Announcements
•Homework: Supplemental Problems
•2nd Project is due at the final exam which is
4:00pm Monday May 2.
•Final Exam will be another hour exam
covering the material we cover since the last
exam. It will be the second thing during the
final exam period (project presentation first)
Black Holes!
Black holes are places where space-time is
warped beyond the breaking point.
The physics of black holes is
Einstein’s General Theory of
Relativity
R

8G 
 4 T
c
For the full form, see the Einstein’s Field Equation link
on the class website
The Schwarzschild Metric
Karl Schwarzschild worked out
Einstein’s equations of general
relativity for a non-rotating point
mass while serving as an artillery
officer in the German army in WWI.
His “metric” describes space-time
near the object. The Schwarzschild
radius is given by
2GM
RS  2
c
 RS
s  1 
R

1
 2 2
 c t 
 RS

1  R





R 2  R 2  2   2 sin 2 

Consequences of the
Schwarzschild Metric
Time Dilation: a clock at a distance R from the singularity
will appear to run slow as compared to a clock at infinity
RS
  1  t
R
Length Stretching: a meter stick placed at a distance R
from the singularity will appear longer as measured by an
observer at infinity
Watch
Spaghettification
video
L
L
RS
1
R
Examples
Determine the Schwarzschild radius of the Sun.
By what factor is time slowed for a clock placed
at twice the Schwarzschild radius from the
singularity of a black hole?
By what factor is a 1.0 m long meter stick appear
lengthened to an observer far away if it placed at
twice the Schwarzschild radius from the
singularity of a black hole?
Example Solution
From Appendix 2 of the textbook Msun = 1.9891 x 1030 kg
2GM
RS  2 
c
  1 

 1.989110
 2.9979 10 
2 6.67  1011
Nm 2
kg 2
8 m
30
kg
2

 2952m
s
RS
R
R

1
t 
 1  S  1  S  1   0.707
R
t
R
2RS
2
The clock near the black hole runs about 70.7% as fast
as the far away clock (or 29.3% slower)
L
L
1.0m
1.0m


 1.41m
RS
RS
1  0.5
1
1
R
2 RS
To a far away observer, the 1.00 m stick would
appear to be 1.41 m long
Rotating Black Holes
A Kerr black hole has two
event horizons
The two even horizons are related
to the angular momentum


Rinner
G
 2 M  M 2  a2
c
Router
G
 2 M  M 2  a 2 cos 2 
c


Lc
a
MG
L is the angular momentum of the black hole
and  is the angle from the rotation axis of
the black hole
The Heisenberg Uncertainty
Principle is one of the
fundamental tenants of
Quantum Mechanics
Et 
E is the energy. Basically the rest mass
energy of the particles (E = mc2). t is how
long a virtual pair exists for and
is
Planck’s constant divided by 2
 1.05457 1034 J  s
A consequence of the
Uncertainty Principle is that
empty space isn’t empty
Steven Hawking applied the
uncertainty principle to the
space near a black hole
Example
An electron-positron virtual pair is created near the
event horizon of a black hole. How long can the pair
last for? If the electron is moving at almost the speed
of light, how close to the event horizon must it be to
fall in before the pair annihilates?
Example Solution
E t   t 
E

1.05457 1034 J  s
2  9.1094 10 kg  2.9979 10
31
8 m
s

2
 6.44 1022 s
Now that we know how long the pair can last we
can find how far they can travel during that time
d
v   d  vt   2.9979 108 m s  6.44 1022 s   1.93 1013 m
t
This is larger than the diameter of a nucleus
(10-15 m) but smaller than the diameter of an
atom (10-12 m)
The Temperature of a black
hole depends on its mass
3
hc
T
8 GMkb
h = Planck constant = 6.626 x 10-34 J-s
Kb = Boltzmann constant = 1.38065 x 10-23 J/K
The emission of a black hole is very similar to that
of a perfect black body at the given temperature
Because they radiate, the will
eventually evaporate away
 BH
5120 G M

4
hc
2
3
The lifetime of a black hole is proportional to its
mass raised the third power
Examples
Determine the temperature and lifetime of a
5.0 solar mass black hole.
Determine the temperature and lifetime of
the supermassive black hole at the center of
the Milky Way which has a mass of 4.1 x 106
solar masses
Example Solution
For a 5.0 solar mass black hole
3

hc
T

8 GMk B 8 6.67 1011


6.626 1034 J  s 2.9979 108 m s
N  m2
kg 2

3
 5 1.989110 kg 1.38065 10 
23 J
K
30
 7.76 108 K
 BH

11 N  m2
kg 2

2
5 1.989110 kg
5120G 2 M 3 5120 6.67 10


4
4
34
8
hc
6.626 10 J  s 2.9979 10


30

1year
68
 4.188 10 seconds 

1.33

10
years
7
3.156 10 s
75

3
Example Solution 2

3
hc
T

8 GMkB 8 6.67 1011


6.626 1034 J  s 2.9979 108 m s
N  m2
kg 2

3
  4.110 1.989110 kg 1.38065 10 
6
23 J
K
30
 9.46 1014 K
 BH

11 N  m2
kg 2

2
4.110 1.989110 kg
5120G 2 M 3 5120 6.67 10


4
4
34
8
hc
6.626 10 J  s 2.9979 10

6

1 year
92
 9.47 10 seconds 

3.00

10
years
7
3.156 10 s
99
30


3
Just for the fun of it!
We now know the temperature and radius of a one
solar mass black hole. Assuming it is a perfect black
body, what is it’s luminosity and peak wavelength?
T = 3.88 x 10-7 K
RS = 2952 m
L  4 R 2 T 4

 4  2952m  5.67 10
2
8
W
m2 K 4
 3.88 10 K 
7
4
 1.411025W
 peak
0.0029m  K
.0029m  K


 7474m  7.47km
7
T
3.88 10 K
The energy of a single photon of this wavelength is
2.658 x 10-29 J so this is ~530 photons per second