14. The Geodesic Equation

14. The Geodesic Equation


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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.

7 thoughts on “14. The Geodesic Equation

  • November 3, 2015 at 4:17 am
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    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.

    Reply
  • September 6, 2018 at 8:35 pm
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    How to find the non zero metric components i.e g_00,g_11,… Of any line element…if anyone know please help me

    Reply
  • November 15, 2018 at 3:00 am
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    I'm impressed with the students that notice errors in the stuff written on the blackboard.

    Reply
  • May 16, 2019 at 12:09 pm
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    great job. pls prof could you give me the possible defintion of the four velocity

    Reply
  • September 9, 2019 at 10:16 am
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    And i thought i was smart

    Reply
  • September 21, 2019 at 3:52 pm
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    @ 17:41: https://www.youtube.com/watch?v=CP61i00VY3c

    Reply
  • September 21, 2019 at 5:25 pm
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    @ 32:59 Greetings from mr Christof(f)el to mr Ellen Gut(t) πŸ™‚

    Reply

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