..
..
..
..
..
.
.
.
.
.
.
Spiral.. Physics
.
Modern Physics
.
.
.
.
.
.
The Special Theory of Relativity:
Dynamics
.
.
.
.
Spiral Physics by Paul D'Alessandris is licensed under a
Creative Commons Attribution-NonCommercial 4.0 International License
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.
2
..
..
..
..
..
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 momentum1
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
Note I’m using a scalar form of this relationship. Thus, this result is only true for motion in onedimension. The full vector treatment of force is more complicated than its worth.
1
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, for motion constrained in
one dimension. (Physicists seldom use a force approach when analyzing motion since a
momentum and energy approach is almost always more useful.)
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(
0
v
KE    3 m(
0
dv
)dx
dt
dx
)dv
dt
v
KE    3 mvdv
0
v
mvdv
v2 3/ 2
0
(1  2 )
c
KE  
This integral can be done by a simple u-substitution,
v2
c2
2v
du   2 dv
c
u  1
v
mc 2 du
KE  
2 0 u 3 / 2
mc 2
[2u 1 / 2 ]
2
1 v
KE  mc 2 [
]0
v2
1 2
c
1
KE  mc 2 [
 1]
v2
1 2
c
2
KE  mc  mc 2
KE  
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
E total  KE  mc 2
E total  80keV  511keV
E total  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.





7
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.
8
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
9
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
10
..
..
..
..
..
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 KE 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
KE (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:
b. 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:
c. 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
The following six particles (with the rest energy E0 indicated) are accelerated through a potential difference
of 10 million volts from rest.
electron
proton
pion
alpha
delta
delta
ep+
++
+
++
E0 (MeV)
0.511
938
140
3727
1232
1232
a. 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:
b. Rank these particles on the basis of their total 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:
c. Rank these particles on the basis of their velocity.
Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest
_____ The ranking cannot be determined based on the information provided.
Explain the reason for your ranking:
14
An electron is accelerated through a potential difference of 10 million volts from rest. Find its kinetic
energy, momentum and velocity. Compare its velocity to that predicted by classical physics.
Mathematical Analysis1
15
A proton is accelerated through a potential difference of 10 million volts from rest. Find its kinetic energy,
momentum and velocity. Compare its velocity to that predicted by classical physics.
Mathematical Analysis2
16
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 Analysis3
17
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 Analysis4
18
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 Analysis5
19
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 Analysis6
20
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 Analysis7
21
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 Analysis8
22
a.
b.
At low speeds, the classical expression, KE = 1/2 mv2, is approximately valid. Below what velocity is
the classical expression accurate to within 5%?
At high speeds, the ultra-relativistic expression, E = pc, is approximately valid even for particles with
mass. Above what velocity is the ultra-relativistic expression accurate to within 5%?
Mathematical Analysis9
23
Etotal  ( pc)2  (mc2 )2 is valid.
2
a.
Show that the relationship
b.
Show that the relativistic expression for kinetic energy reduces to the classical expression in the limit
of low speeds.
Mathematical Analysis
24
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
a.
b.
c.
_______ 4 kg
_
3 kg
after:
3 kg
4 kg
0.6c
_______
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.
Using the classical formula for momentum, calculate the momentum before the collision in the moving
frame.
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 Analysis10
25
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 Analysis11
26
A neutral pion with kinetic energy 1.0 GeV decays into a pair of photons.
0 =>  + 
Both photons travel along the line of the initial pion velocity.
E
(MeV)
before decay
after decay
0


Mathematical Analysis12
27
pcx
(MeV)
pcy
(MeV)
A neutral pion with kinetic energy 1.0 GeV decays into a pair of photons.
0 =>  + 
Both photons travel at an angle from the initial pion velocity.
E
(MeV)
before decay
after decay
0


Mathematical Analysis13
28
pcx
(MeV)
pcy
(MeV)
A charged kaon with kinetic energy 800 MeV decays into an antimuon and a muon neutrino.
+ => + + 
The mass of a neutrino is so small it is essentially zero. Both particles travel along the line of the initial
kaon velocity.
E
pcx
pcy
(MeV)
(MeV)
(MeV)
before decay
after decay
+
+

Mathematical Analysis14
29
A neutral kaon moving at 0.7c decays into a pair of charged pions.
0 =>  + 
The pions each travel at an angle from the initial kaon velocity.
E
(MeV)
before decay
after decay
0


Mathematical Analysis15
30
pcx
(MeV)
pcy
(MeV)
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 an angle from the initial psi-meson velocity.
E
(MeV)
before decay
0
e
after decay
e
Mathematical Analysis16
31
pcx
(MeV)
pcy
(MeV)
An omega-minus with kinetic energy 5.0 GeV decays into a neutral lambda and charged kaon.
- =>  + 
Both particles travel along the line of the initial omega-minus velocity.
E
(MeV)
before decay
after decay


Mathematical Analysis17
32
pcx
(MeV)
pcy
(MeV)
A delta with kinetic energy 3.5 GeV decays into a proton and charged pion.
++ => p + 
Both particles travel along the line of the initial delta velocity.
E
(MeV)
before decay
++
p
after decay

Mathematical Analysis18
33
pcx
(MeV)
pcy
(MeV)
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.
E
(MeV)
before
collision
after
collision
e+
e-


Mathematical Analysis19
34
pcx
(MeV)
pcy
(MeV)
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.
E
(MeV)
before
collision
after
collision
e+
e-


Mathematical Analysis20
35
pcx
(MeV)
pcy
(MeV)
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 along the line of the initial photon velocity.
E
(MeV)
before
collision

ee-
after
collision
ee+
Mathematical Analysis21
36
pcx
(MeV)
pcy
(MeV)
A pion can be produced by a gamma ray striking a stationary proton,
p =>  + p.
Assuming the pion is at rest, what is the initial gamma ray energy?
E
(MeV)
before
collision
after
collision

p

p
Mathematical Analysis22
37
pcx
(MeV)
pcy
(MeV)
An unknown particle decays into a pair of charged pions.
 =>  + 
One pion is detected with energy 550 MeV in the direction of the initial particle’s velocity, while the other
is detected with energy 147 MeV in the opposite direction.
E
(MeV)
before decay
after decay



Determine the unknown particle.
Mathematical Analysis23
38
pcx
(MeV)
pcy
(MeV)
An unknown particle decays into a pair of photons.
 =>  + 
The photons are each detected with energy 180 MeV at an angle of 22 0 from the initial particle’s velocity.
E
(MeV)
before decay
after decay



Determine the unknown particle.
Mathematical Analysis24
39
pcx
(MeV)
pcy
(MeV)
An unknown particle decays into a muon and antimuon.
 =>  + 
The muon and antimuon are each detected with energy 5.0 GeV at an angle of 180 from the initial
particle’s velocity.
E
(MeV)
before decay
after decay



Determine the unknown particle.
Mathematical Analysis25
40
pcx
(MeV)
pcy
(MeV)
An unknown particle decays into a pair of charged pions.
 =>  + 
The pions are each detected with energy 1088 MeV at an angle of 110 from the initial particle’s velocity.
E
(MeV)
before decay
after decay



Determine the unknown particle.
Mathematical Analysis26
41
pcx
(MeV)
pcy
(MeV)
An unknown particle decays into an electron and positron.
 => e + e
The electron and positron are each detected with energy 4946 MeV at an angle of 730 from the initial
particle’s velocity.
E
(MeV)
before decay

e
after decay
e
Determine the unknown particle.
Mathematical Analysis27
42
pcx
(MeV)
pcy
(MeV)
The universe is filled with a gas of microwave photons of average energy 2.35 x 10-4 eV. These photons are
the remnants of the hot plasma present in the early universe. As the universe expanded the wavelength of
these photons also expanded. The wavelength is currently in the microwave region of the spectrum.
These photons limit the maximum energy with which material particles can travel through the universe.
For example, protons with energies above a certain limit will spontaneously produce pions when they
collide with the microwave photons, removing energy from the protons and slowing them down. Any proton
traveling above this energy limit will rapidly lose energy. Your job is to calculate the energy above which a
collision between a proton and a microwave photon can produce a neutral pion.
This calculation is easiest to do in the frame of reference of the initial proton. Basically, rather than
consider a moving proton absorbing a moving microwave photon, consider a stationary proton absorbing a
very high energy photon. Then, once you have the energy of the photon, you can transform your result back
into the earth frame.
A high energy photon with unknown energy is absorbed by a stationary proton, producing a pion.
p =>  + p
a. Assuming the pion is at rest after being produced, find the energy of the incoming photon.
The value calculated above is the energy of the photon in the proton’s frame of reference. The ratio of that
energy to the energy of the photon in the earth’s frame (2.35 x 10-4 eV) gives you the gamma factor between
the frames.
b. Find the gamma factor between the proton’s frame of reference and the earth’s frame of reference.
 
photon energy in proton frame

photon energy in earth frame
c. In the earth frame, the proton moves with this gamma factor. Therefore, in the earth’s frame calculate
the energy of the proton. This is the maximum energy with which protons can travel through the universe
43
..
..
..
..
..
The Special Theory of Relativity:
Dynamics
Projects:
Interstellar Travel – Energy Issues
44
Interstellar Travel – Energy Issues
In a previous investigation, you discovered some of the fundamental difficulties in traveling to even a
relatively nearby star. In this investigation, you will be introduced to a much more important factor, the
incredible amounts of energy needed to accelerate a macroscopic object to relativistic speeds.
I.
Energy Considerations
Consider a spaceship fully fueled of mass M. Some portion of the mass of the spaceship is in the form of its
fuel, such that after “burning” its fuel the ship will have a smaller mass. This mass is the “useful” mass of
the spacecraft, the payload, and we will designate it fM (a fraction, f, of the total loaded mass). For
example, if the ship is 5% fuel, f = 0.95.
1. What is an expression for the amount of fuel on the fully loaded spacecraft?
A. Fusion powered spaceship
The most efficient energy source available using today’s technology is fusion energy, and given that it is
the energy source used by stars, it’s probably a good bet that it may be the most efficient large scale energy
source available in the universe.
Imagine a spacecraft powered by fusion energy. The most efficient fusion reaction effectively combines
four hydrogen nuclei into a single helium nucleus. Since a helium nucleus is less massive than four
hydrogen nuclei, the energy released by this reaction can be calculated using Einstein’s relation E = mc2.
1. Determine the energy released by this reaction by converting the mass difference between the helium
nucleus and the four hydrogen nuclei into a measure of energy.
2. The efficiency of the reaction, , is defined to be the ratio of energy released to energy present in the
initial hydrogen nuclei. Calculate the efficiency of this fusion reaction.
45
3. Using your expression for the total amount of fuel on the spacecraft, and the efficiency of the fusion
reaction, write a symbolic expression (use , not its numerical value) for the total energy released by fusing
all of the fuel.
The energy released via fusion will go toward providing kinetic energy to the spacecraft as well as kinetic
energy to the exhaust helium. Calculating the portion of this energy that goes into moving the spacecraft is
a complicated problem, so we’ll simply approximate that all of the energy goes into kinetic energy of the
ship! (This is a huge over-estimation, but we will find that even with this approximation, a fusion powered
spaceship is not very feasible.)
4. Equate the kinetic energy of the payload (fM) with the total energy released by fusing all of the fuel.
From this equation, solve for f, the fraction of the initial mass of the ship that is payload.
5. In the last investigation, you found that at a cruising speed of v = 0.9992c, a spacecraft could reach Vega
in a little less than 3 years. Using the result above, what fraction of the mass of that spaceship can be
payload, and how much of the ship must be made up of fuel?
You should have found that 99.97% of the mass of your spaceship must be fuel!
A more useful way to express the amount of fuel needed for acceleration is to calculate the fuel to payload
ratio. This is the number of kg of fuel needed to accelerate a single kg of payload.
46
6. Calculate the fuel to payload ratio for your spaceship.
For every 1 kg of payload you want to take to Vega, you would have to supply the ship with over 3000 kg
of fuel. Actually, the situation is far worse. This is simply the fuel required to accelerate the payload up to
the cruising speed. For a complete round trip, one would have to carry enough fuel to boost the payload to
the cruising speed, decelerate it to rest at the destination, boost it to cruising speed again for the return trip
home, and decelerate it upon reaching earth.
7. How much fuel is required for a complete round trip? Express your answer as a multiple of the payload
mass, i.e., the number of kg of fuel per kg of payload. Explain why the answer is not simply four times the
fuel required to boost the payload alone to the cruising speed.
This number borders on the ridiculous. You would need over 10 14 kg of fuel to take a single one kg mass on
a round-trip to Vega.
8. If you aren’t as anxious to get to Vega quickly, you could travel at a cruising speed of 0.5 c. Calculate
the amount of fuel, as a multiple of payload mass, needed for this roundtrip.
9. This should be substantially less fuel, but how long, ignoring acceleration and deceleration, would a
roundtrip journey to Vega (25 c yr away) take at the relatively slow speed of 0.5 c?
47
B. Collecting fuel along the way
One possible way around the problem discovered above is to collect the hydrogen needed for fuel along the
way to Vega. Interstellar space contains approximately one hydrogen atom per cubic meter. With a large
enough collecting “mouth” perhaps this would solve the fuel problem.
In addition to the problem of the size of the “mouth” needed to collect enough hydrogen, a more
fundamental problem remains. The hydrogen is approximately at rest with respect to Vega. However, in the
frame of the spaceship, it is moving toward the ship at the same speed that the ship is moving toward Vega.
Thus, the hydrogen atoms have a large kinetic energy as they are scooped up by the ship. The collision
between the atoms and the ship will effectively slow the ship. When the kinetic energy of the atoms (which
effectively slows the ship) is approximately equal to the energy that can be extracted from the atoms to
accelerate the ship, this technique of collecting fuel along the way becomes relatively useless.
1. Setting the kinetic energy of the collected atoms equal to the energy that can be extracted from the
atoms, calculate the speed above which collecting fuel as you go becomes a losing proposition.
2. Cruising at the speed calculated above, ignoring acceleration and deceleration, how long would it take
you to reach Vega?
48
C. Matter-Antimatter powered spaceship
There is one final possibility. There exists a reaction that has an efficiency of 1.0, where all of the initial
mass energy present is converted into kinetic energy. This is the annihilation of matter with antimatter. If a
proton is combined with an antiproton, the total mass of both protons is converted into the kinetic energy of
the resulting photons. If this reaction does not provide a reasonable fuel source for our trip to Vega, then a
trip to Vega is simply not feasible.
There are many very serious problems with using matter-antimatter annihilation as a fuel source. The first
is that there is effectively no naturally occurring antimatter in the universe! All of the antimatter has to be
“built”. We currently are technologically able to manufacture small amounts of antimatter, so we’ll assume
that in the future this may not be an insurmountable problem. Of course, the energy that is released when
antimatter is annihilated has to be “put into” the antimatter in the first place, but this manufacturing can be
done on a home planet and need not weigh down our spaceship. Storing large amounts of antimatter and
combining it in a controlled manner on the spaceship is a different story, but let’s pretend we’ve mastered
that as well.
1. Imagine a rocket engine that combines matter and antimatter in a controlled way, and focuses the
resulting photons into a tight beam traveling away from the stern of the spaceship. For this engine,
determine f, the payload fraction of the initial mass of the ship, to achieve a cruising speed v. (Hint: The
ship can essentially be considered to be a particle of mass M at rest that decays into a big flash of light and
a smaller particle (the payload) of known mass fM traveling at a known speed v. Conserve both energy and
momentum.)
2. Using the result above, what fraction of the mass of the spaceship can be payload, and how much of the
ship must be made up of fuel?
49
3. Calculate the fuel to payload ratio for a matter-antimatter spaceship. How much fuel, per kilogram of
payload, is needed for a roundtrip journey?
This is less ridiculous, but still requires over a million kg of fuel per kg of payload.
4. If you aren’t as anxious to get to Vega quickly, you could travel at a cruising speed of 0.8 c. Calculate
the amount of fuel, as a multiple of payload mass, needed for this roundtrip.
5. How long would such a roundtrip journey take at the relatively slow speed of 0.8 c.
D. One last concern
1. If relativistic travel were ever possible, we would be wise to avoid bumping into any interstellar dust.
Calculate the kinetic energy of a speck of dust (10-6 g) traveling at 0.8 c. This energy is comparable to how
many 1000 kg cars traveling at 65 mph?
50
Selected Answers
1
K = 10 MeV
pc = 10.5 MeV
v = 0.9988 c
vclassical = 6.23 c
2
K = 10 MeV
pc = 137 MeV
v = 0.145 c
vclassical = 0.146 c
3
a. K = 33.8 MeV
pc = 101 MeV
4
a. K = 63.9 MeV
pc = 632 MeV
5
a. E1 = 3.111 MeV
E2 = 7.808 MeV
6
200 billion $
7
1.3 billion $
8
m = 0.0068%
9
a. v = 0.258 c
b. v = 0.950 c
10
b. -3.78 kgm/s
c. -4.47 kgm/s
11
8
b. 1.5 x 10 s
d. 1.73 x 108 s to reach 0.5c
12
E1 = 1131 MeV
E2 = 4.0 MeV
13
 = 6.8°
14
Emuon = 1247 MeV & Eneutrino = 47 MeV
15
E = 349 MeV
16
 = 37.4°
17
Klambda = 4227 MeV & Kkaon = 836 MeV
or Klambda = 2603 MeV & Kkaon = 2460 MeV
18
Kproton = 3611 MeV & Kpion = 43 MeV
or Klambda = 1930 MeV & Kkaon = 1724 MeV
19
 = 9.8°
20
 = 3.1°
21
E = 2.05 MeV
22
E = 152 MeV
23
K0
24
0
25
0
26
K0
27
0
or Emuon = 106 MeV & Eneutrino = 1188 MeV
 = 40.1°
51
52