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Volume of free-falling dust
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Greg Egan
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Joined: 01 May 2005
Posts: 75

PostPosted: Sat Apr 22, 2006 1:49 pm    Post subject: Volume of free-falling dust Reply with quote

John Baez and others have discussed on this newsgroup the fact that
Einstein's equation can be nicely encapsulated in the relationship
between the second time derivative of the volume of a small cloud of
initially co-moving, free-falling test particles and the energy-momentum
tensor.

The vacuum version of this says simply that the second time derivative of
the volume of the cloud of test particles is initially zero.

Now, this fact holds true in Newtonian gravity as well, and it's easy to
use it to derive the inverse square law for acceleration. If you call
the acceleration due to gravity a(r), then imagine an infinitesimal unit
cube of dust with its z-axis aligned with the radial direction, the two
transverse dimensions of the cube as it falls will both simply be
proportional to r:

delta_x = delta_y = r/r0

where r0 is the distance of the cube from the point mass at t=0, and so:

d^2 (delta_x) / dt^2
= d^2 (delta_y) / dt^2
= a(r)/r0

The second rate of change of the radially-aligned dimension of the cube
will be the rate of change of a(r) with r:

d^2 (delta_z) /dt^2 = a'(r)

So for the second rate of change of the volume of this cube to be zero at
t=0, r=r0, we have:

a'(r0) + 2a(r0)/r0 = 0

which is solved by a(r) = -C/r^2. (It takes a bit more work, but the
Schwarzschild metric can also be derived from the same approach.)

Now, all of this naturally leads one to ask what happens to the volume of
an element of free-falling dust over a *finite* time. We know that to
second order in t when starting from rest the volume is constant, but it
turns out that in the particular scenario that's been described -- free
fall towards a point mass -- the volume *decreases* over time, tending to
zero as it approaches the point mass. The precise relationship is:

v(q)/v0 =

(1/4)(q^2) *
[6 - 2q + 3 sqrt(1/q-1) arctan( sqrt(q(1-q)) / (q-1/2) )]

where q=r/r0, and the range of arctan is chosen to lie from 0 to pi.

This leads to the observation that it's actually impossible to propose
any "alternative" law of attraction such that the volume of a cloud of
test particles would remain constant while falling towards a point mass.
Having declared the initial second derivative zero, we're stuck with
Newton's inverse square law, from which the above formula necessarily
follows. It's mathematically impossible to have a force law that
preserves the volume!

This seems quite counter-intuitive to me, obvious as it is in retrospect.
My gut feeling would have been that somehow you could tweak a(r) so that
the tidal stretching continued to balance the radial narowing. My first
instinct was to say: couldn't you just drop an incompressible fluid down
a funnel, and then declare that its observed motion obeyed the new law of
gravity? But of course there is no universal "law of gravity" for the
elements of that fluid, because the pressure gradient modifies the motion
of each element differently.

Greg Egan

Email address (remove name of animal and add standard punctuation):
gregegan netspace zebra net au
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Greg Egan
science forum addict


Joined: 01 May 2005
Posts: 75

PostPosted: Fri Apr 28, 2006 4:35 am    Post subject: Re: Volume of free-falling dust Reply with quote

In a previous post, I expressed surprise that it was impossible to
"tweak" Newton's law of universal gravity in any straightforward way in
order to ensure that the volume of a cloud of dust free-falling directly
towards a point mass was strictly constant. In reality, the volume is
strictly decreasing, though it is constant to second (actually third)
order in time, starting from rest.

I went away and pondered Raychaudhuri's equation, which implies a very
general association between shear (change of shape without rotation or
volume change) and contraction. Certainly Raychaudhuri's equation makes
it easy to prove that the volume must decrease in the scenario I've
described, but the question remains as to why there is this association
between change of shape and *reduction* of volume.

The simplest way I've been able to come up with to make this association
seem reasonable is to look at the volume of an element in a dust cloud
whose dimensions are changing strictly linearly. The volume in this case
is:

V(t) = (1+at)(1+bt)(1+ct)
= 1 + (a+b+c)t + (ab+ac+bc)t^2 + abc t^3

We can make the volume constant to first order in t by setting

c = -(a+b)

which yields

V(t) = 1 - (a^2+ab+b^2) t^2 - ab(a+b) t^3

= 1 - (1/4)[(a-b)^2 + 3(a+b)^2] t^2 - ab(a+b) t^3

which is strictly decreasing to second order in t unless a=b=c=0.
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John Baez
science forum Guru Wannabe


Joined: 01 May 2005
Posts: 220

PostPosted: Fri Jun 16, 2006 10:36 pm    Post subject: Re: Volume of free-falling dust Reply with quote

In article <20060422122230.C37CC4A5F7@mail.netspace.net.au>,
Greg Egan <gregegan@netspace.zebra.net.au> wrote:

Quote:
John Baez and others have discussed on this newsgroup the fact that
Einstein's equation can be nicely encapsulated in the relationship
between the second time derivative of the volume of a small cloud of
initially co-moving, free-falling test particles and the energy-momentum
tensor.

Yes, popularizing this is my one contribution to general relativity.

Quote:
Now, this fact holds true in Newtonian gravity as well, and it's easy to
use it to derive the inverse square law for acceleration.

Nice derivation! Recently I was challenged to do just this, and being
lazy I handed the challenge over to Ted Bunn, who did it very nicely
here:

http://math.ucr.edu/home/baez/einstein/node6a.html

Unlike you, he had the extra constraint of trying to use the absolute
bare minimum of math, so the derivation could fit nicely in our paper
"The Meaning of Einstein's equations".

I'm not sure his derivation looks vastly simpler than yours, but if
one squeezes his derivation to its essence, it's quite pretty:

To keep things simple, consider a homogeneous planet made of some
rock of constant density D.

Imagine a big ball of little balls of test particles surrounding and
indeed penetrating right into the planet. Start them out at rest:

http://math.ucr.edu/home/baez/einstein/inversesquare1.gif

Then let them fall. The test particles only feel gravity, so
they don't mind falling right through the planet.

The balls of test particles outside the planet (in red) will fall
without changing volume. The balls inside the planet (in green)
will start shrinking according to Einstein's equation:

v''(t)/v(t) = -D/2 at t = 0 (in units where c = 8 pi G = 1)

where v(t) is the volume of any little ball inside the planet.

So, the volume of the big ball will start shrinking at a rate proportional
to the percentage of little balls that are inside the planet.

If you think about it, this means its radius R will start shrinking
at a rate proportional to 1/R^2. Voila!

If someone doesn't follow the last paragraph, they can either do the
math or try this analogy:

Imagine you are sucking air at a constant rate out of a balloon.
The volume of the balloon shrinks at a constant rate. But its
surface area is proportional to R^2. So, its radius must shrink
at a rate proportional to 1/R^2.

Voila again!

You can make this analogy quite vivid by imagining the little green
sphere inside the planet as balloons that are shrinking because the
air is getting sucked out of them:

http://math.ucr.edu/home/baez/einstein/inversesquare2.gif

You may object that in this balloon analogy we're talking about the
*first* time derivative of the volume and radius, while in the gravity
example we're talking about the *second* time derivative. That's true!
But, it doesn't affect the calculation - as the more detailed math
worked out by Ted Bunn shows:

http://math.ucr.edu/home/baez/einstein/node6a.html

If you want to improve the analogy, imagine that you began with
a big balloon full of air and are just *starting* to suck air out
of it, so it's not V'(t) but V''(t) that you are controlling. The
second time derivative of its radius will still be proportional to 1/R^2.

In short:

Gravity doesn't suck - but it always acts like it's *starting* to suck.

Quote:
(It takes a bit more work, but the
Schwarzschild metric can also be derived from the same approach.)

Cool. We didn't go that far.

Quote:
Now, all of this naturally leads one to ask what happens to the volume of
an element of free-falling dust over a *finite* time.

Yes... I've always been scared of that question.

Quote:
This leads to the observation that it's actually impossible to propose
any "alternative" law of attraction such that the volume of a cloud of
test particles would remain constant while falling towards a point mass.
Having declared the initial second derivative zero, we're stuck with
Newton's inverse square law, from which the above formula necessarily
follows. It's mathematically impossible to have a force law that
preserves the volume!

This seems quite counter-intuitive to me, obvious as it is in retrospect.

Yes, is probably part of why I was scared of this question! I'm glad
you tackled it.

I was looking for your posts on 3d gravity and got distracted by this....
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robert bristow-johnson
science forum Guru Wannabe


Joined: 02 May 2005
Posts: 105

PostPosted: Sun Jun 18, 2006 12:19 pm    Post subject: Re: Volume of free-falling dust Reply with quote

John Baez wrote:
Quote:
In article <20060422122230.C37CC4A5F7@mail.netspace.net.au>,
Greg Egan <gregegan@netspace.zebra.net.au> wrote:

John Baez and others have discussed on this newsgroup the fact that
Einstein's equation can be nicely encapsulated in the relationship
between the second time derivative of the volume of a small cloud of
initially co-moving, free-falling test particles and the energy-momentum
tensor.

the Baez-Bunn equation is something a bone-head electrical engineer
like me could understand. i don't know tensor calculus.

...

Quote:
To keep things simple, consider a homogeneous planet made of some
rock of constant density D.

Imagine a big ball of little balls of test particles surrounding and
indeed penetrating right into the planet. Start them out at rest:

http://math.ucr.edu/home/baez/einstein/inversesquare1.gif

Then let them fall. The test particles only feel gravity, so
they don't mind falling right through the planet.

The balls of test particles outside the planet (in red) will fall
without changing volume. The balls inside the planet (in green)
will start shrinking according to Einstein's equation:

v''(t)/v(t) = -D/2 at t = 0 (in units where c = 8 pi G = 1)

this illustrates my wonder why 8 pi G = 1 is considered to be more
natural units than those where 4 pi G = 1, because then you would say

v''(t)/v(t) = -D at t = 0 (in units where c = 1
and 4 pi G = 1)

isn't that more natural to say the Baez/Bunn counterpart to Einstein's
GR equations.

i really think the most natural expression of the Einstein GR equation
is

G_{mu,nu} = 2*T_{mu,nu}

i think that "2" naturally belongs there and that Daryl McCullough in
2000 pointed out its conceptual source:

http://groups.google.com/group/sci.physics.research/browse_frm/thread/f7244928ec125ca7/a0ec7e6c84fc8499

at the risk of getting some of you to dismiss this as "taste", i really
think that the most natural units to do this (along with other physics
that rightly normalize other constants of free space) are those that
make

c = 1
hbar = 1
4 pi G = 1
epsilon_0 = 1 (which, along with c=1, makes mu_0 and Z_0 both
unity, which is really natural)

and then, to repeat pet theory, the fine-structure constant can be
thought of as taking on the value that it does because of the amount of
charge (measured in these units that set epsilon_0 = mu_0 = Z_0 = c =
1) that nature has, for some reason that would be nice to explore if i
were as smart as physicists, bestowed upon charged particles like
electrons, protons, and positrons. that amount of charge, in natural
units, is sqrt(4 pi alpha) = 0.30282212, which is the number physicists
should put on their walls.

sorry to veer off-topic a little, i just don't get why 8 pi G = 1 is
more natural than 4 pi G (and i know it's in the Einstein Eq.).

r b-j
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