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visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: OK. In that case let’s get going. In today’s lecture,

we’re going to be sort of splitting the

lecture of things, if the timing goes as I plan. We’re going to

start by finishing talking about the

geodesic equation. And then if all

goes well, we will start talking about the

energy of radiation– completely changing

topics altogether. I want to begin, as usual,

by reviewing quickly what we talked about last time,

just to remind us where we are. Last time, we at first, at

the beginning of lecture, talked about how to add time

into the Robertson-Walker Metric. And this is the formula that

we claimed was the correct one. For a spacetime

metric, ds squared, the meaning is closely

analogous to the meaning that it would have in

special relativity. The main difference being

that in special relativity we always talk about

what is observed by inertial frames of reference

and inertial observers. In general relativity, the

concept of an inertial observer is not so clear

cut, but we can talk about observers for whom there

is no forces acting on them other than possibly

gravitational forces. And whether or not there

are gravitational forces is always, itself, a

framed dependent question. So it does not have

a definite answer. So observers for which there

is no forces acting on them other than

gravitational forces are called free-falling observers. And they play the role

of inertial observers that the inertial observers

play in special relativity. So if ds squared

is positive, it’s the square of the spatial

separation measured by a local free-falling

observer, for whom the two events happen at the same time. Last time, I think,

I did not really mention or emphasize

the word local. But the point is that

in general relativity we expect in any

small region one can construct an accelerating

coordinate system in which the effects of gravity

are canceled out, as the equivalence principle

tells us we can do. And then you essentially see the

effects of special relativity. But it’s only a small

region, in principle, on an infinitesimal region. So these measurements

that correspond to special relativity

measurements are always made locally

by an observer who is, in principle, arbitrarily

close to the events being measured. If ds squared is negative,

then it’s equal to minus c squared times the square

of the time separation that would be measured by a

local free-falling observer for whom the two events

happen at the same location. I should point out that

a special case of this is an observer looking

at his own wristwatch. His own wristwatch is

always at the same location relevant to him, so it’s a

special case of this statement. So it says that ds squared is

equal to minus c squared times the time that a

free-falling observer would read on his own wrist watch. And If ds squared is 0, it

means that the two events can be joined by a light pulse

going from one to the other. Having said this, we can

go back to this formula and understand why the

formula is what it is. The spatial part is

what it is because any homogeneous and

isotropic spatial metric can be written in this form. And we are assuming that the

universe we’re describing is homogeneous and isotropic. The dc squared piece is really

dictated by item two here. We want the t that we

write in this metric to be the cosmic time variable

that we’ve been speaking about. And that means that it is

the time variable measured on the watches of

observers who are at rest in this

coordinate system. And that means that

it has to be simply minus c squared dt squared. Or else dt would not have

the right relationship to a ds squared to be

consistent with what the s squared is supposed to be. And then we also

talked about why there are no dt dr terms,

or dt d theta, or dt d phi. We said that any such term

would violate isotropy. If you had a dt dr

term, for example, it would make the

positive dr direction different from the

negative dr direction. And that can’t be

something that happens in an isotropic universe. That then is our

metric for cosmology, the Roberrtson-Walker Metric. And another important thing

is what is it good for, now that we decided that’s

the right metric? What use is to us? And what we haven’t done

yet, but it’s actually on the homework, we need

the full spacetime metric to be able to find

geodesics, to be able to learn the

paths of particles moving through this

model universe. So we will be making important

use of this Roberrtson-Walker Metric with its

spacetime contributions. OK. Any questions about that? Now I’m ready to change

gears to some extent. Yes, Ani? AUDIANCE: So in general, the

spatial part of the metric, we can get from the geometry? And in general, can

you just add a minus c squared dt square

for the temporal part? PROFESSOR: It’s

not quite general. Remember we used an argument

based on isotropy here. So I think it’s safe to

say that any metric you’ll find in this class

is likely to have the time entering, and nothing

more complicated than minus c squared dt squared. But it’s not a general statement

about general relativity. Any other questions? Yes. AUDIANCE: [INAUDIBLE], PROFESSOR: OK, the question is,

what would be a circumstance where we would have to deal

with something more complicated? The answer would

be, I think, all you need is to add to this model

universe perturbations that break the uniformity. If we tried to describe

the real universe instead of this ideal universe, where

our ideal universe is perfectly isotropic and

homogeneous, if it said we wanted to describe the lumps

and bumps of the real universe, then it would become

more complicated. And we would probably

need a dt, dr term. OK, next we went on to talking

about the geodesic equation. According to General Relativity,

the trajectories of particles that have no forces acting

on them other than gravity, these free-falling

observers, are geodesics in the spacetime.. So that means we want to learn

how to calculate geodesics, which means paths whose

length is stationary under small variations. So we considered first

just simple geodesics in the spatial metric, because

that’s easier to think about. What is the shortest

distance between two points in a space that’s described

by some arbitrary metric? So first we talked about

how we describe the metric. And we introduced two features

in this first formula here. One is that instead of

calling the coordinates XYZ or something like that. We called them x1, x2, x3, so

that we could talk about them all together in one formula

without writing separate pieces for the different coordinates. So i and j represent

1, 2, and 3, or just 1, and 2, which is the labeling

of the spatial coordinates. And the other important

piece of notation that is introduced

in that formula is the Einstein

summation convention. Whenever there’s an

index, like i and j here, which are repeated

with one index lower and one index upper,

they’re automatically summed over all of the values

that the coordinates take, without writing summation sign. It saves a lot of writing. And it turns out that one always

sums under those circumstances, so there’s no need to write the

sums with the summation sign. Next we want to

ask ourselves, how are we going to

describe the path? Before we can find

the minimum path, we need at least a language

to talk about paths. And we could describe a

path going from some point A to some point B, by giving a

function x supra i of lambda. Well, lambda is an

arbitrary parameter, that parametrizes the path. x supra i are a

set of coordinates. i runs over the values

of all the coordinates of whatever system

you’re dealing with. And you construct such

a function where xi of 0 is the starting point, which are

the coordinates of the point A. And xi of some value lambda f,

where f just stands for final, will be the end of the path. And it’s supposed

to end at point B. So the final

coordinates of the path should be x supra i sub b, the

coordinates of the point B. Then we want to use this

description of the path to figure out what the length

is of a segment of the path. And then the full length will

be the sum of the segments. So for each segment, we

just apply the metric to the change in coordinates. The change in coordinates,

as lambda is varied, is just the derivative of

xi with respect to lambda times the change in lambda. And putting that in

for both dxi and dxj one gets this

formula, relating ds squared, the square

of the length of an infinitesimal segment

to d lambda squared, the square of the parameter

that describes that length. Then the full length is gotten

by, first of all, taking the square root of

this equation to get the infinitesimal length, ds. And then taking the integral

of that over the path from beginning to end. And that, then, gives us

the full length of the path, thinking of it as the

sum of the length of each of infinitesimal segment. OK? Fair enough? Now that we have this

formula for the length, now we have the next

challenge, which is to figure out how to

calculate the path which minimizes that length. And I didn’t use

the word last time, but that what is called

the calculus of variations. And I looked up a little bit of

the history in the Wikipedia. The calculus of variations

dates back to 1696, when Johann Bernoulli

invented it, applied it to the

brachistochrone problem, which is the problem of finding

a path for which a frictionless object will slide and get to

its destination in the least possible time. And it turns out

to be a cycloid, just like the cycloid that

describes our closed universes, closed matter

dominating the universe. And the problem was also

solved by– Johann Bernoulli then announced this

problem to the world and challenged other

mathematicians to solve it. There’s a famous story that

Newton noticed this question in his mail when he

got home at 4:00 AM, or something like that, from

the mint– he was apparently a hardworking guy–

but nonetheless when he seen this problem

he couldn’t go to bed. He went ahead and

solved it by morning, which is a good MIT student

kind of thing to do. So the technique is to consider

a small variation from whatever path you’re hoping

to be the minimum. And we’re going to

calculate the first order change in the

length of the path, starting from our original

path, x of lambda, to some new path,

x tilde of lambda. And we parametrize the

new path by writing it as the old path,

plus a correction. And I’ve introduced

a factor, alpha, multiplying the correction,

because it makes it easier to talk about derivatives. And wi of lambda is just

some arbitrary deviation from the original path. But we want to always go

through the same starting point to the same endpoint,

because there’s never going to be a minimum if we’re

allowed to move the endpoints. So the endpoints are fixed. And that means that

this path deviation, w super i in my notation, has

to vanish at the two endpoints. So we impose these two

equations on the variation wi. Then what do I do is take

the derivative of the path length of the varied path, x

tilde with respect to alpha, and if we had a minimum

length to start with, the derivative

should always vanish. That is, the minimum

should always occur when alpha equals

0, if the original path of the true path, the

true minimum path. And if alpha equals

0 is the minimum, the derivative should always

vanish at alpha equals 0. And vice versa. If we know that this happens

for every variation wi, then we know that our path

is at least an extremum, and, presumably, a minimum. And the path itself is just

written by the same formulas we had before, except for x

tilde instead of x itself. And I’ve introduced an axillary

quantity, a of lambda alpha, which is just what appears

inside the square root. That just saves some

writing, because it has to be written

a number of times in the course of

the manipulations. So our goal now is to

carry out this derivative. And the derivative acts

only on the integrand, because the limits

of integration do not depend on alpha. So just carry the derivative

into the integrand and differentiate

this square root of a of lambda, which is,

itself, a product of factors that we have to use–

product rule and chain rule and various manipulations. And after we carry out

those manipulations, we end up with this expression

in a straightforward way involving a few steps,

which I won’t show again. And the complication is

that what we want to do is to figure out for what

paths that expression will vanish for all wi. We want it to vanish for

all possible variations of the path. And what’s complicated

is that wi appears here as a multiplicative

factor in the first term, but as a differentiated

factor in the second term. And that makes it very

hard to know, initially, when those two terms might

cancel each other to give you 0, which is what

we’re looking for. But the brilliant

trick that, I guess, Newton invented, along

with Bernoulli and others, is to integrate by parts. Integration by parts, I’m sure,

was not a well-known procedure at that time. But if we integrate the

second term by parts, we could remove the

derivative acting on w, and arrange for w to be

a multiplicative factor in both terms. And a crucial thing that

makes the whole thing useful is that when you do

integrate by parts, you discover that you don’t

get any endpoint contributions, because the endpoint

contributions would be proportional to

wi at the endpoints. And remember, wi has to

vanish at the endpoints, because that’s the

condition that we’re not changing the points

A and B. We’re always talking about paths that

have the same starting point and the same ending point. So integrating by parts,

we get this expression, where now wi multiplies

everything, as just simply a multiplicative factor. To write it in

this form, you had to do a little bit of

juggling of indices. The other important trick

in these manipulations is to juggle indices, which

I’ll not show you explicitly. But the thing to remember

is that these indices that are being summed over

can be called anything and it’s still the same sum. So when you want to get

terms to cancel each other, you may have to change

the names of indices to get them to just

cancel identically. But that’s straightforward. So we get this expression. And now we want this

expression to vanish for every possible wi of lambda. And we argued that

the only way it could vanish for every

possible wi of lambda is if the expression in curly

brackets, itself, vanishes. Yeah, if we only know the

values for some particular wi of lambda, then there

are lots of ways it could vanish,

because it could be positive in some places

and negative in others. But the only way it

could vanish for all wi is for the quantity in

curly brackets to vanish. So that gives us our

final, or at least, almost final expression

of the geodesic equation. And that’s where we left off

last time, with that equation. So note that this is

just an equation that would either be obeyed or not

obeyed by the function x super i of lambda. It’s just a

differential equation involving x super i of

lambda and the metric, which we assume is given. OK. So are there any

questions about that? Everybody happy? Great. OK, now we’ll continue

on on the blackboard. OK, the first thing

I want to do is to simplify the equation a bit. This equation is

fairly complicated, because of those square roots

of A’s in the denominators. The square root of A is a

pretty complicated thing to start with, and

the square root of A here is even

differentiated, because it’s got the lambda making

an incredible mess, if you understand all that. So it would be nice

to simplify that. And we do have one trick

which we can still do, which we haven’t done yet. We originally constructed

our path, xi of lambda, as a function of some

arbitrary parameter, lambda. Lambda just measures arbitrary

points along the path. There are many, many

ways to do that, an infinite number of

ways that you can do that. And this formula will

work for all of them, it’s completely general. The formula, when

we derived it, we didn’t make any assumptions

about how lambda was chosen. But we can simplify the formula

by making a particular choice for lambda. And the choice that

simplifies things is to choose lambda to

be the arc length itself. Lambda should be the

distance along the path. And then we’re

trying to express xi as a function of how

far you’ve already gone. And that has the effect, if

we go back to what Ai was, A of lambda really is just

the path length per lambda. So if lambda is the path length

itself, A is just equal to 1. I’m trying to get a formula

that shows that more clearly. Here. If we remember that

this quantity is A, this tells us that ds

squared is equal to A times d lambda squared. So if ds is the

same as d lambda, as you’ve chosen your parameter

to be the path length, this formula makes

it clear that that’s equivalent to A equal to 1. So going back to the

formula, if A is 1, we would just drop it from

both sides of the equation. And all that really matters, I

should point out here, maybe, because we’ll be using it

later, is that A is a constant. As long as A is a constant,

it will not be differentiated, and then it will cancel on the

left side and the right side. So we don’t necessarily

care that it is 1, but we do care that

it’s a constant. And then it just disappears

from the formula. And then we get the

simpler formula. And now we’ll continue

on the blackboard. The simpler formula is

just dds of gij dxj ds is equal to 1/2 times the

derivative of gjk, with respect to xi, times dxj

ds dxk ds, where s is equal to the path length. So I’ve replaced

lambda by s, because we set lambda equal to s. And s has a more specific

meaning than lambda did. Lambda was a completely

arbitrary parametrization of the path. So this one deserves a

big box, because it really is the final formula

for geodesics. Once we write it in terms

of different letters, we will later, but this

actually is the formula. Now I should

mention just largely for the sake of your

knowing what’s going on, if you ever look at some other

general relativity books, this is not the formula that

the geodesic equation is usually written in. Frankly, it is the best form. If you want to

find the geodesic, usually this form of writing

the equation is the easiest. But most general

relativity books prefer instead to

just give a formula for the second derivative, here. Which involves just

expanding this term, and then when we

shuffle things, to try to simplify the expressions. So one can write, to

start, d ds of gij dxj ds. We’re just going to expand it. Now we’re going to be making

use of all the rules of calculus that we’ve learn. Every rule you’ve ever

learned will probably get used in this calculation. So it will be

using product rule, of course, because we have a

product of two things here. But we also have

the complication that gij is not explicitly

a function of s. But gij is a

function of position. And the position that one is

that for any given value of s depends on s, because we’re

moving along the path, x super i of s. So the gij here, is

evaluated at x super i of s. I should give this a new letter. x super k of s. So it depends on s, through

the argument of its argument. So that’s a chain

rule situation. And what we get here is,

from just differentiating the second factor, that’s easy. We get gij DC d

squared xj ds squared. And then, from the derivative

of the derivative chain rule piece, we get the partial of

gij, with respect to xk times the dxj ds times dxk ds. And then to continue, this

piece gets brought over to the other side, because we’re

trying to get an equation just for the second

derivative of the path. So then we get g sub ij d

squared x super j ds squared is equal to 1/2 di– I’ll define

that in a second– g sub jk minus 2 dk gij dxi ds dxj ds. where this partial

derivative with the subscript is just an abbreviation for

the derivative with respect to the coordinate

with that index. So that’s just an abbreviation. Now you could think of this

as a matrix times a vector is equal to an expression. What we like to do is just get

an expression for this vector. So if we think of it as

a matrix times a vector, all we have to do

is invert the matrix to be able to get an expression

for the vector itself. Yes! AUDIANCE: Should that

closing parenthesis be more [INAUDIBLE]? PROFESSOR: Oh, Yeah,

I think you’re right, it doesn’t look right. Yeah. Thank you This has to

multiply everything. Oops! OK, OK. Given enough chances

I’ll get it right. OK, now everybody

happy this time? Thank you very much for

getting it straight. OK, So as I was saying,

we want to isolate this second derivative. We’re hoping to get

just an expression for the second derivative. And this can be interpreted

as a matrix times a vector equals something. We want to just

invert that matrix. Yes? AUDIANCE: Isn’t the ds and

[? the idx ?] [INAUDIBLE]? PROFESSOR: Oh, do I have

that wrong too, perhaps? I think we want j and

k there, that don’t we? OK, attempt number four, or

did I lose count as well? j and k are the

indices and the i matches the free i on the left. And all the other

indices are sound. I think, probably, I finally

achieved the right formula. Thanks for all the help. So inverting a

matrix, the principal is a straightforward

mathematical operation. In general relativity, we give

a name to the inverse metric, and it’s the same

letter g with indices, with superscripts

instead of subscripts. And that’s defined to

be the matrix inverse. So g super ij is defined to be

the matrix inverse of g sub ij. And to put that

into an equation, we could say that if we

take g with upper indices– and I’ll write those

upper indices as i and l– and multiply it by a g

with lower indices l and j, when you sum over adjacent

indices in this index notation, that’s

exactly what corresponds to the definition of

matrix multiplication. So this is just the matrix

g with upper indices times the matrix g with lower indices,

and it’s the i j’th element of that matrix. And we’re saying it should

be the identity matrix, and that means that the i j’th

element should be 0 if it’s off diagonal, and 1 if it’s

diagonal, if i equals j. And that’s exactly

the definition of a chronic or a delta. So this is equal to delta ij. We remember that delta ij is

0 if i is not equal to j, 1 if i is equal to j. That’s the definition. And it corresponds to

that identity matrix in matrix language. So this is the relationship that

actually defines g super il. And it is just the statement

that g with upper indices is the matrix inverse

of g with lower indices. Using this, we can bring

this g to the other side essentially by

multiplying by g inverse. And I will save a little

time by not writing that out in gory detail, but rather

I’ll just write the result. And the result is written in

terms of a new symbol that gets defined, which

is an absolutely standard symbol in

General Relativity. The formula is d

squared x i, ds squared is equal to– we know it’s

going to be equal to stuff times the product

of two derivatives. And the stuff that

appears is just given a name, capital gamma,

which has an upper index i, which matches the left

hand side of the equation. And two lower indices, which

I’m calling j and k, which will get summed with the

derivatives that follow, d x j ds, dx k, ds. And this quantity,

gamma super i sub jk are just the terms

that would appear when we do these manipulations. And I’ll write what they are. Gamma super i sub jk is

equal to 1/2 g super il times the derivative

with respect to j of g sub lk,

plus the derivative with respect to k of g sub lj. And then minus the derivative

with respect to l of g sub jk. And this quantity has

several different names. Everybody agrees how to

define it up to the sign. There are different

sign conventions that are used in

different books. And there are also

different names for it. It’s often called the

affine connection. If you look, for example

in Steve Weinberg’s General Relativity book, he calls

it the affine connection. It’s also very often called

the Christofel connection, or the Christofel symbol. And frankly those are

the only names for it that I’ve seen, personally. But there’s a book

about [INAUDIBLE] by Sean Carroll which

is a very good book. And he claims that it’s

sometimes also called the Riemann connection And

it’s also sometimes called the Levi-Civita connection. So it’s got lots of names, which

I guess means lots of people’s independently invented it. But in any case,

that’s the answer. And it’s just a way of rewriting

the formula we have up there. And for solving

problems, the formula, the way we wrote up

there, is almost always the best way to do it. So this is really just

window dressing, largely for the purpose of making

contact with other books that you might come across. OK, so that finishes

the derivation of the geodesic equation. Now I’d like to give

an example of its use. But before I do

that, let me just pause to ask if there are any

questions about the derivation? OK. So on your homework,

you will, in fact, be applying this formalism to

the Robertson-Walker metric. And you’ll learn how

moving particles slow down as they move through

an expanding universe, completely in an analogy to

the way photons, which we’ve already learned,

lose energy as they travel through an

expanding universe. So particles with

mass also lose energy in a well-defined

way, which you’ll be calculating on the homework. For example, though, I’ll

do something different. A fun metric to talk about

is the Schwarzschild metric, which describes, among

other things, black holes. It in principle,

describes anything which is spherically symmetric

and has a gravitational field. But black holes are the

most interesting example, because it’s where the

most surprises lie. So the Schwarzschild

metric has the form ds squared is equal to minus

c squared d tau squared, which is equal to– this is just a

definition, it defines d tau– but in terms of the

coordinates, it’s minus 1 minus 2 G,

Newton’s constant, M, the mass of the object

we’re discussing– the mass of the black hole, if

it is a black hole– divided by r times c squared, r

is the radial coordinate, times c squared dt squared,

plus 1 minus 2 GM over rc squared times dr squared

plus r squared times d theta squared plus sine squared

theta d phi squared. Now here, theta and phi

are the usual polar angles. We’re using a polar

coordinate system. So as usual, theta

lies between 0 and pi. 0, what we might call

the North Pole, and pi what we might call

the South Pole. And phi is what is often

called an azimuthal angle, it goes around. And the way one

describes coordinates on the surface of

the Earth, phi would be the longitude variable. So 0 is less than or

equal to phi is less than or equal to 2 pi

where phi equals 2 pi is identified

with phi equals 0. And you can go around and come

back to where you started. Now notice that if we set

capital M, the mass of this object equal to 0, the metric

becomes the trivial metric of Special Relativity written

in spherical polar coordinates. So all complications go

away if there’s no mass. The object disappears. But as long as the

mass is non-zero there are factors that

multiply the dr squared term and the c squared

dt squared term. Notice that the factors that

do that multiplying– now one of these should be inverted. Important inverse, it’s a

minus 1 power for that factor. Notice that r can

be small enough so that these factors will vanish. And the place where that happens

is called the Schwarzschild radius after the same person

who invented the metric. So r sub Schwarzschild is equal

to 2 GM divided by c squared. When little r is equal to that,

this quantity in parentheses vanishes, which means

we get infinity here, because it’s inverted,

and we get a 0 there. Now when a term in the metric

is either 0 or infinite, one calls that a singularity. In this case, it’s a

removable singularity. That is, the

Schwarzschild singularity is only there

because Schwarzschild chose to use these

particular coordinates. These are simpler than

other coordinates. He wasn’t foolish to use them. But the appearance

of that singularity is really caused solely by

the choice of coordinates. There really is no singularity

at the Schwarzschild horizon. And that was shown

some years later by other people constructing

other coordinate systems. The coordinate system

is best known today that avoids the

Schwarzschild singularity is a coordinate system called

the Kruskal coordinate system. But we will not be looking

at the Kruskal coordinate system in this class. Leave that for the GR class

that you’ll take some time. OK, now the masses

sum parameter, notice that the

metric is completely determined by the mass. And that’s the same situation as

we found in Newtonian gravity. The metric outside of the

spherically symmetric object, by the gravitational field

in Newtonian Physics outside of a spherical symmetric

object, depends only on the total mass,

which does not depend, at all, on

how it’s distributed as long as it’s

spherically symmetric. And the same thing here. As long as an object is

spherically symmetric, the gravitational field

outside of the object will always look

like that formula. Now there are still

two cases– outside of the object could be

larger than or smaller than this Schwarzschild radius. So for an object like the

sun, the Schwarzschild radius, we could calculate it–

and it’s calculated in the notes– it’s about

two or three kilometers. Hold on and I’ll tell

you more accurately. It’s 2.95 kilometers,

the Schwarzschild radius of the sun. But the sun, of course,

is much bigger than that. And that means that the sun

doesn’t have a Schwarzschild horizon. That is, at 2.95 kilometers

from the center of the sun there’s still sun. It’s not outside the sun. This metric only holds outside

the spherically symmetric object. So it does not hold

inside the sun. The place where this has

the apparent singularity the metric is not valid at all. So there is nothing

that even comes close to anything

worth talking about, as far as the

Schwarzschild singularity for an object like the sun. But if the sun were compressed

to a size smaller than 2.95 kilometers with the same

mass, then these factors would be relevant at the

places where they vanish. And whatever consequences they

have, we would be dealing with. Even though r equals r

Schwarzschild is not a singular point, it is still

a special point. What you can show– we won’t–

but what we can show is that that is the horizon. Meaning that if an object

falls inside this Schwarzschild radius, there is no trajectory

that will ever get it out. Yes? AUDIANCE: Say a star is just

incredibly dense at its core. Is it possible to

have suppression of some fractional life of

a star that’s from that mass that it’s contained? Or like a fusion

reaction that is going on with the net radius? PROFESSOR: OK, could there be

a horizon inside of a star? I think is what you’re

asking, basically. AUDIANCE: One that

actually affects the– PROFESSOR: One that

really is a horizon. AUDIANCE: That’s outside. PROFESSOR: Right. If this were the sun

you were describing, this formula would just

not be valid inside. There would be no

horizon inside. But you’re asking a

real valid question. If a star had, for some

reason, a very dense spot in the middle, could

it actually form a horizon inside the material? And the answer

is, yes, it could. It would not be stable. The material would ultimately

fall in, but it could happen. Yes? AUDIANCE: So like our galaxy

has a super massive black hole in the center. PROFESSOR: That’s right. Our galaxy does have a

super massive black hole in the center. AUDIANCE: Yeah. So you can consider

that as like a larger mass that has black hole, area? PROFESSOR: Right! Right! That’s right. The comment is that

if we go from a star to something bigger

than a star we have perfectly good

example in our own galaxy, where there is a black

hole in the center, but there is still mass that

continues outside of that. And the black hole is

accreting, more matter does keep falling in,

it’s not really stable. But it certainly does

exist, and can exist. Any other questions? Well, our goal now is

to calculate a geodesic. And I will just

calculate one geodesic. I will calculate what happens if

an object starts at some fixed radius at rest and is released

and falls into this black hole. I first want to just rewrite

the geodesic equation in terms of variables that are more

appropriate for this case. When I wrote that, I had a mind

just calculating the geodesics in space, looking

for the shortest path between two points. The geodesic that we’re

talking about when we’re talking about an

object in general relativity moving along the

geodesic is a geodesic that’s a time-like geodesic. That is, any increment

along the geodesic is a time-like interval,

or following a particle. Particles travel on time-like

trajectories in relativity. So the usual notation for

time is something like tau rather than s, which is

why I wrote it this way. ds squared is just defined to be

minus c squared d tau squared. So d tau squared has no

more or less information than ds squared, but it has the

opposite sign and a difference by a factor of c

squared, as well. And another change in notation

which is a rather universal convention is that, when we talk

about space alone we use Latin indices, ijk.. When we talk about spacetime,

where one of the indices might be 0 referring

to the time direction, then we usually use Greek

indices, mu, nu, lambda. So I’m going to rewrite the

geodesic equation using tau as my parameter instead

of s, since we’re talking about proper time

along the trajectory instead of distances. And using Greek letters instead

of Latin letters, because we’re talking about spacetime

rather than just space. So otherwise what

I’m going to write is just identical to that. So really is nothing more

than a change in notation. d d tau of g mu nu,

dx super nu d tau. And it is equal to 1/2 times

the partial of g lambda sigma with respect to x nu dx

lambda d tau dx sigma d tau. Now you might want to go

through the calculation and make sure of the

fact that now we’re dealing with a metric

which is not positive, definite, doesn’t

make any difference. But it doesn’t. It does mean that

now we certainly have possibilities of getting

maxima and stationary points as well as minima, because

of the variety of signs that appear in the metric. But otherwise, the calculations

of the geodesic equation goes through exactly

as we calculated it. And the only thing

I’m doing here, relative to what we

have there, is just changing the notation a bit

to conform to the notaion that is usually used for talking

about spacetime trajectories. Since we’re talking

about radio trajectories, we’re just going to

release a particle at rest and then it will fall

straight towards the center of our spherical object,

we know by symmetry that it’s not going

to be deflected in the positive theta

or the negative theta, or the positive phi or

negative phi directions, because that would

violate isotropy. It would violate the

rotational symmetry that we know as

part of this metric. This Is just the metric of

the surface of the sphere. So theta and phi will just

stay whatever values they have when you drop this object. So we will not even talk

about theta and phi. We will only talk

about r and t, how particle falls in as

a function of time. And then it turns out to be

useful to just first write down what the metric itself tells us. And we’ll divide by d tau. So we could talk about

derivatives with respect to tau. So changing an overall

sign, since everything’s going to be negative

and we’d rather have everything be

positive, we can just rewrite the metric

equation as saying, that c squared is equal

to 1 minus 2 GM over rc squared, times c

squared times dt d tau squared minus 1 minus 2

GM over rc squared inverse times dr d tau squared. So this is nothing more

than rewriting this equation saying d theta is equal

to 0 and d phi will be 0. Written this way,

though, it tells us that we can find dt d tau, for

example, if we know dr d tau. And we also know where we

are, you know, little r. And we’ll be using

that, shortly. To continue a little

further, we’re going to introduce some

abbreviations just so we’re don’t have to write so much. I’m going to define

little h of r as just one minus r

Schwarzschild over r. And this is also 1 minus

2 GM over rc squared. That’s a factor

that keeps recurring in our expression

for the metric. Yes? AUDIANCE: The second

to last equation is supposed to be a c squared

in between the two parenthesis? PROFESSOR: Probably. Yes, thank you. G squared, right? Thanks a lot. In terms of h of

r, we can rewrite that equation

slightly more simply. I’m going to bring

things to the other side and write it as c

squared times dt d tau squared is equal to c

squared h inverse of r plus h to the minus 2 of r

times dr d tau squared. This is just a rewriting

of the above equation, making use of the new notation

that we’ve introduced. And this is the form

we will be using. It explicitly tells

us how to find dt d tau in terms

of other things. So dt d tau is not independent. Since we know dt d tau

in terms of dr d tau. If We get an expression for dr

d tau we’re sort of finished. We could find everything

we want to know about t from the equation we just wrote. So it turns out that

all we need to do to calculate this

radial trajectory is to look at the

component of the metric where that free index, mu, mu

is the index that’s not summed, we’re going to

set mu equal to r. Remember mu is a number that

corresponds to a coordinate. And we’re going to set

mu equal to the value that corresponds to

the r coordinate. And that will be sufficient

to get us our answer. When we do that, the equation

becomes d d tau of g sub r. Now the second index, nu

in the original expression, is summed from 0 to 3

for the gr case, where we have four coordinates,

one time and three spatial coordinates,

but we only need to write the terms where

gr nu variable is non-zero. And the metric

itself is diagonal. So if one index is a

little r, the other index has to also be r,

or else it vanishes. So the only value of nu

that contributes to the sum is when nu is also equal

to the r coordinate. So we get g sub rr d

xr– which, in fact I’ll write it as just dr. x super r

is just the r coordinate, which we also just call r times

d tau is equal to 1/2 dr. And now, on the

right-hand side, we’re summing over lambda and sigma. And lambda and sigma have to

have the property that g sub lambda sigma depends on r,

or else the first factor will vanish. And furthermore,

g sub lambda sigma has be non-zero, for the

values of lambda and sigma that you want, which means that

lambda and sigma for this case has to be equal to

each other, because we have no off-diagonal

terms to our metric. So the only contributions we get

are from g sub rr and g sub tt. So you get the derivative with

respect to r of g sub rr times dr d tau squared. This become squared, because

lambda is equal to sigma. And then plus 1/2 drg sub

tt times dt d tau squared. And note that buried in

here is, if we expand this, the second derivative

of r with respect to time– respect to tau. So we can extract

that and solve for it. And things like dt d tau

will appear in our answer, initially, because

it’s here already. But we could replace dt d

tau by this top equation and eliminate it

from our results. And I’m going to skip

the algebra, which is straightforward,

although tedious. I urge you to go

through it in the notes. But the end result ends up

being remarkably simple, after a number of cancellations

that look like surprises. And what you find in

the end– and it’s just the simplification of this

formula, nothing more– you find that d squared

r d tau squared is just equal to minus Newton’s

constant times the mass divided by r squared. Now this is rather shocking,

and even looks exactly like Newtonian mechanics. However, even though it looks

like Newtonian mechanics, it’s not really the same

as Newtonian mechanics, because the variables don’t

mean quite the same thing. First of all, even

r does not really mean radius in the same sense

as radius is defined by Newton. In Newtonian

mechanics, radius is the distance from the origin. If we wanted to know the

distance from the origin, we would have to

integrate this metric. And in fact, there isn’t

even an actual origin here, because you would have to

go through the singularity before you get there. And you really can’t. That integral is not

really even defined. Although, of course, if we

had something like the sun, where the metric was

different from this small r, then we could integrate

from r equals 0, and that would define

the true radius, distance from the center. But it would not be r. It would be what you got by

integrating with the metric. So r has a different

interpretation than it does for

Newtonian physics. I might add, it still has

a simple interpretation. If you look at this metric, the

tangential part, the angular part, is exactly what you

have for Euclidean geometry. It’s just r squared times the

same combination of d theta and d phi as appears on

the surface of a sphere. So little r is sometimes called

the circumferential radius, because it really does give you

the circumference of circles at that radial coordinate. If we went around in

a circle at a fixed r, the circle would involve

varying phi, for example, over a range of 2

pi, we really would see a total circumference

of 2 pi little r. So r is related

to circumferences in exactly the way as it

is in Euclidean geometry. But it’s not related

to the distance from the origin in the same way

as it is in Euclidean geometry. In addition, tau, here,

is not the universal time that Newton imagined. But rather, tau is measured

along the geodesic. It is just ds

squared, but remember, ds squared is being measured

along the geodesic, which means that it is, in

fact, the proper time as it would be measured

by the person falling with the object

towards the black hole. So tau is proper time as

measured by the falling object. And that follows

from what we know about the meaning of

the metric itself. OK, that said we would

now like to just study this equation more carefully. And since the

equation itself still has the same form as what

you get from Newton, if you remember what you would

have done if this was 801, you can, in fact, do

exactly the same thing here. And what you probably would

have done, if this was 801, would be to recognize that this

equation can be integrated. We can write the equation

as d d tau of 1/2 dr d tau squared

minus GM/r equals 0. I you carry out

these derivatives you would get that equation. And this is just

the conservation of energy version of

the force equation. And that tells us that this

quantity is a constant. If we drop the object from

some initial position, r sub 0, and we drop it with

no initial velocity, we just let go of it at

r sub 0, that tells us what this quantity

is when we drop it. It’s minus GM over r sub 0. This piece vanishes if there

is no initial velocity. And that means it will

always have that value. And knowing that, we

can write dr d tau is equal to– just

solving for that– minus the square root of 2GM

times r0 minus r over r r0 I’ve collected two terms and put

them over a common denominator and added them. So this is not quite as

obvious as it might be. But this is just the

statement that that quantity has the same value as

it did when you started. Now this can be

further integrated. We can write it as

dr over– bringing all this to the other

side– is equal to d tau. And then integrate both sides. Notice when I bring

this to the other side and bring the d tau to

the right., everything on the left-hand side

now only depends on r. So this is just an explicit

integral over r that we can do. And I will just tell you that

when the integral is done we get a formula for

tau as a function of r. And it’s equal to

the square root of r sub 0 over

2GM times r0 times the inverse tangent of

the square root of r0 minus r over r plus the square

root of r times r0 minus r. So when r equals r0, this gives

us 0, and that’s what we want. When we start we’re at r0,

or time 0, or proper time 0. And then as r gets smaller,

as it falls in, time grows. And this gives us the

time as a function of r. We might prefer to have

r as a function of time, but that formula can’t really

be inverted analytically. So that’s the best we can do. Now one thing that

you notice from this is that nothing

special happens as r decreases all the way to 0. Even when you plug

in r equals 0 here, you just get some finite number. So in a finite amount

of time, the observer would find himself falling

through the Schwarzchild horizon and all the

way to r equals 0. I didn’t mention it but r

equals 0 is a true singularity. Our metric is also

singular when r equals 0. These quantities

all become infinite. And physically what

would happen is that, as the object falling

in approaches r equals 0, the tidal forces,

that is the difference in the gravitational force on

one part of the object verses another, will get

stronger and stronger. And objects will

just be ripped apart. And the ripping apart occurs as

being spaghetti-ized, that is, the force on the front

gets to be very strong compared to the

force on the back. So I’ll just get stretched out

along the direction of motion. Now the curious

thing is what this looks like if we think

of it not as a function of the proper time

measured by the wrist watch of the object

falling in but rather, we could try to describe it

in terms of our external time variable. The variable t that appears

in the Schwarzchild metric. And to do that, to

make the conversion, we want to calculate what the

dr dt is, instead of dr d tau. Like maybe an analogous

formula, in terms of t. And to get that, we use

simply chain rule here. dr dt is equal to dr d

tau– which we’ve already calculated– times d tau dt. And d tau dt is 1 over dt d tau. If you just have two variables

that depend on each other. The derivatives are just

the inverse of each other. So this could be written

as dr d tau– which we’ve calculated–

divided by dt d tau. And dt d tau we’ve really

already calculated as well, because it’s just given

by this formula here. So we could write out what

that is and figure out how it’s going to behave

as the object approaches the Schwarzchild radius. So it becomes dr dt

is equal to, I’ll just write the numerator as dr

d tau given by that expression. But what’s behaving

in a more peculiar way is the denominator, which

is h inverse of r plus c to the minus 2, h to the minus

2 of r times dr d tau squared. So now we want to look at

this function h inverse of r. And this just means 1/h of r. It doesn’t mean

functional inverse. That is just equal to r

over r minus r Schwarzchild. And we’re going to

be interested in what happens when r gets to be

very near r Schwarzchild, because that’s where the

interesting things happen, as you’re approaching

the Schwarzchild horizon. And that means that the

behavior of the numerator won’t be important. The denominator

will be going up, and that’s what will

control everything. So we can approximate

this as just r Schwarzchild over r

minus r Schwarzchild. And this is for r

near r Schwarzchild. We’ve replaced the

numerator by a constant. And then if we look

at this formula, this is going to blow up

as we approach the horizon. This is the square

of that quantity. It will blow up faster than the

first power of that quantity. And therefore,

this will dominate, the denominator

of the expression. We can ignore this. When this dominates, the

dr d tau pieces cancel. So that’s nice. We don’t even need to think

about what the dr d tau is. And what we get near

the horizon is simply a factor of c times r

minus r Schwarzchild over r Schwarzchild. It’s basically just h. This becomes upstairs

with a plus sign. And the square root turns it

into h instead of h squared. So this is the inverse of that. OK, now if we try to

play the same game here as we did here, to determine

what our time variable behaves as a function of r, instead

of the proper time variable tau, what we find is that t

of r– this is for r near r Schwarzchild– is

about equal to minus r sub s over c times

the integral up to r of dr prime over

r prime minus rs. This is dr dt. Yeah, this was dr dt

from the beginning. I forgot to write the r somehow. AUDIANCE: Doesn’t

that [INAUDIBLE]? PROFESSOR: Yeah, I didn’t write

the lower limit of integration. I was about to comment on that. The integrand that we’re writing

is only a good approximation whenever we’re near r. So whatever happens near the

lower limit of integration, we just haven’t done accurately. So I’m going to just not write

a lower limit of integration here, meaning that

we’re interested only in what happens as the

upper limit of integration r becomes very near

r Schwarzchild. And everything will

be dominated by what happens near the upper

limit of integration. AUDIANCE: So would you

just integrate over on [INAUDIBLE] for that? PROFESSOR: That’s

right, that’s right. We just integrated over a small

region near, r Schwarzchild. Nu r, which is also about

equal to r Schwarzchild. And the point is, that this

diverges logarithmically as r approaches r Schwarzchild. So it behaves approximately

as minus r Schwarzchild over c times the logarithm

of r minus r Schwarzchild. So as r approaches

r Schwarzchild, this quantity that’s the

argument of the logarithm gets closer and closer to 0. It gets smaller and

smaller approaching 0. But the logarithm of

a very small number is a negative number, a

large negative number. And then there’s

a minus sign here. You get a large positive

number and it diverges. As r approaches r

Schwarzchild the time variable approaches infinity. And that means that

at no finite time does the object ever reach

the Schwarzchild horizon. But as seen from the outside,

it takes an infinite amount of time for the object to

reach the Schwarzchild horizon. As time gets larger and larger,

the object gets and closer to the Schwarzchild horizon,

asymptotically approaching it but never reaching it. So this, of course,

is very peculiar, because from the point

of view of the person falling into the

black hole, all this just happens in a finite amount

of time and is over with. From the outside,

it looks like it takes an infinite

amount of time. And weird things like this can

happen because of the fact that in general relativity time is

a locally measured variable. You measure your time,

I measure my time. They don’t have to agree. And in this case, they can

disagree by an infinite amount, which is rather bizarre,

but that’s what happens. So according to classical

general relativity, when an object falls

into a black hole, from the point of view

of the object nothing special would happen as that

object crossed the Schwarzchild horizon. Everybody believed that

that was really the case until maybe a couple years ago. Now it’s controversial,

actually. At the classical

level, everybody believes that’s still true. I mean, classical

general relativity says that an object can fall

through the Schwarzchild horizon and then

nothing happens. It’s not really a singularity. But the issue is that when

one incorporates, or attempts to incorporate, the effects of

quantum theory, which nobody really knows how to do in

a totally reliable way, then there are

indications that there’s something dramatic happening

at the Schwarzchild horizon. The phrase that’s often

used for what people think might be happening at the

horizon is the word firewall. So whether or not there is

a firewall at the horizon, is not settled at this point. Certainly, though,

classical general relativity does not predict the firewall. If it exists, all the arguments

that say it might exist are based on the quantum

physics of black holes, and black hole evaporation,

and things like that. As you know quantum

mechanically, the black holes are

not stable, either, if they evaporate–

as was derived by Stephen Hawking

in, I think, 1974. But that’s strictly

a quantum effect. It would go to 0 as h bar

goes to 0, and, at the moment, we’re only talking about

classical general relativity. So the black hole that we’re

describing is perfectly stable. And nothing happens if you

fall through the horizon. Except from the

outside, it looks like it would take an

infinite amount of time just to reach the horizon. So we’ll stop there. I guess I’m not

going to get to talk about the energy

associated with radiation. But we’ll get to

that on Thursday. So see you folks on Thursday.

if someone could tell the camera operator to zoom out a little to include the raised blackboard, please do. don't just follow the fellow around the room.

How to find the non zero metric components i.e g_00,g_11,… Of any line element…if anyone know please help me

I'm impressed with the students that notice errors in the stuff written on the blackboard.

great job. pls prof could you give me the possible defintion of the four velocity

And i thought i was smart

@ 17:41: https://www.youtube.com/watch?v=CP61i00VY3c

@ 32:59 Greetings from mr Christof(f)el to mr Ellen Gut(t) π