Download Lecture 1 Assignment 1.1 Binomial expansion

EPGY Summer Institute
Special and General Relativity 2010
Lecture 1 Assignment
SOLUTIONS
1.1 Binomial expansion
In this course we will often examine complex mathematical quantities that have variables that can take very small values.
The typical case is when we want to recover classical, Newtonian, results from relativistic equations when we take the limit
that the velocities involved, v, are very much smaller than the speed of light, c (or put another way when vc << 1). Thus we
need to learn how to approximate mathematical expressions when their arguments are very small. We won’t care to derive
the following relation but in this problem we want to convince you that it is indeed a valid approximation.
First, the full expression to express a polynomial as a series can be expressed as follows. This is called the Binomial
series expansion.
(1 + x)n
=
1 + nx +
n(n − 1) 2
n(n − 1)(n − 2)...(n − m + 1) m
x + ··· +
x .
2!
m!
(1)
In this expression x is restricted to values x2 < 1 and n may be any positive or negative number (it need not be an integer).
Also, this is an infinite series but the form shown here is up to some arbitrary integer m.
Luckily we will not need this full expression. If x << 1 we can get an excellent approximation by using only the first two
terms. This is the approximation that will be used often in this course and it should be memorized (or noted so you can refer
to it later).
(1 ± x)n ≈ 1 ± nx
if x << 1
(2)
(note here we have explicitly shown the result for positive and negative values of the terms).
Before we try to convince you of this expression’s validity let’s explore some examples. For the following expressions the
variable x is much less than 1. Provide the following expansions to first order.
1. (1 + x)2 ≈ 1 + 2x
2. (1 − x)3 ≈ 1 − 3x
3.
1
(1−x)3
= (1 − x)−3 ≈ 1 − (−3)x = 1 + 3x
4. (1 − x3 )3/2 ≈ 1 − 23 x3
5. √
1
1−x2
≈ 1 − (− 12 )x2 = 1 + 21 x2
*(this one will be used by us the most).
6. (1 + 0.1x)−2.4 ≈ 1 + (−2.4)(0.1)x = 1 − 0.24x
7. √
1
a2 −b2
=
1
a
1
p
b )2
1−( a
≈
1
(1
a
+ 12 ( ab )2 ) =
1
a
+
1 b2
2 a3
where b << a.
Verify this approximation
Ok now that you know how to use this expression let’s try to convince you of its validity (and its usefulness!). In the
following table fill in the blanks for the actual function, the first order approximation, and the second order term ( n(n−1)
x2
2!
column B) for each value of x below.
x
10
2
1
0.1
0.01
1 × 10−8
x
10
2
1
0.1
0.01
1 × 10−8
full function
(1 + x)2
121
9
4
1.21
1.0201
1.00000002
√ 1
1−x2
i √199
i √13
∞
1.00504
1.000050004
1.000000005
A: binomial expansion (2)
1 + 2x
21
5
3
1.2
1.0200
1.00000002
1 + 21 x2
51
3
1.5
1.00500
1.000050000
1.000000005
B:
n(n−1) 2
x
2!
2
x
100
4
1
0.01
0.0001
1 × 10−16
3 4
8x
37.5
1.5
3
8
0.000038
0.000000003
3.75 × 10−33
% error of A
82 %
44 %
25 %
0.83 %
0.0098 %
∼0
% error of A
undef
undef
∞%
0.04 %
0.0000003 %
∼0
B/A
4.76
0.8
0.333
0.0083
0.000098
1 × 10−16
B/A
0.74
0.5
0.25
0.0000378
0.000000003
3.75 × 10−33
Examine the values in the last two columns. First, do the results convince you that for values of x much less than 1 that
the binomial expansion is a good approximation to the full function? Second, it should be clear that if x is near or greater
than 1 these expansions break down (for example, using the expression in (1) for general m find the value of the term for
m = 10). Lastly, for values of x much less than 1 it should be clear that, by examining the last column, that higher orders in
the expansion are not needed (or that they do not change the result in column A by much).
Just to convince you of the power of this approximation, compute the following numbers with your calculator.
1
−1=
1 − x2
1
1 + x2 − 1 =
2
√
for x = 3 × 10−9
mine comes up with 0
4.5 × 10−18
for x = 3 × 10−9
Which version gives a more reasonable answer do you think?
1.2 Finding distances between points in 2D and 3D
a) What is the distance between the following pairs of points in the two-dimensional plane? (Points are indicated in
the xy plane as (x, y).)
p
√
√
a.1) (2 m, 3 m) and (5 m, 7 m) Da.1 = (5 − 2)2 + (7 − 3)2 m = 32 + 42 = 25 = 5 m
p
√
√
a.2) (0 m, 4 m) and (5 m, 7 m) Da.2 = (5 − 0)2 + (7 − 4)2 m = 52 + 32 = 34 = 5.8 m
p
√
√
a.3) (-2 m, 3 m) and (2 m, 1 m) Da.3 = (2 − (−2))2 + (1 − 3)2 m = 42 + 22 = 20 = 4.5 m
p
√
√
a.4) (-1 m, 4 m) and (-5 m, 7 m) Da.4 =
((−5) − (−1))2 + (7 − 4)2 m =
42 + 32 =
25 = 5 m
b) What is the distance between the following pairs of points in three-dimensional space? (Points are indicated as
(x, y, z).)
p
√
√
b.1) (2 m, 3 m, 1 m) and (5 m, 7 m, 3 m) Db.1 = (5 − 2)2 + (7 − 3)2 + (3 − 1)2 m = 32 + 42 + 22 = 29 = 5.4 m
p
√
√
b.2) (0 m, 4 m, -1 m) and (5 m, 7 m, 0 m) Db.2 = (5 − 0)2 + (7 − 4)2 + (0 − (−1))2 m = 52 + 32 + 12 = 35 = 5.9 m
p
√
√
b.3) (-3 m, 1 m, 2 m) and (-1 m, 1 m, 6 m)Db.3 =
((−1) − (−3))2 + (1 − 1)2 + (6 − 2)2 m =
22 + 0 2 + 4 2 =
m
√
b.4) (-1 m, 2 m, -12 m) and (-5 m, 1 m, -10 m) Db.4 =
21 = 4.6 m
p
((−5) − (−1))2 + (1 − 2)2 + ((−10) − (−12))2 m =
√
20 = 4.5
42 + 12 + 22 =
1.3 Inconsistency of Galilean Relativity and Maxwell’s Equations
If you have not been introduced to electricity and magnetism, please see the appendix for the relevant concepts and
equations. This material was introduced in the notes that you were sent prior to your arrival.
A very long beam of electrons is formed at SLAC consisting of 1.0 × 1010 electrons per meter, all in a line,
traveling at 2.0 × 108 m/s to the left. In the opposite direction, (to the right), is another beam of positrons with the
same density, traveling at the same speed, and very close to the electron beam. A positron is the anti particle of the
electron. It is identical in every way except it has a charge of +e. Alice stands next to the beam and measures the
current.
a) Does Alice observe any net electric charge of the beam?
No
b) Does Alice observe any electric field at the point P?
No
c) In which direction does Alice observe the net (conventional) current of the beam? The conventional current
for both the electrons and positrons is to the right (in the direction of the positrons.)
~ field, if any, to be in at point P? By the right hand rule, if the
d) In which direction does Alice observe the B
current is to the right you point your thumb to the right (of your right hand) and your fingers curl around the
~
current line. At the point P, above the current, your fingers should point out of the page (screen). Thus, B
point out towards you as you read this.
Alice observes the proton to accelerate away from the beam while Bob sees it to stay at the same distance form
the beam (no net force on it). Inconsistency relies on relativism, facts are different in different frames (lead to
observe Bob dead in one frame (A) and alive in another (B).
A stray proton is traveling near the beam. At a specific time, the proton is at point P, 1.0m above the beam and
traveling at 2.0 × 108 m/s.
e) What direction is the net force, if any, exerted on the proton as seen by Alice? We have the direction of the
~ field, thus the right hand rule gives: fingers point in the direction of ~v , to the left, and rotate
proton and the B
~ field, out of the screen, (and since the proton is positively charged) your
(palm first) to the direction of the B
thumb points in the direction of the force: up, away from the beam.
Now consider Bob who is riding on a very (VERY) fast ship traveling to the left and parallel to the beam at a
speed of 2.0 × 108 m/s. He views the same situation and measures the current of the beam.
The image on the left below represents what Bob sees in his reference frame where he sees himself to be at rest.
Label on the diagram the directions and magnitudes of the velocities of the proton, electron beam, and positron
beam as perceived by Bob.
f) Does Bob observe any net electric charge of the beam?
No
g) Does Bob observe any electric field at the point P?
No
h) In which direction does Bob observe the net (conventional) current of the beam? The current for the positrons
is to the right just as before (and twice as strong since they are moving twice as fast) and the electrons are not
moving -thus they do not contribute to the current. The current actually turns out to be exactly the same.
Thus it is the same as for Alice.
~ field, if any, to be in at point P? Same as for Alice.
i) In which direction does Bob observe the B
j) What direction is the net force, if any, exerted on the proton as seen by Bob? The force law is F = qvB sin θ
but since the proton is at rest (v = 0) there is no magnetic force acting on the proton.
k) Compare the motion of the proton as viewed by Alice and Bob. On each figure above draw the trajectory
of the proton after it goes by point P as it appears to each of them. Discuss the problem with this result provide an example where this result can lead to a logical inconsistency. (Use a trigger that sets of a bomb for
dramatic effect.)
Tomorrow we will ask you how special relativity rectifies this situation.
Problem 1.4: Units of Spacetime
Light moves at a speed of 3.0 × 108 meters
second . One mile is approximately 1600 meters. One furlong is approximately
200 meters.
hr
s
13
a) How many meters of time are there in one day? l = ct = (3 × 108 m
s )(24 day )(3600 hr ) = 2.6 × 10 m
1600m
1mi
−6
b) How many seconds of distance in one mile? t = ( 3×10
s
8 m )( 1mi ) = 5.3 × 10
s
c) How many hours of distance in one furlong? t =
1f ur
( 3×10
8m
s
1hr
)( 3600s
) = 1.9 × 10−10 hr
d) How many weeks of distance in one light-year? 52 weeks
1f ur
s
9
)(3 × 108 m
e) How many furlongs of time in one hour? l = ( 200m
s )(3600 hr ) = 5.4 × 10 f ur
f) How many light-years (ly) of time in one year? 1ly
g) How many light-days (ld) of time in one year? 365 ld
h) Express the speed of light, c, in units of
ly ld
y, y,
and
f eet
nanosec .
3.25f eet
ld
8m
c = 1 ly
)( 101ns
−9 s )(3 × 10 s ) =
y = 365 y ; c = (
m
eet
eet
0.98 fns
≈ 1 fns
Problem 1.5: Traveling to the Andromeda Galaxy I
This problem is Spacetime Physics (Ch 1, #9)
The Andromeda galaxy is approximately 2 million light years distant from Earth as measured in the Earth-linked
frame. Is it possible for you to travel to Andromeda in your lifetime? Lets explore this in steps. In the following
consider the distance from Earth to Andromeda to be exactly 2 × 106 ly and neglect any relative motion of the Earth
and Andromeda.
a) Trip 1: Your one-way trip takes 2.01 × 106 years (as measured in the Earth-linked frame) to cover the distance
of 2.00 × 106 ly . How long does the trip last as measured in your rocket frame?
First find β, β =
v
c
=
2.00
2.01
2
Next find γ = (1 − β ) −
tship =
tEarth
γ
= 0.995 (recall light c = 1ly/y).
1
2
5
= (1 − (0.995)2 )−1/2 = (0.009975)−1/2 = 10.01
= 2.01 × 10 years.
b) What is your speed on Trip 1 as measured in the Earth-linked frame? Express this speed as a fraction of the
speed of light, β = vc .
Found above, β = 0.995.
c) Trip 2: Your one-way trip to Andromeda takes 2.001 × 106 y, How long does this trip last as measured in
your rocket frame? What is the speed of the rocket ship in this case (expressed as before)? First find β,
2.00
β = vc = [ 2.001
] = 0.9995
Next find γ = (1 − β 2 ) −
tship =
tEarth
γ
1
2
= (1 − (0.9995)2 )−1/2 = (0.0009993)−1/2 = 31.6
= 6.3 × 104 years.
d) Trip 3: Now set the rocket time for the one-way trip to take 20 years, you are in a hurry to get to Andromeda.
What is the speed of your rocket (as measured in the Earth-linked frame) in this case? [You will need the
result of the previous problem on binomial expansions]
In rocket frame we want tship =
insignificant).
This gives γ =
2.00×106
20
tEarth
γ
= 20 years. Where tEarth just about equals 2 × 106 y (any difference is
= 100000.
Solve this for β, β = (1 −
1 1/2
γ2 )
∼ 1 − ( 12 )( γ12 ) + ... = 1 − 5 × 10−11 = 0.99999999995
Problem 1.6: Traveling to the Andromeda Galaxy II
This is question 1-10 of Spacetime Physics
In the Star Trek series a so-called Transporter is used to “beam” people and their equipment from a starship to
the surface of nearby planets and back. The Transporter mechanism is not explained, but it appears to work only
locally. (If it could transport to remote locations, why bother with the starship at all?) Assume that one thousand
years from now a Transporter exists that reduces people and things to data (elementary bits of information) and
transmits the data by light or radio signal to remote locations. There a receiver uses the data to reassemble travelers
and their equipment out of local raw elements.
One of your descendants, named Samantha, is the first “transporternaut” to be beamed from Earth to the planet
Zircon orbiting a star in the Andromeda galaxy, two-million light-years from Earth. Neglect any relative motion
between Earth and Zircon, and assume: (1) transmission produces a Samantha identical to the original in every
respect (except that she is 2 million light-years from home!), and (2) the time required for disassembling Samantha
on Earth and reassembling her on Zircon is negligible as measured in the common rest frame of transporter and
receiver.
a) How much does Samantha age during her outward trip to Zircon? 0 years.
b) Samantha collects samples and makes observations of the Zirconian civilization for one Earth-year, then beams
back to Earth. How much has Samantha aged during her entire trip? 1 year.
c) How much older is Earth and its civilization when Samantha returns? 4 million and 1 years.
d) Earth has been taken over by a tyrant, who wishes to invade Zircon. He sends one warrior and has him
duplicated into attack battalions at the receiver end. How long will the Earth tyrant have to wait to discover
whether his ambition has been satisfied? A little over 4 million years.
e) A second transporternaut is beamed to a much more remote galaxy that is moving away from Earth at 87
percent the speed of light. This time, too, the traveler stays in the remote galaxy for one year as measured by
clocks moving with the galaxy before returning to Earth by transporter. How much has the transporternaut aged
when she arrives back at Earth? 1 year. The person ages 1 year by their own clock -that is what determines
how much someone ages.
1
Appendix: Formulas for problem 1.2
For the following question use the following facts which were presented in the “Brief Introduction to EM”.
~ = 1 q12q2 r̂, 1 = 9 × 109 N m2 /C 2
Electric force on a charge: F~ = q E
4π0 r
4π0
12
Electric charge of an electron, −e: −1.6 × 10−19 C, of a positron, +e: +1.6 × 10−19 C.
Magnetic force on a charge moving at velocity v: F = qvB sin θ.
Direction of magnetic force via right hand rule.
For positive charges, point fingers of right hand in
direction of v, rotate, palm first, to the direction of B d, Thumb points in the direction of the force. For negative
charges, the force is in the opposite direction in which your thumb points.
0I
Magnetic field, B, produced by a current I: B = µ2πr
, µ0 = 4π × 10−7 TAm
Direction of magnetic field produced by a current via another right hand rule: Point thumb in
direction of currents (flow of positive particles) fingers curl in direction of magnetic field around wire.
One last fact not presented in that document.
Definition of current in terms of a large number of flowing charged particles: I = nvq, where n =
number of charged particles per meter, v is the speed of these particles, and q is their electric charge. Conventional
current is defined as the flow of positive charges. Even though we now know that electrons flow in wires, they travel
in the opposite direction of the defined current. Again, negative charges traveling in one direction is a conventional
current moving in the opposite direction.
Short Problems
1S.1
List all of the fundamental forces of nature that you know of. How fast are these transmitted?
1S.2
Which of the following are IRFs (ignore air resistance for terrestrial scenarios) ? (what clues tell you that it is/is not
an IRF?)
a) You standing on the surface of the Earth.
b) Jet fighter traveling at mach 3 in a straight line.
c) On board a spaceship, far away from any planets or stars, with its engine turned off.
d) On board a spaceship, far away from any planets or stars, with its engine turned on.
1S.3
You may recall that for an object traveling in a circle at constant speed, the acceleration it experiences, the centripetal
acceleration, is directed perpendicular to its velocity, in towards the center of the circle that it traverses. The
2
magnitude of the centripetal acceleration is a = vR , where R is the radius of the circle and v is the object’s speed.
Find the speed for each and then compare the centripetal accelerations of the following objects.
a) You on a merry go round, 10 m in diameter, and it takes 15 seconds to go around once.
b) A race car goes around a curve of radius 50 m at 200 MPH, (90 m/s).
c) Standing on the surface of the Earth at the equator (compute the speed by noting that it takes 24 hours to
travel once around).
d) The Moon, which orbits the Earth once every 28 days.
e) The Earth which orbits the Sun once a year.
f) Pluto as it orbits the Sun.
g) The Solar System, which orbits the Milky Way Galaxy (at a radius of about 30,000 ly) once in about 220
million years.