velocity. The proof of this relationship is left
as an exercise.
From electrodynamics, the kinetic energy of a charge accelerated through a potential
difference V is simply the product of the charge and the potential difference,
KE  qV
Rather than substituting the numerical value of the charge on an electron (-e = -1.6 x 10-19 C)
into this expression (and obtaining the kinetic energy in joules), we will leave “e” in the
equation and use “eV” as a unit of energy. Note 1.0 eV = 1.6 x 10-19 J.
KE  qV
KE  e(80 x10 3 V )
KE  80keV
Thus, the kinetic energy of the electron is 80 keV.
The total energy of the electron is then
Etotal  KE  mc 2
Etotal  80keV  511keV
Etotal  591keV
6
The momentum is
Etotal  ( pc) 2  (mc 2 ) 2
2
pc  Etotal  (mc 2 ) 2
2
pc  5912  (511) 2
pc  297 keV
(Again, momentum is often easier expressed as “pc” rather than “p”)
The speed of the electron is
v
pc  mc 2 ( )
c
v
297  591( )
c
v  0.503c
Collisions and Decays I
A neutral pion (rest energy 135 MeV) moving at 0.7c decays into a pair of
photons. The photons each travel at the same angle from the initial pion
velocity. Find this angle and the energy of each photon.
Any process that occurs in nature must obey energy and momentum conservation. To analyze
this particle decay, apply both conservation laws to the process.





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First, find the Lorentz factor for the pion.
 
 
1
v2
1 2
c
1
1
( 0.7 c ) 2
c2
  1.4
Applying energy conservation yields:
Ebeforedecay  E afterdecay
E pion  E photon1  E photon2
m pionc 2  2 E photon
1.4(135)  2 E photon
E photon  94.5MeV
The two photons must have the same energy since they travel in the same direction relative to
the initial pion velocity. This is the only way that momentum in this perpendicular direction
can be conserved.
Applying momentum conservation (actually conservation of “pc”) along the initial direction
of travel and using the relationship
pc  Etotal  (mc 2 ) 2 yields:
2
pcbeforedecay  pc afterdecay
p pionc  p photon1c(cos  )  p photon2 c(cos  )
p pionc  2 p photonc(cos  )
E pion  (mc 2 ) 2  2( E photon  (mc 2 ) 2 )(cos  )
2
2
(1.4(135)) 2  (135) 2  2( 94.5 2  (0) 2 )(cos  )
132.3  189(cos  )
  45.6 o
The photons each travel at 45.60 from the direction of the pions initial path.
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Collisions and Decays II
A photon of energy 500 keV scatters from an electron at rest. The photon is
redirected to an angle of 350 from its initial direction of travel. Find the energy
of the scattered photon and the angle and energy of the scattered electron.
’

35°

e-
To analyze, apply energy conservation:
E photon  Eelectron  E ' photon  E 'electron
500  511  E ' photon  E 'electron
1011  E ' photon  E 'electron
note that the electron initially has only rest energy.
pc  Etotal  (mc 2 ) 2 ):
2
Apply x-momentum conservation (and use
pc  pe c  p' c cos 35  pe c cos 
500 2  (0) 2  0  E ' photon  (0) 2 cos 35  E ' electron  (511) 2 cos 
2
2
500  E ' photon cos 35  E ' electron  (511) 2 cos 
2
Apply y-momentum conservation:
pc  pe c  p' c sin 35  pe c sin 
0  0  E ' photon  (0) 2 sin 35  E ' electron  (511) 2 sin 
2
2
E ' photon sin 35  E ' electron  (511) 2 sin 
2
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This yields three equations with the requested three unknowns (E’photon, E’electron, and ).
If you enjoy algebra, solve this system of equations by hand. If you have better things to do
with your life, use a solver to find:
E ' photon  425keV
E ' electron  586keV
  58.1
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The Special Theory of Relativity:
Dynamics
Activities
11
Six particles (with rest energy E0) are detected in the collision “debris” at a particle accelerator. Each
particle’s total energy (E) is measured.
A
B
C
D
E
F
E0 (MeV)
100
200
100
400
200
800
E (MeV)
200
300
400
500
400
1000
a. Rank these particles on the basis of their mass.
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking cannot be determined based on the information provided.
Explain the reason for your ranking:
b. Rank these particles on the basis of their speed.
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking cannot be determined based on the information provided.
Explain the reason for your ranking:
c. Rank these particles on the basis of their kinetic energy.
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking cannot be determined based on the information provided.
Explain the reason for your ranking:
12
The following six particles (with the rest energy E0 and kinetic energy K indicated) are collided with their
antimatter partners traveling in the opposite direction with the same kinetic energy. The resulting matterantimatter annihilation produces a pair of photons (traveling in opposite directions).
A
B
C
D
E
F
E0 (MeV)
100
200
100
400
200
800
K (MeV)
200
100
400
200
400
100
a. Rank these particles on the basis of their momentum.
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking cannot be determined based on the information provided.
Explain the reason for your ranking:
a. Rank these particles on the basis of energy of the photons they create.
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking cannot be determined based on the information provided.
Explain the reason for your ranking:
b. Rank these particles on the basis of speed of the photons they create.
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking cannot be determined based on the information provided.
Explain the reason for your ranking:
13
An electron is accelerated through a potential difference of 10 million volts. Find its kinetic energy,
momentum and velocity. Compare its velocity to that predicted by classical physics.
Mathematical Analysis
14
A proton is accelerated through a potential difference of 10 million volts. Find its kinetic energy,
momentum and velocity. Compare its velocity to that predicted by classical physics.
Mathematical Analysis
15
a.
b.
c.
Calculate the kinetic energy and momentum of a neutral pion ( 0) traveling at 0.6c.
Calculate the velocity and momentum of a neutral pion (0) with kinetic energy 200 MeV.
Calculate the velocity and kinetic energy of a neutral pion (0) with momentum 200 MeV/c.
Mathematical Analysis
16
a.
b.
c.
Calculate the kinetic energy and momentum of a psi-meson () traveling at 0.2c.
Calculate the velocity and momentum of a psi-meson () with kinetic energy 200 MeV.
Calculate the velocity and kinetic energy of a psi-meson () with momentum 200 MeV/c.
Mathematical Analysis
17
Electrons are accelerated to high speed in two stages. The first stage accelerates the electrons from rest to
0.990c. The second stage accelerates them from 0.990c to 0.999c.
a. Find the energy needed for each stage.
b. If the energy needed for the second stage is again applied to the electron, what would be its final
speed?
c. If protons were being accelerated, find the energy needed for the first two stages.
Mathematical Analysis
18
Electrical energy can be sold for approximately 10 cents per kilowatt hour. If there was a way to convert
mass energy directly into electrical energy, how much would my corpse be worth?
Mathematical Analysis
19
To compete with e-mail, the U.S. Post Office has introduced Super Express mail, where a letter is sent to its
destination at a speed of 0.999c using a special Letter Accelerator. Assume that a typical letter has a mass
of about 25 g. What should be the minimum cost of a stamp, assuming the post office only charges for the
energy used to accelerate the letter? Assume the post office powers the accelerator using electricity
purchased at 10 cents per kilowatt hour.
Mathematical Analysis
20
Based on the total power output of the sun, calculate the approximate decrease in mass of the sun per year.
In a one billion year period, by approximately what percentage does the sun’s mass decrease?
Mathematical Analysis
21
Imagine a process in which a portion of an object’s rest energy is directly converted into kinetic energy.
a. For an object initially at rest, find the speed of the object if one-half of its rest energy is directly
converted into kinetic energy. Ignore momentum conservation for this calculation.
b. For an object initially at rest, find the percentage of its mass that must be directly converted into
kinetic energy in order to achieve a speed of 0.99c. Ignore momentum conservation for this calculation.
Mathematical Analysis
22
In classical physics, KE = 1/2 mv2. Below what velocity is the classical expression accurate to within 5%?
Mathematical Analysis
23
For a particle of mass m traveling at very close to the speed of light, E ~ pc. Above what velocity is the
approximation E = pc for a material particle accurate to within 5%?
Mathematical Analysis
24
Prove that the relationship
Etotal  ( pc) 2  (mc2 ) 2
2
is valid.
Mathematical Analysis
25
Imagine a head-on elastic collision in the laboratory between an object of mass 3 kg traveling at 0.8c and
an object of mass 4 kg traveling in the other direction at 0.6c. The objects will rebound with exactly the
same speeds, clearly conserving classical momentum in this frame. However, consider the same collision in
a frame initially moving to the right at 0.8c:
Moving Frame
Laboratory Frame
before:
3 kg
before:
0.8c
0.6c
4 kg
after:
0.8c
_______ 4 kg
_
3 kg
after:
3 kg
4 kg
0.6c
_______
a.
3 kg
4 kg
_______
__
Using the velocity addition formula, calculate the velocities of the two objects in the frame of reference
moving toward the right at 0.8c.
b. Using the classical formula for momentum, calculate the momentum before the collision in the moving
frame.
c. Using the classical formula for momentum, calculate the momentum after the collision in the moving
frame.
The different values in (b) and (c) indicate that the classical formula for momentum does not result in
momentum conservation being valid in all frames of reference. We must either abandon momentum
conservation or abandon the classical formula.
Mathematical Analysis
26
A constant force of 1 N acts on an object of mass 1 kg. According to classical physics, how long will it take
the initially stationary object to accelerate to a speed of:
a. 1000 m/s?
b. 0.5 c?
c. c?
d. Using the relativistically correct form of Newton’s Second Law, determine the actual time needed to
reach each of these speeds. (Hint: Express acceleration as a = dv/dt, separate the velocity and time
variables, and integrate the resulting expression.)
Mathematical Analysis
27
A neutral pion with kinetic energy 1.0 GeV decays into a pair of photons.
0 =>  + 
Both photons travel parallel to the initial pion velocity. Find each photons energy.
Mathematical Analysis
28
A neutral pion with kinetic energy 1.0 GeV decays into a pair of photons.
0 =>  + 
Both photons travel at the same angle from the initial pion velocity. Find this angle.
Mathematical Analysis
29
A stationary charged pion decays into a muon and a neutrino.
- => - + 
The mass of a neutrino is so small it is essentially zero. Find the speed of the emitted muon.
Mathematical Analysis
30
A charged kaon with kinetic energy 800 MeV decays into a muon and a neutrino.
- => - + 
The mass of a neutrino is so small it is essentially zero. Both particles travel parallel to the initial kaon
velocity. Find the possible energies of the emitted muon.
Mathematical Analysis
31
A positron of kinetic energy 2.5 MeV annihilates with an electron at rest, creating two photons.
e+ + e- =>  + 
0
One photon emerges at 90 to the initial positron direction. What is the direction of the other photon?
Mathematical Analysis
32
A positron of kinetic energy 20 MeV annihilates with an electron at rest, creating two photons.
e+ + e- =>  + 
0
One photon emerges at 50 to the initial positron direction. What is the direction of the other photon?
Mathematical Analysis
33
An electron-positron pair can be produced by a gamma ray striking a stationary electron,
 + e- => e- + e- + e+.
If the total energy is divided equally among the three end products, what is the initial gamma ray energy?
Assume the electrons and positron all travel parallel to the initial photon velocity.
Mathematical Analysis
34
A neutral kaon moving at 0.7c decays into a pair of charged pions.
0 =>  + 
Both pions travel parallel to the initial kaon velocity. What is the energy of each pion?
Mathematical Analysis
35
A neutral kaon moving at 0.7c decays into a pair of charged pions.
0 =>  + 
The pions each travel at the same angle from the initial kaon velocity. Find this angle and the energy of
each pion.
Mathematical Analysis
36
A neutral psi-meson with kinetic energy 2.0 GeV decays into an electron and positron.
0 => e + e
The electron and positron each travel at the same angle from the initial psi-meson velocity. Find this angle.
Mathematical Analysis
37
An omega-minus with kinetic energy 5.0 GeV decays into a neutral lambda and charged kaon.
- =>  + 
Both particles travel parallel to the initial omega-minus velocity. What are the possible kinetic energies of
each end product?
Mathematical Analysis
38
A delta with kinetic energy 3.5 GeV decays into a proton and charged pion.
++ => p + 
Both particles travel parallel to the initial delta velocity. What are the possible kinetic energies of each end
product?
Mathematical Analysis
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