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OK, so we were about to embark on a distance journey into,
a journey into a large black hole
Before we do, let me remind you of one or two facts
I just hope we have enough
Oh I actually brought some pens
Anyway let me just go back for a moment to accelerated reference frame
and just to remind you of what the various fact about it are
The accelerated reference frame was a frame
in which every observers is moving on a hyperbola trajectory
As you get closer and closer to the light cone here
the trajectory has an increased acceleration
And you can see that from the fact it has to turn a corner
in a smaller amount of space
So it's not like an accelerated coordinate system
in Newtonian mechanics,
where everybody partake in the same exact acceleration
and all move off simultaneously,
but rather the acceleration gets larger and larger as you move toward
Student: Is it an example that special relativity
Student: is not applied to accelerated reference frame?
No this isn't an accelerated reference frame
Student: But (mumble) meter sticks and (mumble) different acceleration
Yes, that's also true in the vicinity of a black hole
This point or this whole line in fact replaces the black hole horizon
And if you standing if you are supported ten meters
or a hundred meters away from the black hole horizon
You have one acceleration
Let's first take the earth,
if you lower down to the surface of the earth
well from a large distance, in, not lower down but you supported
of course your acceleration is larger
the closer you are to the center of the earth
So it's not that surprising
In that sense, the accelerated reference frame
is not a very precise version of a gravitational field
Because gravitational field varies
Here in special relativity we already have this property
the closest thing to the uniformly accelerated reference frame
already has this fact that the gravitational strength
or the acceleration that you feel
varies from distance from this point over here
OK we also discussed, these are the different observers
They could be characterized by a coordinate r
which at least along here just represent
how far they are from the origin of coordinates over here
Characterized by an r
And they are also characterized by a hyperbolic angle
The hyperbolic angle, did I call it omega last time?
What did I call it, omega?
Sort it out if you want the precise relationship
with the origin coordinates on the blackboard
Call it x and t, then to make the relationship precise
x is equaled to r times a hyperbolic cosine
cosine of the hyperbolic angle omega
and t is equaled to r times the hyperbolic sine of omega
very very much like polar coordinates in which x and y
completely different system now but just by analogy
I don't use x and t, just x and y
We will write that the point is given by an r but not an omega
but theta, I used omega for hyperbolic angle but it's a kind of angle
It doesn't go from zero to pi, sorry
It doesn't go from zero to two pi,
goes from minus infinity to infinity
And it's a time like variable
which is obviously moving more in a time direction
than in a space direction
And it's some sort of time along the trajectory
But the analog here would just be x equals r cosine theta
Ordinary cosine not hyperbolic, and y equals r Sine of theta,
OK, let's just keep in mind the analogy
And if you work out, first let's write the metric in polar coordinates
The ordinary metric dx squared plus dy squared just becomes
dr squared plus r squared d theta squared
There might be other direction sticking out of the blackboard
we would add them in here, I'll do that over here
likewise the metric over here was same and similar
the metric over here just becomes dr squared, well let's be careful
signs, signs are important
dr squared with a minus sign, this is proper time d tau squared
and we have r squared d omega squared
which is the analog of dr squared d theta squared terms here
Omega is like time, so it comes with a plus sign here
r was like space, so it comes in with a minus sign
This is equal to dt squared minus dx squared
Now if there happened to be and of course there are,
in additional coordinates, the additional coordinates
can be thought of coming out of the blackboard
What would those additional coordinates be?
Let's call them y and z
Then we would also write minus dy squared minus dz squared
We will do the same thing here
So it's mainly r and omega which have a new things
And there are kind of hyperbolic polar coordinates
Lines of constant omega, which means lines of constant time
in the accelerated coordinate system look like this
This is omega equal zero for example we also can have negative omega
Omega equals one, omega equals two, omega equals three,
omega equals four, dot dot dot dot
And omega equals infinity, the end of time
from the point of view of the accelerated reference frame
is just this light cone here which is x equals t
x equals t of course means x equals ct
if I put back the speed of light
and so this is just the motion of the light ray
OK that was the hyperbolic polar coordinates
We also pointed out that anybody who find themselves back here
cannot send a message out to the observer sort of out here
Of course and an observer out here can jump in, can just let go of
whatever the accelerated mechanism is
Let go and fall into here
in which case it can make contact but if it stays outside
which means if they stay on this hyperbolic trajectories
but in particular state not cross this line
then, let's give them names, this is Alice over here
Alice to fall through the horizon, and Bob,
Bob over here who can continue the accelerate
After a point Bob cannot see Alice anymore
That's not quite true
Bob looks back, let's ask what Bob sees, what does see means?
See means to look with light
Bob can see everything backward along his own trajectory
Can see everything backward along light like direction
And it takes infinite amount of Bob's time, omega goes to infinity
before he is in the position to actually see Alice
right at that light cone there
So from Bob's perspective,
he sees Alice asymptotically approaching this line
we must not call it a horizon,
it would be improper to call it a black hole horizon
there is not black hole here
Sometimes it is called an acceleration horizon
Or Rindler horizon
So Bob sees Alice asymptotically approach the Rindler horizon
Relative to Bob,
Alice is moving faster and faster particularly
as Bob was doing the moving from our point of view
But Bob is moving outward with an enormous velocity
That means that he sees Alice moving with a large velocity
What happen when somebody with that large velocity?
You get Lorentz-contracted
So Bob would see Alice getting thinner and thinner
in the vertical direction
Not at horizontal direction
And as she gets sort of squeezed on to the horizon
but Alice doesn't see anything out of the ordinary,
she just go right through
So that was a setup which was intended
Student: so would Bob see Alice starts to slow down?
Well, yes yes Bob would see Alice's time slow down
What would Alice see?
Here she just look back and she sees somebody
scooting along the accelerated trajectory
Keep in mind there is some asymmetry here
when you talk about what Bob sees and what Alice sees
Let's draw the diagram again
There is some asymmetry,
the question is where the asymmetry comes from
Let's see what the asymmetry is
OK so here, here is Bob, here is Alice
And for Bob to see Alice,
he has to look backward along light like direction like that
He looks backward, we all look backward, I mean backward in time
We can never see anything at the same moment as we are looking,
we see things in the past
He looks back and he see Alice at that point,
a little bit later he see Alice at that point,
a little bit later he see Alice at that point and so forth
Alice looks at Bob, she doesn't look in this direction
She looks back this way, and she never see Bob disappeared
In fact she just see Bob accelerating away from her
So this means she does see Bob's motion slow down,
because Bob is moving away from her
But, you know something moves away from you, they seem to slow down
But Alice in this picture never sees Bob disappear or anything else
She just keep watching them
as he recedes further and further into the distance
along the accelerated trajectory ya?
Student: I do have a confusion, I see that
Student: the space is a transformed Minkowski?
Yeah, just a coordinates
Student: Just coordinate transformation?
It's coordinate transformation, now we imagine
Student: I don't see the uniform acceleration
Well we are imagining now a collection of observers
This is not just a coordinate transformation
It is a coordinate transformation
But now we're imagining observers who are moving at a fixed r
So imagining a collection of observers each one of which has a fixed r
That means they are accelerating off
and we are asking what they see
Ya, it's a coordinate transformation but at the same time
we're imagining there are people
located at fixed values of these spatial coordinates r
Student: Fixed r means uniform acceleration?
Fixed r means uniform acceleration, ya?
Student: The d omega squared, that's not quite proper time
No it's not proper time
Sorry, again?
Student: It's the d omega squared over r, d
d omega squared is not, let's say r equals fix r, fix r right?
ds, right here
d tau, along one of these trajectories,
dr is zero, dx and dy are zero
so along one of the trajectories
d tau is just equal to r d omega, and r is fixed
but the rate of proper time relative to the rate of omega
differs from r to r
from one r to another OK?
So if you have exactly the same kind of clock over here
as over here as over here,
"namely clock which tick off proper time	"
they would not be ticking off omega
If you want clocks which ticks off omega
Then those clocks have to have different rates
of reading off proper time
But that's just the questions of coordinates
You can imagine clocks which are built to tick off omega
You have to have them built differently at each point here
But so what would that be
For example the proper time between here and here
is longer than the proper time between here and here
So if we want a clock which built to read off omega
They would have to run at a different rate, each one on different rate
Student: Suppose you have observer who was going along one of
Student: those paths, one of those trajectories, only their, actually
Student: their real x component would be displaced, and then
The real, I guess, what was the word real?
Student: I don't know what how to
The x coordinates?
Student: It doesn't change r
Student: If you put it in different point of space
Student: with the same trajectories
Oh you wanna put another trajectories at this point of space
Student: We chart them on the same shape
Yes, right, Same acceleration
Student: so you have (mumble) that different gravitational potentials
Student: because of the fact that one has the displace r back
Remember the gravitational potential is not an invariant thing
You can add constants to the gravitational potential
doesn't change anything
You can ask what they feel different force at the bottom of their feet
They don't,
they feel exactly the same force at the bottom of their feet
They are being supported by
Student: So they don't see each other
Student: moving in one direction or another?
No, they don't stay the same distance apart from each other
in their own frame of reference
Since they are parallel trajectories let's draw them
They correspond to shifting the origin by drawing the same curve
OK, now, in the original coordinates, this distance is equal to
this distance is equal to this distance is equal to this distance
just by, all we've done is translating the hyperbola
But in their own reckoning, they don't measure distance along here
Well, at this point he does measure, but here
Student: All these observers are the inconstant distance from
Each in their own frame of reference sees the one ahead of him
always being the same distance away
Not true here because of Lorentz contraction right
So you could call this an accelerated coordinate systems
in which you just display each one by the same amount
But it would have the property that
an observer moving along here would see the guy in front of him
receding away and not staying at the same distance
get receding of hmm, what's that?
Student: Would the other observer also see the other? You know
Well you got to figure it out, you got to figure it out
I'm not sure what each one sees, but whatever they see
the distance that this person over here sees him
See perhaps is the wrong word
The distance would be measured by a measuring rod along here
would not be the same as the distance along a measuring rod along here
where this line here was chosen to be space like surface
for a moving observer over here, would not be the same
So I believe he would see him recede
But something we got to figure out
The properties of what each one sees
would depend on time in particular
So it wouldn't be a static description
from the point of view of the accelerated reference frame
OK, so that's good
Now let me write down,
I think I did write down the Schwarzschild Black hole metric last time
did I not?
And the many thing I wanna do today is to show you the connection
between this and the Schwarzschild metric
Once you know that connection
between that picture and the Schwarzschild
then there are many many questions that you can answer
simply by thinking the corresponding question here, OK?
So let's write down the Schwarzschild metric
which looks entirely different
Here it is, d tau squared equals one minus two M over r
I'm setting the speed of light as usual to equal to one
G, thank you, d t squared minus one divided by one minus 2 MG over r
d r squared, plus r squared, not plus, minus r squared
times the metric of a two dimension unit sphere
Why? This is,
what we are using is basically a spherical polar coordinates
Ordinary spherical polar coordinates just wouldn't have this MG here
would be dt squared minus d r squared minus r squared
times I think I called it last time d Omega squared
And d omega squared is d theta squared or was it d phi squared?
d phi squared I guess plus either cosine or sine depending on
where you measure the angle from, sine phi squared d theta squared
where theta and phi, just looking around you and there's polar angle
and a azimuthal angle
That is the metric of a Schwarzschild black hole,
it's all there is to it
and analyzing this taking a part in understanding it
It will teach you about, as much about the black holes as,
Well, it will teach you a lot about black holes
OK incidentally, just to remind you,
this is also the metric of the earth's gravitational field,
but only out beyond the surface of the earth
Another word this is the solution, the general solution of the problem
to a spherically symmetric distribution of mass
outside of the place where that mass is
Another words for r larger than the boundaries of the chunk material
As we take, as we take the object smaller and smaller and smaller
Eventually it tends to be everywheres
But of course there are some problem in this metric
Some singularities,
and there are two kinds of singularity in this metric
One of them is a really nasty singularity
where if you encounter it you will not be very comfortable
The other is the coordinate artifact
Another word it's funny in this coordinates
The horizon is one minus 2 MG over r become equal to zero
2 MG equals r that's the horizon
And why do I say something singular is happening?
Well first of all there is something happening to the d t squared
Notice that if you are setting r equals to 2 MG
Then the proper time and the ordinary Schwarzschild time,
this is called Schwarzschild time
have an infinite ratio of the rate at which they run
Any amount of ordinary time correspond to zero, proper time
So that's a little bit funny
But really, we've already seen it, we see it right here,
d tau squared at r equals to zero
which is right at this point over here,
this is a different r incidentally
I think we are going to run into trouble
if I continue to call this r over here the same r over here
No, I'm gonna call it rho, rho
I'm used to calling it rho so I'm gonna call it rho, d rho squared
It's about to happen that we will get into this conflict of notation
I call it r originally
because I want to compare it with ordinary polar coordinates
where we always use r
On the other hand, that r we will see is not quite the same as this r
Related to it, but a little bit different
And this thing has been called r also from time in memorial
So we have to change notation,
we have to give way to some other notation
Either on that side or that side, I prefer to do it over here
OK, now remember though that r and rho are related to each other
They may or may not be exactly the same thing we'll find out
But notice over here, here we have this omega
which is the time from the point of view of this accelerated observers
Notice that rho equals zero, that any amount of d omega
Correspond to no amount of d tau at rho equal zero
Something like that is going on over here at two M G equals r
Arbitrarily large amount of dt correspond to no proper time at all
So these are funny points,
now there is nothing physically important going on here
It's just that we've chosen coordinates
which have a particular center over here
Such that, that's exactly if we want a polar coordinates,
ordinary polar coordinates,
it would just be the statement that at the center
Any amount of angle at all,
right at the center correspond to no circumference, OK?
As you move out, a little angle correspond to
a larger and larger piece of perimeter, distance along the perimeters
So it is kind of a stagnation point in the coordinates over here
but it's not anything physical going on
So you see there are situations where coordinates can do something
funny where there really is nothing going on,
or at least nothing the observer of that point would notice, OK?
We are gonna try to find out is this
Now, first of all, one thing happens at the two M G equals r,
the coefficient of dt squared goes to zero
That's got to do with the fact the clock slow down
as you get closer and closer to the horizon
We'll come back to it I think
But and that wouldn't be so bad,
but this also does something bad at r equals two M G
At r equals two M G, this goes to one
One minus one goes to zero
And even worse thing than going to zero,
the coefficient of dr squared becomes infinite
Hmm? Question? Alright
So it looks like there's something nasty going on at r equals two M G
well, if not from this term over here,
which is doing something fairly similar to what's going on over here
dr squared term is blowing up in our face
So from one point of view that sounds like something nasty there
we will see there isn't
Next, at r equals to zero, there is also something funny happening
At r equals zero, the coefficient of dt squared blows up
But even worse it blows up with a negative sign
When r is equal to zero, this becomes big,
one minus something big is negative something big
So the coefficient of dt squared blows up with a negative sign
And the coefficient of dr squared goes to zero
Why? No, is that right? Yes I think so
At r equals zero, what's in here becomes infinite
and therefore the coefficient of dr squared get smaller and smaller
but also with the wrong sign
So something is happening at r equals to two M G
And in fact it is interchanging the signs of dt squared and dr squared
In other words let's imagine that r is less than two M G
Then this is negative and this one is positive
This term is negative and this one has been
interchanged of what was space like with what was time like
Now does that mean the space has become time
and time has become space?
No it just means the coordinates are doing something funny
We will come back to what it is
Alright let me change let me take apart of this metric a little more
Instead of working with rho, I want to invent another variable
I don't know how to call it, I've run out of radial variables
I hate to introduce some other coordinates,
OK I'm gonna do I have to I have no choice
I'm going to take rho squared and give it a new name, OK let's see
Give me a name for rho squared, capital R OK
rho squared equals capital R
Then we have of course that this is equal to R d omega squared
That goes to zero when R goes to zero that's not surprising,
so rho squared d omega squared
But how about this d rho squared? What is that?
Let's calculate that,
let's calculate the relationship between d rho and d R
What is dR d capital R?
d capital r is equal to twice rho d rho right?
OK, let's divide that, and notice that d rho is d R over twice rho
"Now let's put in d rho squared, d rho squared is minus d r squared 	"
divided by, in this case, it happens that there is a four there,
the four is not the interesting thing
We can get rid of four by a trick
But the rho squared, rho squared in the denominator right?
Everybody see what I did?
I just wrote d rho is equal to dR divided by 2 rho
And I stuck it into there
What is rho squared? Rho squared is R
Forgetting the four, the four we can actually soak up into a
change of unit of R, I simply actually chose the wrong definition
I should have put a one half in here or something,
would have gotten rid of the four
But there it is, let's look at this metric,
this now has d omega squared term goes to zero just like this one does
at R equals zero
At R equals zero, it has the same property as this
namely something goes to zero
On the other hand look what's happening over here
also an R equals zero, the coefficient of dR squared term
is getting infinite
What was in the numerator here appears in the denominator here
What's in the numerator here appears in the denominator over here
There is nothing pathological in the space time over here
I've just use some pathological coordinates which have the property
that the metric has in terms of the time components
it has something which goes to zero at R equal zero
And in terms of the space component
that has something which goes to infinity
The something, uh, should I slow down here?
OK, everybody with me? Alright
So you see in itself,
there's nothing pathological in space just because
its coefficient here goes to zero and
coefficient here goes to infinity in the same way
As I said, the relative four which appears here is not
an important issue
Now there is something else if we look at this metric here
and we thought of R as our coordinates
Notice what happens when R goes from positive to negative
When R goes from positive to negative this term becomes negative
And this term becomes positive
Whatever R less than zero means if it means anything at all
It has this interchange of space and time just as this does
Just as this does
So the question is can we understand with R less than zero?
R is rho squared, how can rho squared be less than zero?
OK, the point is rho squared less than zero is up in here
Let's see if we can understand why that is?
Each one of these hyperboloids or hyperbolas correspond to,
let's see what it is
x squared minus t squared equal rho squared
Each one of these hyperbolas corresponds to a different value of rho
And precisely the same way for a circle
we would have a x squared plus y squared equal rho squared
Here we have x squared minus t squared equal to rho squared
Which happens to be what I called R
Also known as rho squared
What happen when R got negative,
look at this, what happen when R got negative?
Let's suppose R gets to be goes from plus one to minus one
So x squared minus t squared equals to one
That's a hyperbola like this
What happens when it gets to be minus one?
Then it becomes x squared minus t squared equals minus one
Or t squared minus x squared equals to one
That's another hyperbola, where is that hyperbola?
Ya, that's a hyperbola up here
Let's look at it at this point right over here
At that point over here t is equal to one, x is equal to zero
And t squared minus x squared is equal to one
So it's not that anything crazy has happened over here,
The space has become time and time has become space,
so anything like that
It's just that the coordinate that we are using has a funny property,
let's think about what R equals zero
Let's see what R equals zero means
On this side over here, not R equals to zero sorry
Not R equals zero, ya omega equals zero, omega equals zero
Omega equals zero is this line over here
That's what it is when R is bigger, when R is positive
When R is negative it just become this line over here
So it's a crazy coordinates in which we are calling time
makes a right angle turn here, alright?
Here's R, here's omega equals one
omega equals one looks like that, and when it gets to this point
It makes a turn and goes over here
So the main point here is that coordinates,
if they are chosen in some funny ways
can make it look something like very dramatic as happening at a point
While in fact there is nothing dramatic happening over there
It is just that we are choosing coordinates to do funny things
OK, how do we actually go from, ya?
How do we actually go from this metric is there is some approximation
"in which it is just this metric over here	 "
Can we see some approximation in which these two things are the same
And if we can we will learn something important about the black hole
We'll have would have found in an approximation, certain situations
where the black hole is really approximately flat space time
Ya, not, yes that's right
OK, so let's take this metric and fiddle with it a little bit
We are going to be interested in a very near horizon
We are going to try to analyze it very near the horizon
Near r equals to two M G, let's rewrite this, this is r minus two M G
over r times dt squared minus dr squared divided by
or multiplied by r, divided by r minus two M G
I've done nothing, forget the r square
we'll do something with the r squared d Omega squared in a minute
All I've done was to write one minus two M G over r
as r minus two M G over r, that's nothing at all
And now I'm interested in the vicinity of the horizon,
r very close to two M G
while I can't set r equals to two M G over here, that's too extreme
I do want to keep track of the fact
that we are not right at the horizon
But on the other hand, the r over here
There I can set it to two M G near the horizon
This doesn't, as I move near the horizon
a little closer a little further
R doesn't change very much
Let's suppose R is a kilometer
And I move in and out, capital R, R Schwarzschild
Supposing the Schwarzschild radius two M G, it's a kilometer
And I moved in and out by a millimeter, what happens to r?
Well it changes from a kilometer to a kilometer plus a millimeter
A negligible change in r, alright?
So I might as well over here just write r is equal to two M G
and that is a good approximation, I don't wanna do it in the numerator
Because although r doesn't change very much,
percentagewise r minus two M G does change a lot
when I go from r, from near the horizon
It could actually change sign
So I don't want to muddle with this, but in the denominator
not much happens when I change r a little bit near the horizon
Likewise over here two M G,
that's a good approximation to the metric very near the horizon
Notice it still has the property that it goes to zero
when r goes to two M G
And this one goes to zero in the denominator when r goes to two M G
And I made a small approximation near the horizon
Now let's see what is going on over here
Minus, r squared again what we have is ya,
we moved in a little bit closer a little bit further from the horizon,
r squared doesn't change very much
It changes from a kilometer squared
to a kilometer plus a millimeter squared
It doesn't change much, and so very very close to the horizon
We might as well call it also two M G squared d Omega squared
We will come back to this over here
This is what we are doing, let me tell you what we are doing
Here is the horizon of the black hole r equals two M G
We are interested in the neighborhood of a point near the horizon
I'm interested in what somebody sees when moves in and out
Well if they move in they don't move out again
And if they are moving a little bit near the vicinity of the horizon
And they simply want to blow this up
They wanna blow it up, look at them under a microscope
and see what the horizon looks like
So apart from the pieces here
where we have to keep the difference between r and two M G
We simply set r equals to two M G
Now what was going on over here, what was d Omega squared
if I look only in the immediate vicinity of this point
if I look only in the immediate vicinity of this point is that
it's not like looking at the immediate vicinity of a point on a sphere
a point on the sphere, the immediate vicinity
of a point on a big sphere just look like a flat plane right?
This, whatever it is, is a sphere with radius two M G,
that's what it is
it's a sphere of radius two M G as I moved around on here
But a sphere of radius two M G
it doesn't matter how big this sphere is
as long as it's big enough when I move around on it
All I see is a flat plane,
so I can actually replace this by a flat plane
It is just the tangent plane to the sphere
Just like we replace the tangent plane to the earth by flat space,
the flat space approximation
It's not good for the entire horizon
but it is good for the neighborhood of the horizon
So actually we can replace this by dy squared plus dz squared
The two coordinates, this coordinate here, we call, whatever, r
The other two coordinates I'm calling y and z
And this can just be called dy squared plus dz squared why?
Because in the immediate vicinity of a point
that could be approximated by a plane
If we approximate it by a plane, it's just dy squared plus dz squared
So now we are getting there, we have the metric right at that point
at the immediate vicinity of that point,
here it is, r minus two M G over two M G
dr squared two M G over r minus two M G, and the other piece of it
"Very simple, almost trivial, just dx squared plus dy squared 	"
Oh sorry, dz squared plus dy squared
Now I just want to empathize
this is not the exact metric of the black hole
It is the metric of a piece of space very close to a particular point
What we are trying to find out is
what is actually going on in the small neighborhood of that horizon?
Is there something nasty and vicious because r is going to two M G
Or is this more like this over here,
where is in fact nothing vicious and nasty going on
It was just the flat region, flat space, flat space time
OK, so this has vague similarities with this, something in numerator
the same thing in the denominator, a factor of four
which I wasn't very careful to, I could have define things
I could have define things in a way that get rid of it
I didn't so there it is
OK, so let's change coordinates again,
now this whole game is a game of changing coordinates
Until you find the coordinates where you recognize what was going on,
where you recognize with some simplicity what's going on
I don't wanna change to this capital R coordinates
This capital R coordinates was just a contrivance to show you that
there are situation where a numerator of one thing
will become the denominator of the other,
when this goes to zero this one goes to infinity
That was just the contrivance
to make it looks as much as possible like this
But I'm really more interested in the original polar coordinates here
Here is the original polar coordinates, and my claim is,
that by a smart change of variables I can make this metric over here
these two pieces of it look exactly the same as this,
close to the horizon
How could we do that?
Well the trick is to realize what rho is?
rho is the proper distance from this point over here
Spatial distance, it's the proper distance,
how do I know it is the proper distance?
d tau for move out along here,
d omega is equal to zero if I moved horizontally
and d tau is just equal to minus d tau squared is minus d rho squared
That's the definition of proper distance along here, OK?
Let's see if we can find the proper distance
from the horizon to an arbitrary point r
Here is an arbitrary point r
and then use that to replace the coordinate r
The proper distance from the r, from the horizon for that point
I want to use that as a new variable, I will call it rho again
How do we find the proper distance from this point over here
which is r equals two M G to an arbitrary point r?
Well, anybody have a suggestion?
Student: Plug in the zero for the other coordinates?
Yeah, we are plugging in zero for the other coordinates
But then we have to do a bit of work
Yeah, we are gonna move out straight out without changing time
Without changing x without changing z and y
and see how much distance between here and here
The proper distance is determined by this term in the metric
If we move out without changing time and without changing dz and dy
Then the proper distance we moved out, let's call it ds
ds squared is the same as d tau squared except for the minus sign
The proper distance ds squared is just equal to
dr squared times two M G over r minus two M G
OK, we take square root of that
Now how do we find the distance from this point to that point?
We have to do an integral, the distance is the integral of this, OK?
This is the differential distance when you change r by a little bit
If we actually want a distance from r equals two M G
to an arbitrary point we have to integrate
So let's call that integral rho,
it's the proper distance to an arbitrary point
And it's the integral d rho, square root of two M G
that's a constant that comes out
Divided by r minus two M G square root, here we are
This is the proper distance, where do I integrate from?
We integrate, sorry, this is dr
We integrate from r, from r equals two M G to r
I'm gonna stop for question at this point
because this is quite critical to understand what is going on
Ah that's just the point r
OK, let's call this r prime the integration variable, right
And we integrate from r equals two M G to the moving point
or to the arbitrary point r
r outside the horizon to be specific
OK, can anybody do,
let's first of all take the two M G outside the integral
That's easy, that part is easy
Yes, there's a two M G of course
So we take out the two M G in the numerator here,
and we have an integral to do
It's a square root integral, a simple square root integral
One over square root of something
And what is the integral of one over square root?
Who can do one over square root?
x to the minus one half integrated, what is that?
It's x to the three half, oh sorry, x to the plus one half,
726
00:52:57,910 --> 00:52:56,240
and then a two?
Ya two? Right?
The integral of the x to minus one half is twice x to the one half
How do I prove that? I differentiate this,
I get x to the minus one half, one half times two is one, good
So here we are rho,
the distance between the horizon and a point r is given by
Did I lose? I think I did, you are absolutely right
Right, we have to take the square root of this whole thing
Good, thank you, it's square root of two M G that was the numerator
And the square root of r prime minus two M G in the denominator
The square root came from taking the square root of ds squared
dr square root square root
OK so what do we have? We have then that
rho is equal to twice, that's the 2 from here
Then there is a square root of two M G, that's from here
And then the square root in the numerator
square root of r minus two M G
While I plug in the in point into the integral
That's rho, again, what is it?
It's the proper distance from the horizon
to an arbitrary point at the distance r
It's also a change of variables, it's also a change of variables
Where we've, where I am now imagine the change of variables
Which I substitute for the variable r
The Schwarzschild coordinate r, I substitute the coordinate rho
Notice this is a square root, let's see that we have, if you remember
R was equal to rho squared, again there was a variable over here
which is related to a variable over here by a square root
rho is equal to square root of r
Here we have rho related to the square root of r minus two M G
So they do look similar
OK let's rewrite the metric
Let's rewrite the metric, it's a bit of nuisance, it's not very hard
I am taking you through all the steps
I could just write down the answer
but I want to take you through the steps
Here we are, no coordinates,
coordinate transformation to proper distance
which is just equal to square root of eight M G,
I got the 8 by bringing the 2 inside
Times the square root of r minus two M G
That's the proper distance from the horizon
And all we have to do now to rewrite the metric is,
well we have to figure out what it is
First of all, rho over eight M G is r minus two M G,
square root of r minus two M G
So let's square it
Sorry, square root right?
That's rho divided by square root of eight M G
is equal to the square root of what's on this side
Let's square it!
rho squared divided by eight M G is equal to r minus two M G
Well, that's convenient
because we have r minus two M G all over the place in this metric here
so it's convenient that we know what r minus two M G is
Let's now rewrite the metric
First of all it's this term,
I realize that this is a tedious thing to sit through
You don't need to sit through it, you can do it yourself
But let's do it together
And here we are, r minus two M G over two M G
Well, here it is, r minus two M G is rho squared over eight M G
Then I believe there is another two M G downstairs
that makes that 16 M squared G squared right?
Times what?
Times dt squared right?
Now what about this monstrosity here?
There is a quick answer to this so really quick answer to it
What is it in terms of rho? Anybody?
Ya but then we have to also convert d rho d r to d rho
What was the definition of rho?
No no the definition was the proper distance from the horizon
If rho is the proper distance from the horizon
Then this is just d rho squared,
it is the proper distance from the horizon
The way we got rho was basically writing that
d rho squared is dr squared over two M G
It's just equal to this thing here,
so we don't have any work to do to figure out
the rho part of the metric
It is just d rho squared
It is the proper distance minus, minus, minus
Minus good old dz squared and the dy squared
which go along for the ride
We are almost there, what do we comparing this with?
Let's go back, what is my Rindler metric? My polar coordinate metric
My polar coordinate metric was
rho squared d omega squared minus d rho squared
And if there were other coordinates coming out of the blackboard,
if there were other coordinates coming out of the blackboard,
then we will put them in by hand, minus dz squared minus dy squared
Is this the same as this? Not quite, what's the difference?
16MG squared, OK, but now there is a simple thing we can do
We can just make a change, another change of variables namely,
let d omega, I will just let omega equal t over 4MG
omega squared is,
well let's just do d omega, d omega is equal to one over four M G dt
and guess what?
d omega squared is just dt squared over 16 M squared G squared
So I just look at this and say oh my goodness this is nothing but
dt over 4MG squared,
that's all I have to do is change the variable from t to omega
and this becomes rho squared d omega squared blah, blah, blah
the same thing as this here
OK, so let's just go through what I did again
without actually doing it
I changed,
well the first thing I did was I approximate it near the horizon
It's called the near horizon approximation
And it's not just near the horizon,
it's near a specific point on the horizon
We substituted for one minus two M G over r, r minus two M G over r
and set r approximately equal to two M G
So that just gave us a simple expression over here
Same thing upside down over here,
and then this metric on this sphere
was approximated by the tangent space
which just mean it's the same operation
as the one I do on the surface of the earth
I used x and y, for small patch, or a farmer's field
you doesn't need to keep track of the curvature of the earth,
you just need x and y or north and east whatever
And use its flat space metric
OK, that was the first step, simplified this way
Next step was to compute the proper distance that correspond
Another word it really simply stated
it was just setting dr squared of two M G just equal to d rho squared
and figure out what rho is, let's write it that way
Let's just think of it that way
dr square root of two M G over r minus two M G
equals d rho, definition of d rho
I took the square root of this,
square root of this square root of this
And then I integrate it to find out that rho was nothing but,
where is it?
Here it is, that gave me a new coordinate rho which simplify this term
The main point was simplify this term, why do I simplify it, well
dr times the piece of stuff over here is just equal to d rho
so this just becomes d rho squared
This on the other hand was also not too complicated
It just turn out to be rho squared divided by 16 M squared G squared
Why what?
Who said d rho equals dr?
I didn't say that,
d rho squared is equal to dr squared time two M G over r minus two M G
Say that again?
Student: the second term on the right hand (mumble)
Student: Ya ya, that's replace dr squared with that?
No, no I didn't, I replace dr squared divided by this, OK?
Look, here we are
Let's go back one more step, do it once more
here we have this dr squared over 1 minus two M G
Well, two M G over r minus two M G,
OK? Who ever has a question, are you happy with that?
I didn't do anything, I just rewrote it
I'm setting that equal to d rho squaredd
You can't ask me why, afterwards why d rho squared is equal to this
It just is, OK?
Definition of d rho squared, I take the square root of both sides
And I integrate to find out rho is equal to this
Now there is no question that this thing over here
is equal to just d rho squared
It was the way it is defined
It is a trick to redefine the coordinate
to simplify these things over here
That's all
You can always do it, you can always make the change of a variable
to change from dr to d rho,
where by definition, dr times the square root here is equal to d rho
OK so by the time we finish, we get back to here
Well that's pretty interesting, it tells us, what does it tell us?
It tell us that the geometry,
the near horizon geometry very near this point
is really no different than the geometry of flat space time
in polar coordinates
You can't tell the difference
very very close to the horizon of a black hole
You can't tell the difference between flat space and a black hole
number one
Number two, this rho variable
which was proper distance from the horizon
and which enters into the metric in this form over here
is obviously close to the horizon
The same variable as the accelerated coordinate system uses
to distinguish one observer from the other
Just the ra, I don't know what to call it
The variable from one coordinate to the other
So the horizon of the black hole, the horizon of the black hole
and different observer station at different distances
are just like the observers station
at different distances in the accelerated reference frame
Why accelerated? Because gravity and acceleration are the same thing
If I station somebody over here near the horizon of a black hole
By station I mean hold them there,
hold them there by a rope around the neck
They are gonna be feel like they are being pull,
they are gonna feel like being accelerated in that direction
The further out we are,
the less the acceleration, the less stretching of the neck
Same thing here,
to keep somebody at one of these accelerated coordinate systems
You might imagine pulling them along by a rope around their neck
The closer they are to the horizon here,
the harder you have to pull them
So it's exactly the same geometry, the near horizon geometry
is the accelerated coordinate system geometry
Of course it is good old flat space time
But the coordinate in this region here are
Not ten after ninth, no I keep, right, I keep
time direction or black hole or something, it's going too fast
OK, so that, what does that tell us?
It tells us that, as I said repeated over and over again
The horizon of a black hole is not a singular place
Now in particular again,
the fact that relative sign change between here and here
That makes it look like space become time and time becomes space
There's nothing really happening, to see what it correspond to
To see what it correspond to
we have to go back to that capital R coordinate
I will just remind you the capital R coordinate
which was R equals rho squared?
Is that what it was? Ya
The metric in terms of that was R dt squared minus
there was a silly one over four
that doesn't mean anything, R downstairs, d R squared
So that's very much like what is going on here
Student: in this side of the blackboard,
Student: the accelerated objects closer and closer to the horizon
Student: get to there at infinity?
Student: Could I find a time on this side of the (mumble) black hole?
What I can do is point it to this diagram
From somebody looking from outside,
this line here is omega equals infinity right?
omega equals one two three four, omega equals infinity out there
From the reckoning of the accelerated coordinate system,
first of all, nothing passes the horizon until I make it equal to infinity
And furthermore somebody watching, they can wait for ever and ever,
they go way out here
It's a long time in the future according to this fellow here,
he looks back and still doesn't see anything having crossed the horizon
Here it is, here's the horizon
He looks way back, and you're gonna to have a very long time,
you may be close to that horizon, not crossed it,
he looks back and he sees poor Alice having not yet crossed the horizon
Now, exactly the same thing happens here, exactly the same thing
And this is the picture to come to when you
Alright now there is a difference, there is a difference,
the difference is far from the horizon, far from the horizon
the metric is quite different
Close to the horizon it looks like globe flat space,
but flat space in a strange accelerated coordinates,
and strange, I mean they have a singularity over here,
time is this hyperbolic angle
But far away when r is very large, this is small,
two M G over r is negligible,
and it just again looks like flat space
But flat space in which time is time and space r,
when which ordinary coordinates r and t
So there is an transition from very close to the horizon
where looks like flat space and polar coordinates,
to very far away where just looks like flat space and ordinary coordinates
Far away you're not be tempted to call t a hyperbolic angle,
but close in to the horizon, t or at least t over four M G, where is it
T over four M G, omega was T over four M G
Yes, T over four M G close in becomes a hyperbolic angle
So there's a characteristic difference in these coordinates
close to the horizon and far from the horizon
I can give you a
Well, I won't give you another analogy for it, it will just get
Student: Any observer has something enforced towards the event horizon
Student: how far (mumble), cause continuous, now Alice
Student: actually get the event horizon
Nobody who stays outside a black hole
Student: It did, (mumble) did
That's right
Student: So unless you're at the event horizon, inside the event horizon,
Student: and Alice
Student: into the black hole, how could the black hole ever be larger
Student: because nothing it gets into it
It gets larger by passing its energy very very close to the horizon
and causing the horizon to grow a little bit
It's questionable whether you should distinguish
the stuff which is very close to the horizon from the horizon itself
The horizon grows, not because anything goes into the black hole
It just merges with the horizon
Student: What's that distance, it becomes bigger?
Yeah, it's Planck distance
Student: Is this relate
No no, at some point as you watch,
imagine what happens as you watch something fall through the horizon
That something is sending out signals, right?
Sending out radio signals, or sending out,
No, let's say it's sending out light
Let's see if we can see what happens or observe from the outside
We can do it by drawing so we can do it by equations
I prefer doing it by drawings
What's it?
Student: Why don't you take radio signals?
Oh because radio signals, I want to start with something short wavelength
and get them longer and longer wavelength
Radio, there's nothing longer than,
I don't know any name for a thing longer than a radio signal
So if I want to start with something short wavelength, I'll start with,
what would I say, optical light?
Optical light, what come after optical light, infrared
After infrared there's microwave, after microwave and so forth
I want to have a range that I could talk about
OK, so, let's first, yeah, let's um
So here's Alice, Alice is going in,
and she's sending out in her own reference frame,
she's got a little oscillating dipole which is sending out light,
a little atomic dipole which is oscillating,
here are the oscillations, one oscillation per unit of proper time,
in some units, which is appropriate to the emition of light
So away crest from here and trough from here
and crest from here and trough from here and so forth
That's Alice
Here's Bob, and Alice is sending out
I wish I had another color, I don't have another color
Let me see if I do have another color
And I'm not gonna leave those psychologist my pens
OK so let's see what Bob sees
Bob sees a wave, the first wave hit him over here,
the second wave hits him over here,
the third wave hits him over here,
the twenty-fifth wave hits him over here
How much proper time is there between the waves as they hit Bob?
That gets drawn out, longer and longer
One way to see it is just to see that Bob is just receding
and therefore Doppler shift
From this point of view there's no real difference
between Doppler shift and gravitational red shift
As Alice gets closer and closer to the horizon here,
she sends out waves and the last wave that she sends out over here
gets to bob at a very very late stretched out time
If the waves actually have finite frequency,
then there's gonna be a basically a last wave that she sends
This means that Bob sees Alice more and more red shifted
She sending out light signals,
but Bob sees after short amount of time or certain amount of time,
he's receding away, and he sees infrared, then he sees microwave,
then he sees radio waves with a meter,
then he sees radio waves with a kilometer,
then he sees radio waves with million kilometers
And radio waves with million kilometers don't have very much energy
It sending out, you know, one photon,
some number of photons per unit time Alice is sending out,
Bob is receiving those photons slower and slower
at longer and longer wavelength,
so the energy that he is receiving is getting lower and lower
What he sees, is Alice's dipole is slowing down,
that everything about Alice because he sees through light,
he sees slowing down
And you don't need black holes to see it,
you just see it from this picture here
That Bob sees Alice slowing down and slowing down and slowing down,
and finally she sends out her last wave, OK?
Why it's the last wave, while she had higher frequencies,
she could hit some more in there
And yet higher frequencies she could fit even more in there
But any finite frequency oscillator that send out the radiation,
there is gonna be a last wave that she sends out
and then Bob doesn't see it anymore
She's gone, as far as Bob is concerned
What does happen is the remnant of Alice is merged with the horizon
and in fact makes the black hole a little bigger
Now to see that we'll have to solve Einstein equations
within falling material and so forth, that's a little bit hard,
uh but sees, ya?
Student: So what the ways Einstein put the size of the black hole,
Student: did he put some (mumble)
That's, no, that's right,
as long as we should put out as about the size of the black hole
Student: Does this Alice see Bob slowing down?
OK so this Alice see Bob, what Alice sees is Bob receding away,
and moving closer and closer to the speed of light
Now when something goes pass you with the speed of light,
yes it is slowed down, it's slowed down by Lorentz contraction
So yes Alice watching Bob, well, yes, OK
Alice watching Bob, sees Bob recede away and therefore, yes, slow down
There's a problem for Alice though
This is the near horizon geometry,
and we haven't talked about what happens at r equals zero
Something really bad happens at r equals zero,
that is not a coordinates singularity
If we were to calculate the curvature,
in particular let's say the curvature scaler
I don't know not the curvature scaler,
some measure of the curvature of this geometry
We'll find there's nothing radical happening at the horizon,
but at r equals zero
Remember r equals zero was the place where, let's go back to Newton
What happens at r equals zero?
Gravity becomes infinity
Tidal forces become infinity, alright
If we calculated the tidal forces,
associated with this thing here,
we'll find it infinity at r equals zero
OK? So r equals zero is a real genuine singularity,
a place where curvature becomes infinite,
it's a place where tidal forces become infinite
It is not a place which is anything like flat space
But the question is where is it on this diagram?
Let's redraw the diagram and try to figure out where
Student: You said it's on the (mumble) mass inside the
What's it? It's telling you what?
Student: The answer is inside the (mumble)
Ya, but done if you have a completely gravitational collapse
and things collapse to a point, ya
It's collapsed to a point
Student: This mathematics does not (mumble) black hole
No, don't add Alice
Student: But Alice is here
Right, right, right
Ya, ya, Alice, you can let her behind the black hole
Student: I have a question, Alice has some friends and a conversation
Student: is there restriction
Student: on those conversations?
Well Bob, Bob, don't falling freely?
Student: Alice and her friends are all falling freely
Yeah they just told to each other just as if nothing was happening
until they get into the singularity
Ya?
Student: If her friends (mumble)
Oh yes they can
What's it? say it again?
Student: If one is a little heavier, then he can,
Student: when one gets to the event horizon the other still behind
OK, so that's damn like that
Here they're they both falling, OK?
They communicating back and forth freely,
nothing funny happening
Now it is true that once Alice
Here's Alice, and here's Shirley
Alice is at head of Shirley, she's falling in first,
here she is
Just she gets near the horizon she sends a signal out to Shirley,
Shirley gets the signal
Once she passes the horizon,
she cannot get a signal to the Shirley who is before the horizon
There's no way she can get a signal from here to here
But she can get a signal from here to here
Even if there was no black holes,
she couldn't get a signal from here to here,
that would exceed the speed of light
"So Alice sends Shirley signals,	Shirley sends Alice signals,"
nothing funny about them
Now if Shirley would have decided
not to go into the black hole at last minute
Here she is, and then she decides not to go into the black hole,
not to cross the horizon,
but to get onto one of these accelerated trajectories
May be she was connected to the space station by a cable,
and the cable won't allow her to go down so far,
and then got pull her she supported away from the horizon,
Then Alice can see Shirley,
but Shirley cannot see Alice behind the horizon,
there's no way she can see behind the horizon
So if Shirley freely falls,
her relationship to Alice is not unusual,
back and forth they can communicate
If Shirley at the last minute decides not to fall through the horizon,
then indeed Alice falls out of her experience, ya?
Student: So ionization doesn't occur at the horizon?
What kind of thing?
Student: Ions on the ends of the horizon?
Now you're talking, now we get into quantum mechanics,
which we're not gonna do this quarter
Do atoms ionize at the, that depends who's watching them
Student: If (mumble) is moving along these dy or dz,
Student: means that moving by on that surface,
Student: will it continue moving there (mumble)
No it will fall into the horizon and
Student: But the space is flat there, it should keep in flying
Yeah the problem is these trajectories that move along here
are going to be space like
Then why is that, supposing,
supposing we have a trajectory very close to the horizon, OK?
That's moving with a dx, sorry, a dy or a dz, OK?
So let's see, so that means dy or dz is not equal to zero, right?
But when we extremely close to the horizon,
let's say practically at the horizon, then this term is zero,
and this term is zero
Oh sorry, well this term is zero because dr is equal to zero
We're not moving in and out, we're moving along the horizon, right?
OK? So this one not there because dr is equal to zero along that motion,
and this one is zero because the coefficient of dt is equal to zero
Student: Second is zero over zero
Well we're slightly away from the horizon, slightly away the horizon,
just outside, OK?
But the main thing is dr is exactly equal to zero
We're skirting the horizon at a fixed r,
that was the question, if you skirt the horizon at a fixed r,
that means this is exactly zero, alright?
And r minus two M G is just a small number
On the other hand this is not zero but it's very very small,
in particular the coefficient here is so small that if d,
if we get close enough, we get close enough,
this would eventually get smaller than dz
So if we make a little excursion along dz here by a given distance,
no matter how long we take to do it, if closed enough to the horizon,
this term is zero
That means the trajectory here is space like,
and it means it's exceeding the speed of light
So anything that tries to skirt along the horizon
is exceeding the speed of light
And that's because time has slowed down so much
or the coefficient of dt squared has slowed down so much,
that the trajectory becomes space like
Do you understand the difference between,
do you understand what a space like trajectory is, right
So skirting along the horizon is forbidden motion,
it's motion faster then the speed of light
Supposing, alright, let's talk about a thing
Well I think that will do it, right
So you can't skirt along a horizon like that
Student: (mumble) into the singularity
Let's get to the singularity now
The singularity is that r equals zero here
Now let's remember the connection
r equals zero or r equals small, let's take small values of r
For small values of r,
that's where the interchange between positive and negative is taken place
For small values of r, we're in here
Small values of r don't correspond, fixed values of r,
don't correspond to time like trajectories, trajectories of observers,
but they correspond to space like surfaces like this
Just to see what's going on, let's go back to this metric over here
Supposing R if R is positive, then we're on the outside
If R is negative, we're on the inside
Alright here's the inside, here's the outside
this case R being negative,
that's different than little r being negative here,
it's like r minus two M G being negative
Big R is like r minus two M G, alright
When big R changes sign,
the hyperboloids x squared minus t squared equals R
become hyperboloids like this
So fixed R outside corresponds to the motion of an accelerated observer
Inside it corresponds to a kind of space like surface,
another way to say it is fixed R inside the black hole,
is some time, not some place
It's a time rather than place, fixed R, OK?
Now what about r equals zero?
Well r equals zero is sort like capital R equal negative
As I said, r minus two M G is sort like capital R
What does R equal something negative correspond to?
It corresponds to a surface like this
So little r equal zero is a surface like this,
this is little r equal zero like that
That's the singularity
And it's not a place so much, it is a time
Once you fall through here, notice once you fall through here,
you have no way of avoiding the singularity
The speed of light is forty-five degree angles,
you cannot move faster than the speed of light
When you out here, and you send a light ray in the outward direction,
it might escape
if you send it the inward direction, it might hit the singularity,
it will hit the singularity
If you're inside,
it doesn't matter which direction you send a light ray in
you got it in singularity, ya?
Student: If you free fall
Student: then there's no reason you use this thorny coordinates
Well free falling just means you moves in straight lines
You don't have to use any coordinates, here's the geometry
Student: I think the whole idea that the time and x z coordinate switch
Student: are because this thorny coordinates
The coordinates are funny, but the geometry, this is the geometry
This picture doesn't require
Let's take the r equal zero away
That's it, that's the geometry
That's the geometry, that's what it looks like
I don't need to put any coordinates on it,
that's what it looks like
And once you in here, you can't escape falling into the singularity here
Student: (mumble) time and space switching
Well I keep telling you, time and space don't interchange,
it's just you draw on coordinate axes which make right angle turning
Student: So basically the opinion is, other,
Student: except time little facts, the free falling observer doesn't
Nothing Student: Yeah
Student: So at certain proper time does it
At a certain proper time you've cross the horizon,
but crossing the horizon is not itself a dangerous event
You doomed but you not died
Student: Take (mumble) proper time, then you will reach
Student: the signal might not important, (mumble) some angle
If you're just falling then you pass the horizon you have some
particular amount of time left over, ya, ya, ya, that's right
The maximum amount of time that you have
is basically approximately equal to the time that it takes
to cross the horizon of the black hole,
it's just a numerical fact
Uh if you measure distances and time in the same units, c equals one,
then the Schwarzschild radius becomes a time
Schwarzschild radius of a star would be about a meter
Kilometer, excuse me
Light takes blah blah blah seconds to cross kilometer,
one over three times ten to the fifth
It doesn't matter, order of magnitude, order of magnitude,
order of magnitude it takes about one trans time across the black hole
and into the singularity
Student: Once you're there
I don't know what to say, once you in there
Student: It will see inward
No, no
Everything become infinitely dense, infinitely distorted, infinitely um
Quantum mechanics becomes important,
everything is torn apart into its constituents and worse,
and we're into a real Terra Incognita
But really you're never gonna get far enough to
experience that singularity and so, ya?
Student: Does the singularity actually,
Student: does the infinity actually exist in real life?
Student: Is there something by before you get there
Student: sort of, maybe it's the factor, quantum mechanics or something
Well nobody really knows, nobody really knows
My own feeling is no it's ultimate endeavor things
but let's put this way
Let's just say nobody really knows, but we know how to follow it,
for example into a region where the curvature might be,
the radius of curvature might be no bigger than the radius of a proton,
easy to follow it there, nothing,
we know very very well what physics is like down to those distances
So you would not like to be squeezed into a radius
equal to the size of a proton
Actually ya, tidal forces would be effective magnitude, they would
Student: So is it the thing that (mumble) will grow large as
With the horizon, one sense is to impact the horizon
There is something dreadful that happens at the singularity
But the singularity as I said is not so much a place
A singularity, if a singularity was a place spot over here,
and you flowing in, while you can go around it
If the singularity is the end of time, as I've drawn here,
there's no way to avoid it, uh, so
There's something I should tell you about real, about genuine black holes
You would not survive
the trans across the horizon of a stellar mass black hole
The tidal forces are just too large
If you're to estimate, this is a Newtonian calculation,
if you're to take a stellar mass and compress it down to the size of a
of a, compress it to the size of a black hole,
which would be about a kilometer
Well a kilometer doesn't seem very small,
it seems like you could pass through it safely
But the tidal forces, just the gradient of the gravitational field would,
you know, squeeze you out like a toothpaste through a outlet
So that would be,
it would be very dramatic going through the horizon of a
On the other hand if you're as small as a paramecium,
then the gravitational gradients across your body would be negligible,
and you would fall right through the horizon without any trouble
Uh, similarly, if we took a black hole which was,
I don't know, a billion solar masses
A billion solar masses, the horizon is so big and so flat
that the tidal forces are not very strong on your body
at a billion solar masses
It's a little bit comp, how should I say
Make the account intuitive,
you might think the bigger the mass of the black hole,
the worse it is to go through the horizon of the black hole,
but that's not true
The point is, that the bigger the mass of the black hole,
the bigger the horizon, and the less curved and the less tidal forces,
and the less you would be squeezed going through the horizon
So if a black hole were a billion solar masses,
you would go through the horizon pretty easily
I think I once estimated the, I can't remember whether, let's see,
billion solar masses so uh
It's a billion kilometers,
how long does it take to go a billion kilometers?
About an hour, right?
Is that right? A billion kilometers is like the radius of the solar system
Ya, it's about the radius of the solar system which is about a light hour?
So alright, so the biggest black hole is founded
in the centers of the biggest galaxies,
You could probably last for half an hour or something like that
after falling through the horizon
Not longer than that, that's how much time we take a light to go across it
Take a black hole of a trillion solar masses
There are no such things, probably there're no such things
You would last another thousand time longer
So going through the horizon of a black hole,
for a stellar mass black hole is not as innocent as I may have sound
But the principle is there, a sufficiently small living creature
would survive going through the horizon of a solar mass black hole
Would last very long, hit the singularity in a fraction of a second,
but still, it would be at the singularity not at the horizon
OK, um, I don't know, is there anything left to, ya?
Student: I have a question about the coordinates
Ya
Student: The coordinates we've been using,
Student: coordinates onto a series of rods and clocks
Yes
Student: How to solve a thing
That are what?
Student: That are takes location outside a black hole, not moving,
Student: So it looks they see accelerated coordinates as series of
Student: rods and clocks and which're undergoing acceleration
Relative to what? Relative to what?
OK, I know
Accelerated relative to what?
Ya, relative to Alice, OK
Now for Alice fall through the horizon of the black hole,
believe me, Bob is accelerated relative
Student: I have a point about the understanding of the coordinates
Student: near the black hole, we're assuming that the
Student: we have a bunch of points scatter around and small
Student: cube and clock there, and they're running,
Student: telling, a point that inside the black,
Student: you can't actually do that on the surface of a black hole and so
You can't stand there
Student: Yeah, but you could extend that coordinates across the
Student: boundary of the black hole
Student: And you still get a series of points and clocks,
Student: and the clocks are still measuring time, aren't they?
There's nothing to prevent you in this picture over here
from using a set of coordinates, and related to real clocks and rods
which look like that
Alright?
This is um, Alice is a freely falling reference frame
OK it's just a reference frame
which freely fall through the horizon of a black hole
Student: OK, what we're talking about, is the t-s and r-s
Student: is what we studied in the freely fall through black hole
That's right, that's the one looks like this
Student: And it looks like that at outside the black hole
Alright, and inside the black hole,
we just extinct a coordinate which look like this
Student: OK
Right
Student: So we can imagine series of rods and frames
Student: which actually extend the coordinate
You can imagine
But they have to be falling
You can not imagine anything standing still behind the horizon, why not
Student: What the variables are plug in
Anything that's standing still, that means a fixed R, fixed little r
Student: You can't move along one of those coordinates (noise)
Student: like these coordinate because those coordinates are fixed ones
That's right
If you try to make something at fixed r here,
it will give exceeding speed of light
Something at a fixed r behind a horizon is exceeding speed of light
Student: Isn't that the r in the first equation?
Student: exceeding the speed of light
Yes, wait, behind the horizon it's exceeding the (speed of light), yeah
So no there can't be rods which are go through here with these rods
just standing still statically
There they're going
Student: So basically, notional coordinates system you can't imagine
Student: actually constructing it
No, you can imagine constructing it, but in here,
the things that would measure r would be clocks,
and the thing which were measure, let's see, this is,
this was time, the time coordinate becomes coordinate like that
So behind the horizon, this coor, but the only coordinates, keep in mind,
they're rather arbitrary
We've just drawn some coordinates, behind the horizon,
if you want to measure r, you'll use clocks,
and if you want to measure t, you'll use meter sticks, alright?
t flows along here, the same coordinate,
the same mathematical coordinate flows along here
You measure things in here with meter sticks
Student: Take an Einstein's clock and (voice too low)
Well, keep in mind this is not a problem of general relativity,
this is a problem of special relativity,
this is just ordinary flat space time, no gravitating objects,
except as seen by an observer who is
been pulled along at a constant acceleration
The one thing which is not there in that case would be that singularity
The singularity is really the gravitating object, right, that,
that's the real signal that there's tidal forces for example
Student: Can we go back to the way back to where the metric constitute?
Student: There's a solution to Einstein's equation,
Student: start from some distribution of energy
Any, that's right
Student: what this distribution start from?
Any, any as long as it's enclosed within here, right
So we solve Einstein's equations,
we make some assumptions, I'll tell you what assumptions
We first start with Einstein's equations
R mu nu minus one half g mu nu equals T mu nu
Now, or with some four phi-s and G-s and it doesn't matter
Now we're studying a region outside of place where there isn't any
energy distribution
And that means that we're studying solutions to this equation,
knowing full well that inside here there may be some material,
but outside here, sorry, g mu nu R equals zero
Those are the Einstein's equations in the absence of any matter
It's exactly the same with Newton
Newton you'll write Del squared of a potential
equals the energy distribution, matter distribution, mass distribution,
but outside the mass distribution you'll write zero
You know that inside something is happening,
but outside, there's nothing happening
So you asked what are the general solutions of this equation,
under certain conditions?
Alright I'll tell you what are the conditions are
What are the general solutions of this equation
which are radially symmetric
Another words which are only a function of r
Let's start with this symmetric
What is the general solution of the equation Del squared Phi equals zero?
The Phi is only the function of r
and not a function of theta phi and time, for example
Well there' general, what is it?
Ah, one other restriction, that's a solution
One other restriction, Phi equals zero far away
Phi equals zero far away
You just told me in fact what the general solution is
x squared plus, well, quadratic formula
No, no no no, linear formula,
quadratic formula has constant on the right hand side, linear formula
But assuming also it goes zero far away
Well the answer is there're none
There are no solution of this equation where phi goes to zero far away
But now supposing you say what are the solutions of this equation,
outside a certain sphere
I'm not caring whether you solve the equation on the inside,
not worrying about whether you solve the equation on the inside,
but only caring about whether you solve the equations on the outside
then there're solutions
Gener, only, one form solution
the one form solution is that Phi is a constant over r,
the only mathematical solution
Now what you carrying this constant in the real world?
The constant is the mass contained within that sphere
So it's M G, a minus sign
The physics tells you it's a minus sign, and the M G is Newton
Minus M G is Newton
That's the most general solution of this equation of Del squared Phi zero
if you don't care if the equation is solved on the inside, OK?
Now supposing we take a sphere
Student: At first I understood the answer, but don't understand the mass
Student: Was, what happened to the mass inside
Well that's, you can follow it,
we can take, we can ask what happens when we take a very small sphere
Well the answer was exactly the same
set the mass inside that small sphere some place
What happens if you (mumble) a point particle?
You (mumble) a point particle
and the equation satisfied everywheres except
at the point where the point particle is
Where the point particle is, there's a singularity as r equals zero
OK so what, and if you like, you can make mass as falls in
and all accumulating a point
A point of space
Where is the material of all equals black hole?
Assume here, it falls in, and automatically arise on singularity
The singularity is also r equals zero
So all the material that fill in
is located in some sense at r equals zero
But r equals zero is not space like, and not time like
So where's the material? It's here
That being r equal zero
Everywheres outside here, which means everywhere out of here,
Einstein's equations is solved with zero right hand side
The only place where there's source is sort of here
But that's the end of time anyway, that's just the singularity
For Newton it's also a singularity, but with a very different count
Is there any question?
Student: So what prevent the Newtonian singularity (mumble)
Stydent: eventually things repel
What prevents the Newtonian thing?
Student: Yeah (mumble)
Ya, that's a fair question
Student: So is there something, is it possible something in here
Student: that prevent the singularity the singularity the same one?
There's a very interesting history to this question, it was no,
there was no, I don't want to (mumble), you know
that a bunch of mass points
in Newtonian physics will not collapse to a point,
will not collapse to a singularity
Or let's put it this way
It's a real set of measure zero additional condition
which allows it to collapse to a singularity
And the reason is just a tiny angular momentum,
a one particle relative to the other will have missed each other
Just a little bit of angular momentum will provide some
centrifugal force, they will
So unless they're really very very symmetrically distributed,
with no angular momentum at all,
there was it, go pass each other and come and miss each other,
and not form a singularity
There was none, it was a theorem about cosmic mechanics
The cosmic, Einstein assumed, that a black hole could never formed
He assumed that somehow the same thing would happen,
and it would be impossible
Student: Is that the same thing that makes Globular cluster
Student: basically stable? As a whole? (mumble)
(voice too low), ya
There's a different here, I'll come here
Just a single point particle, supposing there was,
suppose there was from (mumble),
a singularity in Newtonian physics, point particle
And then you throw another one there
Unless you have
It's a point, a point particle
Unless you have the alignment with (cough) presume
it will miss this point and just go back out
What happens here, what happens here?
Once the particle passes the horizon,
there's no way that it can stay,
that's because the singularity has a different character,
it's the end of time, not a point of space
So Einstein was wrong, his logic was wrong,
he had the picture that the singularity is a point,
and that anything he tries to add some mass to it,
it will miss and go back out
Whereas singularity is a real trap,
he don't understand the idea of the horizon,
and the horizon is really the point of no return,
that once you pass it, you'll
Student: You mean it's not a collapsing star provide a lot of pressure
Student: and guide everything inward?
Classically, I mean the Newtonian way
Student: Yeah
Ya, no
Student: I don't understand this all, this is presuppose that
Student: this is a singularity, begin with that, (mumble)
Well light cone are singularities
Student: I mean is that clear that if there is a singularity, (mumble)
I was saying even there was a singularity, you couldn't add to it
even if there was, so even if you can get on started and tag one,
you couldn't (mumble)
If you can't (mumble)
obviously you can't (mumble) make one
Student: That looks, says in terms of phase space,
Student: picture change from classical picture
to Liouville (mumble) picture
Student: if you look at the incompressibility of phase space,
Student: incompressibility is there because the Planck constant,
Student: which suggest
No the incompressibility is there because you leave those thing,
not because of Planck constant
The incompressibility of phase space is a classical theorem
Well keep going, stressing your question
Student: Alright, but wouldn't the Planck constant,
Student: basic uncertain relation prevent further compression,
Student: so to speak, in the phase space, in the black hole and in the pha
OK, so the problem is,
that in order to have a resistance against compression,
you have to have a reasonable density of matter,
and a reasonable amount of kinetic energy and so forth,
and then you can squeeze it, but you can't squeeze it in phase space
But if you look at how dense something has to be,
in order to form a horizon, the answer is if it's heavy enough,
the density can be negligible
Um, the kind of force you're talking about which prevent collapse
It dose have to do with the incompressibility of phase space
But just think of the incompressibility of phase space
providing spacial force which keep things from collapsing
And you get to trouble typically in compressing
when push is too big, when density is too large
How large you expect the density to make a black hole? Let's (mumble)
You have a given amount of matter rho
Don't get it this way
Mass is equal to two M G over R, Mass
I get tired, uh, stick to this way
Uh, two M G is equal to R
Student: Is there c squared?
Yeah, good, c squared, OK
Let's divided this by R cubed, OK?
Why? Because I'm interested in M over R cubed, which is density
The bigger the black hole, the more mass the black hole,
the smaller the density of a collection of material have to be
before it forms a black hole
Why? Take another mass, you put it in a volume,
the density only has to be of an order,
the R's order is equal to two M G
Two M G squared
So for a given amount of mass, in order to make it form a black hole,
the density only has to be one over two M G squared
This is why people are very very certain a black hole's form,
they're very certain before the astronomical observations
You don't have to make, have enough mass, arbitrarily low density,
arbitrarily low density, let's say it in other way
Let's take an arbitrarily low density rho, rho is as small as you like
Particles every sixty meter or whatever
Now take a big enough volume
take enough of them, take a big enough radius of them
How much mass? Let's take a radius R
That's a radius R, how much mass is there?
Well, rho times R cubed, third four pi and so forth
No, this is, ya, four third pi R cubed
This is order of magnitude, the amount of mass is there, right?
So R, even for this very low density,
R, the radius of the full mass is M over rho
The rho is fixed, rho is just the very low density, to the one third power
No matter how small the rho is, if you make M big enough,
this will be less than twice M G
Why? Because this increases linearly,
and this only increases to the one third power
So for a fixed but tiny rho,
you can eventually make M over rho to the one third
smaller than the Schwarzschild radius
Once it's smaller than the Schwarzschild radius, it's a black hole
So no matter how rarefied the material is,
if you create enough uh
Student: If you take the ah, a certain energy density,
Student: what would be associated with for Schwarzschild radius,
Student: to create the black hole with the density?
You mean with the normal cosmological
The radius of the universe, the radius of the horizon, ya
Ya, ten billion light years
I think we could be pretty sure that in our universe
there cannot be a black hole of size bigger than ten billion light years
Student: So can we say our universe of inside a black hole
No, not inside a black hole, but inside a inside-out black hole
It's a inside-out black hole and surrounded by a horizon
but the the analog of the inside of the horizon is outside the region
We are inside of this sphere, as if we're outside the horizon
Things fall toward, things fall away from us
They don't fall toward us, for some big enough scale
Because of dark energy, things fall away from us,
they don't fall toward us
That's the, that's the accelerated expansion of the universe
Things fall away from us and because they fall away from us,
eventually if you go far enough out,
you'll find they moving faster than the speed of light rolling of you
and that becomes spherical horizon,
but it's not that we are inside of a black hole
Being in a black hole, you crash into a singularity eventually
Being inside a cosmic horizon, you just stay and grow up
Student: The singularity is the end of time,
Student: how (mumble) happen that you do (mumble)
Uh, end the time is probably wrong,
it's the end of time in the sense of definitely end of time
Student: Just the proper time changes in the Schwarzschild
Ya, proper time in singularity is (voice too low)
Student: We have fixed r in that point and that's
What's it?
Student: At singularity you know r is zero you still have a fixed r
Student: which will put you travel faster than c at singularity?
Say it again
Student: When r is zero at the singularity, and you also have a fixed r
Student: which would
You can't fixed r because fixed r you have to exceed speed of light
Student: Right, so you wouldn't have a fixed r at r is zero,
Student: at singularity, it will be a fixed r
What I've said, I mean physics of what happens
when something is in singularity is obscure at the best
There's no sense which it stops, it just,
I don't know what to say about it
Physics doesn't or haven't get there
Densities become so large, pressures become so large,
(mumble) become so large that no matter how much physics you know,
you'll always go beyond that
There's no physic, temperature will become infinite,
pressures will become infinite, densities will become infinite
So if you want to understand physics, some desired density
that means that you could understand what happens
near the singularity some shell
But, the lower the worse, you wait enough to, not very long, it get worse
And you always be driven beyond what we know about the matter
So, um, is it the problem that we don't know enough to figure out
what happens in the singularity?
Well I don't think so, I don't think that's the point, I think um,
nothing survives on (mumble)
Student: Since as long as you just stay outside it's also irrelevant
That's true
Student: Because nothing goes on the other side in effect
That's right, that's right
But you could jump in to find out, you could jump in, I'm curious
If you outside the black hole, stay outside the black hole
the singularity (voice too low)
You jump into the black hole you try to find out
what happens in the singularity, you're not going to find out either
Student: You can't tell you (noise)
It's not only that, it's not only that, eventually,
let's look at it, let's look at the (voice too low)
Let's take two atoms, fall into the singularity
Here's a singularity, now atoms can send back and forth messages,
messages can simply be the photons to make a Coulomb field, but
Effect of this, this atom is influenced by that atom,
well you know atoms influence each other and so forth
When they get too close to the singularity,
this atom over here cannot send the signal to the other one
Why? There is not enough time to send the signal
They fall and out of course lose contact with each other
They can no longer influence each other
And so as you get closer and closer to the singularity,
the atoms of your body fall out of contact with each other,
and you just disintegrate into a collection of uncoupled atoms
that have no connection between them
But it's worse than that, the electrons of the atom fall out of contact,
out of communication with nuclears,
and protons and neutrons inside,
nuclears fall out of contact with each other
So everything just disintegrate in a completely uncoupled somethings,
bits which don't talk to each other
Student: Is it really a
Student: You approximate much to make a metric,
Student: to make a correspondence to Rindler space,
Student: only could work when r is near the Schwarzschild radius
Yes
Student: So by going near the singularity
Student: you're going far from the Schwarzschild radius and some up
I've drawn the picture of what that metric doesn't look like,
within the horizon, so that's what are you seeing
Student: You have sort of set up approximations
Student: that make you by on the other side
Ya, ya
Student: So that's some of like Oppenheimer diagram
Say Christoffel diagram
Student: So is that the singularity (mumble)
Singularity what?
Student: (voice too low) but on the other side of it
The other side of singularity?
Student: Ya, the other side (mumble)
The metric are just undefined on the other side of it
it's just undefined on the other side of it
Or if it is not fine, nobody knows
You simply go through a physics which is totally unknown
Student: But (mumble) the coordinate is exist, r is t,
Student: that's more than that
Well how do you, how do you, alright, you might have
you could ask, can you analogically continue the metric pass that point
To where, up to um, to what, to r, you can't
Student: let's just, let's just pick up (mumble)
Student: and the other question is
Let's see can you um, r, so we want r to be more
Where we want to continue it to, we want to continue it to where?
Student: r be negative
r be negative
So what happens if you try to continue this to r negative?
Uh this become positive, this become negative
Student: And this obviously says that uh,
Student: there's no mass (mumble)
There's no what?
Student: There's no matter (mumble)
Student: It's obvious that light can't pass that surface
Well
Student: You put that t started from one side
Look you could draw light trajectories through it (mumble)
Student: That because you draw that already
But we can't find out
Even who willing to jump into the black hole he can't find out,
Student: (mumble) information is severely lost
That's quantum mechanics
Let's not getting to that now, that's beyond my (mumble) class
Student: When Alice arrives the horizon, and she's carrying meter stick,
Student: is that meter stick suddenly become a clock?
No
Alice, here's Shirley,
Shirley is one end of the meter stick and Alice is the other end
Here is a meter stick, here is a meter stick, nothing happens to it,
just go straight through
There's nothing fancy happen near the horizon
No, a meter stick is not become a clock
Moreover, Alice's clock does not become a meter stick, and moves this way
Just using stupid coordinate
Using stupid coordinate, with you called r,
should have called something else about the horizon
Just because you call it r, doesn't mean it is the radius
Student: Is there any difference between clocks and meter sticks
Say it again?
Student: There's no difference between a clock and a meter stick
Really?
Student: Because how could we measure time?
Student: You take how we measure time, you (voice too low)
You can use meter sticks to help you measure time,
but a meter stick is a meter stick, the clock
Student: There's nothing Alice, there is nothing Alice that is (mumble)
There's nothing, nothing dramatic happens that Alice and Shirley,
absolutely not
Student: Bob has a reflector, right?
Student: And Alice was sending signals (mumble), right?
Student: Was she getting anything change as she cross over the horizon?
Bob has a reflector
Student: She is sending message and she is receiving message
Bob has a reflector
Here's Bob, now you don't have to keep track of the fact
that Bob is accelerating away from Alice, that's OK
Alice is here
Sorry, what is it, Alice
Student: Keep free falling
She's still outside, she's outside, she sends a light beam,
it oops, here she is, alright?
Press the singularity over here, Alice sends a message,
and the message comes back, yep?
Alice sends a message over here, and the message comes back,
behind the horizon
But from here Alice can no longer send a message
So she can send a message
Student: She will know, she will know that something happened
No, no no no no no
All she knows that, she knows that she gets closed to the singularity
Student: Oh, She wouldn't know that she's passing the horizon?
No, that's right
That's right, so, so what she sees is when she sends a message,
"the message is taking longer and longer to get back	"
She send a message from here and just take a very very,
let's forget the singularity,
she sends a message from here and it takes a very very long time
for it to come back
It's more or less it's back very very visually
So as she gets closer and closer to the horizon,
and sends out her just usual radio waves,
those radio waves have to get to Bob which takes a very long time
and it have to come back which takes the same time
And they have, the moving mirror has red shifted the photons,
So the photons get very very red shifted
So Alice could see would be a very very red shifted version
of what she sent
But, and, but she cannot see that the version from here
Student: So why is Maxim Braverman says the black hole (mumble)
We're only need some quantum mechanics so that,
so that's from quantum mechanics so (voice too low)
Student: How the gravitons (mumble) the black hole and (noise)
Who said that?
Student: You take a gravitational force from the black hole,
Student: You use that to see gravitons?
You're taking to literally the idea
the gravitational force comes from gravitons (voice too low)
That's, that's
Or if you like you could take the gravitons being emitted
from the surfaces of the black hole
You can, it's perfectly OK
to imagine that all the material of the black hole accumulates,
since it navigate to the horizon, it must accumulate at the horizon
It must sinking in, closer and closer,
asymptotically get closer and closer
But all of the mass is concentrated in the little little thin shell
Point of view of somebody outside,
it's concentrated in a very thin shell in the horizon
If you like to think of gravitons being jumping back and forth,
then you can do it
They jump back and forth from the surface, just through the horizon
So outside just take the materials collecting, kind of symmetric layers,
which through every falling, closer and closer,
this is a classical statement, which through every falling,
closer and closer and thinner and thinner to a symmetry layers,
never quite getting there
But the mass is there, you can tell mass is there
because you put a surface around there and gravitation field out there
And you can always think of it as a mass being localize,
just a very very thin layer
Student: About (mumble) which is not spherically symmetric,
Student: that might put (mumble) be rotating,
Student: So it would be flatten down in something
Student: The spinning flat them down
There is a numeric on how fast it can rotate
and how spherical symmetric it can get
Student: So basically they can't (mumble)
They don't
That's right
Rotating black holes don't (?)
Student: And they could have parts of them like
Student: parts rotating around each other
Student: for example the cllap, a binary star collapse and
Student: some kind binary object that fill inside the black hole,
Student: the piece will collapse
Student: and the singularity will stop that
Ya, we have a couple of binary things going around each other
inside a black hole, OK?
That's really not very different then the atom with electrons and protons
Student: So there's no need for example for (mumble) to escape the
Student: black hole to tell us what's going on
There's no need, and it's not possible
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