Transcript Slide 1

HW
The Nature of Space and Time-General Relativity Black Holes
READ: Chapter 8 and Section 12.4 (Binaries with Black Holes)
Chapter 8: Questions (HW#28)1, (HW#29)3
Problems (HW#30)2, (HW#31) 3, (HW#32)
6, (HW#33)7, (HW#34)8, (HW#35) 10
We will do this later, not for chapter 7.
Chapter 12 (HW#36) Problem 12.9(note, inclination is the angle
between the plane of the orbit of the two objects and a perpendicular
plane to our line of sight ..see Fig 5.7)
Curved Space time
• Shortest distance between two points is called the
geodesic (light path in GR?)
• We normally think in terms of Flat or Euclidean space,
geodesic = a straight line. Consider sum of angles of a
triangle =?o
• But consider spherical space: geodesic = great circles,
hence sum of angles of triangle =? Ex?
90o
•What curves Space time…Einstein said MASS!
90o
90o
General Relativity
A new description of gravity
Postulate:
Equivalence Principle:
“Observers can not
distinguish locally
between inertial forces
due to acceleration and
uniform gravitational
forces due to the
presence of massive
bodies.”
Another Thought Experiment
Imagine a light source on board a rapidly
accelerated space ship:
Time
Light
source
Time
a
a
a
a
g
As seen by a
“stationary” observer
As seen by an observer
on board the space ship
Thought Experiment (2)
For the accelerated observer, the light
ray appears to bend downward!
Now, we can’t distinguish between
this inertial effect and the effect of
gravitational forces
Thus, a gravitational force
equivalent to the inertial force
must also be able to bend light!
Thought Experiment
This bending of light by the gravitation of massive
bodies has indeed been observed:
During total solar
eclipses:
The positions of
stars apparently
close to the sun
are shifted away
from the position
of the sun.

New description of gravity as
curvature of space-time!
Principle of Equivalence
“ A uniform gravitational field in some direction is indistinguishable from a uniform
acceleration in the opposite direction.”
Note: if you are in an accelerating reference frame pseudo-forces appear in the
direction opposite to the true acceleration.
Example 1. Car brakes (a true force: friction accelerates the car backwards)
but you accelerate forward and think a force (pseudo-force) pushes you forward.
You can accelerate out the windshield but no true force is on you!
Example 2. To travel in a circle you need an acceleration toward the center of the
circle (Centrifugal force) but you feel a force (pseudo-force) outward.
Exs. Amusement park rides, pail of water in a circle, Centrifugal devices to separate
out blood serum or heavy elements in solution.
Example 3. Person in elevator on a scale see Fig 8.4 in text.
Reminder Newton’s second law SF=ma
Case I: elevator is not accelerating -mg+Fs =0
Fs
or Fs =mg! Third law-> force of scale on person is
same as person on scale=reading of scale!
mg is called the Weight or W=mg
mg
s
g
Equivalence principle illustrated
Example 3 Case II Elevator is in deep space: no gravity. The elevator with the
person on the scale is accelerating at a! (Rocket engine)
Set a=g…What does the scale read?
Scale has to push person to g acceleration or Fs =ma=mg reaction on the scale
is mg or scale reads Fs =mg! Person does not know whether they are on earth
or in deep space accelerating at g! Hence…equivalence!
Case III. Elevator cable breaks and is in free fall.
What is Fs =?
Free fall also when in orbit
A DEEPER MEANING!
Inertial mass, m, ie a=F/m or m acts like a resistance to F.
Inertial mass
Formally, Inertial Mass = Gravitational Mass!
DEEPER!
Gravitational mass
Inertial Mass = Gravitational Mass!
This is the reason all bodies near the surface of the Earth have the same
acceleration!
Given a mass, m, near the Earths surface.
Gravitational Force on the mass m from the Earth ME is
FG =GmME /RE2
here m is the gravitational mass acted upon.
Since F=ma (with m the inertial mass).
and Inertial Mass = Gravitational Mass,
the acceleration experienced by mass m from gravity is:
a =FG /m =GME /RE2
or all bodies of any mass m have same “a”
near the surface of the Earth as found by Galileo and proved by Newton
above.
It is not obvious that the Inertial mass and the Gravitational mass are the same.
There must be something special about Gravity!
Other Effects of General Relativity
• Perihelion advance (in particular, of Mercury)
Closer to the mass creating the
curvature the greater the
effect! Hence Mercury! Since
curvature changes, this causes the
elliptical orbit not to close, but to
advance as follows:
sun
Advance is 5600 sec as per Newton but 43
sec of arc/century cannot be accounted for.
Einstein’s results
were 43 sec/century
from GR!
See text about
challenges to this!
GR implies:
Planets follow
the curvature of
space-time
due to the mass
of the Sun( or star)
Try rolling marbles
in a gently sloping
salad bowl.
Record your
observations
for EC!
Bending of EM radiation
q
Radius of sun=b
q
y
A Classical argument for the bend to get a
feel for the GR solution.
Consider a photon whose E=hu is
equivalent to a mass =mphc2
mph grazing the sun will experience a
gravitational force =GMsmph /b2 =mpha
or a=GMs/b2 (b=radius)
The displacement d under acceleration a
is given by d=at2 /2 and the distance y=ct
d
with t equal to the time the mph is under
the acceleration a.
No sun
Tanq ~ q = d/y =(GMs /b2 )( t2/2) /ct =tGMs/2b2c
path
An estimate of time is in the order of t=b/c
So q=1/2(GMs /bc2 ).
GR via Einstein’s curved space time formulation predicts q=4GMs /bc2
which predicts an effect 8 times larger than a classical approach!
Even so the angle with b as the solar radius is at most only 1.74” of arc,
a difficult measurement which has been tested (Eddington 1914,1919) but with an error
factor. Radio time delays via interplanetary satellites in the extra path around
the sun confirmed GR effect to 0.1%! The bending due to galaxies curving space
causes a lens effect!
Another manifestation of bending of light:
Gravitational lenses
A massive galaxy cluster is bending and
focusing the light from a background object.
See also fig 8.1 in your text!
Gravitational red shift: Light from sources near
massive bodies seems shifted towards longer
wavelengths (red). Logical explanation:
• Remember a photon has Energy E=hu
and since ul=c u=c/l so E=hc/l.
A photon has its energy change
as it moves away from a massive body.
That is, it loses some energy as the
gravitational field does work on it. This
means that l must change. Since E goes
down l gets bigger.
• We have a RED shift!
GR Redshift from a static and spherically
symmetric mass distribution
From a solution from the Schwarzchild solution (very complicated Tensor Mathematics!)
to Einstein’s Space-time field equations: This solution takes a mathematical form called
a “metric” that describes
the Geometry of space-time around a mass. One can derive the red shift of a
photon escaping a mass starting at r1=R and at a distance r2 =r: similar to Eq. 8.4
in Kutner.
r
R
l2
-l1
What happens to
l2
l1
l2
=
[
(1-2GM/rc2)
---------(1-2GM/Rc2)
]
1/2
if (1-2GM/Rc2)->0! ?
What is the value of R that does this? R=? Solve now!
This R is called the Schwarzchild radius: Welcome to a BLACK HOLE!
Gravitational red shift: a classical
approach to get a feel for GR solution
l2
r2
r1
When a particle, mass m, is in a gravitational field it has
gravitational potential energy = -GMm/r.
We again can consider the photon has an effective mass
Ie Eph =hu = hc/l = meff c2 or meff =h/cl
Total Energy =photon energy +gravitational Potential energy for meff
l1
Etotal =
Eph –GMmeff /r = hc/l –GMh/rcl.
At each point r1 and r2 the total energy is the same(conservation of
energy) and using the last expression at each point we get
hc/l1 –GMh/r1cl1 =hc/l2 –GMh/r2cl2
->
1/l1 (1-GM/r1c2) = 1/l2 (1-GM/r2c2) or the nonrigorous classical result->
l2/l1 = (1-GM/r2c2)/ (1-GM/r1c2)
GR result-> l2/l1 = ( (1-2GM/r2c2)/ (1-2GM/r1c2) )1/2
For small shifts (1-x)1/2 ~= 1-x/2
eq 8.3
eq 8.4
8.4-> 8.3 (effect of approximation)
Gravitational red shift: for photon reaching
us r2-> infinity from an object radius r
• GR-> l2/l1 = ( (1-2GM/r2c2)/ (1-2GM/r1c2) )1/2
• r2-> infinity with r1 =r we get
 l2/l1 = (1/(1-2GM/rc2)1/2 = (1-2GM/rc2)-1/2
• Expanding as before l2/l1 =1+GM/rc2 or
 (l2-l1)/l1= Dl/l =GM/rc2
eq 8.6 Kutner
• l=l1 for photon leaving surface of radius r!
• When is the GR redshift the most for values of r?
How big are these shifts? be sure to do Example 8.1 in the
text…very small Sun not measurable but White Dwarfs ok.
Best Dl= 3 x 10-4 Sirius B and =6 x 10-5 40 Eridani
Other GR considerations
The Mossbauer effect
is used to verify GR red shift on Earth Kutner p 146.
Essentially, a crystal emits a well defined Gamma ray and an identical crystal
can absorb that gamma ray provided it is the same wavelength. Put a crystal
emitter in the basement and a crystal to absorb on the roof. The gamma ray
photon moving through earths gravitational field undergoes a red shift and
cannot be absorbed on the roof (different l). Moving the crystals together
one changes the wavelength and can calculate the shift which matches GR
very well.
Time dilation
In a gravitational field can also be derived and expression of change
is similar to GR red shift
t2/t1 = ( (1-2GM/r2 c2)/ (1-2GM/r1c2) )1/2
Airplane and rocket tests with atomic clocks follow this formula after
taking into account SR time dilation!
GR PREDICTS GRAVITATIONAL RADIATION!
=BRIEF DISTORTION OF SPACE-TIME!
Escape Velocity
Velocity needed
to escape Earth’s
gravity from the
surface: vesc ≈
11.6 km/s.
Now, gravitational
force decreases
with distance (~
1/d2) => Starting
out high above the
surface => lower
escape velocity.
vesc
vesc
vesc
If you could compress Earth to a smaller radius
=> higher escape velocity from the surface.
Non-relativistic!
=(2GM/R)1/2
Examples Vescape
•
•
•
•
Earth Fg =GM1M2/R2 units 1 dyne=G g2/cm2
But F=ma ->1 dyne =g cm/s2 or
units of G = cm3/s2 g or
G=6.67 x 10-11 cm3/s2 g
• ves=(2GM/R)1/2 
•
•
•
•
=(2 x 6.67 x 10-7 cm3/s2 g x5.98 x 1027 g/6.37 x 108 cm)1/2
=1.12 x 106 cm/s =11.2 km/s ~ 7mi/s
A NEUTRON STAR Ves = 2 x 1010 cm/s ~ 2/3 c
If a star hits the Schwartzchild Radius we solved on slide 17,
namely:
• Rs =2GM/c2 WHAT IS Ves = for this case? Solve now!
The Schwarzschild Radius
So we see => This is a limiting
radius where the escape
velocity reaches the speed of
light, c:
2GM
Rs = ____
c2
G = Universal const. of gravity
M = Mass
Rs is called the
Schwarzschild Radius of
a Black Hole!
And the Red shift is????
Vesc = c
What are the values for the
Schwarzchild Radius
• Black holes are determined by the Mass and
Size (Angular Momentum and electric charge is
also considered but we will ignore these effects
for now).
• How big is the RS for an object of one solar
mass in km?. Msun=2 x 1033g
• C=3 x 1010 cm/s G=6.67 x 10-8 dyne cm2/g2
• Calculate NOW!! RS =?
Rs = 2GM/c2
• That was example 8.2!
Schwarzschild Radii
Since, Rs is directly proportional to M
One can see Rs =3km (M/Msun)
Recall, density, r =M/Vol : Vol =4/3pR3
Calculate r for one solar mass Rs =?
Equation 8.11 for black hole r ~=1017g/cm3 (M/Msun)-2
Derive for extra credit!
Schwarzschild Radius and Event Horizon
No object can
travel faster than
the speed of light
=> nothing (not
even light) can
escape from inside
the Schwarzschild
radius
• We have no way
of finding out what’s
happening inside
the Schwarzschild
radius.
 “Event horizon”
Black Hole Tidal Effects I
Tides depend on the
difference of the force
of gravity. Or the rate
of change of gravity.
Tides from the moon on
the earth depend on
the Force on a Mass
(m) at a distance r being
different at each point
on Earth.
Low
Forces
High
Differentials
From center
force
Low
High
Black hole Tidal Effects on an
unfortunate Astronaut
Consider that the force of Gravity on a mass m leads to an acceleration
g, derived from
F=mg =GMm/r2 or g(r) =GM/r2 g at earths surface~ 10m/s2~103cm/s2
The tidal acceleration is the differential in the gravitational
acceleration. Or Dg =-2GM/r3 Dr
-> How did I get this?
Example 8.3-What is the difference in the acceleration of gravity at the
feet to the head on an astronaut whose height is 2 meters drifting
into a black hole of 1 Solar mass ->r = 3km and how does this compare
With the gravitational acceleration at earth’s surface:
Calculate now!
As black holes get more Massive what can we say about this
tidal effect???? Hint how does the Radius go!
An astronaut descending
down towards the BH will be
stretched vertically and
squeezed laterally (tidal
effects).
This effect is called
“spaghettification”
General Relativity Effects Near
Black Holes : an “Astronut” travels
in to it!
Time dilation
Clocks starting at
12:00 at each point.
After 3 hours (for an
observer far away
from the BH):
Clocks closer to the
BH run more slowly.
Time dilation
becomes infinite at
the event horizon.
Astronaut far away
never sees falling
Event Horizon
astronaut reach the Rs
since the time is
General Relativity Effects
Near Black Holes
Gravitational Red Shift
Dl/l =GM/Rs c2 from the surface!
All wavelengths of emissions
from near the event horizon
are stretched (red shifted).
Or as the “astronut” travels
inward he emits a wave. As he
gets closer each wave
received gets longer!
See text sec 8.4.2
Event Horizon
Traveling
toward
R
s
Since space-time is sharply curved near the Black hole as Astronut
approaches and sends light out the photons bend with the
curvature
Photons escape here
Exit Cone
Captured
Gets smaller
Nothing escapes
passed the
Event Horizon
SEE THE EFFECTS OF ROTATION Fig 8.13
At r=3/2Rs photons
emitted horizontally
orbit in
photon sphere!
You can see
the back of your
head here!
General Relativity
:Visualizations
At a distance, the
gravitational fields of a black
hole and a star of the same
mass are virtually identical.
At small distances, the much
deeper gravitational potential
will become noticeable.
World line = path of the photon
Movies to Visualize Space time
• Movies from the Edge of Spacetime
HW
REMINDER!
The Nature of Space and Time-General Relativity Black Holes
READ: Chapter 8 and Section 12.4 (Binaries with Black Holes)
Chapter 8: Questions (HW#28)1, (HW#29)3
Problems (HW#30)2, (HW#31) 3, (HW#32)
6, (HW#33)7, (HW#34)8, (HW#35) 10
We will do this later, not for chapter 7.
Ch 8 Answers
• Q. 8.1 What do we mean when we say gravity alters the geometry of
space and time?
• Q. 8.3 Why don’t you need a solar eclipse to measure the bending of radio
waves past the edge of the sun?
Ch 8 Answers
• Q. 8.1 What do we mean when we say gravity alters the geometry of
space and time?
• Q. 8.3 Why don’t you need a solar eclipse to measure the bending of radio
waves past the edge of the sun?
• P. 8.2 Neutron star with 1 Msun and radius of 10 km, what angle is
light bent as it passes the solar edge? Use equation 8.1: θ = 4GM/Rc2
• M = 1 solar mass = 2.0 x 10^33 gm and R=10 km=10^6 cm (CGS)
•
= (4)(6.67 x 10^-8 dyne-cm^2/gm^2)(2.0 x 10^33 gm)/(10^6 cm)
•
(3 x 10^10 cm)^2 = 0.59 rad = 33.8=340 (2π rad = 360)
Ch 8 Answers
• Q. 8.1 What do we mean when we say gravity alters the geometry of
space and time?
• Q. 8.3 Why don’t you need a solar eclipse to measure the bending of radio
waves past the edge of the sun?
• P. 8.2 Neutron star with 1 Msun and radius of 10 km, what angle is light
bent as it passes the solar edge? Use equation 8.1: θ = 4GM/Rc2
• M = 1 solar mass = 2.0 x 10^33 gm and R=10 km=10^6 cm (CGS)
•
= (4)(6.67 x 10^-8 dyne-cm^2/gm^2)(2.0 x 10^33 gm)/(10^6 cm)
•
(3 x 10^10 cm)^2 = 0.59 rad = 33.8=340 (2π rad = 360)
• P. 8.3 White dwarf of 1 Msun and a radius of 5x10^3 km find l that Ha
is shifted when seen by distant observer. Use 8.5 gives
• λ’ = λ(1 + GM/rc^2) = (656.47 nm)[1 + (6.67 x 10^-8 dynecm^2/gm^2)(2.0 x 10^33g)/{(5 x 10^8cm)(3 x10^10cm)^2}]
• = 656.66 nm
Not much !!!
Ch 8 Ans
• P. 8.6 Compute your Schwarz radius and density for BH of your mass.
– If an object shrinks to the Schwarzschild Radius it will ultimately
collapse to a singularity (basically vanish from our universe). A
singularity is like what happens to 1/x when x0. Eq. 8.8 is: RS
=2GM/c^2 (Example 8.2 did RS we can use ratio)
• For 75 kg person, RS = (3 km)(M/ MS) = (3 x 10^5cm)(7.5 x 10^4g)/
(2.0 x 10^33 gm) = 1.13x 10^-23 cm Really shrink!
• Your density , ρ =Mass/Volume(sphere) ; ρ = M/(4πR^3/3) =
75g/4π(1.13 x 10^-23 cm)^3/3 = 1.2 x 10^70
gm/cm^3 : or
using equation 8.11 ρ = 7 x 10^73 gm/cm^3! Why a difference?
•
Ch 8 Ans
• P. 8.6 Compute your Schwarz radius and density for BH of your mass.
– If an object shrinks to the Schwarzschild Radius it will ultimately
collapse to a singularity (basically vanish from our universe). A
singularity is like what happens to 1/x when x0. Eq. 8.8 is: RS
=2GM/c^2 (Example 8.2 did RS we can use ratio)
• For 75 kg person, RS = (3 km)(M/ MS) = (3 x 10^5cm)(7.5 x 10^4g)/ (2.0 x
10^33
gm) = 1.13x 10^-23 cm Really shrink!
• Your density , ρ =Mass/Volume(sphere) ; ρ = M/(4πR^3/3) = 75g/4π(1.13
x 10^-23 cm)^3/3 = 1.2 x 10^70
gm/cm^3 : or using equation 8.11
ρ = 7 x 10^73 gm/cm^3! Why a difference?
• P. 8.7 What mass BH does density equal 1 gm/cm^3? Using
equation 8.11 1gm/cm^3 = (1 x 10^17 gm/cm^3)(M/ MS )^-2; M/ MS =
3.2 x 10^8, M =6 x 10^41 gm
•
Ch 8 ans
• P. 8.8 What mass BH does g between feet and head of astro =
g of Earth (1000 cm/sec^2)?
• Use example 8.3: dg/dr = -GM/r^3; (dg/dr) = 1000 cm/s^2/ 200cm
(astronauts size=2meters) we get M/R^3 = 7.5 x 10^7 gm/cm^3
the density. Eq 8.11: 7.5 x 10^7g/cm^3 = (1 x 10^17g/cm^3)( M/
M)^-2
; so M/ M = 3.7 x 10^4
• P. 8.10: Compare rate of clocks very far away to one 1.5 RS from a
3 MSUN BH?
• Use eq. 8.7 with r2 at infinity. r1 = 1.5 (2GM/c^2);
• T2/T1 = (1-1/1.5)^-1/2 = 1.73
OLD AGE Stellar Collapse: Review: The
Remnants of Sun-Like Stars: White
Dwarfs
Sunlike stars build
up a Carbon-Oxygen
(C,O) core, which
does not ignite
Carbon fusion.
He-burning shell
keeps dumping C
and O onto the core.
C,O core collapses
and the matter
becomes
degenerate.
 Formation
of a
White Dwarf
White Dwarfs
Degenerate stellar remnant (C,O core)
Extremely dense:
1 teaspoon of WD material: mass ≈ 16 tons!!!
Chunk of WD material the size of a beach ball
would outweigh an ocean liner!
White Dwarfs:
Mass ~ Msun
Temp. ~ 25,000 K
Luminosity ~ 0.01 Lsun
Core has high pressure!
Degenerate Matter Pressure!
• Depends on the Pauli exclusion principle: no two
electrons can be in the same state:
• Regular gas pressure cannot stop the collapse in this
case.
• Electrons packed in tight have a higher energy than
normal and exert a pressure higher than normal ideal
gas. Hence, this degenerate situation stops the collapse:
• P ~ r (density) about 100 x P (ideal gas)
• Electron energy and momentum become relativistic as
calculated by S. Chandraseker (Nobel prize in ‘83 for
stellar structure)
The Chandrasekhar Limit
The more massive a white dwarf, the smaller it is.
Pressure becomes larger, until electron degeneracy
pressure can no longer hold up against gravity.
The relativistic electrons create a pressure that has a
maximum support!
WDs with more than ~ 1.4 solar
masses can not exist!
If the core mass
is greater than
1.4 SM, star
collapses.
We get
a Neutron star->
PULSARS:
A later chapter!
Black Holes beyond Neutron stars
Just like white dwarfs (Chandrasekhar limit: 1.4 Msun),
there is a mass limit for neutron stars:
Neutron stars can not exist
with masses > 3 Msun
We know of no mechanism to halt the collapse
of a compact object with > 3 Msun.
It will collapse into a single point – a singularity:
=> A Black Hole!
Black Holes in Supernova Remnants
Some
supernova
remnants with
no pulsar /
neutron star in
the center may
contain black
holes.
Observing Stellar Black Holes
Because no light can escape a black hole,
=> black holes can not be observed directly.
Material falling into a black hole emit x-rays
If an invisible
compact object is
part of a binary, xrays can pulsate
and then we can
estimate its mass
from the orbital
period and radial
velocity.
Mass > 3 Msun
=> Black hole!
O star =15 Msun Read sec 12.4 for details
Stellar black hole
Supermassive black holes
are found at the core of galaxies
Matter falling in accelerates and spirals out in two jets:
The exact mechanism is still unknown and various models have
been proposed.
Checked to here