Hello.
I want to talk a little bit about cosmology
and the shell theorem.
The universe is isotropic and homogeneous.
That's called the cosmological principle.
That means, if I look out at the universe
from this point-of-view, I could envision
that the universe is composed of uniform density
shells.
These are concentric shells, and they continue
out for infinity, away from me, the observer.
On the other hand, if I have an object out
here, it also has the experience of seeing
the universe as composed of uniform density
shells, concentric shells, going out for infinity.
So what is your immediate reaction to this?
If I asked what is the force on this particle,
and what is the force on this particle.
Obviously since everybody sees an equivalent
bunch of symmetric concentric shells, the
force on each observer in this system is going
to be zero.
I would say that was "obviously" except that
a large number of really well-trained cosmologists
disagree with this.
Okay, they say they disagree with this.
I think I can actually prove it.
We can integrate but instead of integrating
with the shells, which could lead to some
kind of possible error involving partial sums,
I'm going to use concentric cylinders.
But first, let me talk about what could be
an error involving partial sums.
The partial sum in question here is a partial
sum of the variety of an infinite sum as n
equals negative infinity to infinity of negative
one to the nth power, and it looks a little
bit something like...
It looks a little something like this.
negative one plus one minus one plus one,
and um, each of these minus one's and plus
one's cancel out, so that given any point
on the system you've got a zero net force,
because you've got an equal amount of matter
pulling either direction.
Now, I would like to kind of argue that the
sum of (-1)^n is basically zero.
There is argument about that in some places.
I've read arguments that if we call this "s"
or no... sorry, they define it as s=sum(-1)^n
of n equals zero to infinity, and they actually
end up saying that 1 minus "s" equals "s"
and then, solving that to get 1 equals 2s
and they come up with s equals one-half, which..
you know, you're working with infinities...
But my point is going to be a different nature,
that when you sum up together negative one
this sum can be anything you want... anything
you want!
And how is that done?
Well, let's say you ask this guy how much
force is on the rope?
What he's going to do is he's going to say
I see me and four other guys in front of me
pulling on that dot over there.
So that's minus five men's force or probably
about 500 pounds pulling to the left.
And then, he's going to say that this guy
and this guy are going to cancel out, and
this guy and this guy are going to cancel
out, and that guy and that guy are going to
cancel out.
And so we can get an answer of a force of
five-hundred pounds to the left on that guy
right there.
And similarly If you ask this guy, he's going
to see himself and these two other guys contributing
300 pounds to the right on that object.
But he'll say, everybody else, forward and
backward is going to cancel their forces around
him, so he'll say there's 300 pounds to the
right on that object.
Now of course in the situation I've drawn
here, you're going to have basically an infinite
tension in the rope, um, probably everywhere,
and somewhere the rope is going to break,
if not on a fairly regular basis.
But, if you sum up the forces on this point,
does it really make sense, just to pick an
arbitrary---for each person just to decide,
I'm going to count the people between myself
and that point, or would it make more sense
for them to try to imagine the situation from
the perspective of that point, to calculate
the forces?
So I'm going to kind of invoke a principle
of "Force Feeling".
I'm not sure if I have a proof for this(?)
but I think that in order to calculate the
force FELT by an object, you must do it from
the perspective of that object.
In other words, don't use a method that leads
to asymmetric partial sums.
Or more rationally, IF you use...
You only need to do this...
Let me erase that.
You only need to do this if you're using a
method that leads to asymmetric partial sums.
So now I want to show how Newton's shell theorem
leads to asymmetric partial sums of an infinite
series.
Here's the basic idea: We have an observer
right here.
And then we have the point around which we
want to calculate the force over here, and
so what we're going to do is imagine the shells
around the observer.
Shel here; shell here, shell here, and I'll
keep drawing shells out until I get out to
there.
And basically, we say, this guy all of these
shells contribute a force to the left.
Now, after that, (You can do integrals to
show that-that all of those contribute to
the left) but then after that, the next shell
around it.
I drew this too big to get it all on the map,
so I'll just kinda draw it like that.
Well, there is, this mass of the shell pulls
to the right, but the mass over here, pulls
to the left, and it just happens that the
symmetry of this is interesting, and it involves
an integral, if I make the big radius R, and
this radius little r from here, it involves
an integral from...
Well we'll (incorrectly labeled) call this
L, and it involves and integral of L going
from capital R minus little r, to let's call
this 
square root of Big R squared plus little r
squared, and then we start over here and this
exactly cancels out with the integral from
L equals square root of Big R squared plus
little r squared to Big R plus Little r.
I don't know exactly how to write that, but
anyway, the force to the right ends up canceling
exactly with the force to the left, and that
happens with all of those spheres, so by doing
it this way, we basically get minus one, plus
minus one, plus minus one so a bunch of forces
contributing to the left, and then you start
getting those cancelings.
Let me just make sure that that is... that
was a dash, not a minus sign. and that was
a dash, not a minus sign.
Anyway...
So obviously Newton's shell method leads to
this problem of asymmetric partial sums.
But we can do the integral using another geometry,
such as an infinite series of concentric cylinders.
Here,
I envision a situation where the mass, m is
separated from the observer, by a radius,
little r.
I also envision an infinitely long cylinder
with density sigma, aligned with its center
along the axis between the observer and the
mass.
The whole mass of the cylinder is chopped
up into tiny differential masses, called dM,
Each differential mass creates a differential
gravitational field at the central mass, according
to the universal gravitational constant, G,
times that differential mass, divided by the
square of the distance between the point mass
and the differential mass.
So I'll call that L, the distance from m to
the differential mass.
The differential mass of the system is going
to be equal to the area-density of the cylinder,
sigma, times the differential area element
around the cylinder.
Now we can use a symmetry argument to say
that all of the x and y components of the
force on this mass are going to cancel out,
so that all we're left with is a force in
the z-direction, so I'm going to say dg sub
z, dF sub z over m, and I'll put a note there...
dot-dot-dot, and what I'm going to do to get
that F sub z component is multiply that total
force from the differential mass (dM) by sine
of theta, here, which is the angle, which
is (the length) no the proportion.. no the
component of this triangle in the direction
of z.
So I'll just copy those z's down, and the
sine theta down.
Now we can talk about a differential area,
that is going say from here..
No, I should put it up above...
We'll start here, where I've labeled the dM,
but instead of the dM being just a tiny mass
at that tiny point, I'm going to have the
differential mass go all the way around the
thing, since we've already invoked this symmetry,
and that dM--that dA--sigma dA.. that sigma...
well, it's just going to stay there.
the dA is going to be 2 pi r, which is the
circumference of the circle times dz, which
is the tiny differential distance around that
cylinder.
Now we can replace that sine of theta with
the opposite (z-r) divided by the hypotenuse
(L) and we will have an integral we can work
with.
Pull all of the constants out.
The universal gravitational constant is a
constant.
And the radius in this situation is the radius
of this cylinder... it's a constant throughout
this context.
Sigma is the density of that cylinder, which
will be a constant, and we will apply an integral
of (z-r) dz over square root of (z-r) squared
plus R squared, and that integral comes out
to be ZERO.
And notice that I've got rid of the problem
that comes from using the sphere method--the
Newton's shell method.
There was a problem with partial sums before,
but now we're taking care of that partial
sum...
There's no partial sum left because we're
going all the way from negative infinity to
infinity here.
And then we'll notice that since this infinite
cylinder contributes nothing to force on this
mass we can see that none of the cylinders...
because we're going to make an infinite number
of concentric cylinders around here to do
the calculation, and we will find that it
always comes out to be zero, and so the total
sum is going to be zero, and you can see that
since this force is zero, it has no dependence
on what the mass is.
or who the observer is who is looking.
It's always going to be zero.
There's NO force on that mass due to all of
those infinite uniform density shells.
It always comes out to be zero.
But if you prefer to do it with partial sums,
instead of drawing those cylinders, you'll
draw spheres around this mass and you will
find that these spheres, done this way, all
of these spheres will contribute a certain
amount of force on that mass and then you
keep going outward and none of these spheres...
the amount of force in that part exactly cancels
out with the amount of force in this direction.
So that one would be zero.
So you'd get a net total of a force in that
direction toward the observer.
So if you agree with my principle of "Force
Feeling" in order to calculate the force felt
by an object, you should do it from the perspective
of that object if you're using any method
that leads to asymetric partial sums around
the distant mass, but if you think that you
should do that, then you should have a look
at "Relativity, Gravitation, and World Structure"
by Edward Arthur Milne, because most people
criticize Milne's model because they claim
it describes an empty universe.
That's NOT what Milne's model does.
It describes a model where all of the forces
cancel out, and you get zero... like this...
Now this whole argument is described on the
essence that the amount of material in the
universe is actually infinite, and that's
the awy Arthur Milne described it, too.
He said that the amount of mass in the universe
is infinite.
If we assumed that the universe was not infinite
then there actually would be a center of mass
of the universe, and you could distinctly
say that one of these guys is getting the
correct result, because the matter IS symmetric
around him, and this guy might be wrong, because
he's counting matter in the universe over
here, that doesn't exist.
Or vice-versa.
So, if the amount of matter in the universe
is finite, then the equations of modern cosmology
actually become fairly reasonable, except
that you got to choose the correct center.
So last week, when I was writing this article
about hypothesis vs Impression I said that
my current impression of the standard model
was that the universe was a finite uniform
density sphere out to a certain radius, and
is empty beyond that radius.
The reason that I assumed that was because
the math works out if you do assume that.
but if you assume that the actual radius is
infinite, and it is not empty beyond any radius,
then all you're doing is creating this artificial
asymmetry, and all of these results are different,
wherever there is an incomplete partial sum
of an infinite series.
