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Spiral.. Physics
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Modern Physics
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Spacetime and General Relativity
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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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Spacetime and General Relativity
Minkowski Metric
Two observers both measure the time separation and spatial separation of two
explosions that occur in interstellar space. One observer finds the explosions to
be separated by 22 s and 5.5 x 109 m.
a. Is it possible that the other observer detects the two explosions as
simultaneous? If so, how far apart are the events in this second frame?
b. Is it possible that the other observer detects the two explosions as occurring
at the same point? If so, how far apart in time are the two explosions in this
second frame?
In classical three-dimensional space, the distance between two points can be determined by
Pythagoras’ Theorem,
(dr ) 2  (dx) 2  (dy) 2  (dz ) 2
This distance is the same for any two observers, even if their individual measurements of x-,
y-, and z-separation are different. This relationship forms a method of calculating distance in
three-dimensional space and is referred to as the metric of three-dimensional flat space.
In Special Relativity, Pythagoras’ Theorem is not a valid way to calculate the distance
between two points. Hermann Minkowski discovered that if the temporal (dt) and spatial (dx,
dy, dz) separation between two events are combined in the following way,
(ds) 2  (cdt ) 2  ((dx) 2  (dy) 2  (dz ) 2 )
the resulting quantity, the spacetime interval, is the same for all observers. This result is the
metric of the four-dimensional flat spacetime that obeys Special Relativity. This metric is
referred to as the Minkowski metric.
Since this combination of spatial and temporal separations is the same for all observers, we
can use it to answer the above question. Label the two observers #1 and #2, and, if the events
are simultaneous for observer #2, dt2=0.
(cdt1 ) 2  ((dx1 ) 2  (dy1 ) 2  (dz1 ) 2 )  (cdt 2 ) 2  ((dx2 ) 2  (dy 2 ) 2  (dz 2 ) 2 )
((3x108 )(22)) 2  ((5.5 x10 9 ) 2 )  (0) 2  ((dx2 ) 2  (dy 2 ) 2  (dz 2 ) 2 )
1.33x1019  ((dx2 ) 2  (dy 2 ) 2  (dz 2 ) 2 )
 1.33x1019  (dx2 ) 2  (dy 2 ) 2  (dz 2 ) 2
Since the distance between events in frame #2 cannot be the square-root of a negative
number, it is impossible for any other observer to see these two events as simultaneous.
3
For part b, if the two events are to be located at the same point,
(cdt1 ) 2  ((dx1 ) 2  (dy1 ) 2  (dz1 ) 2 )  (cdt 2 ) 2  ((dx2 ) 2  (dy 2 ) 2  (dz 2 ) 2 )
((3x108 )(22)) 2  ((5.5 x10 9 ) 2 )  (cdt 2 ) 2  (0 2 )
1.33x1019  (cdt 2 ) 2
3.65 x10 9  cdt 2
dt 2  12.2s
Therefore, it is possible that another observer can see the two explosions as occurring at
exactly the same point in space, separated by 12.2 s of time.
4
Minkowski Metric in Polar Coordinates
We are free to express the Minkowski metric in
whatever coordinate system is most useful for the
problem under investigation. For example, the
metric expressed in polar coordinates is:
(ds) 2  (cdt ) 2  (dr ) 2  r 2 (d ) 2
Notice (in the diagram at left) that each small
“step” in the radial direction, dr, is exactly the
same length. The metric reflects this fact because
there is no multiplicative factor in front of the dr
term.
However, the length of a “step” in the tangential
direction, d, depends on how far you are from
the origin. The farther you are from the origin,
the longer a single step in dis. This is reflected
by the multiplicative factor of ‘r’ in the tangential
part of the metric. (As r gets big, a single step in
d gets big.)
What would it mean if there were similar multiplicative factors in front of the temporal and radial parts of
the metric? A factor in front of the dt part of the metric would mean that steps in dt (i.e., clicks of a clock)
would be of different duration in different places. A factor in front of the radial part would mean that radial
steps were of different lengths in different places. It would be like taking the coordinate grid shown above
and stretching it by different amounts in different places, resulting in a coordinate system (and underlying
space that the coordinate system is trying to represent) that is no longer flat, but rather warped or curved.
For example, if the radial part of the metric was
multiplied by a factor that got large as r got small,
this would result in a coordinate system that
looked something like the one at left. Notice that
the radial steps get larger and larger as the radial
distance gets smaller and smaller. Also notice that
there is really no way to adequately draw this
coordinate system on a flat surface; the
coordinate system is intrinsically curved.
5
Defining the Schwarzchild Metric
In General Relativity, the flatspace Minkowski metric cannot be used to describe spacetime.
In fact, the metric depends (in a very complicated way) on the exact distribution of mass and
energy in its vicinity. For a perfectly spherical distribution of mass and energy, the metric is
(ds) 2  (1 
2GM
)(cdt ) 2 
2
c r
(dr ) 2
 r 2 (d ) 2
2GM
(1  2 )
c r
This metric is referred to as the Schwarzchild metric, and describes the shape of space near a
spherical mass such as (approximately) the earth or the sun, as well as the space surrounding a
black hole. There are a number of subtle points you need to understand to use this metric.
1.
This is the metric for a slice of spacetime that contains the mass center. Since the
mass is spherical, all slices through the mass center are identical. The metric is
expressed in polar coordinates, (r, ), with the mass center at the origin.
2.
Notice that the tangential component of the metric is unchanged from the Minkowski
metric, meaning there is no deformation in that direction. However, both the
temporal and radial portions are deformed by multiplicative constants, so radial
lengths and time intervals are different in different locations of space.
3.
Notice that as M goes to zero, or r gets very large, the metric approaches the
Minkowski (flatspace) metric.
4.
t is the flatspace time, the time measured on clocks very far from the central mass,
where spacetime is assumed to be flat.
5.
r is the reduced circumference. There are several ways to measure your distance
from an object, such as physically traveling to the object or bouncing a signal off the
object. If you tried to measure your distance from a black hole in either of these
manners, you would have a very tough time, because either you (or your signal)
would never return. Therefore we need a different method of determining radial
distance. To do this, we will imagine wrapping a tape measure around the black hole,
measuring its circumference, and then dividing the circumference by 2. The
resulting number is termed the reduced circumference, and, in flat space, would
actually equal the value of our radial distance. (It won’t equal the “real” radial
distance from the black hole because the “real” radial distance is undefined (and
undefinable!).)
6.
Notice that the metric diverges (becomes infinite) at r horizon = 2GM/c2. Thus a single
radial step at this location is infinitely long (and it appears that a single clock tick has
no duration)! This “radius” (actually reduced circumference but we’ll be sloppy and
call it radius from now on) is termed the Schwarzchild radius and forms the event
horizon of the black hole. At, or within, this radius, events are “beyond the horizon”,
meaning they are unseen and unseeable from radii greater than the Schwarzchild
radius. Basically, once you pass over the horizon, you are no longer in contact with
the rest of the universe. Ever.
6
Using the Schwarzchild Metric: Time
How close to a black hole of 5 solar masses can you approach before your
spaceship’s clock differs from time measured in flat spacetime by no more than
1%?
Regardless of where you are in space, if you make your measurements over a small enough
region of spacetime that region of spacetime will appear locally flat, just like a straight
tangent line can be drawn at any point on a smooth curve. Therefore, the ship’s measurements
are made using a standard Minkowski metric while the faraway observer must use the
Schwarzchild metric since the spaceship’s clock is far from her location.
(ds ) 2ship  (ds) 2faraway
2GM
(cdt )  ((dx)  (dy )  (dz ) )  (1  2 )(cdt ) 2 
c r
2
2
2
2
(dr ) 2
 r 2 (d ) 2
2GM
(1  2 )
c r
We’ll assume your spaceship is at rest, in both frames of reference, so dx = dy = dz = dr = 0
and d = 0.
2GM
)(cdt faraway) 2
c2r
2GM
 (1  2 )1 / 2 dt faraway
c r
(cdt ship ) 2  (1 
dt ship
The above expression plays a similar role in general relativity that the time dilation
relationship plays in special relativity. It relates the time interval measured by a “special”
observer (one at rest in curved space) to another observer’s time measurements.
Mathematically, it even has a similar structure, with the term “2GM/r” playing the role of “v 2”
in the time dilation formula.
Continuing with the question:
dt ship
dt faraway
 (1 
2GM 1 / 2
)
c2r
2(6.67 x10 11 )(10 x10 30 )
)
(3 x10 8 ) 2 r
14800
0.9801  1 
r
r  744,000m  744km
(0.99) 2  (1 
Since the event horizon is at rhorizon = 2GM/c2 = 15 km, you are about 50 event horizons away
from the black hole.
7
Using the Schwarzchild Metric: Length
Two astronauts are creating a (metric) football field near a 10 solar mass black
hole. The reduced circumference between the two astronauts is 100 m, and the
astronauts lie along the same radial line. What is the radial separation between
the astronauts as measured by the inner astronaut, if the inner astronaut is at
twice the event horizon?
Again, we’ll assume the spacetime immediately surrounding the inner astronaut is locally flat,
allowing the astronaut to use the Minkowski metric. Since the separation between the
astronauts is expressed in terms of reduced circumference, this can be incorporated into the
Schwarzchild metric. Thus,
(ds ) i2nner astronaut  (ds ) 2faraway
2GM
(cdt )  ((dx)  (dy )  (dz ) )  (1  2 )(cdt ) 2 
c r
2
2
2
2
(dr ) 2
 r 2 (d ) 2
2GM
(1  2 )
c r
To measure the distance between two points, the location of the two points must be
determined at the same time, so dt = 0 in both reference systems. Additionally, since the
points lie along the same radial line, d = 0. Calling this line the x-axis allows us to set dy =
dz = 0. Thus,
(dr ) 2
2GM
(1  2 )
c r
dr
dx astronaut 
2GM
(1  2 )1 / 2
c r
 (dx) 2astronaut  
The above expression plays a similar role in general relativity that the length contraction
relationship plays in special relativity. It relates the spatial interval measured by a “special”
observer (one at rest in curved space) to another observer’s spatial measurements.
Mathematically, it even has a similar structure, with the term “2GM/r” playing the role of “v 2”
in the length contraction formula.
Substituting r = 2rhorizon = 4GM/c2 and dr = 100 m yields
100 m
1
(1  )1 / 2
2
dx astronaut  141.42 m
dx astronaut 
8
Thus, even though the two astronauts only differ by 100 m in reduced circumference, the
distance between the astronauts, measured by the inner astronaut, is 141.42 m. Thus,
spacetime is stretched by a factor of 41% compared to flat spacetime. This provides a measure
of how “warped” spacetime is at this location.
Are you upset with my sloppy use of calculus in the previous example? You should be. The
metrics relate differential changes in time and space (dt and dr) and I just plugged in 100 m
for dr. Is 100 m infinitesimally small? It depends …
More carefully, I should integrate the expression for dxastronaut between the two limits, from
2rhorizon to (2rhorizon + 100 m).
x astronaut 
2 rhorizon100m

2 rhorizon
dr
2GM
(1  2 )1 / 2
c r
This integral is ugly for two reasons: the variable is in the denominator of a fraction that’s in
the denominator of the expression, and the integral has a bunch of constants. It’s easy to get
rid of the constants by using the definition of the event horizon,
x astronaut 
2 rh 100m

2 rh
where rh 
dr
r
(1  h )1 / 2
r
2GM
c2
To solve the more difficult problem, multiply the numerator and denominator by a skillfully
chosen factor:
x astronaut 
2 rh 100m

2 rh
x astronaut 
2 rh 100m

2 rh
r
( )1 / 2 dr
rh
r
r
( )1 / 2 (1  h )1 / 2
rh
r
r
dr
rh
r
1
rh
Next, notice that the combination of terms that appears in the integral is dimensionless,
meaning it has no units. It is always a very good idea to try to simplify complicated integrals
in terms of dimensionless factors.
9
Perform a u-substitution where u is equal to this dimensionless factor and simplify:
with u 
du 
r
,
rh
1
dr ,
rh
lower limit : u 
(2rh )
2
rh
(2rh  100 m)
100c 2
upper limit : u 
 2
 2.03385
rh
2GM
1
x astronaut 
rh
2.03385

2
u
du
u 1
2GM
2.03885
[ u (u  1)  ln( u  u  1)]2
2
c
x astronaut  140.83 m
x astronaut 
Thus, the actual distance between the astronauts, measured by the inner astronaut, is 140.83
m. For this problem, 100 m is “small enough” to be considered infinitesimally small, since the
correct answer differs from the approximate answer by less than 1%. The correct answer is
less than the approximate answer because the correct answer takes into account that space is
less stretched as you move out toward the second astronaut, while the approximate answer
approximates the stretching of space as being constant between the astronauts.
10
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Spacetime and General Relativity
Activities
11
The merry-go-round gedanken experiment illustrates how Einstein came to believe that acceleration curves
space (and time). Three students wrestle with the meaning of this gedanken experiment.
Otis:
Initially, the merry-go-round is rotating at constant speed. Then, to speed it up energy has to
be transferred to the merry-go-round. This energy goes into the merry-go-round, causing its
shape to warp or curve. This lead Einstein to conclude that the energy needed to accelerate an
object must result instead in the curving of the space surrounding the object.
Tina:
When the merry-go-round is spinning, its outer edge, or circumference, must contract due to
length contraction. Since the merry-go-round is not moving in the radial direction, the radius
of the merry-go-round is constant. If the circumference of a circle decreases without a
decrease in radius, the object cannot remain a flat circle. The object must warp or bow, kinda
like a portion of a hollow sphere. Since even a constant speed merry-go-round is accelerating,
this lead Einstein to associate acceleration with spatial curvature.
Brett:
Since the merry-go-round is moving, it must shrink in size due to length contraction.
However, since different parts of the merry-go-round are moving at different speeds, each
piece shrinks by a different amount. If different parts of an object shrink by different amounts,
the object becomes warped. Thus, when you have varying velocity (or acceleration) you get
warping of space.
Which, if any, of the students are correct? For each incorrect response, provide a short explanation why it is
incorrect. If no one is correct, provide a correct answer below.
12
The man-in-the-elevator gedanken experiment illustrates how Einstein came to believe that the effects of
acceleration and the force of gravity are indistinguishable. The three students also wrestle with the meaning
of this gedanken experiment.
Otis:
If the man in the elevator steps on the bathroom scale, and the bathroom scale reads his
weight, the man can conclude one of two things: Either I am at rest in a gravitational field or I
am accelerating through space at exactly 9.8 m/s2. Since there is no way for him to
experimentally distinguish between these two possibilities, he must accept that there is no
fundamental difference between acceleration and the effects of the force of gravity.
Tina:
It’s actually much simpler than that. Since the force of gravity always results in a dropped
object experiencing an acceleration, the two concepts, gravity and acceleration are
fundamental intertwined.
Brett:
The point of the man-in-the-elevator gedanken experiment is to show that if we don’t know
what reference system we are using when we measure something, like our weight, then the
conclusions we draw can appear to be relative. Just as in Special Relativity, were the time
between events is different for different observers, in General Relativity we learn that the
force of gravity acting on us (our weight) is different for different observers.
Which, if any, of the students are correct? For each incorrect response, provide a short explanation why it is
incorrect. If no one is correct, provide a correct answer below.
13
Imagine that you and a friend are riding in trains that are moving relative to each other at relativistic
speeds. As you pass each other, you both measure the time separation and spatial separation of two
firecracker explosions that occur on the tracks between you. You find the firecracker explosions to be
separated by 1.0 s of time and 500 m of distance in your frame. By radio, your friend reports that the
explosions were separated by only 0.80 s of time. Find the spatial separation of the events in your friend's
frame.
Mathematical Analysis
14
Two observers both measure the time separation and spatial separation of two explosions that occur in
interstellar space. One observer finds explosions to be separated by 5 hrs of time and 7.5 light-hours of
distance.
a. Is it possible that the other observer detects the two explosions as simultaneous? If so, how far apart
are the events in this second frame?
b. Is it possible that the other observer detects the two explosions as occurring at the same point? If so,
how far apart in time are the two explosions in this second frame?
Mathematical Analysis
15
Two observers both measure the time separation and spatial separation of two explosions that occur in
interstellar space. One observer finds explosions to be separated by 7.3 minutes of time and 6.5 lightminutes of distance.
a. Is it possible that the other observer detects the two explosions as simultaneous? If so, how far apart
are the events in this second frame?
b. Is it possible that the other observer detects the two explosions as occurring at the same point? If so,
how far apart in time are the two explosions in this second frame?
Mathematical Analysis
16
The nearest star to our Sun is Proxima Centauri, located 4.28 light-years away (measured by a very large,
stationary meterstick).
a. I would like to build a spaceship that will allow me to get to Proxima Centauri in 10 years (as
measured by a clock I leave behind in my office at work). Using only the metric for flat spacetime, how
much time will elapse on my dashboard clock?
b. I would like to build a spaceship that will allow me to get to Proxima Centauri in 10 years (as
measured by a clock on the dashboard of my ship, right under the fuzzy dice). Using only the metric for flat
spacetime, how much time will elapse on the office clock?
Mathematical Analysis
17
Imagine an alien spaceship traveling so fast that it travels from earth to the nearby star Vega (a distance of
25 light-years measured by earthlings) in only 2.0 yrs of spaceship time. Using only the metric for flat
spacetime, how much time will elapse on earth during this journey?
Mathematical Analysis
18
Imagine a particle traveling in a particle accelerator of length 2000 m at a constant speed of 0.9994c (as
measured by laboratory observers).
a. How long does it take the particle to travel the length of the accelerator in the laboratory frame?
b. Using only the metric for flat spacetime, how much time will elapse from the particle’s point of view?
Mathematical Analysis
19
How close to the center of the sun can you approach before your spaceship’s clock differs from time
measured in flat spacetime by no more than
a. 1.0 s per year? Is this larger than or smaller than the earth’s orbital radius?
b. 1.0 s per week? Is this larger than or smaller than the sun’s radius?
c. At what rate would your clock run slow on the surface of the sun?
Mathematical Analysis
20
By how many seconds per year do clocks on the surface of the earth run slow compared to clocks in flat
spacetime due to
a. the curvature created by the sun?
b. the curvature created by the earth?
Mathematical Analysis
21
How close to the black hole at the center of our galaxy (approximately 2.5 x 106 solar masses at a distance
of 4.7 x 1020 m) can you approach before your spaceship’s clock differs from time measured in flat
spacetime by no more than
a. 1.0 s per year?
b. 1.0 s per week?
c. At what rate do clocks on earth run slow due to the curvature created by this black hole?
Mathematical Analysis
22
How close (in multiples of the event horizon) can you get to a black hole before time is slowed by
a. 1%?
b. 10%?
c. 100% (i.e., time measured near the hole is twice flatspace time)?
Mathematical Analysis
23
Imagine two stationary observers near the sun, differing in reduced circumference by 1.0 m. What is the
radial separation between these observers, as measured by the inner observer, if the inner observer is at a
reduced circumference
a. equal to 10,000 times the radius of the sun?
b. equal to 100 times the radius of the sun?
c. equal to the radius of the sun?
Mathematical Analysis
24
How close (in multiples of the event horizon) can you get to a black hole before space is stretched by
a. 1%?
b. 10%?
c. 100% (i.e., radial separations measured near the hole are twice the difference in reduced
circumference)?
Mathematical Analysis
25
Imagine two stationary observers near a black hole, differing in reduced circumference by 1.0 m. What is
the radial separation between these observers as measured by the inner observer, if the inner observer is at
a reduced circumference
a. equal to 100 event horizons?
b. equal to 10 event horizons?
c. equal to 1.001 event horizons? (Hint: Are you too close to the event horizon to let dr = r?)
Mathematical Analysis
26
No real test of gravitational time dilation can involve a clock measuring flatspace time, although the
concept of flatspace time does allow you to compare two clocks at different locations in a gravitational
well. Imagine a clock at rest on the earth’s surface and another at the top of a building, a distance H above
the first clock. Ignore the rotation of the earth.
a. Write an expression for the time measured by the clock a distance h above the surface of the earth in
terms of flatspace time.
b. Write an expression for the time measured by the clock on the surface of the earth in terms of flatspace
time.
c. Divide your two expressions to find the ratio of the two clock readings. This expression does not
depend on flatspace time. (If you like, you can use the binomial expansion to simplify this expression
quite a bit.)
d. At what rate do clocks on the surface of the earth run slow compared to clocks at the top of the Empire
State Building (381 m)?
Mathematical Analysis
27
Imagine a photon of energy E shot directly upward from the surface of the earth. In classical physics, as
the photon increases its height above the surface of the earth its gravitational potential energy (mgH)
would increase causing its kinetic energy to decrease. Of course, since the photon is massless this is a silly
discussion, but let’s pretend that the photon has an “effective mass” given by E = mc 2, meaning m = E/c2.
a. Write an expression for the kinetic energy of the photon when it reaches a height H above the surface
of the earth, using the above expression for the effective mass of the photon.
b. Since the (kinetic) energy of a photon is directly related to its frequency (E=hf), set your expression
equal to hfH, where fH is the frequency of the photon when it is H above the surface of the earth.
c. Replace the initial energy of the photon with hf0, and solve the resulting expression for the ratio of f H/
f0.
Although we used dubious classical methods, this expression correctly quantifies gravitational redshift, the
observed phenomenon that the frequency of light decreases (is shifted toward the red) as light travels away
from the earth’s surface.
d. What is the fractional change in frequency as light travels from the base to the top of the Empire State
Building (381 m)?
Mathematical Analysis
28
Consider a 6 solar mass black hole. A stationary observer at reduced circumference r shines a laser of
wavelength 400 nm radially outward. What is the wavelength of the light detected by a remote observer if
a. r = 100 event horizons?
b. r = 10 event horizons?
c. r = 1.001 event horizons?
Mathematical Analysis
29
A spaceship exploring a 20 solar mass black hole travels very slowly radially inward from a reduced
circumference of 50 event horizons to dangerously close to the “edge”, 1.01 event horizons.
a. What is the change in reduced circumference, expressed in meters?
b. What distance, in meters, does the ship’s odometer record for this journey? (Hint: dr ≠ r)
Mathematical Analysis
30
A spaceship returning to earth after a reconnaissance mission to a 50 solar mass black hole travels very
slowly radially outward from very close to the “edge”, 1.001 event horizons, back to earth, which is very
far away. By how much does the ship’s odometer reading differ from distance measurements made on
earth?
Mathematical Analysis
31