SPCS Summer Institute 2014
Special and General Relativity
Lecture 8 Assignment
Problem 8.1:
This is Spacetime Physics 7-5 a) (in groups)
The rocket observer measures the energy and momentum components of a particle to have the values E 0 ,
p0x , p0y and p0z . What are the corresponding values of energy and momentum measured by the laboratory
observer? We will use the Lorentz transformations to answer the question.
The moving particle emits a pair of sparks closely spaced in time as measured on its wristwatch. The
rocket latticework of clocks records these emission events; so does the laboratory latticework of clocks. The
dt
, p0x = m dx
rocket observer constructs components of particle momentum and energy (E 0 = mc2 dτ
dτ , etc.)
0
0
0
0
from knowledge of particle mass m, the spacetime displacements dt , dx , dy , and dz derived from the event
recordings, and the proper time dτ computed from these spacetime components. [Recall dx = x2 − x1 is
∆x with the two points extremely close together]. Laboratory components of the energy-momentum 4vector come from transforming the spacetime displacements. The Lorentz transformations for incremental
displacements are,
dx
vdx0
) = γ(cdt0 + βdx0 )
c
= γ(dx0 + vdt0 ) = γ(dx0 + βcdt0 )
dy
= dy 0
dz
= dz 0
cdt = γ(cdt0 +
a) Multiply both sides of each equation by the invariant mass m and c then divide through by the
invariant proper time dτ . Recognizing the components of the energy-momentum 4-vector, show that
the transformation equations for energy-momentum are, (see text)
E
= γ(E 0 + βcp0x )
cpx
= γ(cp0x + βE 0 )
cpy
= cp0y
cpz
= cp0z
Performing the suggested operations we get
dt
dτ
dx
mc
dτ
mc2
mvdx0
dt0
+
) = γ(E + vp0x )
dτ
dτ
dx0
dt0
= γ(mc
+ vmc ) = γ(cp0x + βE 0 )
dτ
dτ
= γ(mc2
(1)
Problem 8.2: Fast electrons at SLAC
This is Spacetime Physics 7-6
The two mile Stanford Linear Accelerator accelerates electrons (and positrons) to a final kinetic energy of 47
GeV (47×109 electron-volts; one electron-volt (eV) = 1.6×10−19 Joules). The resulting high-energy electrons
are used for experiments with elementary particles. Electromagnetic waves produced in large vacuum tubes
(“klystron tubes”) accelerate the electrons along a straight pipelike structure 10,000 feet long (approximately
3000 meters). Take the rest energy of an electron to be mc2 ≈ 0.5M eV = 0.5 × 106 eV .
a) Electrons increase their kinetic energy by approximately equal amounts for every meter traveled along
the accelerator pipe as observed in the laboratory frame. What is the gain in M eV /meter? Suppose
the Newtonian expression for kinetic energy were correct. In this case how far would the electron travel
along the accelerator before its speed were equal to the speed of light?
47GeV
4.7 × 1010 eV
eV
=
= 1.57 × 107
3000m
3 × 103 m
m
Now we want to find when 12 mv 2 → 12 mc2 = 12 (.511M eV ) = 0.256M eV . It only takes
0.0163m or 1.6 cm to achieve this energy.
(2)
0.256M eV
15.7M eV /m
=
b) In reality, of course, even the 47 GeV electrons that emerge from the end of the accelerator have a
speed v that is less than the speed of light. What is the value (1 − vc ) between the speed of light
and the speed of these electrons as measured in the laboratory frame? [Hint: For vc very near unity,
2
1 − vc2 = (1 + vc )(1 − vc ) ≈ 2(1 − vc ). Let a 47 GeV electron from this accelerator race a flash of
light along an evacuated tube straight through the Earth from one side to the other (Earth diameter
12,740,000 meters). How far ahead of the electron is the light flash at the end of this race? Express
your answer in millimeters.
We want to find 1 − β, thus we need to find β. Note the following for motion in one direction,
p=m
pc
dx
dx
p
pc
=
= γm
= γmv = γmcβ → β =
=
dτ
dt
γmc
γmc2
E
(3)
This last relation is a very useful on to find β. But let’s find it a different way,
E = γmc2 −→ (47 × 109 eV ) = γ(0.511 × 106 eV ) −→ γ =
47000
= 91976.
0.511
(4)
That’s extremely relativistic, we can now just express this in terms of β,
γ=p
1
1−
β2
−→ 919762 =
β=
p
1
1 − β2
1 − (1.087×−5 )2
(5)
We will need the binomial expansion here,
β ≈ 1 − 5.9 × 10−11
(6)
That’s very close to c! How far ahead is the light beam after traveling across the diameter of the Earth?
We will use Llight = ct and Lelectron = βct = βLlight .
∆L = Llight − Lelectron = ct(1 − β) = LEarth (1 − β) = 12, 470, 000m(1 − β)
= (12, 470, 000m)(5.9 × 10−11 ) = 7.35 × 10−4 m
(7)
This is 0.7mm! That’s fast!
c) How long is the “3000 meter” accelerator tube as recorded on the latticework of rocket clocks moving
along with the 47 GeV electron emerging from the accelerator?
Since we already found γ it is easy L0 =
Problem 8.3: Photon Integrity
L
γ
=
3000m
91976
= 3.2cm.
(in groups) This is Spacetime Physics 8-10
Show that an isolated photon cannot split into two photons going in directions other than the original
direction. (Hint: Apply the laws of conservation of momentum and energy and the fact that the third side
of a triangle is shorter than the sum of the other two sides. What triangle?)
Conservation of momentum is a vector relation,
p~initial
= p~1 + p~2
the x component is,
pinitial
= p1 cos θ1 + p2 cos θ2 .
(8)
Conservation of energy states simply that Einitial = E1 + E2 but for photons we can express this in terms
of momentum via E = pc. Thus,
cpinitial
The only way this can hold is for θ1 = θ2 = 0.
=
cp1 + cp2
Problem 8.4: A Stanford P070 midterm question
What is the speed of an electron whose relativistic kinetic energy equals its rest energy?
The kinetic energy is E − mc2 or (γ − 1)mc2 . If this
is equal to the rest energy mc2 , then we have
√
3
2
E = 2mc or γ = 2. We may now solve for β to find β = 2 . I find it easier to use
so p =
√
E2
=
4mc2 = m2 c4 + p2 c2
β
=
√
pc
γβmc2
3
=
=
E
γmc2
2
3mc. Now
Problem 8.5: Another Stanford P070 midterm question
A particle of mass m has an energy E = 4mc2 . What is the momentum of this particle in units of mc?
What is the energy of this particle in a reference frame in which p = 2mc?
Use E 2 = m2 c4 + p2 c2 to find,
16m2 c4
−→
= m2 c4 + p2 c2
√
p = 15mc.
For the second part, use the same equation with p = 2mc:
E2
E
= m2 c4 + 4m2 c2
√
=
5mc2
Problem 8.6: Yet another P070 midterm question: Electron pair to several
An electron with relativistic kinetic energy K = (γ − 1)mc2 and relativistic spatial momentum p (say in
x direction) collides with another electron that is at rest. As a result, 4 particles emanate in a reaction
consisting of 3 electrons and one positron. Electrons and positrons are identical in every way except they
have opposite charges −e and +e respectively. Of importance here is that they have the same mass m.
Ignore any electromagnetic interactions in the following.
a) Briefly explain (little calculation, if any, is needed) why it can not be that three of the resulting
particles are at rest and the other moves off with some momentum p̄.
Solution: a)
If only one particle is moving after the interaction, then it must have the same momentum as the initial
incoming electron. thus it has the same energy. Then we would have Ei + m2 = Ei + 3m2 , which can not
be true.
b) If the resulting particles all emanate with some momentum p̄, find the minimum relativistic kinetic
energy K of the initial electron in terms of mc2 (or m if using c = 1).
In order for K to be minimum, we want the final energy to be minimum while conserving momentum.
This occurs when all of the resulting particles same momentum p̄ (or else more initial energy must be supplied
to provide the transverse momentum). Note that K = (γ − 1)mc2 = Ei − mc2 The two conditions are (the
following uses c = 1),
p
Cons. of momentum :
p = 4p̄ −→ p̄ =
4
p
p
2
2
Cons. of Energy :
Ei = m + p + m = Ef = 4 m2 + p̄2
r
p
p2
2
2
m + p + m = 4 m2 +
simpler to keep in terms of Ei
16
r
p2
Ei + m = 4 m2 +
square both sides
16
p2
Ei2 + 2Ei m + m2 = 16(m2 + ) = 16m2 + p2 = 15m2 + Ei2
16
2Ei m + m2 = 15m2
2Ei m
=
14m2
−→ Ei = 7m
Thus using K = Ei − m we get that K = 6m .
Problem 8.7: Special Relativistic Beauty Tips
(in groups)
As we have learned, the theory of special relativity is a font of beauty tips. Let’s revisit two beauty tips and introduce
a third beauty tip.
# 1 By moving quickly by your friends, you will age slower than your friends. (That is, if you accelerate up to some
speed and later slow to rest.)
# 2 By moving quickly past your friends, they will see you as thinner in the direction of travel.
# 3 By accelerating up to some speed, you will expend energy, thereby reducing your mass*.
So, let’s envision that you can run as fast as the fastest human has ever run (as known by Wikipedia) and assume
you can run a marathon at that speed -this speed is about 12 m/s. At this rate (27 MPH for roughly 26 miles, it
takes about 1 hour to complete a marathon), it should take 60 minutes, or 3600 seconds to complete the marathon.
Thus at 12 m/s (or just 10 to 1 sigfig), the questions are,
a) How much thinner do you appear while running? (via length contraction)
b) By how many seconds (or nanoseconds) would you be younger? (via TD -twin paradox).
c) How much mass would you have lost. Use the following assumptions*, you neither gain nor shed mass during
the run, and the metabolic rate is 1000 W (an assumed value based on values given on wikipedia -we only want
an order of magnitude estimate)
You will need to use the binomial expansion to figure these values (to only 1 sigfig please!).
d) Repeat the above (not c) using the fastest speed ever obtained by humans relative to Earth, about 11,000 m/s
(by the Apollo 10 crew, via wikipedia). (Do not consider part c) here.)
.
Problem 8.8: Some random, basic, questions
(in groups)
a) What is the difference between an invariant quantity, a conserved quantity, and a constant quantity?
b) How much mass does a 100 Watt light bulb dissipate in one year?
c) The total energy consumption in 2005 for all nations of the Earth is estimated to be about 5 × 1020 kilowatt hours
(1 kWh = 1000 W × h). How much mass is this equivalent to?
W
d) The solar constant, 1360 m
2 , is the total power received on Earth per unit area. Consider the side of Earth that
receives light to be a disk of radius 6.37 × 106 m. How much mass does the Earth gain by absorbing this light in one
year. (We are assuming all light is absorbed, clearly an overestimation).
e) Reversing the solar constant (non-trivial calculation), determine the amount of energy the Sun loses per second.