Professor Charles
Bailyn: Last time,
I wrote down the Schwarzschild
metric, which was the sort of
culmination of the theoretical
part of the--of this section of
the course.
And I realized,
as I was talking to some of you
afterwards about some of the
implications of this,
that the way I--one of the ways
in which I wrote it down might
have been a little confusing.
So, let me write it down again
and I'll say a couple things.
And what I'm going to do is I'm
going to write it down with a
little r instead of a
capital R and I'll tell
you why in just a minute.
Obviously, it doesn't matter.
I've just changed the notation
slightly.
And the reason I want to do
this is because I want to
emphasize that this r,
the little r,
what I've now written down as a
little r in this
equation,
is not the radius of the object;
r is just a coordinate.
It can be anything.
It's not the radius of the
object for which the mass is
giving you the Schwarzschild
radius.
The Schwarzschild radius is a
very specific radius.
r_s =
2GM / c^(2) so
that's a specific number.
The little r, here,
is just a coordinate and so is
φ and θ, which combine into
this thing which I've called
Omega [Ω]
and T.
And so if you have an object of
mass M and radius--I'll
give--I'll call that
R_0 just to
make sure that you understand
that this is the radius of the
object.
The radius of the object
doesn't come into this metric.
The mass does,
because the mass turns into the
Schwarzschild radius and that's
part of the equation.
But the radius of the object
doesn't matter.
And so, if you then ask,
well, what is a black hole?
A black hole is something in
which the radius of the object
is less than the Schwarzschild
radius.
Because if that's true,
then there is some coordinate
r, some little r
where little r =
R_S.
And then, if you go up into
these equations here very--these
very strange things happen when
little r is equal to
R_S.
Yes, question?
Student: Yesterday,
the big group had [Inaudible]
Professor Charles
Bailyn: Oh you're right,
you're right,
it has to be the other way
around, I'm sorry.
Student: [Inaudible]
Professor Charles
Bailyn: This has to be
squared.
You're right on both counts,
good.
R_S / r.
Yes, it better be that way
because R_S /
r in the normal part of
space is less than 1.
So, thank you very much,
sorry about that.
Okay, I think we're all right.
Are we all right?
Yeah, good.
Okay, so there is some r
where r is equal to
R_S,
and then these very bizarre
things start to happen.
If R_0,
the radius of the actual
object, is bigger than its
Schwarzschild radius,
then that isn't true.
There's no r where
r is equal to
R_S--then no
r equal to
R_S.
Now, you might think that
somewhere inside the object,
there would be a little
r, which is equal to the
Schwarzschild radius.
But that isn't--that turns out
not to be right either,
because the relevant M,
the relevant mass-
Student: Can you lower
the top slide?
Professor Charles
Bailyn: Oh sure.
Student: The whole--the
slide not the - Professor
Charles Bailyn: Oh yeah,
okay.
So, yeah.
Let me put this back.
The relevant mass is the mass
inside r.
So if you're--if you've got
some mass M,
like this, and you're outside,
if you're here,
then the whole of the mass
counts.
But if you're sitting inside
this thing, then the only mass
that counts is this mass here.
And so, for r--for
coordinates for positions inside
the object, then the relevant
mass that you put into the
equation is less than the mass
of the whole object.
And the consequence of that is
that, as you go further and
further inside the thing,
the mass becomes less and less,
and the Schwarzschild radius
becomes less and less.
And so, if the total size of
the object is greater than the
Schwarzschild radius,
then there is no position
inside the thing which is less
than the Schwarzschild radius,
because the relevant mass
decreases, and so does the
Schwarzschild radius as you go
in toward the center of the
object.
So, I just wanted to make sure
you understood that the key to
using this metric is that
the--this r term in it
tells you where you are.
It has nothing to do with the
size of the object.
And you become a black hole
when the size of the object is
smaller than the Schwarzschild
radius.
Because under those
circumstances there's some
position, or set of positions,
where you can be inside the
Schwarzschild radius,
and therefore all of these very
exotic effects start to happen.
Questions on this?
Or have I just baffled you
further?
Well, we'll see.
Student: Is that the
radius of the Sun?
Professor Charles
Bailyn: Sorry.
Student: Is that the
radius of the Sun?
Professor Charles
Bailyn: Oh no,
this is the radius of the
object.
Sun is--Sun has a dot in the
middle--not that you could tell
with my handwriting.
Okay, I don't know if--whether
that was helpful or not,
but I felt the need to say it.
Yes, please.
Student: Could you
please explain the diagram that
you [Inaudible]
Professor Charles
Bailyn: This right here?
Yeah, so the situation is
that--what is the relevant
M to put into this
equation?
This is true,
by the way, in Newtonian
physics also.
If you're inside an object then
the mass outside of your
position, this stuff,
doesn't count toward the
M.
And you only want to count the
mass inside of where you are.
Student: Okay,
so that whole circle is an
object?
Professor Charles
Bailyn: Yeah,
so this whole circle is an
object--object with mass and
some--with some mass and some
radius.
But now you're asking the
question, "What happens if
you're somewhere inside this
thing?"
Yes?
Student: Does that mean
that the dotted line is
R_S?
Professor Charles
Bailyn: No,
the dot--well,
the dotted line could be
anything.
You're at some position inside
and this--the dotted line is the
radius at which your particular
position is.
Student: [Inaudible]
Professor Charles
Bailyn: No,
it's just wherever you are
inside the thing.
Less mass counts than you think
it's going to because it's only
the stuff inside your own radius
that counts toward the metric,
or toward the gravitational
force in the Newtonian thing.
You can be at--at whatever
point you choose to be inside
here, you can draw this same
thing.
And the point of this is that
if you are now down in here,
where you thought the
Schwarzschild radius was going
to be,
for the whole of this mass--in
fact, the Schwarzschild radius
is still going to be inside your
position because much less mass
will count toward it.
This isn't helping,
I can tell by the look on your
face.
Okay, let me try again.
Let me try again.
Okay, here's an object.
And this object has a radius
bigger than its Schwarzschild
radius, and it's got some mass.
Okay, so, now I'm going to
draw, in dotted lines,
its Schwarzschild radius.
That's inside here, okay?
Now, supposing you're inside
this object.
Supposing you're living here.
Then, at that point,
you--I draw this sort of
interior radius,
here.
And if I'm at this point,
then I ask, what is the
Schwarzschild radius that goes
into the metric for something
that's sitting inside for person
X, here.
And the answer to that is it's
actually smaller than the
Schwarzschild radius of the
entire object,
because less mass counts.
Only this mass counts towards
the Schwarzschild radius.
So, there's some part of the
object whose mass no longer
counts, because it's outside
your position.
And so, as I move in here to
some interior point,
the Schwarzschild radius that
is appropriate for me to use
becomes smaller.
Does that make more sense?
Student: Can you draw
that if it were a black hole?
Professor Charles
Bailyn: Ah,
so if I was black--if it was a
black hole, right.
Okay, so now,
let's do the opposite case.
Good question--is less than the
Schwarzschild radius.
So, now here's my object,
and here's the Schwarzschild
radius.
That's the difference, right?
And therefore--and the
consequence of this,
by the way, is that this whole
object sinks down to a point.
Because, remember,
things inside the Schwarzschild
radius have to contract down to
a point.
So, I can't be part way through
it, no matter what I do.
And, you know,
if I'm outside this tiny little
thing, if I'm here,
I have the same Schwarzschild
radius,
because all of the mass is
interior to my position.
So, the key thing is,
is the object smaller or bigger
than its own Schwarzschild
radius?
Student: So,
in that case,
all the mass is going to be
closer than-- Professor
Charles Bailyn: All
the--yeah,
all--well, all the mass--
Student: [Inaudible]
is going to [Inaudible]
Professor Charles
Bailyn: If the radius of the
object is smaller than its
Schwarzschild radius,
then all the mass is inside the
Schwarzschild radius.
And any point in our Universe
where, you know--and so,
all r greater than
R_S is out
here,
somewhere, and it doesn't
matter where you are.
All that mass counts because
it's all inside the
Schwarzschild radius anyway.
And then, when you get to this
point, you have this magic
moment where r =
R_S and
everything blows up.
Right?
And that can't happen here.
Because as you move in,
in the middle of the object,
the Schwarzschild radius
shrinks down away from you as
you get in,
because less and less mass
counts towards it,
therefore it gets smaller.
Yes?
Student: Would it be
possible to have an object
[inaudible]
density--like such that there
would be a point at which you
would [inaudible]
Professor Charles
Bailyn: Ah,
okay.
So, could you have some
incredible density gradient
where you're actually gaining
more--you're actually catching
up to the Schwarzschild radius
as you go in.
In principle, yes.
But if you work out what that
density gradient would have to
be, it would be incredibly
steep.
Remember, there was a problem
set at some point where we
discovered that the less mass
there is in a black hole,
the denser it has to be in
order to be inside its
Schwarzschild radius.
So, as you're losing mass as
you get down in here,
the amount by which it has to
be denser on the inside goes up
really sharply.
But yes, such a thing can be.
Imagine, for example,
a black hole with an
atmosphere.
Imagine there's some--or some,
just, gas floating around out
here.
If you think of that as one
object, then when you're outside
the gas, if you think of the
outer edge of the gas as the
edge of the object,
then that's not a black hole.
And if you come on--if you go
down--sink down into the
atmosphere, eventually you'll
find it.
So, that's an incredibly steep
density change because you've
gone from an atmosphere to some
object.
So, in principle,
that's possible.
In any kind of real object
that's holding itself up,
probably not,
because the density gradient
would have to be so severe.
Yes?
Student: How would this
work for non-spherical objects?
Professor Charles
Bailyn: How does it work for
non-spherical objects?
Excellent.
Okay.
So, let's see.
What happens is,
if you have a
non-spherical--first of all,
the gravitational forces are
sufficiently strong near a--if
you're anything close to a black
hole,
that you're likely to end up
spherical.
The only way you don't is if
you're spinning.
Because then,
you know, you bulge out at the
center.
Spinning black holes exist.
They have a different metric.
And the angular term becomes
important because you're
spinning around and it's a
different metric.
And the shape of the event
horizon where r is equal
to R_S or the
equivalent in this other metric,
is non-spherical.
And so, you can have
non-spherical event horizons if
the thing is spinning,
because that has the effect of
making the mass distribution
different.
And this is called Kerr black
hole, it has a Kerr metric,
K-E-R-R.
And that's a different metric,
and it's got a whole other set
of terms that have to do with
the rotation of the angular
component.
But again, the same thing
happens.
When you get to some place,
all the terms blow up but it's
a non-spherical surface where
that occurs.
And you have to do it in
cylindrical coordinates.
It becomes a mess--not that it
isn't a mess already.
Other questions?
All right, let me go back to
the effects of general
relativity.
We've already done one of these.
So, what I'm going to talk
about is the post-Newtonian
gravitational effects.
So, these are things where it
just barely starts to deviate
away from what the predictions
of Newton's laws would be.
So, just to go back,
if you take that metric and you
kind of expand it in small
quantities where the
Schwarzschild radius is much,
much smaller than wherever you
are.
And you take the first term of
the expansion.
Mathematically,
it turns into the same things
as Newton's laws.
And so, when you're out there
in more or less flat space,
or close to flat space,
then Newton's laws still apply,
because the expansion of the
metric gives you the exact same
result.
As the gravitational forces
become larger,
you start to get post-Newtonian
terms, and things start to be a
little bit different from
Newton.
And we've already had one of
these--so, post-Newtonian effect
number one is the precession of
the perihelion,
where, unlike the prediction
for Newton, the orbit doesn't
exactly repeat.
It sort of moves on a little
bit each time.
And you get one of these nice,
sort of, Spirograph patterns,
as an elliptical orbit
gradually precesses,
or gradually moves around.
So, that was the first one.
And that was the first one that
was seen.
That was seen in Mercury,
which is the planet you would
expect to see such a thing,
because it's closest to the Sun
and therefore,
R_S / r
is the largest number for
Mercury--for any planet in the
Solar System.
So, Mercury.
And, that actually turns out to
have been known before Einstein
actually worked out the general
relativity.
So, now, post-Newtonian effect
number two is the deflection of
light.
The whole idea,
remember, is that there's
the--is that gravity works
because it curves space and
objects moving along this curved
space appear to have curved
trajectories.
That's true of light also.
Light is also moving across
this curved space.
So, there's a prediction in
general relativity that if you
shine light near a massive
object, it's the path that that
light will take.
It's curved.
It's as if the light's being
gravitationally attracted by
this object, So,
let's see here.
Imagine the following situation.
Here are you,
you're observing something.
Here's a star.
And the path that light takes
to this star,
from the star to your eye,
is some kind of straight line.
And so, that's where you
perceive the star to be.
Now, supposing that along--that
between you and this star,
there's some object.
So, here's an object with mass
M, some massive object.
Now, because the path is going
to curve, the path that the
light takes from the star to
your eye is going to now look
like something like this.
And it's deflected by the
presence of the mass,
which curves the space that
it's in.
That means, you think,
because you're interpreting
this as if the light is coming
straight towards you,
you think the star is here.
Or it appears to be there.
Because, you say,
all right, what is the angle at
which the light is coming into
my eye?
It's coming from there.
You don't realize that the star
is actually over there and the
light has gone up toward the
ceiling and bent back down to
your eye.
So, it's as if I'm looking up
there and seeing the exit sign.
Because the light from the exit
sign has gone up to the ceiling,
been bent around by some black
hole or something,
up at the top of the classroom,
and come back down to my eye.
Yeah?
Student: Well,
I mean, wouldn't the light on
the other side of the planet
showing the massive thing also
be [Inaudible]
Professor Charles
Bailyn: Ah,
it is possible, yes.
It is possible.
In this particular case,
probably not,
because there wouldn't be
enough bend in order to get it
in here.
So, light going this way,
for example--let me take
another color before this gets
too complicated.
Light coming this way would
bend like that and it would miss
your eye.
However, there are situations
in which you see two images of
the same object,
one from each side.
These are called gravitational
lenses.
Student: Can you see
many, many images?
Professor Charles
Bailyn: In principle,
if it's lined up exactly right.
Imagine that it's perfectly
aligned.
Then you see this way,
you see that way,
and you see out of the paper
and you can see a whole ring.
This has actually been observed.
Not the whole ring,
but large parts of the ring,
because it's never perfectly
aligned.
But you can see these things.
These are called Einstein's
rings, and you can actually see
them.
You can see these kinds of
multiple images in cases where
this is a very distant galaxy or
quasar.
Quasars are very bright,
very small, extragalactic
objects.
And this is an intervening
galaxy or galaxy cluster.
And you can see these multiple
images.
I'll bring some pictures in
next time.
They're very--they're
amazing-looking things.
It's often true that you can
see four images.
That comes about when the
intervening mass is not
spherical.
And so, you get weird
deflections and you--they're
multiple directions.
You see the same object--a very
spectacular thing.
These were discovered only
recently but it turns out that
you can observe this effect even
in our own Solar System.
Because what you can do--I'll
bring in some pictures of
gravitational lenses,
those are amazing things.
And what also happens is,
if you're looking at something
big, like a galaxy,
that has a shape,
it distorts the shape.
It stretches the whole thing
out, and you get these amazing
pictures.
All right.
In any case,
in our own Solar System,
you see this in the following
way.
Here's the Sun,
and here's a star right behind
the Sun.
And the light from the star
comes like this,
and so you think the star is
here.
So, if you look at a star when
it's right next to the Sun,
right at the edge of the Sun,
it seems to be in a different
position relative to stars
further--relative to other stars
up here.
Here's another star.
Then it would--if you looked at
it at night when the Sun wasn't
in the way.
Yes?
Student: Then how do you
do astronomy during the daytime?
Professor Charles
Bailyn: Excellent question,
I'm getting to that.
Okay, so--but the principle
is--supposing the Sun were dark,
okay.
Look at star behind the Sun,
or right next to the Sun,
really.
You can't see through the Sun.
I'm not that crazy.
Star behind Sun.
And it appears to be in a
different place,
position, than when observed at
night, without the Sun in the
way.
Okay, so how do you do
astronomy during the daytime?
That's a good question.
Student: Eclipses.
Professor Charles
Bailyn: Eclipses,
yes.
You wait until there's a solar
eclipse and then you do your
observation.
Do this by observing during an
eclipse.
And if you do this,
the position,
the change in position of the
star, if it's right at the edge
of the Sun--and you've now
blocked the Sun,
because here's the Moon, right?
And the light from the Sun
doesn't come through.
This change in position,
that's an angle,
is around 1 arc second.
And that's a positional change
you can observe.
Remember the size of--the
apparent size of a star,
when you look at it,
is about 1 arc second.
So, it seems like it shifted a
significant amount.
And so, you go,
you take a picture of the sky,
of the stars behind the Sun,
during a total eclipse of the
Sun.
Now, the problem with this is a
practical problem.
Total eclipses of the Sun
happen only for very short
amounts of time,
and only occasionally,
and only in a small--over a
small portion of the Earth at
any one time.
How many people have seen an
eclipse of the Sun?
Pretty spectacular thing, huh?
But, you know,
you have to travel to get
there.
It doesn't come to you.
So, here's what happened.
In 1917, Einstein publishes his
theory of general relativity.
And then, in 1919,
there's going to be an eclipse
in Brazil, I think it was.
Eclipse in Brazil.
And it seems like a good idea
that you should actually go and
test this theory,
which has made a very specific
prediction about what ought to
be going on with these stars.
So Eddington,
we've heard of him before,
he's the guy who trashed his
student Chandrasekhar some years
later.
But at the time he was young,
up-and-coming British
scientist, mounts an expedition
to Brazil to test this theory.
And it works.
Goes down there,
he takes these pictures.
It does just what Einstein said.
And this is the event that
makes Einstein into a great
international figure.
This is front-page headlines
around the world.
And part of the reason of why
it was is, look at these dates.
Einstein is a German scientist.
This is the middle of World War
I.
And so, in the middle of World
War I a German scientist--he's
later thought of as a Jewish
scientist,
but at the time,
Hitler was not yet on the
scene.
And so, Einstein was thought of
as a German scientist.
So, this German scientist
publishes a grand,
bizarre, new theory of the
Universe right after the war.
The Brits mount an operation to
go and check and see if this is
right, and it turns out to be
correct.
So, this is now right after
World War I, in the era when
everybody thought,
well, we're never going to have
a war again.
There's going to be peace and
justice and brotherhood
throughout the world for the
rest of time.
And as part of this,
the British mount an expedition
to confirm a German theory of
science.
And this is regarded as a huge
sign of the great new future
that we're moving into of
international cooperation.
And a consequence of this--this
is front-page news all over the
world, and it makes Einstein
famous.
This is what makes Einstein
famous.
So, the fable,
here, is the 1919 eclipse
expedition.
And the moral that was drawn at
the time is, the great
international,
is how science is this
wonderful international--maybe
even universal--quest.
And, in retrospect,
what it is, is,
it's a great demonstration of,
you know, how science is
supposed to work.
New theory makes a prediction.
You go out and measure the
effect, and it works.
So, this is,
science works as advertised.
Except, you know,
maybe it doesn't.
Because subsequently,
in subsequent years,
people looked back at the
apparatus that Eddington and his
team had put together and said,
you know, you can't actually
make measurements that accurate
with this stuff.
So, how were they able to make
this measurement?
And, oddly enough,
for about fifty years,
the photographic plates –
what they had done was taken
photographs of the sky – had
disappeared.
And so, nobody could go back
and check the original data.
At some point,
subsequently,
they kind of reappeared
inside--just recently,
actually,
inside the personal effects of
a British astronomer who had
gone to live in Chile.
And nobody quite knows how he
got a hold of them.
And, I actually have to say,
I'm not sure what the outcome
of this is--whether people have
actually gone back and tried to
measure this again.
So--and at the time,
you know, Eddington,
who, in his subsequent life,
demonstrated a real propensity
for believing his own theories,
if you know what I mean--which
served him poorly when he was
dealing with Chandrasekhar,
who was thinking of other
things.
Eddington had said before he
went on the expedition that he
was absolutely certain this was
going to work because he had
studied Einstein's theory and
had come to believe it,
because it was so beautiful and
elegant.
And so, there's a famous story
where somebody said,
well, Mr.
Eddington what's--or said to
somebody else on the expedition,
what will Eddington do if it
comes out wrong?
And they said,
well, Eddington will just go
insane.
And so, there's a little bit of
suspicion attached to this
particular experiment.
However, subsequent to that
this has been measured
repeatedly, many times,
during many eclipses,
and it's very clear that this
works.
And as I said,
there's now evidence from other
astronomical objects for the
deflection of light,
all of which confirms
Einstein's theory.
Yes?
Student: What does the
second line say?
Professor Charles
Bailyn: What does the second
line say?
Let's see, eclipse expedition,
scientists and international,
universal quest,
science works as advertised,
maybe.
That's what I was trying to
write down.
So, that's the second
post-Newtonian test.
And the fact that it works in
1919 is what makes people really
believe relativity,
and what makes people think
that Einstein's this intense
genius.
Okay, there are a couple of
other post-Newtonian effects.
So here's, now,
post-Newtonian effect number
three.
And this is something called
the gravitational redshift.
You've heard of a redshift.
Redshifts and blueshifts.
They come about because of the
Doppler shift.
Something moves towards you,
its light gets shifted to the
blue.
Something moves away from you,
its light gets shifted to the
red.
Turns out, the wavelengths of
light also change due to
gravity.
Or due to being emitted in a
curved space-time,
if you want to use the
relativistic formulation.
Basically, here's how it works.
You have some object with
radius R,
mass M,
and you have a light source on
the surface of that object.
And you look at it from a
distance.
You're out here at infinity
watching this light.
And the light,
if you emit green light here,
by the time it reaches you,
it'll have shifted into the
red.
And there's an equation.
Let me write down the equation.
So, this is delta lambda
[Δλ]--oops,
that one's not working anymore.
Δλ over λ_0.
That's the shift in light.
And remember that the Doppler
shift has some equation,
here.
Well, here's the equation for
this shift, due to the
gravitation of an object.
R_S /
r, where this is the
distance to the given mass.
Or if you're emitting light
from the surface of an object,
the distance to the mass is the
distance to the center of the
object.
So, in this particular case it
would be the radius of the
object, minus 1.
So, you can see that if there's
no mass, then the Schwarzschild
radius is zero,
and this is 1 - 1,
and there's no shift.
Okay, now why does this happen?
This has to happen.
Imagine that you're standing on
the surface of an object and you
throw an object up into the air.
I have an object here.
I'm going to do this.
What will happen?
As I throw it up into the air,
it's going to slow down.
If I throw it faster than the
escape velocity and I don't have
a ceiling and stuff,
then it'll go--it'll keep going
away forever,
but it will,
nevertheless,
slow down.
If I throw it up slower than
the escape velocity,
then it's going to rise to some
height, stop,
turn around and fall back.
Right?
So, that's what happens with an
object.
Why is this happening?
The reason it happens is
because as it moves--as it tries
to move away from the Earth,
it has to lose energy,
because it's pushing its way
out of the gravitational field.
And so, it gradually slows
down, because its kinetic energy
decreases, as it goes further
away,
to balance the change in the
potential energy,
for those of you who remember
high school physics.
It balances the change in
potential energy,
so the total energy is zero.
And so, the kinetic energy has
to slow down.
And whether it slows down
enough to stop it or not depends
on whether it's moving faster
than its escape velocity or not.
Supposing I do this with a
flashlight.
And instead of throwing a
keychain up in the air,
I throw a photon up in the air.
The photon starts out at the
speed of light.
Now, that, too,
has to lose energy as it moves
away from the Earth.
But light, as we know,
always travels at the speed of
light.
It doesn't slow down.
It can't slow down.
So, what it has to do is,
it has to lose energy in some
other way.
And the trick is that the
energy of a photon is equal to a
couple of constants,
h times c, over
the wavelength.
So, as the energy gets less,
the wavelength gets longer--has
to get bigger.
And so, what is happening here
is, as you shine light off a
gravitating object,
as it gets further and further
away from the object,
it loses energy.
But it loses energy not by
slowing down,
as an ordinary mass of objects
would, but by changing its
wavelength.
And now, go back and look at
this equation again.
Supposing you're shining your
flashlight from the event
horizon of a black hole--from
something where the distance to
the object is equal to the
Schwarzschild radius.
Then, you get Δλ equals
infinity.
And you've redshifted your way
out to an infinitely large
wavelength.
The consequence of that is that
the energy of the light is equal
to zero.
And so, this is how it happens
that light can't escape from a
black hole.
Because, you know,
if you have the Newtonian
vision of the escape velocity in
mind--the thing goes up,
stops, turns around and comes
back--and you imagine shining a
flashlight from inside the event
horizon of a black hole,
you imagine that the light
would go up, stop,
turn around and fall back.
But light doesn't do that,
because light always goes at
the speed of light.
So, what happens instead is,
it redshifts itself out of
existence altogether.
And so, that's the way light is
prevented from leaving a black
hole.
Because if you shine from
either inside,
or at, the event horizon,
it'll redshift itself to
infinity.
Yeah?
Student: Everything you
just said still makes perfect
sense in the context of the
Newtonian concept,
why do we need to [Inaudible]
Professor Charles
Bailyn: Okay,
so there is a Newtonian version
of this.
You can imagine how you expand
this thing.
But this particular equation is
relativistic.
Oh, the other thing is that,
of course, in Newtonian
physics, there's no reason for
light not to slow down.
This business that light always
goes at the speed of light,
regardless of the observer,
is a relativistic effect.
So, you know,
if you take a photon and throw
it in the air,
in a Newtonian concept,
there's no reason it can't slow
down and turn around and come
back.
Yes?
Yeah?
Student: [Inaudible]
Professor Charles
Bailyn: Δλ over λ.
Δλ over λ.
Student: Sorry,
next page.
Professor Charles
Bailyn: This?
Student: [Inaudible]
Professor Charles
Bailyn: Okay,
so this is speed of light -
Student: Yeah.
Professor Charles
Bailyn: And this is another
famous constant called Planck's
Constant.
I don't think we're going to
use that equation for anything
in particular.
But it's a couple of constants.
So, basically,
the energy is proportional to 1
/ λ.
Student: Okay.
Professor Charles
Bailyn: Yeah.
Other questions?
Okay.
If you turn it around--if you
sit on the surface of,
I don't know,
a neutron star or something,
and look at distant starlight,
that light will pick up energy
as it comes towards you.
So, if you sit on a gravitating
object and look at a distant
light, that light will actually
be blueshifted,
because it will gain energy
falling down toward you.
And so, while this is always
referred to as the gravitational
redshift, because we're always
imagining that we're some
distant observer in a flat space
looking at some gravitating
object,
in principle,
if you could live on a neutron
star, all the stars would look
bluer than they do to us because
of this effect.
So, it works both ways.
This, by the way,
has now been tested in the
laboratory.
Here's how you do it.
You set up--you sit on top of a
ladder and you look at light
coming from the surface and
coming up toward you.
And then, you have
somebody--you have some
confederate, or graduate
student, or something,
who sits on the ground and
looks at light coming sideways.
And you determine that the
wavelength detected here is
slightly larger than the
wavelength detected here.
That's a really small effect,
because the Earth's
gravitational field--you know,
if you do this little
calculation and you try and
figure out R_S
/ r for the surface of
the Earth,
it's a really,
really small difference from 1.
Nevertheless,
because you're in the
laboratory and we know how to
measure the wavelengths of light
with incredible accuracy,
this effect can actually be
measured in the lab,
and has been.
Yes?
Student: So,
if you're on a black hole or
whatever, shining a light upward
and instead of doing the arc
thing just--see through
the--stretches out and winks out
of existence at the top,
then winks back into existence,
falls back?
Professor Charles
Bailyn: Well,
it can't fall back,
right.
It can't fall back because it
can't stop, turn around and come
back.
Student: So,
it just winks out of existence
up here?
Professor Charles
Bailyn: Well,
notice what happens.
Supposing you go from the event
horizon only to slightly above
the event horizon.
The difference in the
wavelength you see is this
quantity at the event horizon,
minus this quantity a little
bit above the event horizon,
because it's still going to be
redshifted.
But this is infinite at the
event horizon.
So, by the time it moves even a
tiny bit off the event horizon,
it's already gone.
Student: Okay.
Professor Charles
Bailyn: Yeah,
it's because this goes--this
term becomes infinity large.
Oh, but I should say,
this brings up a good point.
Supposing you're here and it's
not a black hole.
So, it's not going to--it's not
going to be disastrous.
But you're observing it,
not at infinity,
but close.
So here's--this is at some
R, and this is at
R_1 a little
bit further away.
So, what you would observe,
here, is Δλ / λ = Δλ at
R_1 minus
Δ--let me write this so that
there's some hope of reading it.
So, this gives you the Δλ you
observe at R_1,
over λ_0,
is equal to Δλ at
R_1,
calculated by that formula,
minus Δλ at R.
Because, it redshifts a little
bit from here to here.
It would redshift more from
here out to infinity.
And so, the way you calculate
that is you calculate redshift
from here to infinity,
minus redshift from here to
infinity.
And since--for something that
is at the event horizon,
the redshift from there to
infinity is infinite,
even if you're observing it
just a little bit higher,
it's already gone.
Student: Okay.
Professor Charles
Bailyn: Does that make
sense?
How are we doing?
Other questions?
Yes?
Student: [Inaudible]
Professor Charles
Bailyn: Oh,
the photons,
remember--there's a
relationship between the whole
way wave particle duality works
for photons.
There's a relationship between
the energy of the photon and the
wavelength.
That was this E =
hc / λ thing.
So, as the wavelength becomes
infinite, the energy in that
photon goes to zero.
So, it's still a photon.
But if you've got a photon with
zero energy, you're not going to
notice it.
Yes?
Student: Just as we,
like, you know,
use the gravitational field of,
like, Mercury,
Jupiter, or whatever to kind of
slingshot probes into outer
space, could light,
like, pick up the energy by,
like, you know,
going around massive objects?
Professor Charles
Bailyn: Well,
light picks up energy as it
falls towards a massive object.
That's the inverse
gravitational redshift.
You sit on the gravitating
object.
You watch light coming at you
from a distance.
So, it does that.
That slingshot effect only
works if you add energy to the
thing and kick yourself back out
of orbit at just the right
moment.
You can't do that with light,
because you can't give it
extra--you can't give it an
extra velocity kick,
at any moment,
to push it back out.
But it is true,
that if you sit--you can do
this experiment backwards,
right?
You have your light source up
here, and you observe it from
down here, and then you'll see
it blueshifted.
Other questions?
How are we doing?
Okay.
Let me mention one last
post-Newtonian effect.
This is post-Newtonian effect
number four.
This is the existence of
gravitational waves.
And the way this works is--let
me put these over here.
Remember the bed sheet and the
basketball, over here,
from a few days ago?
All right, now,
imagine that that basketball,
which is some massive object,
is moving back and forth on
the--it's orbiting some other
thing on that bed sheet.
What will happen?
Well, the bed sheet will ripple
in different ways as the thing
moves around.
And this creates,
essentially,
a wave of distortions of
space-time that propagates
outward, as it turns out,
at the speed of light.
So, the way this works is,
as a mass moves back and
forth--perhaps because it's in
orbit,
or for some other reason--it
generates ripples in space-time.
These propagate outward at the
speed of light.
And we'll talk later,
maybe, about how one--what it
means to have a ripple in
space-time, and how one might
detect it.
But there's a more immediate
effect, which is that the energy
in these so-called gravitational
waves--they carry energy with
them.
And the energy comes from--is
extracted from the orbit of the
object that's moving back and
forth.
So, the orbit gradually loses
energy.
And that means that the orbit
gets smaller,
gradually.
So, the orbit gets gradually
smaller.
And so, in the Newtonian
approximation,
things stay in orbit and
they're perfectly stable forever
and ever,
and they just go orbiting
around, and around,
and around.
But, in general relativity,
there's a gradual,
slow loss of energy,
and objects will gradually
spiral in.
Now, this happens really slowly.
It has not been detected in the
Solar System.
That is to say,
even the orbit of Mercury,
we can't measure its orbital
period accurately enough to see
this slow down.
Certainly all the planets in
the Solar System have been
happily going in their orbits
for four-and-a-half billion
years with no sign of this
effect.
But you can see this in certain
kinds of binary stars.
Stars which happen to have very
short orbits.
And in particular,
there's something called the
binary pulsar.
Pulsar is a kind of neutron
star.
And these gravitational wave
effects, the gradual decrease in
the orbit, has been observed in
binary pulsars.
And we'll talk about that,
I guess, next time.
 
