Transcript Document
Today in Astronomy 102:
black holes and how to prevent them
The Schwarzschild singularity and the sizes of black
holes.
Degeneracy pressure: a quantum-mechanical effect that
might stop matter from collapsing to form a black hole,
when gas pressure or material strength aren’t enough.
9 October 2001
Astronomy 102, Fall 2001
1
Karl Schwarzschild’s Work
In 1916 Schwarzschild read Einstein’s
paper on general relativity. He was
interested in the physics of stars, and had
a lot of spare time between battles on the
Russian front, so he solved Einstein’s field
equation for the region outside a massive
spherical object.
His solution had many interesting
features, including
prediction of space warping in strong
gravity, and invention of embedding
diagrams to visualize it.
verification gravitational time dilation,
just as Einstein had pictured it
prediction of black holes, though this
was not recognized at the time.
9 October 2001
Astronomy 102, Fall 2001
Karl Schwarzschild
2
Einstein’s field equation
The field equation is the ultimate mathematical expression of
Einstein’s general theory of relativity.
Astronomy 102 version:
“Spacetime, with its curvature, tells masses how to move;
masses tell spacetime how to curve.”
Physics 413 - Astronomy 554 version:
2 l
x h 2 a x s 2 a
2 g ml
2 g mk
2 g lk
x h 2 a x s 2 a
1 gl
+ g hs
+
- a
a
l
a
m k
k
a
m l
2 x k x m x k x l x l x m x l x l
x xl
x x
x xl
x x
2 gl
h
h 2a s 2a
2 a s 2 a
2 g lm
2 g lm
1
1
x
x
x
x
l
+ g hs
- g mk
- m
- a
2 m
a
l
a
m
m
a
l
x l x m
x x l x m x
2
x x l x m x
2 x x m x x l
= -8pG
p
n
9 October 2001
nm
dx nk 3
d (x - xn )
dt
Astronomy 102, Fall 2001
3
What you get when you solve the field equation
In case you’re interested (i.e. not on the exam):
The solution to the field equation is a function called the
metric tensor. This function tells how much distance or
time is involved for unit displacements in the spacetime
coordinates.
Accordingly, the metric tensor is related to the absolute
interval. Each different solution to the field equation
corresponds to a different absolute interval. The absolute
interval corresponding to Schwarzschild’s metric turns
out to be given (in spherical coordinates) by
2
r
2GM
2
2
2
2
2
2
2
s =
+ r + r sin - c 1 - 2 t 2
2GM
rc
1- 2
rc
F
H
9 October 2001
Astronomy 102, Fall 2001
I
K
4
Calculated from Schwarzschild’s solution for an
object with M=c2/2G. (From lecture on 13 September)
20p
16p
Distances between the circles:
1.185
1.074
1.057
1.106
1.065
1.087
1.135
12p
8p
4p
1.300
All the distances would be 1
if space weren’t warped.
y
x
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Astronomy 102, Fall 2001
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One way to visualize warped space: “hyperspace”
(Also from lecture on 13 September)
To connect these circles
with segments of these
“too long” lengths, one
can consider them to
be offset from one
another along some
imaginary dimension
that is perpendicular
to x and y but is not z.
(If it were z, the circles
wouldn’t appear to lie
in a plane!). Such additional dimensions
comprise hyperspace.
9 October 2001
1.057
1.065
1.074
1.087
1.106
1.185
1.135
1.300
y
x
Hyperspace
Astronomy 102, Fall 2001
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Schwarzschild’s view of the
equatorial plane of a star
Figure from
Thorne, Black holes
and time warps
(equator)
Center of star
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Astronomy 102, Fall 2001
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Schwarzschild’s solution to the Einstein field
equations
Results: the curvature of spacetime outside a massive star.
For a given, fixed star mass M, he examined how space
and time are curved if the star is made smaller and
smaller in size.
Singularity: if the star is made smaller than a certain
critical size, the gravitational redshift of light (time
dilation, remember) predicted by his solution was infinite!
(See next slide.)
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Astronomy 102, Fall 2001
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Figure from Thorne, Black
holes and time warps
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Astronomy 102, Fall 2001
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Implications of “Schwarzschild’s singularity”
If a star is made too small in circumference for a given
mass, nothing can escape from it, not even light.
• This would be a black hole, and the critical size is the
size of the black hole’s horizon.
This is similar to an 18th century idea: “dark stars”
(Michell, Laplace...); that if light were subject to
gravitational force, there could be stars from which light
could not escape.
• The critical size of Schwarzschild’s singularity turns
out to be the same as that for the 18th century dark
star.
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Astronomy 102, Fall 2001
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The Schwarzschild singularity
According to Schwarzschild’s solution to Einstein’s field
equation for spherical objects, the gravitational redshift
becomes infinite (i.e. time appears to a distant observer to
stop) if an object having mass M is confined within a sphere
of circumference CS, given by
CS=
4pGM
c2
Schwarzschild
circumference
where G =6.67x10-8 cm3/(gm sec2) is Newton’s gravitational
constant, and c = 2.9979x1010 cm/sec is, as usual, the speed of
light (and p = 3.14159....).
You need to understand this formula.
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Astronomy 102, Fall 2001
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The Schwarzschild singularity (continued)
Any object with mass M, and
circumference smaller than CS, would not
be able to send light (or anything else) to
an outside observer -- that is, it would be
a black hole.
The sphere with this critical
circumference - the Schwarzschild
singularity itself - is what we have been
calling the event horizon, or simply the
horizon, of the black hole.
9 October 2001
Astronomy 102, Fall 2001
CS
12
Examples: calculation using the horizon
(Schwarzschild) circumference
Example 1: what is the horizon circumference of a 10M
black hole?
4p GM
CS =
c2
3
cm
4 3.14 6.67 10 -8
sec 2 gm
2.0 10 33 gm
=
10 M
2
1M
3.00 1010 cm
sec
(Compare to the
= 1.86 107 cm
discussion of the
km
black hole Hades,
= 1.86 107 cm 5
= 186 km
10 cm
pg. 29 in Thorne)
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Examples: calculation using the horizon
(Schwarzschild) circumference, continued
Example 2: what is the horizon circumference of a black hole
with the same mass as the Earth (6.0x1027 gm)?
C S=
=
4pGM
c2
3
cm
4 3.14 6.67 10 -8 2
s gm
2
cm
3.00 10 10
s
= 5.6 cm (!!)
9 October 2001
Astronomy 102, Fall 2001
6.0 10 27 gm
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Examples: calculation using the horizon
(Schwarzschild) circumference, continued
Example 3: what is the mass of a black hole that has a horizon
circumference equal to that of the Earth (4.0x109 cm)?
First, rearrange the formula:
C S=
4pGM
c2
4pGM c 2
c2
C S=
4pG
c 2 4pG
CSc 2
=M
4pG
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Examples: calculation using the horizon
(Schwarzschild) circumference, continued
Then, put in the numbers:
10 cm
9
10
00
.
3
cm
10
4
CS c 2
sec
=
M=
3
4p G
cm
4 3.14 6.67 10-8
sec2 gm
2
= 4.3 1036 gm
36
= 4.3 10 gm
9 October 2001
1M
33
2.0 10 gm
= 2.15 103 M
Astronomy 102, Fall 2001
(!!)
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Mid-lecture Break
Exam #1 is still having its grades recorded. It’ll be
available tomorrow in recitation, and Thursday in lecture.
Einstein and
his violin
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Astronomy 102, Fall 2001
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Singularities in physics, math and astronomy
A formula is called singular if, when one puts the numbers
into it in a calculation, the result is infinity, or is not well
defined. The particular combination of numbers is called the
singularity.
Singularities often arise in the formulas of physics and
astronomy. They usually indicate either:
invalid approximations -- not all of the necessary physical
laws have been accounted for in the formula (no big deal),
or
that the singularity is not realizable (also no big deal), or
that a mathematical error was made in obtaining the
formula (just plain wrong).
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Singularities in physics, math and astronomy
(continued)
Example of a classical physics law with a singularity:
Newton’s law of gravitation.
r
F
m
GMm
F= 2
r
M
r is the distance between the centers of the two spherical
masses. A spherical mass exerts force as if its mass is
concentrated at its center.
Clearly, if r were zero, the force would be infinite!
This formula will not appear on homework or exams. It is
used only because it is a good example of a singularity.
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Singularities in physics, math and astronomy
(continued)
This singularity is not realized, however:
the mass really isn’t concentrated at a point.
a spherical shell of matter does not exert a net
gravitational force on a mass inside it.
m
r
.
M
9 October 2001
Consider mass m inside mass M:
outer (yellow) matter’s forces on
m cancel out, and only inner
(green) exerts a force. As m gets
closer to the center (r 0), the
force gets smaller, not larger.
No singularity!
Astronomy 102, Fall 2001
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Reaction to the Schwarzschild singularity
Schwarzschild’s solution to the Einstein field equation was
demonstrated to be correct - the singularity is not the result of a
math error.
Thus most physicists and astronomers assumed that the
singularity would not be physically realizable (just like the
singularity in Newton’s law of gravitation) or that accounting
for other physical effects would remove it.
Einstein (1939) eventually tried to prove this in a general
relativistic calculation of stable (non-collapsing or exploding)
stars of size equal to the Schwarzschild circumference.
He found that this would require infinite gas pressure, or
particle speed greater than the speed of light, both of which
are impossible.
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Reaction to the Schwarzschild singularity
(continued)
Einstein’s results show that a stable object with a
singularity cannot exist.
From this he concluded (incorrectly) that this meant the
singularity could not exist in nature.
Einstein’s calculation was correct, but the correct inference
from the result is that gas pressure cannot support the
weight of stars similar in size to the Schwarzschild
circumference.
If nothing stronger than gas pressure holds them up, such
stars will collapse to form black holes -- the singularity is
real.
• Stronger than gas pressure: degeneracy pressure.
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Degeneracy pressure
This involves a concept from quantum mechanics called the
wave-particle duality:
All elementary particles from which matter and energy
are made (including light, electrons, protons, neutrons...)
have simultaneously the properties of particles and
waves.
Which property they display depends upon the situation
they’re in.
Degeneracy pressure consists of a powerful resistance to
compression that’s exhibited by the elementary constituents
of matter when these particles are confined to spaces small
enough to reveal their wave properties.
In more detail….
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Astronomy 102, Fall 2001
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Particles and waves
Particles exist
only at a point
in space.
Waves extend over a region of space.
Electric
Field,
for
instance
Location in space
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Light can be either a particle or a wave
Particle example: the photoelectric effect -- the 1905
explanation of which, in these terms, won Einstein the 1921
Nobel Prize in physics.
Light (in the
form of
photons)
Electrons
Metal slab
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Light can be either a particle or a wave (continued)
Wave example: the Doppler effect.
Lasers
Observer sees:
To observer
V (close to c)
V (close to c)
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Electrons can be particles or waves
Particle example: collisions between free electrons are
“elastic” (they behave like billiard balls).
Wave example: electrons confined to atoms behave like
waves.
Nucleus
Atom
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Electron “cloud”
(extends over space as a wave does)
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How to evoke the wave properties of matter
All the elementary constituents of matter have both wave and
particle properties.
If a subatomic particle (like an electron, proton or neutron) is
confined to a very small space, it acts like a wave rather than
a particle.
How small a space?
The size of an atom, in the case of electrons (about 10-8 cm
in diameter).
A much smaller space for protons and neutrons (about
10-11 cm diameter).
Generally, the more massive a particle is, the smaller the
confinement space required to make it exhibit wave
properties.
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