>> In the last lecture, I
derived the Schwarzschild metric.
So that's ds squared equals minus 12 minus 2m
over r dt squared plus 1 divided by 1 minus 2m
over r dr squared plus r
squared d omega squared.
And this is a solution of the
vacuum Einstein equation --
-- where by vacuum we mean the stress
energy momentum tensor vanishes.
If we let our spacetime be described by
this metric for all values of little r --
that is, for all values of the areal radius --
then this is a spherical black hole of mass m.
Now one question people often have is the
following: If this represents an object
of mass m, a black hole, then how can that be,
since we solved the vacuum Einstein equations?
Vacuum means t mu equal 0, so there
are no matter fields anywhere,
so how can there be mass, or said another
way where is the mass in this spacetime?
The answer is that we're too used
to thinking of mass as a property
that only material particles can
have, but remember, in relativity,
mass is the same thing physically as
energy, and fields can have energy.
For example, the electromagnetic field can carry
energy, and therefore it can add mass, and also,
the gravitational field itself can have
energy associated with it, or mass,
and that's what's happening here.
The black hole, the mass, is just
a property of the spacetime itself.
And recall how we define m. We define
it using a Kepler type experiment
where we release a particle or a satellite
at large R and measure its period and relate
that to the mass m. So the measurement or the
identification of the mass has nothing to do
with the presence of any material particles.
Now sometimes you'll hear someone say that
the mass of a black hole is all concentrated
at the singularity, and there may be
some sense in which that's correct,
but it leads you to view the
mass as being concentrated
at a physical location inside the
black hole, in a physical center,
and that's not the right point of view.
The singularity is actually
not a location in space.
It's a set of events in the future.
So instead of saying, "Where's the mass?"
we should be asking, "When is the mass?"
But now I'm getting a little bit ahead of
myself in talking about it the singularity.
In order to understand these
statements, we need to go back
and examine carefully the causal structure
of the spacetime described
by this Schwarzschild metric.
Let me begin by making a
few simple observations.
We're considering the spacetime
described by this metric for all values
of the aerial radius little r, so
we need to worry about what happens
when little r is equal to either 2m or 0.
When r equals 2m, this metric component
vanishes, while this metric component goes
to infinity, and when r is
equal to 0, this component goes
to infinity and this component vanishes.
So what is happening --
-- at r equals 0 and r equal 2m?
The first thing we need to
understand is that just
because some metric components
are going to 0 or infinity --
-- this does not imply that there's anything
strange or abnormal about the geometry.
So, for example, let's just consider flat space
in two dimensions using polar coordinates.
All right.
The metric is dr squared plus r squared d theta
squared, and here we have a metric component,
the theta-theta component, goes to 0 at
one point, namely when r is equal to 0.
That doesn't mean there's anything strange
or abnormal about the geometry at that point.
What it's really telling us is that
we have a bad coordinate system.
So if you draw the coordinate
lines, the r equal constant lines,
and the theta equal constant lines, all of these
radial coordinate lines converge at the origin.
So what that means is this point
has multiple coordinate labels.
So r theta could be 0 pi over 2, or it
could be 0 minus 3 pi over 4, and so forth.
So there's a breakdown between the
one-to-one correspondence between pairs of --
sets of coordinate labels
and points in the space.
In this sense, polar coordinates
are bad coordinates,
and that shows up in the
vanishing of this metric component.
So let me give you another example.
Let's let r equal 1 over rho.
So we're defining a new radial coordinate
rho, and dr is minus 1 over rho squared d rho.
So the metric in these coordinates
is ds squared equals 1 over rho
to the fourth power d rho squared plus
1 over rho squared d theta squared.
Now in these coordinates, these metric
components blow up at rho equals 0, but again,
that's not a reflection of anything
funny or abnormal about the geometry.
It's just a reflection of
our choice of coordinates.
In fact, we can convince ourselves that there's
nothing wrong with the geometry described
by either of these metrics by computing
the curvature tensor, and if we did that,
we would find the curvature is 0.
After all, this -- both of these metrics
just describe a flat, two-dimensional plane.
So the real test for anything
strange going on with the geometry is
to compute the curvature, Riemann curvature --
-- r mu nu alpha beta, and in this example,
we would just find that's equal to 0,
which implies the geometry is flat.
So let's do the same thing for Schwarzschild.
Let's compute the curvature.
When we do that, we find that
the Riemann tensor has components
like r t tr equals minus 2m divided by
r squared times 2m minus r and so forth.
Now in this case, with the
Schwarzschild spacetime,
we need to go a step beyond what
we did for the flat space case,
where the Riemann tensor was just 0.
We need to compute curvature and variance.
In other words, we need to compute things that
don't depend on the coordinate system at all.
So we need to compute scalar quantities.
Scalars are functions that have a particular
value at each point in the spacetime,
and their value doesn't depend on
the coordinate system being used.
So, for example, we can compute
the square of the Riemann tensor,
r mu nu sigma rho r mu nu sigma rho.
This is a scalar, and its value is equal
to 48m squared over r to the sixth.
Now there's another way that we
can construct invariant quantities,
and that is to introduce an orthonormal basis.
So, for example, let's introduce basis vectors
et with components 1 divided by the square root
of 1 minus 2m over r comma 0 comma 0
comma 0 and another basis vector er
with components 0 comma square root
of 1 minus 2m over r comma 0 comma 0,
and you can check that each of
these basis vectors is normalized.
The et basis vectors are normalized to minus 1,
and the er basis vectors are normalized to 1,
and now we can compute the scalar quantity r
mu nu alpha beta e sub t mu e sub r nu e sub t
alpha e sub r beta, and this is
equal to minus 2m over r cubed.
And we can go on to construct other curvature
and variance quantities that are scalars built
out of the curvature tensor, and in every
case, what we find is a similar result.
Namely, the curvature invariants
are well-defined at r equal 2m
but blow up to infinity at r equals 0.
So for Schwarzschild --
-- the curvature is just fine.
It's finite at r equal 2m, but it's infinite
at r equals 0.
So the fact that the gtt component
of the metric goes to 0 at r equal 2m
and the grr component blows up at
r equal 2m, that's just a result
of the coordinate system we've chosen.
On the other hand, r equal 2m does
have a special global meaning,
and that is it's the boundary that divides
spacetime into two regions, an interior
and an exterior region, with the
difference being that light rays
in the interior region can
never reach infinity --
can never reach asymptotic
infinity -- where r is infinite.
So r equal 2m is referred
to as the event horizon.
and r equal 0 is referred to as the singularity.
The curvature blows up at the singularity,
and there's nothing that we can do about that.
That has nothing to do with coordinates.
So a change of coordinates can't change the
geometry or the property of the spacetime,
but different sets of coordinates can be
useful for understanding different features
of the spacetime, so what I'd like to do now is
to change from the Schwarzschild coordinates --
in particular, from the Schwarzschild
time coordinate --
to a new time coordinate
called the Kerr-Schild time.
So let me just remind you in Schwarzschild
coordinates the Schwarzschild geometry is minus
1 minus 2m over r dt squared
plus 1 divided by 1 minus 2m
over r dr squared plus r
squared d omega squared,
and now we'll introduce the Kerr-Schild time.
And I'll just put a little
subscript KS here for Kerr-Schild.
It's related to the Schwarzschild
time coordinate t
by Kerr-Schild time equals Schwarzschild
time plus 2m times the natural log
of absolute value of r over 2m minus 1.
Then dt Kerr-Schild is equal to dt Schwarzschild
plus 1 divided by r over 2m minus 1 times dr,
and we can of course solve this for dt and then
plug that into here to find the following result
for the metric: ds squared equals minus
1 minus 2m over r times dt squared,
which is dt Kerr-Schild minus 1 over r over 2m
minus 1 dr, all of that squared, plus this term,
1 minus 2m over r dr squared
plus r squared d omega squared.
Now we multiply this out.
We have minus 1 minus 2m over
r dt Kerr-Schild squared.
Then there's a cross term here.
That would be plus 1 minus 2m over r, and then
divided by r over 2m minus 1 times a factor
of 2 dt Kerr-Schild dr, and then the
term coming from squaring this term --
we have minus 1 minus 2m over r times 1
divided by r over 2m minus 1, quantity squared,
and we also have another term
that's proportional to dr squared.
That's plus 1 over 1 minus 2m over
r, all of that times d r squared,
and then finally we have this
term, r squared d omega squared.
Now there are a few more steps of
algebra to go through to simplify this,
and I won't go through it in detail, but let
me just write down the answer, and while I --
when I do, that let me change notation.
Instead of calling the new time coordinate with
a sub KS, let me just drop the KS and just call
that t. so we'll understand that
t is now the Kerr-Schild time.
The result is minus 1 minus 2m over r dt
squared -- that's Kerr-Schild time squared --
squared plus 4m over r dt dr plus 1 plus 2m over
r dr squared plus r squared d omega squared.
So this is the metric for a
Schwarzschild black hole written in terms
of the Kerr-Schild time coordinate.
This Kerr-Schild coordinate system
has some nice properties in common
with our original Schwarzschild
coordinate system.
For example, r is the areal radius.
So r measures the area of the spheres that
are defined by the rotational symmetry
of the spacetime and represented
by this part of the metric.
So the area is 4 pi r squared.
The other property it has is that
the Killing vector field is d by dt.
You can see because the metric components
in these coordinates are t-independent.
This might seem surprising that d
by dt is the Killing vector field,
because d by dt was also
the Killing vector field
when we were using our original
Schwarzschild coordinates.
Let me remind you of the relationship
between the Schwarzschild coordinate,
which here is called t, and the new
time coordinate, here called t sub KS.
This transformation doesn't actually
change the t-coordinate lines --
in other words the lines of
constant r, theta, and phi.
What it does is it changes
the t equal constant surfaces.
So let me draw a picture.
These are the t equal constant
curves or surfaces
in the original Schwarzschild coordinates.
And these are the integral curves
of the Killing vector field.
So that's d by dt Schwarzschild,
but it's also d by dt Kerr-Schild,
and now the t equal constant surfaces in the
Kerr-Schild coordinate system look like this.
So these are t Kerr-Schild equal
constant, and in both cases,
we're carrying the spatial coordinates
-- the r, theta, phi coordinates --
along the integral curves
of the Killing vector field.
In this case, though, with
Kerr-Schild coordinates,
the t equal constant surfaces are not
orthogonal to the Killing vector field.
Our next task will be to investigate
the causal structure of the black hole.
And we do this by examining the light columns.
So we need to compute the null directions.
So let's consider a radial curve at some curve
t of lambda, r of lambda, theta of lambda equal
to a constant and phi of
lambda equal to a constant.
Now we want this curve to be null, so we
want ds squared to be 0 along the curve,
and ds squared is minus 1 minus 2m over r dt
squared plus 4m over r dt dr plus 1 plus 2m
over r dr squared plus r squared d
omega squared, and now if we plug this
into the line element, we
have minus 1 minus 2m over r
of lambda times t dot squared times d lambda
squared, but since this is equal to 0,
I'll drop the d lambda squared, plus 4m over r t
dot r dot plus 1 plus 2m over r, r dot squared,
and d omega squared is 0 because
theta and phi are constants.
Now this can be rewritten as minus
t dot squared plus r dot squared --
that takes care of this 1 and this 1 --
plus 2m over r times t dot squared plus
r dot squared plus 2 times t dot r dot.
So these first terms are t dot
plus r dot times t dot minus r dot,
and these next terms can be written as 2m over
r times t dot plus r dot quantity squared.
So now we can factor out the
factor of t dot plus r dot.
That's times minus t dot minus r dot, so
let's just write that as r dot minus t dot,
plus the remainder here is 2m
over r times r dot plus t dot.
Now we're looking for all the ways
in which this expression can equal 0.
So one way for this to equal 0 is for t dot plus
r dot to equal 0, and this of course tells us
that t dot equals minus r dot, and
we can now integrate this to find
that t equals minus r plus a constant.
Now another way that this expression can
equal 0 is for r to equal 2m, a constant.
So if r is equal to 2m, then r dot is 0
here and here, and 2m over r is just 1,
so the minus t dot cancels the plus t dot, and
this whole expression in square brackets is 0.
Now the final set of null directions
comes from setting the factor
in square brackets equal
to 0 for r not equal to 2m.
So for r not equal to 2m, we set this equal to
0, and we have r dot plus 2m over r times r dot.
That's this term and this term.
Let's bring the t dot terms to the other side.
We have t dot minus 2m over r times t dot.
Now we can rearrange this as --
let's write this term on the left.
This is t dot times 1 minus 2m over r equals r
dot times 1 plus 2m over r, and now we'll divide
by this factor to obtain t dot equals 1 plus
2m over r times r dot divided by 1 minus 2m
over r. Now this can be integrated.
The solution is t equals r
plus 4m times the natural log
of absolute value of r minus 2m plus a constant.
So these are all the radial null
curves in the black hole spacetime,
and we can plot them in a tr diagram.
That's what I've done here.
So in this diagram, the vertical axis is
the t-axis, horizontal axis is the r-axis.
So keep in mind each point in this
diagram represents a 2-dimensional sphere
of areal radius little r. Now the straight lines
in this diagram are the null directions
t equals minus r plus constant.
And these curves that fan outward here
and inward here are the curves t
equals r plus 4m times natural log
of absolute value r minus 2m plus constant.
The remaining null curve in
this diagram is r equal 2m.
Now r equal 2m is of course
the black hole horizon.
So it's represented by this heavy black line in
the diagram, and r equal 0 is the singularity,
so I'm going to put a little squiggle
line along r equal 0 to remind ourselves
that the curvature is singular there.
Now let's sketch some of the light cones.
So for an event -- say an event right
here -- these are the future null rays,
so the light cone looks something like this.
For an event here, we have --
here's the light cone, and so forth.
This event.
And you'll notice as we get closer
and closer to the horizon, 2m,
the light cones are becoming more and more
tipped to the left, towards small values
of r. Inside the horizon, here's
a light cone, and so forth.
Now the horizon itself, 2m, is a null curve.
So here's a light cone that
sits right at the horizon.
Here's another one.
Let me draw one more light cone
outside the horizon, say right here.
Now keep in mind that the light cones
represent the causal future of events.
So nothing can travel faster than
light, so any material particle
or observer must follow along a time-like
curve that stays inside the light cone.
So, for example, here's an observer, observer's
world line always staying inside the light cone
at each event, and notice this observer
is staying outside the black hole,
outside referring to the
region r greater than 2m.
The interior or inside of the
black holes are less than 2m.
Now let's draw the world line of an
observer who crosses the horizon.
So this is a time-like world line, so it always
stays within the light cone at any given point.
Let's say the observer crosses the
horizon right here at this event.
The observer must stay inside the light cone,
and you can see because the light cones
are all tipped towards the singularity.
This observer has no choice but to
continue moving to smaller and smaller r
until they hit the singularity at r equal to 0.
We can see what's happening more clearly
in the lower part of the diagram.
Let's say we have an observer who
reaches this event right here.
That observer has no choice but to move within
the light cone, so that observer must remain
between this null curve and this null curve.
So that means the observer's going to hit
the singularity somewhere in this region,
and no amount of force, no amount
of acceleration can change that,
because this is really the
future for this observer.
So just keep in mind that this
t-coordinate is merely a coordinate.
It has no physical meaning, okay?
In particular, we shouldn't think of
time as running up in this diagram.
In fact, for any observer in
the interior of the black hole,
time is moving to the left,
towards the singularity.
The singularity you can see
is not a location in space.
It's actually a future time.
It's not a place or location in space,
and this is why it can't be avoided.
It's the future for any observer or
any particle that crosses the horizon
into the interior of the black hole.
So this diagram I think is a good way to
understand the causal structure of a black hole
and the origin of the familiar
statement that nothing can escape
from a black hole, not even light.
You can see in this diagram that even
all of the light rays are eventually drug
into the singularity and no light
rays escape across the horizon at 2m.
Let me comment on one last feature of a black
hole that people sometimes find confusing.
Here we have a particle that's falling
into the black hole across the horizon,
and here we have an observer on the outside.
So let's pretend the particle is glowing.
It's giving off light, and these
are some of the light rays that pass
from the particle to the observer.
So when the particle -- so this is the light
that's released from the particle at this event
and received by the observer at this event.
Here's the light that's emitted at this event
and received at this event and so forth,
and you can see that what's happening is
that as the particle moves closer and closer
to the horizon, the light received by the
observer is getting more and more spread out.
So that's as these light rays, these null rays,
become more bunched up close to the horizon,
they take longer and longer to escape out to
the observer near infinity, and in the limit,
as the particle approaches the horizon, it takes
infinitely long amount of time from the point
of view of the observer for these
light rays to reach the observer.
So the observer never actually sees
the particle cross the horizon.
In fact, the event at which the
particle reaches the horizon --
light released from that event
just travels along the horizon.
It never escapes out to infinity,
out to where the observer can see it.
So although the observer never sees
the particle cross the horizon,
it's not the case that the observer sees
the particle just hover above the horizon.
What happens is these light rays
are becoming more and more spread
out as the particle approaches the horizon,
and that means they're becoming
more and more red-shifted.
So we can think of these light rays representing
the world lines of the peaks and troughs
of a light wave, and they're
becoming more and more spread out,
which means the light is becoming
more and more red-shifted,
so no matter how sensitive the observer's
equipment might be, there comes a point in time
when the observer can no longer see the particle
because the light has red-shifted
below the threshold of the instruments.
So what the observer sees is the particle just
faded to blackness as part of the black hole.
So the statement that an outside observer
will never see a particle cross the horizon is
technically correct, but misleading.
A particle certainly can cross the horizon
of a black hole, and of course once it does,
it has to hit the singularity, and we can even
calculate the time it takes the particle to move
from the horizon to the singularity.
It's not too difficult of the calculation.
I think I'll save the details
for a practice problem,
but just let me quote some of the results.
If we have a particle that falls freely --
-- so in other words, along a geodesic,
and we'll make it a radial geodesic --
starting from infinity and
it falls through the horizon
and into the singularity, the time it takes
from the horizon to the singularity of
course depends on how big the black hole is.
For a black hole of mass 1
solar mass, the mass of the Sun,
the time is approximately 6.6 times 10 to the
minus 6 seconds, so about 6.6 microseconds.
That's very fast on a human time scale.
For a large black hole, the
time scale can be much longer.
For example we can consider the black
hole at the center of the galaxy M87.
This was recently imaged --
-- by the Event Horizon Telescope.
This black hole has a mass of
about 6.5 billion solar masses,
and the time it would take a particle to fall
from the event horizon of this black hole
to the singularity is about 11 hours.
