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Let me just tell you what the story is with dark energy,
big rips,
Whether dark energy will have a tendency to tear atoms apart and so forth
Uh, we're not,
In past classes we studied a little bit about dark energy,
so, for those who were there, I'll just remind you,
for practical purposes,
let's forget the deep story
But for practical purposes,
dark energy or cosmological constant,
or whatever you like to call the vacuum energy,
you observed uh, accelerated expansion of the universe
That is equivalent to a small repulsive force,
a component to gravity
They say gravity is not repulsive,
well suspend judgment about that until we get to it in this class
Uh, and imagine a small repulsive component to gravity,
which is not proportional to one over R squared
like Newtonian gravity,
but in fact where the force increases with distance
It increases with distance,
roughly the same way well how does the how about a uh, springs
Supposing you have a particle on the spring connected to the wall,
or two particles connected by springs
how does the force depend on the separation between them?
The force grows linearly with the separation between them, of course
until you break the spring but let's not break the spring,
The force grows linearly,
and it's proportional to the distance between uh, the
uh, particles at the end of the spring
Force grows linearly
There are some constants there,
usually called the spring constant let's put it in
And because it's attracted,
pulling two things together,
there is a minus sign there
Uh, and imagine that we have such a spring,
okay now what and there were two particles attached to it
What does the cosmological constants do?
What the cosmological constant the repulsive cosmological constant,
what it does or positive cosmological constant or positive dark energy,
what it does is it adds a little bit of force of just this type,
proportional to distances, but repulsive not attractive,
this is attractive that's a minus sign there
Repulsive
So it is a little bit a repulsive force to the otherwise attractive force
between the two particles on the end of the spring
Now I'm not really thinking about two particles on the end of the spring,
but it is a good model for us
What I'm really thinking of for example is uh, an atom
an atom is a system which is composed of
two more particles held together by forces
The forces are not entirely different than spring forces,
but what would happen if you add to this
a little bit of, much smaller than the spring constant
The spring constant which hold atoms together is quite strong,
the dark energy forces are minute, they're incredibly small,
until you get very very far away,
notices this force grows with distance
Forget this one
We are going to add now a little bit of component
the coefficient in front of it is called the cosmological constant,
or the dark energy,
also times R
What would it do to the spring if there was a
tiny tiny coefficient here and that coefficient is truly small
for ordinary laboratory experiments
All it would do, would be to change the equilibrium position
of the particles at the end of the spring a little bit
It would effectively change the spring constant here,
make it are a little bit smaller
But what will happen
if you made this spring constant the little bit weaker?
All those would happened is that the
spring with the particles on it
would be a little bit bigger
That's all
Atoms will grow by an entirely negligible amount,
they wouldn't be torn apart
They wouldn't be torn apart,
it would just change the equilibrium position of the electrons
in the atom by a tiny completly insignificant amount,
and that would be the net effect on the atoms,
it would be the net effect on almost anything,
that was otherwise bound together
The reason is that this lambda here,
how small is it it's so small,
that this force doesn't become significant, until you get out to the
full measure of the entire or global universe,
that's where it begins to get big
But if you ask about it on the scale of the solar system,
it's negligible it's tiny by comparison with the other forces
in the solar system gravity and other things
And so what it would do is it would change
the equilibrium position of the earth a little bit
a tiny tiny bit,
or it would change the size of a galaxy a little bit,
and expanded out a little bit
But it wouldn't overcome the forces which hold things together
Now there's a theory called the big rip,
it violates every principle of physics [laughter]
It does, it violates every principle of physics
Uh, why anybody would take it seriously I don't know
but when seriously
work out the equations,
and say there are limits between this and such-and-such cannot be bigger
than this so such-and-such cannot be small than that,
for deep fundamental principles
Somebody's gotta come along and say yea,
but what if those principles are wrong?
and such-and-such could be a little bit bigger or a little bit smaller
than the bound
that uh, theoretical principles tell you?
Somebody will come along and say that,
they'll write a paper about it,
the paper will last for
till it gets to the New York Times
And then they will slowly fall into the junk heap of bad physics ideas
The big rip is one of those ideas
But I'm not going to try to tell you right now
what is wrong with it
We will eventually come to it,
but it's basically the idea that this constant lambda here,
is time-dependent,
that it grows with time,
now there is no reason on earth to believe that
Not only there's no reason on earth to believe it,
there were very very strong reasons on earth to disbelieve it
But you can say who cares what theory says,
let's make a prediction of what would happen if lambda grew with time
Well if it grew with time,
eventually it would become stronger than the forces
which bind together other objects
At first it would become stronger than the force
which hold galaxies together
the galaxies would fly apart
and it would become stronger than the forces which hold the solar system
the solar system would fly apart
and eventually will become stronger than the forces
which hold the spring together, springs flies apart
As I said,
that's called the big rip,
that this constant lambda the dark energy is not a constant at all,
but that it increases with time
Decreases with time is allowable,
increases with time violates uh, some strong, yeah?
Student: I interpreted what you're saying
Student: that lambda is just a function of mass or whatever?
No, lambda is just a numerical constant, period
Student: Well, the question I'm getting is,
Student: what if there is no mass and is lambda still in effect?
It's a force that would be there between any pair of particles
if the particles are there
every particles aren't there so there would be the particles
there will be no force on them so uh
Student: lambda evolves with time? Or change?
No, according to most current thinking
it's called the cosmological constant for good reasons,
because all of the good theories says that it's gotta to be a constant
Now, could it not be a constant?
Uh, it doesn't violate any
very very deep principles for the decrease with time
It violates big principles to increase with time
so ummm
But we will come to this,
I just wanted to assure you to make you feel better about the future
Student: I heard that dark energy have a huge effect on particles
Student: If it's so small, how can that be huge?
Well the dark energy density in this room is tiny,
It's extremely small, I'll tell you how small it is
Take a cubic meter,
and how many,
how many protons, well proton is certain mass equals mc squared,
so there's a certain amount of energy in every proton
Uh, roughly speaking
I think the dark energy in this room per cubic meter
uh, would be
I'll have to work it up but roughly it is a order about
a thousand protons in the cubic meter,
it is an negligible amount of energy
Its gravitational effects on things are really negligible,
but if you had it smear out through the entire universe that way,
then as efficiently big distances away
Now imagine now
imagine now we live in the world,
which has an extra bit of energy,
that's causing repulsion,
but it spread out throughout the entire universe in a smooth distribution
Then eventually at sufficiently large distances,
it would make an effect that repulsive affect
on things which would become big
but those distances are cosmological in size,
so it has an effect on the global universe
But it doesn't have an effect on
certainly not on laboratory physics,
it doesn't even have a significant effect
on any kind of astronomical physics
Uh, by astronomical I mean things smaller than the entire universe
In fact for whatever reason
this number happens to be such, that only that this was
only become significant and comparable to other forces
at something like the radius of the entire universe
So it's only becomes important under those circumstances,
People who want to tap the vacuum energy to get vacuum energy to do work
and as a solution to the energy crisis,
have to deal with the fact,
that if they wanted to get a tank full of gasoline,
they would have to extract the energy of a volume something like the
orbit of the moon um, or something like that
So there's not a lot of energy there,
and no way to tap it but
Student: let's suggest the dark energy would be a relative energy?
Relative to what? What is that mean?
I'm not sure what you mean by relative energy
You mean, well it gravitates,
it gravitates that has uh, has the effect of the uh,
distribution of mass in the universe
What's, I'm not, uh,
you mean the only differences over the importance cause we comment?
I'm not sure if relative was being used
Student: so I may check the library on that
Ya that's probably
Student: So is there some certain particles associate with dark energy?
No, no, there aren't dark energy
Dark energy doesn't have the form
of any energy that's associated with particles
We'll come to it, I hope we will come to it,
in the course of this lectures
But no, there's no particles
Now dark energy is to be contrasted with what is called dark matter,
dark matter is an entirely different thing totally different thing,
which is made up out of particles,
or at least it is believed to be made up out of particles,
but uh, this is not where I was going to my .. I just wanted to
for those who were involved in these exchanges of emails
I thought I would give you my view of it
Let's come back
to the Newtonian laws of gravity
We need to spend a little more time with Newton,
before we can move on with Einstein
When I say Newton of course many of the things I'm going to say,
Newton might not have really recognize,
but they are forms or expressions of Newtonian physics
First a little bit of mathematics it's not really mathematics,
it's just formalism, ways of expressing the equations, symbols
First of all I told you last time,
we use the symbol Del
last time, it's called Del
Upside down Delta,
and it sort of a kind of vector but it's not really a vector,
with uh, definite magnitude in definite directions
It stands just like every vector if I wrote any vector over here V,
we could say that is the same thing as
giving its components,
the x component the y component and z component
So one way of viewing a vector is just that
it stands for three numbers,
The x y and z components of
v could be velocity for example
The symbol del with an arrow above it
stands also for three objects
but those three objects are really derivatives,
derivative with respect to x,
derivative with respect to y,
derivative with respect to z
Whenever you see a del it always means you differentiating something,
you are always doing a derivative of some kind
A partial derivative or a set of partial derivatives
of uh, uh, of something
Now for example how do you use it
It's really called a differential operator,
it's not a set of numbers it's a set of operations,
but it also has three components
So let's say,
let me the first example,
the operation del,
if it acts on a scaler, now a scaler just means a function of position,
function of x y and z
I'll stop writing xy and z soon enough, and just called x
But take a function of x y and z,
and applied to it the operation, del,
what does it make?
it makes a vector phi is not a vector,
but del times phi is a vector,
and one of the components of the vector,
the components of the vector are just,
derivative of phi with respect to x,
derivative of phi with respect to y,
derivative of phi with respect to z
So it creates a vector from a scaler,
that's one example of the use of del
Another example,
supposing it act on a vector field, it can also act on a vector field
What does it mean, let's take
del dot A, let A be, oh let's use V, del dot V
That's called the divergence of V we've talked about last time,
but just to give you a little more feel for how the symbols are used,
dot stands for dot product
If I have two vectors ,let's call one W ,for one moment let's call one of
them W and let's call the other one V,
and I will take the dot product of the two vectors
What it stands for?
Stands for the x component of W,
time the x component of V,
plus the y component of W times the y component of V,
plus third term,
which is similar,
that's the meaning of W dot V, the standard old-fashioned dot product
between vectors which most of you have seen
Aright, by the same rule,
we simply pretend that the symbol del stands for a vector,
it's like W,
but it's not it's not a real vector it's just the set of operations,
and what this gives is
the derivative of Vx with respect to x,
plus the derivative of Vy with respect to y,
plus the derivative of Vz with respect to z,
but otherwise it has the same form,
except replace W by this set of derivative operators
That's also neat notation
It's a notation or device,
for doing certain kinds of derivatives, partial derivatives
on various kinds of things
If phi is a scaler,
then Del dot phi is a vector
What if V is a vector?
What is Del dot V?
It's a scaler
In the same way that the dot product between two vectors as a scaler,
the divergence of a vector,
is a vector(scaler)
These notations are pervasive throughout
all the physics electricity magnetism,
gravity,
and without them,
we would not be able to,
it would not be able to function very well
Okay that uh, that's just the preliminary warm-up definition,
it's really just definition
There's not any deep mathematics, the deep mathematics is of course in
the notion of a derivative or partial derivative,
derivative along a direction
Now I
mentioned of course,
that phi here is a function of position,
a function of position like that is called the scaler field
It's a scaler,
and it depends on position
depending on position like that makes it a field
V, the components of V may also be
functions of position,
and in that case V is called a vector field
Alright, so anything which depends on positions is called a field,
and that's our basic notation
All right, let's come to the field of the gravitational field
The gravitational field is a property every point of space,
and let me define it for you
First of all let's imagine that it is created by some masses
So let's imagine some masses,
some mass points,
and I'll label them as usual i j k
i j k run from one and to whatever
let's call this over here the ith mass
Now what we do to find the gravitational fields we invent another mass,
we imagine one more mass which we call the test mass,
the test mass is a little tiny mass that we move around in space,
we examine the forces on it,
and in terms of the forces on a different places in space,
we define a field
And that field is a force field,
or strictly speaking a field of acceleration
So here is what we do,
we take our little test mass,
we hold it, let it go, we see how much acceleration it accelerates with
and in what direction, acceleration is a vector,
it has a direction it has a magnitude
And that acceleration,
the acceleration
experienced, let's draw a little test mass
I'll just draw a little crossing on it there
It may or may not be real,
it may just be in our imagination or
may be a very small little mass that we move
around and explore the vicinity
of the real masses
And that acceleration is a vector,
it depends on position I'm not going to write xy and z now,
I'll just write x,
x now standing for x y and z
That acceleration,
and let me write it down for you,
let's call the vector,
from the ith mass to the test mass,
vector pointing from the ith mass to the test mass,
let's call that Ri
Last time I defined the vector as the vector,
from the test mass to i
Now I'm defining it to go from ith to the test mass,
though two vectors are equal and opposite to each other
But by defining the vector Ri,
as the vector from i to test mass,
I will save myself a minus sign
Which direction,
ah, sorry
I think I want it to go the other way,
if I want to save myself the minus sign, I make it go the other way
Ri goes that way,
from the tast mass,
to the particle which is creating the gravitational field
Now which way is the force on the subject
due to the ith mass?
It's pointing toward the ith mass, right?
The gravitational force on the test mass is pulling towards the ith mass,
and so it's along the vector Ri,
not opposite to it,
and that would save me having to write a negative sign
Alright, so now
What do we know about this field of acceleration?
Well it gets a contribution from every test, from every real mass here
and the contribution in particular from the ith mass is a sum
A is a sum over all the masses, all the real masses
it depends on the gravitational coupling constant the Newton's constant,
it depends on the distance from the test mass to i that is Ri,
Ri squared,
and what else?
And it depends on the mass of the ith particle
Each particle has a mass,
the acceleration on this particle due to the ith particles,
is proportional to the mass, is proportional to G,
and it's divided by Ri square but everything is written here
None of the things that are written here are vectors,
G is a number, m is a number,
and R, R is the length of a vector,
it's square is just a number
There's no vectors on the right hand side,
so it's got to be something wrong with this equation
What's wrong is I have to remind myself
what direction the acceleration is,
and it's along the direction Ri
So, the way to deal with that is to put another R vector here,
but then we have too many R enumerated,
we have to put another one in the denominator, I have to make this cubed
Here the length of this is R
Another way to think about it is to leave it as R squared,
and think of this as a unit vector the symbol for unit vector,
a vector of unit length,
is a vector of a little hat on top of it,
little hat on top of it means unit vector,
vector of unit length
Alright, so this vector here,
is a vector,
it tells you the direction of things,
it tells the direction of the acceleration,
but the magnitude of it all comes from here,
and we add them all up
The acceleration on the test mass is the sum of the vectors,
all different uh, contributions
This depends on position,
for example the further away we take it,
from the distribution of mass over here the smaller it is
If we bring our test mass right up to one of the particles here,
there's a great big force on it
And we bring it on the other side the direction switches
So this A here is a field,
it depends on position,
and it's called gravitational field
Alright, there's a way to summarize this
which we have talked about last time
I tell you anything I didn't tell you about last time
but uh,
the way to summarize it
is in terms of the divergence of A
let's look at the uh,
del dot A,
that's the divergence of A,
and let's look at let's just take a simple case
The simple case let's suppose is only one massed point,
that has one mass point here,
then let's look at the gravitational field,
in the neighborhood of that mass point
Well it's pointing inward
If I take my mass if I take my test mass,
and move around in the vicinity of the of the uh, the heavy mass here,
I'll discover a gravitational field which varies from point a point,
always pointing toward that mass
So the gravitational field,
the vector field point inwards toward that point,
The pointing the inward pointing this out there,
is called the divergence of a field
The divergence of the acceleration field here,
is due to masses,
and it's proportional to the mass
At every point, there's a divergence which is
proportional to the amount of mass at that point
To express it mathematically,
we have to invent the concept called mass density
Now we all know what we mean by mass density,
it's not a new concept for most of you
Mass density simply means,
take a volume of space,
ask how much mass ,and now imagining
that uh,
either the masses continuously distributed,
that  not in the form of point particles,
or that there are so many point particles tiny tiny point particles,
though we might as well effectively think of it is being distributed
Then the amount the mass in the unit volume,
or the amount of mass per unit volume,
is called rho,
equals mass per volume
We take a small volume,
count all the mass in it,
delta M over delta V is called the mass density,
uh, the mass per unit volume
And of course a varies in general from place to place,
here there's no mass per unit volume,
here there's mass per unit volume
So the mass per unit volume,
is itself a field,
it's the density field
On the left hand side here,
we have the divergence of A
A is a vector field,
but its divergence is a scaler
The relationship between masses and gravitational field,
which is basically the relationship that Newton wrote down here it is,
can be re-expressed in the form,
that del dot A,
is equal to minus,
four Pi, I'll tell you in a moment,
I'll remind you where the 4 Pi comes from, why it is there,
times,
the mass density,
and times,
Newton's constant
The strength of gravity is always proportional to Newton's constant,
so the effect of a mass is always proportional to newton's constant
It's proportional to the amount of mass,
that's the vou
This is the basic equation,
that you can think of as a field equation,
it relates two fields,
the field of acceleration,
to the field of mass density, and the constant G is just a constant
in the relationship
This is called Gauss' law,
why is there a minus sign here incidently?
Well, the minus sign simply takes into account,
that if you have a mass point there,
the acceleration field is pointing inward
So you don't really have a divergence,
you have a convergence of the gravitational field
But a convergence is just a negative divergence
So that's why the negative sign is there
This is called Gauss' law
But there's another thing which is named after Gauss,
and it's Gauss' theorem
This is a law of nature,
this is equivalent to Newton's relationship
between the acceleration field and the mass density,
it's something that you get from experiment
or if not well from experiment, the observation
But the mathematical theorem which is Gauss' theorem,
which also has to with divergence of vectors
So let me remind you about Gauss Theorem,
it is so central to everything in uh,
and certainly gravity theory
I feel justified in spending another 15 minutes or 10 minutes on it
Gauss' Theorem says,
that if you have a field,
which has a divergence,
that divergence is a scaler,
and you take some region of space any regional space that has a boundary
could be a sphere
doesn't have to be a sphere,
it could be a sphere with two ears sticking out of it, uh
it should have the topology of a sphere,
it shouldn't be a donut,
it should not have holes in it or anything like that
even then it's OK but let's just take it to be a chunk of space like that
Then what Gauss' law says is that if you integrate
the divergence over the volume, dx dy dz, inside the region
How do you do that, what does that mean?
That means you break up the region into little cubes,
three-dimensional cubes
You break up in the tiny three-dimensional cubes, in each cube,
you take the divergence of A,
you multiply it by the volume of the little cube and you add them all up
That's integrating the divergence of A
over the interior region here
What it says? What Gauss' theorem says
is that integral is equal to a surface integral
are sum over the surface, so let's draw a piece of suface,
and also now break up the surface,
the bounding surface,
also into little squares little cells
And each cell, the field A,
has a component perpendicular to the cell it may also have the components
in the directions parallel to the surface,
but A has a direction perpendicular to the surface
Let's call that A perpendicular,
or A sticking a component of A sticking out of the surface
It says that Gauss' theorem
says that the integral of the divergence of A over the interior
is equals
to the integral over the boundary
That's usually called the sigma, a sigma stands for a little surface area
stands for little surface here
The integral d sigma
of a component of A sticking out of the surface
In other words I was just add up all the components
of A sticking out of the surface
I told you what the significance of this would be for the flow of water,
if uh, the divergence of A stood for the divergence of the velocity
field of water
They would just be the amount of water
being pumped in,
to a region here
is equal to the amount going out through the boundary
But this is Gauss' Theorem, let's just isolate Gauss' Theorem now,
the integral over a volume of the divergence of A,
is equal to the integral over the surface
on the component of A perpendicular
Now, that's Gauss' theorem
Now, let's put them together,
let's put them together to see how they work together
Let's imagine first that we have a distribution of mass
which is spherically symmetric,
that means it doesn't have any dependence on angle in space
Uh, that the uh, distribution of mass has the symmetry of the sphere
symmetry of rotational vaviance
Another word, there's another word for having symmetry
of the sphere, anybody know this?
Isotropic,
isotropic, it's the same in every direction,
about a central point here
Alright, so uh,
it could be a ball of material but it doesn't have to be a solid
ball of material,
but it should be shaped like a sphere
So we have some mass in here,
and I want to know,
what the gravitational field is
What I do is I surrounded
nicely spherically symmetrically concentrated with a sphere
I apologize for the fact that my drawings are two-dimensional,
if I have some way to make them three dimensional I would but I can't
Uh, so they're two-dimensional versions of surrounding a
spherically symmetric distribution a material
by a shell, by an imaginary mathematical shell
Can the red be seen?
Alright so there is a red shell around here
Now let's see what that says
First of all we integrate up del dot A,
but what is Del dot A?
From Gauss' law not from Gauss' Theorem but from Gauss' law,
that Del dot A is minus four Pi
times
mass density,
times G, so let's work that out
So we're gonna get minus four pi,
we're gonna get G,
and then we're going to get the integral of the mass density
over the interior of the sphere
The minus four pi G can come on the outside of the integral,
since it's only a constant
and the integral itself just gives us this integral was just,
what you see here
Now what is the integral of rho over volume?
That's the mass
Integrating up the mass density,
just gives you the mass enclosed
So there is mass enclosed within the red sphere,
Okay? so let's rewrite that, that's minus
four Pi G,
times the mass within the sphere
If the red sphere is bigger than, let's call it a sub planet,
If the red sphere is bigger than the planet,
and that is the entire mass of a planet in there,
minus four pi M G minus four pi G M is the left hand side
What about the right-hand side of the equation
The right-hand side of the equation says,
take the perpendicular component
of the acceleration of the gravitational field
Now if the mass distribution is isotropic,
the gravitational field will also be isotropic,
in other words,
the component, the perpendicular component
of the field will be the same everywhere
It'll be pointing out with the same component
everywhere is on the sphere
So that means that the perpendicular component of A doesn't vary
from point to point on the sphere,
and that means that this integral here
is just the perpendicular component, whatever happens to be
times
the integral of the d sigma, what is the integral of the d sigma means?
It is the area of the sphere,
it's just adding up all area cells,
the total area of the sphere
And that's the area of the sphere which is
four Pi R squared
R here not being the radius of the planet
but being the distance from the center of the planet to the red sphere
Okay so now you see why the four pi
was in Gauss' Law,
why Gauss put it into his law
It was in order to be able to cancel out out at this point
It's arbitrary you could have defined things differently
Now, we solve this for the gravitational field and what is it tell us?
It tells us that the gravitational field of a mass,
is minus M G over R squared
Something we already knew,
that's Newton's law
That the acceleration due to a massive object,
is the mass of the object,
times G divided by R squared
But it gives us a nice,
we talked about this last time, we are just going back over
It also tells us that it doesn't depend on
mass having all been concentrated at a point,
or depended on was the mass was spherically symmetrically distributed
Okay so that,
that's I did this again to point out to you that
Del dot A equals minus four pi rho G,
is equivalent,
to A being a sum
of all the mass points,
of newton's characteristic expression for, uh, for gravitational field
So that's cool, huh?
It's a nice way to express things,
this summarizes the very very complicated equation on top of it
It doesn't look so complicated,
but if you wrote it out for lots of mass points,
and so forth that would become complicated
Here's a way to express it where it's rather simple
Uh, while we were at it,
somebody asked me last time,
about the gravitational field within the earth,
what is it like within the earth?
Here we did gravitational field outside the earth,
we can now carry out a calculation of
what's going on inside the earth if we like,
it's kind of interesting
All we have to do now,
is take a red surface same calculation,
but now take the red surface inside the earth
But now we have to be careful,
because the amount of mass within the red circle in red sphere
is not the total mass of the planet anymore,
it's only the mass within the red sphere
So let's do the calculation over again,
the right hand side stays the same,
the right hand side is still 4 Pi,
the perpendicular component of the field comes 4 Pi times R squared
Well this R is now,
the radius of the inner sphere here
What about the mass within the red sphere,
how much masses there within the red sphere?
Well for that we have to make some kind of assumption,
about how the masses distributed throughout the red sphere
Let's make the assumption which is not so bad for the earth,
it's not a bad approximation for the earth,
that the mass density within the earth is uniform,
The mass densitie is uniform inside the earth,
and then when you get to the surface of the earth it drops
abruptly to zero
That's not exactly true it's denser
near the center and leaner near the outside
But let's take that as a model
A uniform mass per unit volume with in, uh, with in the earth
Then how much mass is inside this red region here?
That's not the total mass of the earth now,
it's different, so let's calculate it
It's the volume of the sphere,
times the mass density,
which is equal
minus
the mass within the sphere,
and mass within the sphere,
is volume of the sphere,
four thirds Pi R cubed,
that's the volume of the sphere times rho,
four thirds Pi R cubed I think I got it right
Oh, on this side G, good ,thank you
All right we can cancel out some things we can cancel out the four Pis
And then we can divide it by R squared,
if I divide by R squared when I get is
A perpendicular is equal to minus one third rho times G,
one third vou times G times R
What does this say?
This says that the acceleration of a test mass
Of course, we would have to drill a hole through the earth,
if we wanted to drop a particle,
we can't drop particles if it stuck in the rock
But if we had some sort of particle,
well either we drill a hole through the earth and drop particles down,
or we drop particles which don't interact with the material of the earth
Neutrinos don't interact much with the material of the earth
We drop a particle what is it do,
well it experiences an acceleration,
which is toward the center,
that's a minus sign here
The perpendicular component of A,
is negative that means towards the center,
it's got some constants in it,
rho G and three those are just numbers,
but it grows linearly with the distance from the center of the earth
So it's an acceleration,
which means a force,
if we multiply the acceleration by the mass of the test particle
on both sides
That I would have the mass on the test particle,
the mass on the test particle is toward the center,
and it's proportional to the distance away from the center
Uh, there's a name for such a system,
it's a harmonic oscillator
A force which is proportional to distance,
another words,
a particle displaced from the center will be pulled back to the center,
exactly as if it was on the spring
With a spring constant,
that depends on the mass density depends on the gravitational constant,
uh, and three
So if we dropped a particle if we drilled a hole through the earth,
and drop the particle,
it would just oscillate up and down, up
and down, like the harmonic oscillator ,I forget the
the time constant ,I think it takes about ten minutes
go through the earth or come back or something oo that order of magnitude
So this is not important, to the general theory of relativity,
it is just another example of how to use
Gauss' law and Gauss' Theorem
to solve what would be a very hard problem,
if we have to add up the contribution of every particle in the earth
I mean just think about it,
uh, how hard that would be
yet this uh, this method allows us to do it easily
Any questions? Student: Is there some reason why the distribution
of the matter outside the shell is negligible?
It's not eligible, zero, that's a good point, okay,
uh, alright
so let's talk about this,
That's right, that was one of the consequences are one of the curious
consequences, I'm sure gave newton some pause there when he realize it
but that's right,
it is only the material inside the sphere,
which contributes to the gravitational field
The material on the outside,
all cancels out
The gravity from here in the gravity from here in the gravity from here,
they all cancel out, Newton knew it,
and he proved it without calculus which is quite uh, the reason he
prove all these things without calculus is because none of his uh,
audience knew any calculus,
and so he have to do everything geometrically,
but we see it from Gauss' law
Gauss' law tells us among other things,
that the gravitational field of a spherically symmetric mass,
only depends on the mass within the radius R,
and doesn't depend upon all the mass outside the radius R
A consequence of this uh, I think we talked about last time,
but I might just say it again,
is that if the mass was entirely distributed on a shell,
imagine the mass was on a shell, thin shell,
then inside the thin shell there would be no gravitational field at all
Particles will move around as if there was no gravity at all,
On the other hand particles outside the shell,
would see the gravitational field of the shell exactly the same
as if there's a point at the center
So that's a rather remarkable fact,
and it's special to the one of R squared law,
noticed it tells you
Assuming Gauss' law and Gauss' theorem
you uh, get one of R square off
Student: I'm having a hard time equating those two,
Student: I figure that the surface of the earth and I got a four but ...
Uh, I may lost a four somewhere, uh, where?
Which are where?
So I go back I need to go back a step in
Student: No, the expression there,
Student: on your right, the lower there
Student: breaking to the expression on the board to your left,
Student: at the surface of the earth, those two gravitational
Student: acceleration should be the same on the surface
What should be the same as what?
Student: A on the right board should be the same as A perpendicular
Student: on the left board at the surface of earth.
Oh at the surface of the earth
Good, that I make a mistake do I lost a factor somewhere?
I may have lost a factor, is it a factor 4 that is missing?
4 Pi
From which one are
Student: When you cancel the 4 Pi of that ...
You are absolutely right it should agree on the surface of the earth
so someplace I lost a 4 Pi
this one this one of course is correct
that's just Newton
Suduent: [UNINTELLIGIBLE]
One person,
does somebody see where I have missed?
Student: You cancel the four pi at both side of the equations
Student: No it should be a function of the mass
Student: One you have vou One you have mass
So let me to go back to do with it, okay
All right here is a buzz
Well, okay, one is rho and one is mass yeah yeah yeah
That was right,
it was right,
yap yap yap yap yap it related that's right,
that's right, now
to those who didn't follow the last discussion forget it, it was right
One more concept to which is important before we get onto relativity
is the equivalence principal, the equivalence principal in particular
is the idea of gravitational potential,
just uh, one point
the gravitational field is a field of acceleration, ya?
Student: At the shell, is the gravitational field continues or
No it jumps at the shell
It jumps at the shell as you go from outside the shell to inside the shell
[Voice too low]
Student: well because the finite thickness
If the shell have finite thickness, of course that's smooth,
but the shell is literally and infinitely thin shell,
then it jumps, the gravitational field jumps
Student: The equations you had said the the force, the acceleration
Student: varies linearly from the center of the earth
Until we get to the boundary of the earth
Student: you also said that if we consider a series of shell
Student: the force on an object
Student: and the shell can be neglected once in the inside, but the mass
Student: the sub sphere whatever
Student: is not linearly depended on earth
No
that's right, that's right, we canceled out yup, okay,
so let's let's go back, let's go back
We had on the left hand side we have on the left hand side
of the equation, the gravitational field times the area,
R squared
That was one side of the equation the other side of the equation,
have the mass which you correctly say,
was proportional to R cube
Right by R square,
and we got A proportional to R
That's what we did,
the R square the um,
the area grows as r squared
the mass grows as r cubed
and therefore the gravitational field
it's like one of a R
Student: Can you derive the equation again?
Which one?
Student: The one you raised the board behind you?
This one? Okay, it is Gauss' law again
We take a sphere,
which is smaller than the size of the earth,
okay
One side of the Gauss' Law
gives us the total mass within that sphere, is that clear?
Who has a question? Ya, is that clear? Alright,
one side of the equation is the mass with and that sphere
What is that? That is four thirds Pi R cube times the density,
well we also have to throw in a G, the G, yeah
Four thirds Pi R cubed,
times rho is the mass within that sphere
Now we multiply it by G,
that's one side of the Gauss' equation here
The other side of the Gauss' equation
is the integral of the component of A sticking out from the surface,
that's equal to the surface area,
4 Pi R squared,
times A perpendicular there is a minus sign there
So we have on the left side something like R cubed,
on the right hand side R squared
I can cancel the four Pi,
and then divide it by R
By cancel four Pi I get G over three R cubed rho
is equal
to minus R squared A perpendicular
And then divided by R squared,
divided by R squared,
leaves one power of R there
So it tells me that the gravitational field is linear just like a spring
Student: On the surface of the earth
Student: you can equate that to M G divided by R squared
Student: M G and four thirds and it is four thirds Pi R cubed
Student: You missing a four pi
This one is missing a four Pi? No, no.
Student: You miss a four pi multiplying G
Student: The mass of the earth is four thirds pi R cubed
Four thirds pi R cubed and rho,
Student: Right, okay so on one side you have four thirds Pi R cubed,
Student: you have a four pi and the other side you don't
Is that good or bad? Okay.
I thought I did it carefully but maybe I made a mistake
Let's take a vote, how many think I'm right?
Let do it during the break and I will go through it with you
It is not entirely possible that I make numerical
Student: How you get to M by doing the intergal?
Student: That's wierd, you miss the cause
Yeah, so we will work it out on the detail during the break
Okay I was going to talk about gravitational potential but I think I'll
do that next week and I want to get onto the equivalence principle
Little bit of some new physics,
this of course was all old physics and take us another hour to
go through Newtonian mechanics
Just repetition of what we did last time,
but let's move on now to the equivalence principle
Now, equivalence principle of course is the principle that says
that a field of acceleration,
that acceleration and gravity are the same thing
But to make it specific,
uh, from
The elevator analogy is a good analogy
Now for reasons that are
a little strange I'm going to have my elevator accelerate horizontally
It's just because I want to use the coordinate encoded X instead of Y
And you'll see why, we will get to it
So we have an elevator,
somebody stands in the elevator
I want to emphasize that this has nothing to do with this room
It could be in any direction that the elevator is accelerating
It could be accelerating by means of a cable or whatever,
Rockets I don't care, whatever,
uh, pulling it
And it's being pulled with an acceleration
Let's call that acceleration little g
What about
an object that's dropped by the passenger within the elevator
That object,
according to Newton's laws,
in a frame of reference outside the elevator,
inertial frame of reference, watching this,
that object stands still
But what does it do relative to the person in the elevator
Well the elevator person gets accelerated that way with acceleration g
But he doesn't see he's accelerated, he just see I'm standing still,
and he sees the little thing to drop,
accelerate toward the floor,
and of course accelerate toward the floor
with exactly the acceleration g,
but not toward the right, he sees it accelerates toward the left
So he says there is a field of acceleration inside the elevator
There's A field inside the accelerator pointing downward
Everything he drops
gets accelerated downward exactly, downward meaning to the left,
exactly as if uh,
there was a gravitational field,
A field of acceleration pointing to the left
How big is it?
Well it's just exactly equal in magnitude
to the acceleration that the elevator's moving with, little g
And so he says is an acceleration field of strength g pointing toward
the floor of the elevator
Ok, looks like gravity, sounds like gravity, smells like gravity
are the conclusion that Einstein came to in this
You might say, why take Einstein to say this?
Almost any small child who's been in elevator
could have recognized it,
that gravity,
the effects of gravity,
the effects of gravity on everything that this person does,
drop a ball does whatever he wants to do,
is exactly identical to what
you'll experience if you are in the elevator being accelerated
Gravity and acceleration are the same thing according to Einstein
No, no, no, the physical effects
of being in a accelerated elevator are identical
to the effects of gravity
Now, as I said that seems almost trivial
I'll tell you, let's take a break and then come back to some
some discussion of its non-triviality
Why Einstein was called Einstein instead of uh, Joe Palooka?(Comic Icon)
All right, so let's talk a little bit about how Einstein
use this idea of gravity being the same as acceleration
We've been just say that,
but he used it, and uh
Let's go through one application of it,
and then start
to discuss uh, some ideas about the general theory of relativity
So now let me put, now let me returned to uh
Before I do that,
there was some reason why I put the elevator horizontal
It had to do with the graphs that I want to draw
Let's talk a little mathematically
about what it means to be in an accelerated frame of reference
The accelerated frame of reference was the um,
was the frame of reference, as seen
by the passenger in the elevator
This frame is his or her frame of reference
Okay, so,
x's horizontal that's why I had the elevator going horizontally,
because I want to use x
And I want to plot x along the horizontal axis
and along the vertical axis I'm going to plot time T
this is all the space-time, x, well
I haven't plot of the other directions, y and z
but time and x
and that's the world inside uh, inside and outside
Now, let's first talk about
what it means for a frame of reference to be uniform velocity,
before we get to acceleration, let's just suppose the
elevator were moving with uniform velocity not accelerated
Let's call the bottom of the accelerated floor,
let's call that
x prime equal zero in other words we have two frames of reference,
me standing on the outside
and I have a coordinate
which I measure by meter sticks
and I measure it from right over here
one meter two meters three meters four meters and so forth
All right, so that's my frame of reference,
and I call my coordinate x,
and here are the ends of the meter sticks,
meter sticks
And the first meter stick, this is x equals zero
x equals one
x equals two
and so forth
All right, now what about an observer,
somebody in the elevator,
who's moving with uniform velocity
He has his coordinates
he stacks up his meter sticks in here,
his meter sticks are not the same as my meter sticks, my meter sticks
are standing still, and my friend, he is moving in my frame,
and he will call the bottom of the floor of the elevator,
he will call that x prime equals zero
prime for passenger
So there's also an x prime coordinate
let's take x prime equal zero that's the floor of the elevator,
the floor of the elevator is moving,
and so the floor of the elevator looks like that
As time goes on, the floor or
elevator moved further and further to the right
Not only does the floor of the elevator but x' equals one follows along,
x' equal two follow along
Those were intended to be parallel, actually they are
So the moving frame of reference,
it measures distances relative to the floor of the elevator over here,
whereas a stationary frame of reference measures distances
relative to x equal zero over here
Now what's the relationship between x and x prime,
well let's take a any point over here
This is a point in space and time, it's a point of space at some time,
and let's ask what its x value is
it's x value is the distance to here
Another words it's the coordinate that the stationary observer sees
Just call it x
What about x prime?
X prime is the distance to the floor of the elevator,
the elevator is moving
So x prime is this distance
What's the difference between them
the difference between them is just the distance
that the elevator has moved, velocity of the elevator times time
That's so far the elevator has moved in the mount of time t,
it's moved with its velocity, it's moving with the uniform velocity,
v times t
So what is x prime, here is x prime over here,
x prime is equal to x minus vt
That's the relationship between the primed coordinate
and the un-primed coordinate
And this is about to used to this idea of
relating frames of reference like this
That's the relationship between the moving frame and the stationary frame
Now there's one other thing in the Newtonian physics
We are not yet doing relativity, in the Newtonian physics,
all observers,
measure the same time,
time is the same for everybody
And so we can write another equation,
that t prime is the same as t
This is Newtonian physics, this is not yet relativity,
this is not yet Einstein, this is modified Einstein
But, this is the relationship between
the coordinates measured by the moving observer
and the coordinates measured by the stationary observer
That's coordinate transformation
X prime equals x mines vt, t prime is equal to t
Now, what, let's suppose now
that there was an object which is standing still
in the outside frame of reference
This guy dropped his ball here, the ball stands
Let's say it stand still in my frame of reference just for fun
How does it move in his frame of reference?
Uh
It's standing still in my frame of reference,
what's it doing in his frame of reference?
Uh, and I'm particular interested in the velocity
All right, if a ball standing still in my frame of reference
it means x is equal to a constant
It's just standing there are still, the elevator is going by
the person in the elevator is looking at the ball,
sees its moving
Alright if x is a constant,
then x prime is of course just minus vt
but what I wanted to do is differentiate this equation
with respect to time, what dose it say?
It says that x prime dot,
I'm using dot to indicate time derivative,
standard notation we've used it before,
a dot indicates a derivative with respect to time
that's equal to x dot
But if x is constant, it's standing still in my frame of reference,
That's zero
What's the derivative of minus vt with respect to t?
Just minus v
So this was a long winded way to say
that the velocity of the ball in the,
as seen from inside the elevator is minus v
Of course it is
The object is standing still in my frame of reference,
the observer is moving pass it with velocity v,
and of course sees it's moving backward relative to him,
with velocity minus v, that's all it says
What does it say about the acceleration
as seen in the moving frame of reference
Let's differentiate one more time
Differentiating it one more time that gives us
the acceleration in the prime frame of reference
What's the derivative of v with respect to t?
If v is a constant?
Zero
So if it's not accelerating in my frame of reference,
it's not accelerating in his frame of reference
Very simple
Okay? Trivial in fact
But now let's suppose that,
the elevator is not moving with uniform velocity,
but that it's moving with uniform acceleration
Then the distance that the elevator moves is not the vt
Supposing it's moving with uniform acceleration,
a, acceleration equals little g,
acceleration due to gravity
Then, how was the floor of the elevator move relative to me?
It's accelerating like that
Well it doesn't move on a straight line, it moves on a curved line,
it moves on a parabola
it's uniformly accelerated relative to me, it's moving on a parabola,
and that parabola is x equals one half g t squared
Is this known to everybody? This is a uh,
standard formula for uniform acceleration
Yell out if you don't know it
Okay this is uh, this is the elementary
calculation of a thing moving with uniform acceleration
So the moving observers, the floor move to the right
One half gt squared, along with the floor all the other
meter sticks, the end of all the meter sticks to the same thing
They accelerate off on a curved path in space-time,
and acceleration means a curved path in space-time
So let's uh, let's see what the relation between x prime and x axis
Again the same thing,
here's x prime
here is x
What's the difference between them?
The difference between them is x minus
one half gt squared
This distance is one half gt squared
Distance from here to here
X prime is x minus gt squared
This is the new transformation,
for transforming to a accelerating frame of reference
Okay let's say again
that, the passenger drops a coin ,whatever it is
and that the coin happens to be standing still,
he throws it just such a way,
that it's standing still in my frame of reference
That again means x is a constant,
What does it say about x prime? It says that x prime
What does it say about the velocity
If x is standing still and we differentiate this we get the velocity
in the prime frame of reference
Zero from differentiating x
standing still in my frame of reference but what happens
if we differentiate minus one half gt squared?
Minus gt
But I'm interested in the acceleration
that the observer in this sees so I differentiate it again,
x prime double dot,
the acceleration seen from inside here
What I get if I differentiate once more?
Just minus g
This is a mathematical statement of the simple trivial observation
that the acceleration works the same way as gravity
The accelerated elevator gives a rise to an acceleration
for anything that you drop, which is just minus g,
just like an ordinary gravitational field of the earth
So anything which is dropped in this elevator
accelerates toward the floor,
not toward, not in direction of the elevator's acceleration,
but in the backward direction, that's the minus sign here,
and accelerates backward,
with the acceleration minus g
So that's a mathematical statement,
but what I want to bring out is that it really has to do
with using a set of curved coordinates
It has to do, or the acceleration is really the choice of a
curvilinear set of coordinates to describe spacetime
That's the real lesson here
Acceleration,
an accelerated reference frame is described by a coordinate transformation
to a set of curved coordinates
So we we need to keep that in mind,
the connection between gravity and curvature were beginning to develop
But having done that let's now come back to some simple physics
One of the things Einstein became instantly famous for
We can work it out qualitatively, it's the bending of light
Okay so we first, the first step, was to say I really mean it
when I say that the laws of physics in a gravitational field
are the same as the laws of physics in a accelerated reference frame
and particular not just the laws of Newtonian mechanics
but the laws of everything
Nobody at that time had the faintest idea,
how gravity affects electromagnetic fields,
they didn't know
In fact I don't even know the question was asked at that time, serious way
And so it wasn't known for that reason
it wasn't known how gravity affects the motion of light rays
What Einstein said is I now know how gravity should affect the motion
the motion of light rays
So here is an elevator, now I'm drawing
the elevator vertical, the vertical
elevator would accelerate upward
Let's suppose a light beam is shining from one side to the other
First supposed the elevator is not accelerating,
let's suppose the accelerator was standing,
uh, elevator was standing still
and you shine a light beam, what does it do? It gose in a straight line,
it goes right across the elevator, at this point over here
Now what happens if the elevator is accelerating upward?
Well let's send the light beam just arbitrarily, let's send the light beam
at exactly the instant when the elevator begins to accelerate
It's not yet moving, it's just beginning to accelerate,
So the light beam starts moving off horizontally,
but now the accelerate, the elevator is accelerating upward
and so where does the light beam hit the other side of the elevator?
Down here somewheres
And it travels in a curve path
In fact that curve path will be a parabola again
It will travel on a parabolic orbit
And that parabolic orbit,
you can work out easily,
just from the motion of the elevator
What do you find?
You find that the light ray has a downward component of acceleration,
which is equal to g,
the same acceleration
If the elevators is accelerated upward
with the acceleration g,
then the light ray will have a downward component
of acceleration which will also be g
So by that little argument,
nothing more than that Einstein deduced the fact that light falls in the
gravitational field
As I said this was unknown,
nobody even suspected it,
that the existence of heavy masses which create gravitational fields
will cause lights to do anything other than travel in a straight line
The fact light doesn't travel in a straight line,
where it passes through the earth,
passes across the earth,
why? Because the gravitational field of the earth can be thought of
as an upward acceleration
So, light will fall
Uh, let's go through
Student: question?
Yeah
Student: Suppose you shine a light in the direction of the acceleration?
Yeah,
and so that it's a little more complicated it still it's still uh,
it still has an effect, it effect on the energy of the light ray
Student: And will it exert a effect to speed?
No
Student: Cause it depends on
Right, the acceleration in the vertical direction will correspond to
change of energy of the, yeah
It won't affect speed
Student: one more question?
Yeah
Student: They have the experiment,
Student: to verify general theory of relativity
Student: if you take Newtonian mechanics,
Student: and assume the photon is a particle of mass, E over c squared,
Student: and just use Newton's gravity equation
Yeah, let's do it
Student: That, does general relativity predict exactly the same thing?
By a factor of two
But let's see if we can roughly worked out
the bending of light by the sun
The question is this, for light ray passes the sun,
and clearly Einstein has already proved that light falls
in a gravitational field,
So he wanted to estimate
The first thing he did before we had the full apparatus of the
general theory of relativity
He had as much as I've explained up till now,
and he said I will try to estimate how much the light bends
another words what angle does the uh, does the light ray,
get uh, accerated, what's the right word
deflected
By what angle does the light ray get deflected
Alright we can make an estimate, at this point he was reduced to
estimates, because uh, he didn't know the details of how
this spherical earth with its varying gravitational field
would really affect things
But in particular he was interested in a light ray which just skim
the surface of the sun
Got it close to the sun, why?
Because the effect would be maximal the closer the lightray got to the sun
Of course he wasn't interested in light rays
that went through the sun light rays don't go through the sun
A light ray just barely skim the sun
Uh, how much it got deflected?
Okay, let's see if we can figure it out
Uh,
first of all in passing, passing through the
when it passes close to the sun
it has an acceleration toward the sun
which is just the acceleration field of the sun
So when it's in the vicinity of the sun let's break it up,
when it's out beyond,
the distance comparable to the radius of the sun,
it's too far from the sun to feel anything
When it's within some region comparable to
twice the radius of the sun here,
it feels the gravitational pull of the sun
What's the acceleration that it feels downward?
Well for that we just have to work out the acceleration due to gravity
We'll have to work out is how much acceleration is there
at the surface of the sun
I think we already had a formula for that, right?
The acceleration is, uh, M, mass of the sun, G, over R squared
That's the
Do I have it right?
MG over R squared,
downward, alright?
Everything gets accelerated downward, light as well as everything,
and all with the same acceleration
So M G divided by R squared is the acceleration,
the vertical component of the acceleration, alright
How much velocity does the,
how much vertical component of velocity, now vetical doesn't mean
toward the sun, it means vertical on the blackboard here
How much vertical component of velocity does it have
after it passes the sun?
Well, to calculate the change in velocity,
you multiply the acceleration by time, acceleration times time is
the change in velocity
So you take this acceleration
and multiply it by the amount of time,
let's call a delta t,
that it takes for the light beam
to cross the radius of the sun
now we can estimate that because we know how fast light moves,
it moves with the speed of light
Uh,
so how long does it take? It takes time delta t,
which is approximately twice the radius,
this distance here,
divided by the velocity
That's velocity times time equals distance
So time is distance that it travels
divided by the velocity, what velocity?
Velocity of light,
the velocity of light going across here
So that's 2R divided by c
So let's plug that in
that gives us another two
Two is not important
R divided by c
Then what is that? That's twice MG over RC, alright?
that's the downward component of velocity here
But that's not the angle,
the angle is basically the ratio of the,
if we have a small angle relative to the horizontal,
here's horizontal small angle relative to the horizontal
That's the velocity,
that's the direction of the velocity after it passes the sun
Then the downward component here, the uh, the angle
is the ratio of the downward component to the horizontal component
The downward component divided by the horizontal component
is essentially the angle,
at least the angle which is small
So, the angle, Let's call it theta,
is equal to the ratio of the downward component, here it is,
two MG over RC
that's the downward component of the velocity
divide that by the horizontal component of velocity,
and that gives you the angle
Horizontal component of the velocity as one over c
and here's the answer
2MG over R C squared
That's a very small angle typically,
mainly because c is a big number,
and G is a small number
So G is a very small number,
c is a very big number, c squared is even bigger
And so that's typically a small angle,
when the uh, when the light ray go across the sun
So that was the crude estimate that Einstein made
Uh, I can't remember what the uh, what correct answers from the
general theory of relativity, it's within a factor of two of this
I think maybe it's just exactly
is the either twice or half, I can't remeber
Uh,
so every light beam that passes by the edge of the sun,
gets deflected by that much
Alright, that was Einstein's prediction, it was verified by looking at,
uh, stars passing across the sun during an eclipse
we don't need to go through that,
most of you know the story
Uh,
but the main point here
is that Einstein used the idea
that gravity and acceleration are the same,
something that was known to everybody
but he used it in the context which nobody had ever thought of,
namely he applied it to the motion of light,
and to electromagnetic phenomena in general
Okay so let's uh, now
Yes, gravity is like acceleration, but that idea is
pretty much limited to a uniform gravitational field,
uniformed vertical gravitational field
It doesn't really make sense when you try to think about it globally
in terms of the
earth's gravitational field, when you stop thinking about
flat earth approximation
So here's the earth
the affect of the earth's gravity on anything over here, on here
Einstein said is exactly the same,
as if you weren't in elevator over here, accelerating upward
with uh, with a certain acceleration
Well first of all,
you can
The acceleration has to be different in different places,
different direction in different places,
different magnitude in different places
And so there's no overall sense
in which you can't take the gravitational field of the earth
and replace it by a single accelerated frame of reference
the acceleration has to be this way over here,
this way over here, this way over here, this way over here
So there's certainly no sense in which
the full gravitational field of the earth can be replaced,
by saying you are in a accelerated reference frame
What you can say, is for a small amount of time
and over a small amount of different distance,
distances which are too small to detect the fact of the gravitational
field has different directions in different places,
another words on a small elevator,
for a small amount of time,
the effect of the earth's gravitational field is the same as
the effect of an accelerated elevator,
but only for a small amount of time, why?
Because, unless you allow the acceleration to change with time
But for a small amount of time that you can pretend that
A small amount of time and a small amount of space,
not enough space to feel the change in the direction of gravity,
you can say that gravity has the same effect as acceleration
But if you really think about it that mean you have to
think much more deeply about what the uh,
what the gravitational field of the whole around the earth is
Now this is related to something else,
if you're freely falling near the surface of the earth,
you don't feel the effect of gravity
Uh you feel you can tell the difference between free fall or not
but there's no difference between falling at the surface of the earth
and being at the middle of space with no gravitating object at all
What about somebody falling over here?
Could somebody freely falling, just freefall, can they or can they not
tell the difference between
falling and being uh, in an elevator so to speak, being uniformly
accelerating, what's the right word
Can a person tell a falling
as opposed to being in outerspace? Can he tell the difference between
gravity and acceleration? And the answer is yes
They can tell the difference because the gravitational field of the earth
is not uniform
It points in different directions and it varies with distance
So that means an observer,
here is an observer, a cubic observer,
a cubic observer has a different gravitational field on them
in different places,
the bottom of his feet is being pulled harder than the top of his head,
he was being stretched that way
Uh, the gravitational field, is sort of pinched in
toward the center here and the some sense of being squeezed this way
You know what does it call, those are tidal forces,
we talked about title forces last time
tidal forces are real, you can feel them
But in order to feel them, you have to be large enough
to sense the variation in the gravitational field
So the right statement is if you are small enough,
if you are small enough and for a small enough amount of time,
then the gravitational field, you don't feel it, and falling
and it can be completely replaced
by being in an accelerated frame of reference
Gravity and acceleration are the same
But if you big enough to feel the curvature of the earth's surface,
and if you big enough, uh, you wait long enough, enough time
and you could feel the variation of the radical component
of the gravitational field
then you can definitely tell the difference between being in outer space
and uh, being in freefall
Freefall isn't so free,
you get squished one-way stretch the other way
These are the tidal forces, the effect of tidal forces are in obstruction
to eliminating the gravitational field by replacing it
by an accelerated frame of reference
You can't really replace gravitational field
by an accelerated frame of reference
except locally and over small times
so the equivalence principal is a limited idea
that makes sense only in this limited sense
There's an obstruction to getting rid of the gravitational field
and replacing it by free fall
By uh,
by a field of acceleration
And that obstruction is tidal forces
There's no way that you can construct an accelerated frame of reference
which will completely get rid of gravitation
The way that we would if the gravitation were uniform
Gravitation uniform you go to an accelerated frame of reference
falling down with uniform acceleration,
and gravity is gone in that frame of reference
No way to do it, for the real earth
The real earth is a gravitational field which is created by mass,
and mass correspond to divergence of gravitational field
that's the divergence of the gravitational field which gets in the way
of trying to replace it by a uniform acceleration
So uniform acceleration is,
how to say it,
eliminating gravity by replacing it by a accelerated frame
of reference there's an obstruction and the obstruction is
the masses in the universe creat a divergence of the gravitational field
Uh,
I begin to get tired, so I think I should probabaly
quit any minute now
Uh,
we need to ,you need to review some special theory of relativity
Perhaps I'll pick another ten minutes and just remind you of
the things we learned about special relativity in previous courses,
and which I want you to know
Let's see
Yeah, question?
Student: Inside the elevator that's special relativity holds ...?
We of course are studying in the approximation at the moment
of the Newtonian physics
I don't think I wanted to do special relativity tonight
I think I want to do that, instead,
I want to talk about geometry a little bit
Geometry and curvature
Curvature is a concept we're gonna need,
and I think we can just uh,
discuss it without discussing physics for a little bit
How do I described the geometry of this blackboard?
The geometry of the blackboard is flat space
Just flat two dimensional space
Uh, how do I described its geometry
The way a mathematician would describe its geometry
is by specifying the distance between any pair of neighboring points
If you know the distance between any pair of neighboring points
you can rebuild the entire geometry from that
Knowing the distance between any pair of neighboring points nearby points
is enough to rebuild the geometry of this black board,
and to discover that it's flat
Let's take a pair points,
let's characterize them by coordinates, x and y
or just let's called x sub i, i runs from 1 to 2
and here's another point at position x plus dx and y plus dy
two points separated by a differential distance dx and dy
What's the different, what's the distance between them?
The distance between them is just dx squared plus dy squared
Oh that's the square of the distance between them
The square of the distance between them, let's call it ds squared
is dx squared plus dy squared,
that's just Pythagoras theorem,
dx dy ds
Ds squared is dx squared plus dy squared
Now that is something which is true
if you use for your coordinates, nice rectangular Cartesian coordinates
What would happen,
if I've replaced the Cartesian coordinates on the blackboard
by some crazy set of coordinates, coordinates for which
the axises will curve in some highly complicated kind of way?
There are lots of different coordinate systems
that you can use to describe the blackboard
They don't have to all be drawn as a rectangular grid
We can use polar coordinates, polar coordinates look like this
All kinds of other coordinates,
and in general, the lines of constant coordinate
lines of constant x and lines of constant y
would be curves on the blackboard,
that would be a general description of the blackboard,
not tied to nice rectangular mesh like this
Alright, so let's suppose now this could be x equals one, x equals two
x equals three
Here is y, and here's x euqals zero,
here's y equals one, y equals two, y equals three
I have a grid
if I tell you the value of x and y
you can tell me exactly where the point is on the blackboard
but the grid is no longer a rectangular grid
the distance between two points is no longer given by dx squared plus
dy squared
There's a general formula for any kind of curve coordinates like this,
looks like this,
that's ds squared,
has some coefficients,
whatever it is,
it will involve these little differential displacements,
and involve them quadratically,
for the second powers, we're going to go over this again,
I want to lay  it out for you now, we're gonna to improve it
But it's going to be some quadratic form
This is called a quadratic form, dx squared, dx times dy and dy squared
It'll have all of these things in it,
and furthermore these gs will depend on where you are
They'll be numbers, they'll be numbers, but they'll really be fields,
they'll depend on x and y
I'm really getting tired so I should probably quit in next five minutes
I don't know how many of you have seen this formula,
how many haven't seen it? How many have not?
Alright we are going to work it out next time
and I'll show you why this is true
But before I do, let's just go through what some of these things mean
What does it mean to have a g12 here? In the original formula you have
nothing multiplying dx times dy
This is dx squared, this is dy squared
I say more generally, if you think about general sets of coordinates
with curvilinear coordinates
then there will be terms of dx times dy, you know what dose it mean?
When it has terms of dx times dy,
it means the coordinates axises are not perpendicular
That's what that means
What about g11 and g22?
Those have to do with the relatives spread
If we are to make coordinates like this,
which were much denser in one direction than the other direction
then the g11 and g22 would be different than each other
If the coordinates curve and things change from place to place,
then these gs depend on position
But the general formula for general curve coordinates
for the distance between two points looks something like this,
and it contains three functions of position
g11 g12 and g22
those functions are called the metric, we're going to go through them
I'm gonna show you why that's true
But we're still just describing the blackboard,
We're describing blackboard, and the blackboard could always be described
by these coordinates here
When it's true that you can describe the geometry
in the simple way like this, the geometry is called flat
Even if you decide to describe it in a more complicated way,
if you can find new coordinates that simplify in this form,
it's called flat
This is the hall mark of flat geometry, Pathagoras theorem if you like
There are geometries which are not flat,
there are geometry that no matter
how do you choose the coordinates,
you cannot reduce it to this form
Those geometry are called curve,
now a example of curve geometry,
the surface of the sphere
The surface of the sphere,
you cannot choose coordinates on the surface of the sphere,
which makes the distance between points dx squared plus dy squared,
it's always going to be more complicated than that
no matter how you choose the coordinates
What about the surface of the cylinder?
Is surface of the cylinder curved?
No, it's not, it's flat
Because you can always chose exactly the same coordinates
that you would choose if you unroll it
You can't unroll a sphere and spread it out flat on the surface,
but you can spread out the cylinder make it flat on the surface
Cylinder is flat, a sphere is not
How about uh, let's see what other kind of uh,
a saddle surface, a saddle surface？
Shaped like a saddle is also not flat,it can't be flattened out
uh, cone,
what about a cone?
A cone is flat everywhere except that the tip
A cone is flat everywhere except the tip
because you can always take a cone
Here is a cone, right? Here is a cone,
at the tip, something funny is going on
But anywheres else you could unwrap it and spread out on the table,
so you can use coordinates on a cone, except for the tip
which look like this, cones are not flat but only at the tip
Cylinders are flat, plane is flat, spheres are not flat
How do you tell, what's,
whether a surface or given metric like this
describe something flat or not?
Well, the answer is curvature
which I haven't defined, we're going to define it
The answer is curvature
But curvature is the obstruction to flattening out the coordinates
Tidal forces are the obstruction
to getting rid of the gravitational field
by changing coordinates uh, to accelerated reference frames
That connection is not accidental
Curvature is the obstruction to flattening a surface out
or to thinking of it as flat
Tidal forces are the obstruction
to removing a gravitational field
by coordinate transformation to an accelerated frame of reference
Uh, that's the connection we want to build up
The curvature of space and time is
essentially the same thing as the effect of a real gravitational field,
a real gravitational field are tidal forces
And that's we are going to find
A curvature in space-time is the same as tidal forces
Okay let's quit now before I fall over uh, exhaustion
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