Download Black Holes and Special Relativity

Astronomy 114
Lecture 24: Black Holes & Relativity
Martin D. Weinberg
[email protected]
UMass/Astronomy Department
A114: Lecture 24—06 Apr 2007
Read: Ch. 24
Astronomy 114—1/15
Announcements
PS#5 solutions posted today.
PS#6 posted on line (due next Wednesday)
Exam next Friday covering: Stars, Stellar Evolution,
Stellar remnants
A114: Lecture 24—06 Apr 2007
Read: Ch. 24
Astronomy 114—2/15
Announcements
PS#5 solutions posted today.
PS#6 posted on line (due next Wednesday)
Exam next Friday covering: Stars, Stellar Evolution,
Stellar remnants
Today:
Black holes
Relativity
Black Holes, Chap. 24
A114: Lecture 24—06 Apr 2007
Read: Ch. 24
Astronomy 114—2/15
Announcements
PS#5 solutions posted today.
PS#6 posted on line (due next Wednesday)
Exam next Friday covering: Stars, Stellar Evolution,
Stellar remnants
Today:
Black holes
Relativity
Black Holes, Chap. 24
Next up (Wed?): Optics and Telescopes
Optics and Telescopes, Chap. 6
A114: Lecture 24—06 Apr 2007
Read: Ch. 24
Astronomy 114—2/15
Conservation of Energy
Example: potential energy is energy that can be
converted to kinetic energy by moving in a force field
v=0
v=vf
Potential
energy is
positive to start
Kinetic energy
is gained as ball
rolls down hill
Ground level
Potential energy
is zero at bottom of hill
A114: Lecture 24—06 Apr 2007
Read: Ch. 24
Astronomy 114—3/15
Conservation of Energy
Example: potential energy is energy that can be
converted to kinetic energy by moving in a force field
v=0
v=vf
Ground level
v=0
Kinetic energy
is gained as ball
rolls into well
Potential energy
is negative at
bottom of well
v=vf
A114: Lecture 24—06 Apr 2007
Potential
energy is zero
at top of well
Read: Ch. 24
Astronomy 114—3/15
Conservation of Energy
Example: potential energy is energy that can be
converted to kinetic energy by moving in a force field
v=0
v=vf
Ground level
v=0
Potential
energy is zero
at top of well
Kinetic energy
is gained as ball
rolls into well
Potential energy
is negative at
bottom of well
Total energy (kinetic plus potential) is conserved
v=vf
A114: Lecture 24—06 Apr 2007
Read: Ch. 24
Astronomy 114—3/15
Escape velocity (1/2)
Need a critical velocity to escape the surface of a
gravitating body
To escape the gravitational pull of a planet, need an
initial velocity that will make your total mechanical
energy be unbound
Total mechanical energy of a blob (the energy of
motion) added to the energy given up by falling in:
E = KE + P E
1 2 GM m
mv −
=
2
R
A114: Lecture 24—06 Apr 2007
Read: Ch. 24
Astronomy 114—4/15
Escape velocity (2/2)
Energy must be greater than zero to be unbound:
1 2 GM m
mv −
0 =
2
R
s
2GM
vesc =
R
Examples:
Earth: vesc = 11.2 km/s
Moon: vesc = 2.4 km/s
A114: Lecture 24—06 Apr 2007
Read: Ch. 24
Astronomy 114—5/15
Schwarzschild Radius
For a given velocity and mass, we can compute the
radius we need to just escape:
2GM
R= 2
vesc
A114: Lecture 24—06 Apr 2007
Read: Ch. 24
Astronomy 114—6/15
Schwarzschild Radius
For a given velocity and mass, we can compute the
radius we need to just escape:
2GM
R= 2
vesc
Nothing moves faster than the speed of light. . .
Light cannot escape from a Black Hole if it comes from a
radius closer than the Schwarzschild Radius, Rs :
2GM
Rs =
c2
where M is the mass of the hole.
A114: Lecture 24—06 Apr 2007
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Astronomy 114—6/15
Sizes of black holes
2GM
Rs =
c2
A black hole with a mass of 1 M⊙ has Rs =3 km
0.6 M⊙ White Dwarf: radius of 1 Rearth = 6370 km
1.4 M⊙ Neutron Star: radius of 10 km
A114: Lecture 24—06 Apr 2007
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Astronomy 114—7/15
Implications for the Laws of Physics
Need to modify laws of motion to think about
velocities near the speed of light (Special Relativity)
Newton’s theory gravity implies that the force
propagates infinitely fast (action at a distance)
Also need to modify law of gravity to incorporate new
laws of motion
General Relativity:
Einstein’s theory of gravitation (1915)
First solutions by Karl Schwarzschild (1916)
Not taken seriously until the 1960s
A114: Lecture 24—06 Apr 2007
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Astronomy 114—8/15
New theory of motion
Thought experiment. . .
Q: Two cars have a head-on collision. Each is moving at
45 mph. What is their relative speed?
A114: Lecture 24—06 Apr 2007
Read: Ch. 24
Astronomy 114—9/15
New theory of motion
Thought experiment. . .
Q: Two cars have a head-on collision. Each is moving at
45 mph. What is their relative speed?
A: 90 mph
A114: Lecture 24—06 Apr 2007
Read: Ch. 24
Astronomy 114—9/15
New theory of motion
Thought experiment. . .
Q: Two cars have a head-on collision. Each is moving at
45 mph. What is their relative speed?
A: 90 mph
Q: Two spaceships have a head-on collision. Each is
moving at 0.9 c. What is their relative speed?
A114: Lecture 24—06 Apr 2007
Read: Ch. 24
Astronomy 114—9/15
New theory of motion
Thought experiment. . .
Q: Two cars have a head-on collision. Each is moving at
45 mph. What is their relative speed?
A: 90 mph
Q: Two spaceships have a head-on collision. Each is
moving at 0.9 c. What is their relative speed?
A: 1.8 c? No, it is 0.99 c (according to special relativity)
A114: Lecture 24—06 Apr 2007
Read: Ch. 24
Astronomy 114—9/15
New theory of motion
Thought experiment. . .
Q: Two cars have a head-on collision. Each is moving at
45 mph. What is their relative speed?
A: 90 mph
Q: Two spaceships have a head-on collision. Each is
moving at 0.9 c. What is their relative speed?
A: 1.8 c? No, it is 0.99 c (according to special relativity)
Why do we need a new theory?
A114: Lecture 24—06 Apr 2007
Read: Ch. 24
Astronomy 114—9/15
Special Relativity
What is the medium that propagates light?
c. 1900: scientists look for differences in the speed of
light in different directions
Michelson-Morley experiment: no difference
Einstein showed that this can be explained assuming
the speed of light is constant for all observers
A114: Lecture 24—06 Apr 2007
Read: Ch. 24
Astronomy 114—10/15
Special Relativity
Two principles of Special Relativity:
The laws of physics must be the same in all
reference frames
(True in Galilean relativity, Newtons Laws, too)
Speed of light must be constant in all reference
frames (New!)
A114: Lecture 24—06 Apr 2007
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Astronomy 114—10/15
Consequences of Special Relativity (1/2)
Consider two observers moving at near the speed of
light relative to each other
At the instant that they are at the same place, one
observer fires a flash bulb or strobe
Sphere of light propagates from this point
c
c
A
B
v=0.8 c
c
What does each observer see?
c
A114: Lecture 24—06 Apr 2007
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Astronomy 114—11/15
Consequences of Special Relativity (2/2)
Each observer must see a sphere of light.
How can this be?
Stationary observer A sees
q
B’s length contracts: L = Lo 1 − v 2 /c2
q
B’s clock dilates: T = To / 1 − v 2 /c2
q
B’s energy increases: E = Eo / 1 − v 2 /c2
From B’s point of view, the same thing happens to A.
Both observers A & B see light sphere from the flash
bulb.
A114: Lecture 24—06 Apr 2007
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Astronomy 114—12/15
Photon clock
Photon Clock
1.5 m
Clock ticks when pulse
arrives, sends new pulse
Pulse from laser travels 3
meters to receiver
t = 3m/3 × 108 m/s = 10−8 s
Laser
source
Photocell
detector
A114: Lecture 24—06 Apr 2007
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Astronomy 114—13/15
Photon clocks in moving frames
As seen by spaceship pilot (B) at v = 0.8c
A114: Lecture 24—06 Apr 2007
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Astronomy 114—14/15
Photon clocks in moving frames
v=0.8 c
As seen by mission control (A)
A114: Lecture 24—06 Apr 2007
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Astronomy 114—14/15
More consequences
Lengths and clocks
Simultaneous events in one frame are not
simultaneous in another
Pole and barn paradox
A114: Lecture 24—06 Apr 2007
Read: Ch. 24
Astronomy 114—15/15
More consequences
Lengths and clocks
Simultaneous events in one frame are not
simultaneous in another
Pole and barn paradox
80 m pole, 40 m barn with doors at both
ends
Runner runs through barn at v = 0.5c
√
In barn frame: L = 80 1 − 0.52 = 34.9 m
√
Runner: barn is L = 40 1 − 0.52 = 17.4 m
Barn doors close and open simultaneously in barn’s frame but not in Runner’s
frame
A114: Lecture 24—06 Apr 2007
Read: Ch. 24
Astronomy 114—15/15
More consequences
Lengths and clocks
Simultaneous events in one frame are not
simultaneous in another
Pole and barn paradox
Twin paradox
A114: Lecture 24—06 Apr 2007
Read: Ch. 24
Astronomy 114—15/15
More consequences
Lengths and clocks
Simultaneous events in one frame are not
simultaneous in another
Pole and barn paradox
Twin paradox
One twin travels at 0.99c to star 10 ly away
Travelling twin sees length
to star
√
contracted: L = 10 ly 1 − 0.992 = 1.41 ly
Time to get there and back:
T = 2 × 1.41 ly/c = 2.82 years
Twin on Earth sees: 2 × 10 ly/c = 20 years
A114: Lecture 24—06 Apr 2007
Read: Ch. 24
Astronomy 114—15/15