[SQUEAKING]
[RUSTLING]
[CLICKING]
SCOTT HUGHES: So in
this final lecture,
I want to think a bit
sort of with an eye
towards thinking about
how one might actually
make measurements that prove
the nature of the black hole
spacetime that was discussed
the previous lecture.
I'm going to discuss motion
in a black hole spacetime.
We touched on this a little
bit in the previous lecture,
where we discussed the motion
of radial light rays, OK?
We, in fact, used radial
light rays as a critical tool
for describing the
properties of the spacetime.
We use that to help us
understand the location
of events horizons.
But I want to think a
little bit more generally.
What might it look
like if I have
material orbiting in
the vicinity of one
of these black holes?
What if it's not light?
What if it's made out of matter?
And so what this is
going to boil down to
is understanding the behavior
of geodesics in a black hole
spacetime.
And the naive approach to doing
this is not wrong, but naive.
What you do is you would
just take the spacetime--
take your Schwarzschild or
take your Kerr spacetime--
and turn a very large crank,
grind out all of the connection
coefficients, evaluate
the geodesic equation,
integrate it up.
Solve for the geodesics.
Boom, you got yourself
your motion, OK?
And that is absolutely correct.
You can do that using
your Kerr space time
or your Schwarzschild spacetime.
In fact, if you do that for--
you can get the connection
coefficients describing this.
Those are relatively
easy to work out.
I think they are listed in
Carroll, in equation 5.53,
according to my notes.
That may just be
for Schwarzschild.
But at any rate, they're all
listed there, and have a blast.
This approach is
not wrong, but--
my notes say, but
it is not useful.
That's not really true.
I'll just say that there is a
more useful approach to this.
A more fruitful approach is to
exploit the fact that these are
highly symmetric spacetimes.
Exploit the symmetries
and the Killing
vectors, and see
how they can be used
to reduce the number
of degrees of freedom
that you need to describe.
In this lecture, I'm
going to go through this
in quite a bit of detail
for Schwarzschild.
The concepts that I'm
going to apply work
for Kerr and for
Kerr-Newman as well, OK?
Schwarzschild is just a
little easier to work with.
It's something that I can
fit into a single lecture.
In particular, one of the nice
things about Schwarzschild--
so let's go ahead and
write down that spacetime.
So one of the nice things
about Schwarzschild
is it is spherically symmetric.
This means I can always
rotate coordinates such that--
well, when me think
about it-- let's
just back up for a second.
Imagine that I have some kind of
a body orbiting a Schwarzschild
black hole.
Spherical symmetry
tells me that there
must be some notion of a
conserved angular momentum such
that that orbit always
lies within a given plane.
Put it another way.
Because it is fairly symmetric,
there cannot exist a torque.
The black hole cannot
exert a torque that changes
the orientation of
the orbital plane.
In particular, because
it is fairly symmetric,
there is no unique notion of
an equator to this object.
And so you might as well define
any orbits to live in the theta
equals pi over 2 plane.
You can always rotate
your coordinates
to put any orbit in the
theta equals pi over 2 plane.
It's actually a pretty
simple exercise.
I won't do this,
but if you start
with an orbit that is in the
theta equals pi over 2 plane,
and it is moving such that its
initial velocity would keep it
in a theta equals
pi over 2 plane,
it's a very simple exercise
using the geodesic equation
to show that it will
always be in the theta
equals pi over 2 plane.
So spherical symmetry
says, you know what,
let's just forget about the
theta degree of freedom.
I can always define my
coordinates in such a way
that it lives in the theta
equals pi over 2 plane.
Boom, I have reduced my
motion from, in general,
being three spatial dimensions
to two spatial dimensions.
That's another one
of the reasons why,
for pedagogical purposes,
it's nice to start
with Schwarzschild.
For Kerr, this is
not the case, OK?
Kerr is a little bit
more complicated.
You have to treat the
theta motion separately.
It's not strictly symmetric,
and so you can't do that.
I have actually spent
a tremendous amount
of my career studying
the orbits of objects
around Kerr black holes.
And I do have to say that the
additional complications that
arise from this
lack of sphericity,
they're really beautiful, OK?
There's an amazing amount of
fun stuff you can do with it.
You know, there's
a reason why I just
keep coming back to
this research problem,
and part of it is
it's just bloody fun.
But if you're teaching this
stuff for the first time,
it's not where
you want to begin.
All right, so
Schwarzschild allows
us to reduce it from a
three-dimensional problem
to a two-dimensional problem.
And Schwarzschild also
has two Killing vectors.
The time derivative of every
metric component is equal to 0.
That means p downstairs
t is constant.
So there is a timelike
Killing vector.
So p downstairs t is constant
every level along the orbit.
It means there exists a
timelike Killing vector.
And so what we do
is we associate this
with the energy of the orbit.
We call this up to a minus sign.
And we choose that minus
sign because if we imagine
orbits have very,
very large radius,
the spacetime is
nearly flat, and we
want to sort of clear out
the minus sign associated
with lowering our index here.
We're going to call
that constant negative
of the energy of the orbit.
The spacetime is also
independent of the angle phi.
And so p sub phi is a constant.
This means that the spacetime
has an axial Killing
vector, something associated
with motions around a symmetry
axis.
So I'm going to
call this L-- well,
actually, I will tend
to call it L sub z.
You can kind of think
of this as, after I
have put everything in the
theta equals pi over 2 plane,
this is like an angular
momentum on the--
angular momentum parallel to
the axis normal to that plane,
and I call that the z-axis.
It's worth noting that
all three of these things
are also true for
Reissner-Nordstrom black holes.
This one is not true for
Kerr, but these two are, OK?
So although the details
change, many of the concepts
will carry over when you look
at different, more complicated
classes of black holes.
Let's now think about
the forward momentum
of a body moving in a
Schwarzschild spacetime.
So let's say it's
got a rest mass m,
then it will have
three components.
Remember, I have put
this thing in a plane
where there is no theta motion.
It will have three theta
motions describing its motion
with respect to the
time coordinate,
the radial coordinate, and
the actual coordinate phi, OK?
So before I do
anything with this,
let's take advantage of the fact
that we have these quantities
that are constants.
So using the fact
that p downstairs mu
is what I get when
I hit this guy--
that be a mu, pardon me.
This is what I get when I
hit this guy with the metric.
I can write out a p
sub t and p sub phi.
So let's do that
on another board.
Well, let's do a p sub t first.
So p sub t is going
to be gtt, p sub--
whoops.
OK, this is minus.
There's an m from that--
minus sign from my metric.
And this whole thing I define
as the negative energy, e.
OK, so that combination--
we'll do something at the
rest mass in just a moment.
But you should
basically look at this
and saying that dt d
tau, which tells me
about how the body moves with
respect to the Schwarzschild
time coordinate--
that complement of the
four velocity times 1 minus
2gm over r is a constant.
Let's look at p downstairs phi.
So this is m r squared
sine squared theta.
Ah, we've chosen our orbital
plane, 1 times d phi d tau.
This is equal to [INAUDIBLE]
momentum L sub z.
In my notes, I might flip around
a little bit between L sub z
and L.
So let's massage these
a little bit more.
So I can take these
two expressions
and I can use them
to write dt d tau
and d phi d tau in terms of
these conserved quantities.
So dt d tau is equal to--
I'm going to call it e hat
divided by 1 minus 2gm over r.
d phi d tau is going to be
Lz hat over our squared.
And so these
quantities with a hat
are just the conserved values
normalized to the rest mass,
OK?
They all are proportional
to the rest mass,
but it's actually if you
want to know what the--
this is essentially telling me
about how clocks on the orbit
tick relative to clocks that
are infinitely far away.
And that can't depend on the
mass of the orbiting object.
This is telling me about
how this small body is
moving according to the clock
of the orbiting observer.
And again, that can't depend
on the mass of the object.
OK, so that's kind of cool.
So I've now managed
to relate three--
excuse me, two of
the three components
of the forward momentum to
functions of r and quantities
that are known to be constant.
That's good.
We have one more
overall constraint.
We know that if I take p dot p,
I get negative of the rest mass
squared.
So this, when I write it out--
OK, notice every single term
is proportional to m squared.
So I can divide that out.
I can insert dt d tau.
I can replace for this
e divided by this guy.
I can replace d phi d tau by
my Lz hat divided by r squared.
Doing so, manipulating
a little bit, I get--
we get something like this.
And let me rearrange
this a tiny bit.
All right, I have
managed to reduce this
to a one-dimensional
problem, OK?
So going from that line over
to there, basically all I did
was insert the relationship
between dt d tau and E, d phi
d tau and L, cleared out some
overall factors of things
like 1 minus 2GM/r, manipulate,
manipulate, manipulate,
and what you finally get is
this lovely equation here
that tells you how the radial
velocity, the radial velocity
with respect to
proper time, how it
depends as a function of r
given the energy and the angular
momentum.
I have written it in this
form because the problem
is very strikingly reminiscent
to the Newtonian problem
of understanding the motion of
a particle in a 1/r potential,
which we often describe as
having an effective potential
that has a gravitational term--
Newton's gravity,
or if you're doing
things like quantum mechanics,
the Coulomb potential,
and a Coulomb barrier
associated with the angular
momentum of an orbit.
So one way to approach
what we've got now,
one thing that we could
do, is essentially just
pick your energy and your Lz--
pick your energy and
your angular momentum--
pick an initial position--
let's imagine you synchronize
the clocks at t equals 0--
and then just integrate.
You've got your dr
d tau given here.
Don't forget, you also have d
phi d tau and dt d tau related
to the energy and your
angular momentum, like so.
Boom.
It's a closed system.
You can always do just sort of
a little numerical integration
of this.
In a certain sense,
this completely
specifies the problem.
But there's so much
more we can do.
In particular, what
we see is that all
of the interesting behavior
associated with this orbit
is bound up in this
function V effective.
So something that's
really useful for us to do
is to take a look at what
this V effective looks like.
So suppose you are
given a particular value
for E hat and Lz hat, and you
plot V effective versus r.
Well, what you typically
find is that it's
got a behavior that looks kind
of like this, where this value
right here is V
effective equal to 1.
Notice as r goes to
infinity, you get 1 times 1.
So this asymptotes to 1.
Notice E hat squared has the
same dimensions as V effective.
In fact, in the unit
choices I've used here,
they are both dimensionless.
So what we can do is plot--
let's imagine that we have--
you know what?
I'm going to want to sketch
this on a different board.
Let me go over here.
So I'm going to want to look
at a couple of different values
of V effective--
excuse me, a couple of
different values of E hat.
OK.
Here's an example.
This guy is asymptoting at 1.
So since E hat, as I said,
has the same units as V
effective, let's
plot them together.
So example one-- imagine
if E hat lies right here.
OK.
So let's call this E hat 1.
So this is some value
that is greater than 1.
What is the point of doing this?
Well, notice-- I'm
going to flip back
and forth between these
two middle boards here.
Let's look at the
equation that governs
the radial motion of this body.
dr d tau has to
be a real number.
This has to be a real number.
So we have to have E
hat squared greater than
or equal to V effective
in order for dr d tau
to have a meaningful solution.
So let's look at my
example here, E hat 1.
E hat 1 is greater than
my effective potential
everywhere at all radii
until I get down to here.
Let's call that r1.
So in this case, dr d tau,
if I think about this thing--
so note that defines
dr d tau squared.
Let's suppose we take
the negative square root.
dr d tau is positive
and inward--
or, well, it's negative--
negative and real, negative
and real, negative and real,
negative and real,
negative and real-- boom.
It's 0.
Can it go into here?
No.
It cannot go into there, because
there E hat squared is less--
sorry.
That should have been squared.
Inside here, E hat
squared is less
than the effective potential.
dr d tau is imaginary.
That doesn't make any sense.
The only option is for this
guy to change sign and trundle
right back out.
So E hat greater
than 1 corresponds
to a body that comes
in from large radius,
turns around at particular
radius where E hat is
the square root of the
effective potential,
and then goes back out to
infinity or back out to-- let's
not say infinity-- goes
back out to large radius.
OK?
In Newtonian gravity, we would
call this a hyperbolic orbit.
This corresponds
to-- so remember,
when I did this I have not--
I'm not actually-- I'm only
computing the radial motion.
I'm not looking
at the phi motion.
So this actually, when you
look at both the radial
motion and the phi
motion, what you see
is that this is a body that
comes in and then sort of whips
around that small radius
and goes right back out.
Let's look at another example.
Let's call this E2--
some value that is less than 1.
OK.
Well, for E hat 2,
it's a potential.
The potential is underneath
E hat 2 squared only
between these two radii, which
I will call r sub p and r sub a.
What we expect in this case is
motion of this body essentially
going back and forth
and turning around
at periastron and apoastron.
This is a relativistic
generalization
of an elliptical orbit.
In general relativity,
they turn out
generally not to be ellipses.
So we call this an
eccentric orbit.
In the weak field limit, if
you imagine r being very, very
large, it's not hard to show
that the motion is nearly
an ellipse, but it's an ellipse
whose long axis is slowly
precessing.
This actually leads to the
famous perihelion precession
of Mercury that Einstein
first calculated.
And this is an
exercise that I am
asking you to do on one of the
final P-sets of this course.
Using what I have set
up here, it's really not
that difficult to do.
Let me go to another board.
And note that one could imagine
an energy such that dr d tau
is exactly 0.
So if you choose your
energy so that you
sit right here at the
minimum of the potential--
I will label this as point s--
the energy that
corresponds to exactly that
point is what would be a--
that is, there is a single point
at which dr d tau equals 0.
Anywhere away from
that, dr d tau
would be imaginary, so
that's not going to work.
But right at that point,
dr d tau equals 0,
and you get a circular orbit.
Notice there's a second point
at which that can happen.
Let's call this point u.
Perhaps you can guess why I
called these points s and u.
If you imagine that you
add a tiny amount of energy
right here, well,
what it will do
is it will execute small
oscillations around the point
s.
It will sort of move
it up to something
that's similar to what
I drew up there as E2,
but with a very
small eccentricity.
So if I slightly disturb a
circular orbit down here at s,
I essentially just
oscillate in the vicinity
of that circular orbit.
S stands for stable.
If I have an orbit up here at
u and I very slightly perturb
it, well, it'll either go in
and eventually reach r equals 0,
or it'll go out, and then
it's completely unbound,
and it will just keep trundling
all the way out essentially
forever.
This guy is unstable.
Stable orbits are particularly
interesting and important.
So let's look at these orbits
with a little bit more care.
So the very definition
of a circular orbit
is that dr d tau equals 0.
Its radius does not change.
If dr d tau equals
0, then it must
have E hat equal the square
root of the effective.
Both of these orbits
happen to live
at either a minimum or a
maximum of this potential curve.
So I'm going to require
that the partial derivative
of that potential with
respect r be equal to 0.
So let's take a look
at this condition.
My effective potential
is given up here.
Do a little bit of algebra with
this, set this guy equal to 0.
What you'll find after
your algebraic smoke clears
is that you get this condition
on the angular momentum.
Notice as r gets really large--
oh, shoot.
Try it again.
Notice as r gets
really large that this
asymptotes to plus or minus
the square root of GM r.
That is indeed exactly what you
get for the angular momentum
of a circular Newtonian orbit.
OK.
So it's a nice sanity check.
It appears to be
somewhat pathological
as r approaches 3GM, though.
Hold that thought.
OK.
So now let's take
that value of L,
plug it back into the
potential, and set
E equal to the
square root of that.
A little bit of algebra ensues.
And what you find is that
this equals 1 minus 2GM/r
over, again, that factor under
a square root of 1 minus 3GM/r.
Again, we sort of see
something a little bit
pathological happening
as r goes to 3GM.
Let me make two
comments about this.
So first of all, notice that
this energy is smaller than 1.
I can intuitively-- you can
sort of imagine-- remember,
E hat is the energy
per unit rest mass.
You can think of
this as something
like total energy over M--
so the energy associated
with the orbiting body.
It's got a rest energy,
a kinetic energy,
and a potential
energy divided by m.
For an orbit to be bound,
the potential energy,
which is negative, must
have larger magnitude
than the kinetic energy.
So for a bound orbit, E
kinetic plus E potential
will be a negative quantity.
So the numerator is
going to be something
that, when normalized
to m, is less than 1.
So this is exactly
what we expect
to describe a bound orbit.
Notice, also-- so if
you take this formula
and look at it in the large r
limit, it goes to 1 minus GM
over 2r.
This is in fact
exactly what you get
when you look at the energy
per unit mass throwing in--
sort of by hand--
a rest mass.
The minus GM/2r
exactly corresponds
to kinetic plus potential for
a Newtonian circular orbit.
So lots of stuff is
hanging together nicely.
So we've just learned that we
can characterize the energy
and angular momentum of circular
orbits around my black hole.
Let's look at a couple other
things associated with this.
So these plots where I
look at the radial motion,
this effective potential, as
I mentioned a few moments ago,
there's additional
sort of degrees
of freedom in the motion that
are being suppressed here.
So this thing is also
moving in that plane.
It's whirling around
with respect to phi.
We've lost that information in
the way we've drawn this here.
Let's define omega to be the
angular velocity of this orbit
as seen by a distant observer.
OK.
Why am I doing it as seen
by a distant observer?
Well, when things
orbit, there tend
to be periodicities that imprint
themselves on observables.
It could be the
period associated
with the gravitational wave
that arises out of this.
It could be the
period associated
with peaks and a light curve
if this is a star orbiting
around a black hole.
It could be oscillations
in the X-ray flux
if this is some kind of
a lump in an accretion
disk of material
orbiting a black hole.
So if this is the angular
velocity seen by distant
observers-- remember, the time
that distant observers use to--
the time in which the
distant observers' clocks run
is the Schwarzschild time t.
So this will be d phi dt, which
I can write as d phi d tau--
this is the angular velocity
according to the orbit itself--
normalized to dt d tau.
Now, these are both
quantities that are simply
related to constants of motion.
What I've got in the numerator
here is L hat over r squared.
And what I've got
in the denominator
here is E hat over 1 minus--
I dropped my t.
It would happen at some point--
1 minus 2GM/r.
So let's go ahead and
take our solution here.
My E hat is 1 minus 2GM/r
divided by square root of blah,
blah, blah.
The 1 minus 2GM/r cancels.
My L hat is square root
GM r divided by, again,
that square root 1 minus 3GM/r.
Notice, the square root 1
minus 3GM/r factors all cancel.
So this becomes plus or minus
1 over r squared square root GM
r.
Looks like this.
Plus and minus basically
just correspond to
whether this motion
sort of is going
in the same sense as
your phi coordinate
or in the opposite sense.
There's really no
physics in that.
It just comes along for the
ride that both behave the same.
If you guys get
interested in this,
and you do a similar calculation
around a Kerr black hole,
you'll find that your
prograde solution gives you
a different frequency than your
retrograde solution because
of the fact that the
dragging of inertial frames
due to the spin of the
black hole kind of breaks
that symmetry.
Something which is
interesting and--
well, I'll make a comment about
this in just a second, which
is kind of interesting.
And here's what I'll say--
sometimes people think this is
more profound than
it should be--
is this is, in fact, exactly
the same frequency law
that you get using
Newtonian gravity.
This is actually exactly
the same as Kepler's law.
That seems really, really cool.
And it is.
It's actually really useful.
It makes it very easy
to remember this.
But don't read too much into it.
More than anything,
it is a statement
about a particular quality
of this radial coordinate.
So remember, in Newtonian
gravity r tells me
the distance between--
if I have an orbit at r1
and an orbit at r2, then I know
that the distance between them
is r2 minus r1.
In the Schwarzschild spacetime,
the distance between these two
orbits is not r2 minus r1.
However, r2 labels a sphere of
surface area 4 pi r2 squared.
And r1 labels a sphere of
surface area 4 pi r1 squared.
It's easy to also show that
the circumference of the orbit
at r2 is 2 pi r2,
the circumference
of the orbit at r1 is 2 pi r1.
That, more than anything,
is why we end up
reproducing Kepler's
law here, is
that this areal
coordinate is nicely
amenable to this interpretation.
So is this orbit--
so I described over here
an orbit that is unstable
and an orbit that is stable.
I have described
how to compute--
if I wanted to find a
circular orbit at a given
radius, those formulas
that I derived over there
on the right-most
blackboards, they tell me
what the energy and
the angular momentum
need to be as a function of r.
Is that orbit stable?
Well, if it is, I can
check that by computing
the second derivative of
my effective potential.
So my orbits are stable if the
second derivative with respect
to r is greater than 0,
unstable if this turns out
to be negative.
Let's look at the crossover
point from one to the other.
What if there is a
radius where, in fact,
the stable and the
unstable orbits coincide?
In fact, what one
finds, if you look
at the effective potential--
you imagine just sort of playing
with L sub z.
So let's say we
take that L sub z,
and we just explore it for
lots of different radii
of the orbits.
You find that as the
orbits radius gets
smaller and smaller, your
stable orbit tends to go up,
and this minimum sort
of becomes flatter,
and your unstable orbit
just kind of comes down.
They sort of
approach one another.
There is a point just
when they coincide--
this should have an
"effective" on it.
My apologies.
Right when they
coincide, this defines
what we call the
marginally stable orbit.
I may have put this
one on a problem set.
But it might be one of the
ones I decided to drop.
So I'm just going to go
ahead and do the analysis.
When you compute this, bearing
in mind that your angular
momentum is a constant--
so take this, substitute in
now your solution for L sub z,
which I've written
down over there--
what you find is that the
marginally stable orbit--
let's call it r sub ms--
it is located at
a radius of 6GM.
This is a profoundly
new behavior
that doesn't even
come close to existing
in Newtonian spacetime--
spacetime-- doesn't come
close to existing in Newtonian
gravity, excuse me.
[SIGHS] I'm getting tired.
The message I want
you to understand
is that, what this tells us is
that no stable circular orbits
exist inside r equals 6GM.
So this is very, very
different behavior.
If I have-- let's just say
I have a very compact body
but Newtonian gravity rules.
I can make circular
orbits around it,
basically go all the way
down until they essentially
touch the surface of that body.
And you might
think based on this
that you would want to
make orbits that basically
go all the way down, that
sort of kiss the edge of r
equals 2GM.
Well, this is telling
you you can't do that.
When you start
trying to make orbits
that go inside-- at least,
circular orbits-- that go
inside 6GM, they're not stable.
If someone sneezes
on them, they either
are sort of blown
out to infinity,
or they fall into
the event horizon.
And in fact, one of the
consequences of this
is that, in
astrophysical systems,
we generically expect there
to be kind of-- if you imagine
material falling
into a black hole,
imagine that there's like a
star or something that's just
dumping gas into orbit
around a black hole, well,
it will tend to form a disk
that orbits around this thing.
And the elements of
the disk are always
rubbing against each
other and radiating.
That makes them get hot.
They lose energy because
of this radiation.
And so they will very
slowly sort of fall in.
But they make this kind of--
it's thought that in
most cases they'll
make this kind of
thick disk that
fills much of the spacetime
surrounding the black hole.
But there will be a
hole in the center.
Not just because the
thing is a black hole.
I don't mean that
kind of a hole.
There'll actually be something
surrounding the black hole's
event horizon, because there
are no stable circular orbits.
Once the material comes in and
hits this particular radius,
6GM in the Schwarzschild case,
it very rapidly falls in,
reduces the density of
that material tremendously.
And you get this much
thinner region of the disk
where essentially things fall
in practically instantly.
It should be noted that
this 6GM is, of course, only
for Schwarzschild.
If you talk about
Kerr, you actually
have two different
radii corresponding
to material that goes around
parallel to the black hole's
spin and material that
goes anti-parallel
to the black hole's spin.
And it complicates
things somewhat.
There's two different
radii there.
The one that goes parallel tends
to get a little bit closer.
The one that's
anti-parallel tends
to go out a little bit farther.
But the general
prediction that there's
an innermost orbit beyond which
stable orbits do not exist,
that's robust and holds across
the domain of black holes.
So let me conclude
by talking about one
final category of orbits--
photon orbits.
So let's recall that
when we were talking
about null geodesics, we
parametrized them in such a way
that d--
sort of the tangent
to the world line
has an affine parameter attached
to it, such that we can write p
equals dx d lambda.
These guys are null.
So p dot p equals 0.
For the case of
orbiting bodies, that
was equal to minus mass squared.
Mass is 0 here.
So we're going to follow
a very similar procedure.
The spacetime is still
time independent,
so there is still a notion
of a conserved energy.
It is still actually symmetric,
so there is still a notion
of axial angular momentum.
But because p dot p is 0 now,
rather than minus m squared,
when we go through the exercise
of-- that sort of parallels
what we did for our
massive particle,
we're going to derive
a different potential.
So let's go ahead and
evaluate this guy again.
And I get 0 equals minus
1 minus 2GM/r dt d lambda
squared plus 1 minus
2GM/r dr d lambda squared.
I'm still going to
require the thing
to be in the theta
equals pi/2 plane
so that we know theta term.
And I have set theta to pi/2.
So my sine squared
theta is just one.
So I get this.
I'll remind you
that I can relate--
so the relationship between
the time-light component
of momentum and energy, the
axial component of momentum,
and the angular momentum, it's
exactly the same as before.
There's no rest mass appearing.
And so now I find--
so my energy, I don't
put a hat on it.
It's not energy bringing it
mass, because there is no mass.
That looks like so.
And my L looks like so.
Using them, I can now
derive an equation
governing the radial
motion of my light ray.
And sparing you the line or two
of algebra, it looks like this.
Pardon me just one moment.
OK.
OK.
So kind of similar to what we
had before, if you look at it
you'll see the term involving
the angular momentum
is a little bit different.
As you stare at
this for a moment,
something should
be disturbing you.
Notice that the
equation of motion
appears to depend on the energy.
OK.
That should really bug you.
Why should gamma rays
and infrared radiation
follow different trajectories?
As long as I am truly in
a geometric optics limit,
I shouldn't.
Now, it is true that if you
consider very long wavelength
radiation, you might need
to worry about things.
You might need to solve the
wave equation in the spacetime.
But as long as the wave
nature of this radiation is--
if the wavelength is small
enough that it's negligible
compared to 2GM--
I shouldn't care
what the energy is.
Energy should not
be influencing this.
So what's going on is
I need to reparametrize
this a little bit.
What I'm going to do to
wash away my dependence on--
wash away my apparent
dependence on the energy here,
is I'm going to redefine
my affine parameter.
So let's take lambda--
so L divided by lambda.
And I am going to define
b to be L divided by E.
This is the quantity that I
will call the impact parameter.
And I'll describe
why I am calling it
that in just a few minutes.
So I'm going to take this entire
equation, divide both sides
by L squared.
And I get something
that looks like this.
What I'm going to do is say
that the impact parameter, b,
is what I can control.
It's the parameter
that-- all I can
control that defines the
photon that I am studying here.
And everything after this,
this is my photon potential.
Notice that the photon potential
has no free parameters in it.
If I plot this guy
as a function of r,
it kind it just looks like this.
Two aspects of it are
worth calling out.
This peak occurs
at r equals 3GM.
Remember the way in which our
energy-- oh, still have it
on the board here-- things
like the energy per unit mass
and the angular momentum
per unit mass all
blew up when r equal 3GM.
3GM is showing up
now when I look
at the motion of radiation,
look at massless--
radiation corresponding to a
massless particle, so to speak.
If I look at the energy per unit
mass, and the mass goes to 0,
I get infinity.
So the fact that I
was actually seeing
sort of things blowing
up as r goes to 3GM
was kind of like the equations
hinting to me in advance
that there was a hidden solution
corresponding to radiation that
could be regarded as
a particular limit
that they were
sort of struggling
to communicate to me.
The other thing
which I want to note
is that the potential, the
height of the potential right
here, it has a peak value of
1 over 27 G squared M squared.
Hold that thought
for just a moment.
Actually, let me write it
in a slightly different way.
This equals 1 over
3 root 3 GM squared.
So now, to wrap this up I need
to tell you what is really
meant by this impact parameter.
So go back to
freshman mechanics.
And impact parameter
there is-- like, let's say
I have a problem where I am
looking at an object that
is on an infall trajectory.
There is a momentum p
that is describing this.
And it's moving in such
a way-- it's not moving
on a real trajectory, right?
It's actually moving
in a sense that
is a little bit off of true
to the radial direction.
This guy actually has an angular
momentum that is given by--
let's say that this
is equal to px x hat.
Let's say this is b y hat.
This guy's got an angular
momentum of b cross p.
That's the impact parameter
associated with this thing.
It sort of tells me about
the offset of this momentum
from being directly
radial towards the center
of this object.
Well, I'm going to
use a similar notion
to give me a geometric sense of
what this impact parameter here
means.
Suppose here is my black hole.
It's a little circle
of radius 2GM.
And I'm sitting way the heck
out here, safely far away
from this thing.
And what I'm going to
do is shoot light--
not quite radial.
What I'm going to
do is I'm going
to have sort of an
array of laser beams
that kind of come up here--
an array of laser beams.
And I'm going to shoot them
towards this black hole.
And what I'm going to do is I'm
going to offset the laser beams
by a distance b.
And I'm going to fire it
straight towards this thing.
Go through sort of a
careful definition of what
angular momentum
means, and you'll
see that that definition
of impact parameter
gives you a very nice sense
of the energy associated
with the trajectory of this
beam and a notion of angular
momentum associated with this.
So let's flip back and
forth between a couple
of different boards here.
There are three cases
that are interesting.
Suppose b is small.
In particular, suppose I
have b less than 3 root 3 GM.
So I start over here.
Let's take the limit.
What if b equals 0?
Well, if b equals 0,
this guy just goes, zoom,
straight into the black hole.
As long as it is anything
less than 3 root 3 GM,
what will happen is, when
I shoot this thing in,
it goes in, and
it bends, perhaps,
but it ends up going
into the black hole.
Let's look at this
in the context
of the equation of motion,
the potential, and the impact
parameter.
If b is less than 1 over 3 root
3 GM, then 1 over b squared
will be higher than this peak.
And this thing is
just going to be,
whoo [fast motion
sound effect], it's
going to go right over
the top of the peak.
And it shoots into small radius.
OK.
Remember, L and E are
constants of the motion.
So that b is a parameter
that defines this light
for the this entire trajectory.
1 over b squared is less
than V phot everywhere--
excuse me, greater
than V phot everywhere.
A rather crucial typo.
And as such, it
shoots the light ray,
and it goes right over the
peak into the black hole,
eventually winds
up at r equals 0.
If b is greater than 3 root
3 GM, then 1 over b squared
is less than V phot at the peak.
This corresponds,
in this drawing,
to a beam of light that follows
a trajectory that kind of comes
in here at this point.
dr d tau, if we were to continue
to go into smaller radii,
it would become imaginary.
That's not allowed.
So it switches direction
and trundles right back
out to infinity.
On this diagram, that
corresponds to a light ray
that's perhaps out here.
This guy comes in, and he
gets bent by the gravity
a little bit, and then just,
shoo [light ray sound effect],
shoots back off to infinity.
The critical point,
b equals 3 GM,
is right where the impact
parameter hits the peak.
And so what happens in this
plot is this guy comes in,
hits the peak, and just
sits there forever.
You get a little bit
more of a physical sense
as to what's going on by
thinking about it here.
This guy comes in,
[INAUDIBLE] this,
and then just whirls
around, and lies on what
we call the photon orbit.
What's the radius of
that photon orbit?
r equals 3 GM.
Now, astrophysically, a
more interesting situation
is, imagine you had some
source of light that
dumps a lot of photons in
the vicinity of a black hole.
Some of those photons
are going to tend to--
some of them are going to
go into the black hole,
some are going to
scatter a little bit
and shoot off to infinity.
But if you imagine
that there will
be some population of
them that get stuck
right on the r
equals 3GM orbit--
now, that is an unstable orbit.
And in fact, if you look at
it, what you find is that
your typical photon is likely to
whirl around a bunch of times,
and then, you know, if
it's not precisely at 3GM
but it's 3.00000000000000001GM,
it will whirl around maybe 10
or 15 times, and then
it'll go off to infinity.
So what we actually expect
is if we have an object that
is illuminated
like this, then we
will actually see these
things come out here,
and we will see--
so bear in mind, this is
circularly-- this is symmetric.
So take this and rotate
around the symmetry axis.
We expect to see a ring
whose radius is twice
the critical impact parameter.
So it would have a
diameter of 6 root 3 GM.
It would be essentially a ring
or a circle of radius 3 root 3
GM.
This is, in fact, what the Event
Horizon Telescope measured.
So last year when I was
lecturing this class,
this was announced almost--
I mean, they timed it well.
They basically timed it to
about two or three lectures
before I discussed this
aspect of black holes.
So that was-- thank you,
Event Horizon Telescope.
That was very nice of them.
And of course, we don't expect--
so Schwarzschild black holes
are probably a
mathematical curiosity.
Objects in the real
universe all rotate.
We expect Kerr to be
the generic solution.
And so there's a
fair amount of work
that goes into how you
correct this to do--
so doing this for
Schwarzschild is,
because of spherical symmetry,
it's beautiful and it's simple.
Kerr is a little bit
more complicated.
But, you know, it's a
problem that can be solved.
And a lot of very smart
people spend a lot of time
doing this to sort
of map out what
the shadow of the
black hole looks like,
what this ring would look
like in the case of light
coming off of a
spinning black hole.
So that's it.
This is all that I'm
going to present for 8.962
in the spring of 2020 semester.
So to everyone, as
you are scattered
around the world attempting
to sort of stay connected
to physics and your
friends and your classwork,
I wish you good health.
And I hope to see you again
at a time when the world is
a little less crazy.
In the meantime, enjoy our
little beautiful black holes.
