Riemann Awakes
Oz
<References: <4gb15v$j...@pipe12.nyc.pipeline.com>
<4gggi6$2...@guitar.ucr.edu> <4gj3sc$7...@agate.berkeley.edu>
The wizard laughed. "Yes, you want INSIGHT. But are you prepared to
pay the price? You want to understand the laws of the universe --- but
would you know yourself any longer if you did?" The wizard exploded
into light, Oz staggered over, and everything went black.
Oz awoke to find himself in his cave, lying on the straw-filled bunk.
He didn't remember going there... he had a splitting headache... the
last thing he remembered was opening a book on something or other in 26
dimensions, and the wizard catching him. What a fool! He propped
himself up, and it felt as if his head was about to burst. He looked
around and saw a piece of paper lying on the ground near his bed, with
glowing writing on it, which faded and dissolved as he read:
"Dear Oz --
If you wish to learn more general relativity I am afraid you will need
to pass a test of your valor. So: answer me the following questions:
1. Explain why, when the energy density within a region of space is
sufficiently large, a black hole must form, no matter how much the
pressure of whatever substance lying within that region attempts to
resist the collapse.
2. Explain how, in the standard big bang model, where the universe is
homogeneous and isotropic --- let us assume it is filled with some fluid
(e.g. a gas) --- the curvature of spacetime at any point may is
determined at each point.
3. In the big bang model, what happens to the Ricci tensor as you go
back in past all the way to the moment of creation.
I have taught you enough to answer these questions on your own. Slip
the answers under my door --- I'm busy.
Oz fell back into bed with a groan.
"Absolutely typical." though Oz
"The deal was that I sweep the floors, and the Wiz answers the
questions. Now I sweep the floors AND answer the questions."
Oz muttered dark thoughts under his breath and kicked angrily at his
bunk, which promptly collapsed in a cloud of dust. Oz looked at it
gloomily.
"Trouble is that all the Wiz's notes fade away after a week or two so he
can reuse the paper. Hmmmm. OK, so I can go and look at my Hard Drawn
Descriptions (or HDD's) that I carved on the rocks outside, but it's
blowing a blizzard. In any case some have been taken by the trolls who
used the rocks for Gremlin Pancake Fodder (or GPF's) so they aren't what
you could call complete." muttered Oz despairingly to himself.
"Oh, well, I suppose I had better try" he thought, and walked over to
the cavern's mouth. The snow was blowing a force 12 and the temperature
was 250K. No way was Oz going out there! Oz returned glumly to his hole
in the rock and started to think. This unexpected mental activity was
not nice. In fact the pain was pretty bad, but he tried hard. Synapses
that had been asleep for years complained bitterly about being woken up
by a sharp electric shock, and out of condition neurons puffed around
frantically trying to make sense of the unfamiliar impulses.
OK, so why must a black hole always form when the energy density within
a region of space becomes large enough. Hey, this is practical stuff. We
haven't done any practical stuff! The bast**d! Now if we had seen some
worked out expressions for the Schwarzchild metric, then it would have
been much easier. It would also have been nice if someone had given some
relationship between force and curvature, no matter, doubtless it will
come up in due time.
Well, Oz knew the old fashioned reason, where photons leaving a
spherical shell would be redshifted to zero frequency at infinity, and a
few other explanations too, but a curvature description? Well, obviously
ALL geodisics from the spherical shell must end up back on the shell.
No, that's not really what's being asked. Force, at least locally, is
still a relativistic f=ma. So obviously the force required to hold the
shell at constant r must become infinite or at least strange. Presumably
the curvature could become so great that ... ahhh ... could it be that
spacetime is so curved that there are no timelike paths away from the
point to 'outside'. In which case the only way to go is 'down', or at
least 'up' becomes 'down'. Well, a useful guess I hope.
Oz despondantly went out to do the only thing he knew well. Picking up
his broom he started to sweep the floor.
Suddenly, with an embarrassed 'pop', a small sheaf of papers
materialised out of thin air and fluttered to the floor. "Freshman GR,
lecture #1" was at the top of every sheet.
"Typical Wiz." thought Oz "Questions first, crib sheet after. Humpf!"
Oz gathered them all up and went off to his hole in the rock and started
to read them.
Right at the end he found something.
Wiz said: R_{00} = T_{00} + (1/2) T^c_c
and later: T^c_c = -T_{00} + T_{11} + T_{22} + T_{33}
to give: R_{00} = (1/2) T_{00} + T_{11} + T_{22} + T_{33}
which didn't look right, Oz thought it should be
R_{00} = (1/2) [T_{00} + T_{11} + T_{22} + T_{33}]
Oz wondered what a curvature of 1 would be. He rather guessed that 1 (or
maybe 2) meant that a vector in the R_{00} direction would come back as
having no component in the R_{0} direction. More than 1 would have it
pointing 'backwards'. So if T_{00} is momentum in the time direction and
is equivalent to energy which is equivalent to mass then once the mass
equivalent exceeded 2 then R_{00}=1 whatever T_{xx} was and spacetime
would be curved back on itself. Indeed one might guess that if we were
talking about a solid (or gaseous) body then the pressure (T_{xx}) might
have to somehow equal T_{00} anyway to stay 'in equilibrium' so the mass
density would probably only need to reach +1/2 to cause spacetime to
curve back on itself.
A quick bit of algebra putting in useful constants actually gave:
R_{00}= 1/2[(16pi.k/c^2-1)T_{00} + T_{11} + T_{22} + T_{33}]
which would mean that .....
SQUARK!!!
[A large and fearsome troll stormed into Oz's cave and said:
"We are supposed to be going away for the weekend and you have to take
the dogs for a walk, pack, get ready and GO!. You have to move NOW!!!
Oz left hurriedly. Determined to do some more on Monday.]
-------------------------------
'Oz "When I knew little, all was certain. The more I learnt,
the less sure I was. Is this the uncertainty principle?"
Brian J Flanagan
Thank you very much for your recent (unusually amusing) bit of whimsy.
You know the one I mean.
Warm regards,
BJ
Oz
On a quick look through he has a couple of questions:
Baez:
7. The RICCI TENSOR. The matrix g_{ab} is invertible
and we write its inverse as g^{ab}. We use this to cook up some tensors
.....
Okay, starting from the Riemann tensor, which has components R^a_{bcd},
we now define the RICCI TENSOR to have components
R_{bd} = R^c_{bcd}
where as usual we sum over the repeated index c.
The physical significance of the Ricci tensor is best explained by an
example. So, suppose an astronaut taking a space walk accidentally
spills a can of ground coffee.
.....
More precisely, the second time derivative of the volume of this little
ball is approximately
R_{ab} v^a v^b
times the original volume of the ball. This approximation becomes
better and better in the limit as the ball gets smaller and smaller.
Oz******************************************
So are we take it that 'smaller and smaller' means that contributions
from the Weyl Tensor becomes even more smaller and smaller. Or/and are
we to take it that local spacetime becomes flatter and flatter in the
absence of T so that we only need consider curvature due to T. Hmm, I
suppose that doesn't make sense. Perhaps I mean that spacetime can be
considered flat at a reasonable distance away from our coffee grain of
momentum density. Well, what are you neglecting by taking a smaller and
smaller volume?
********************************************
Baez: ......
geodesics. Remember, if v is the velocity vector of the particle in the
middle of a little ball of initially comoving test particles in free
fall, and the ball starts out having volume V, the second time
derivative of the volume of the ball is
R_{ab} v^a v^b times V.
Oz******************************************
I expect the above is obvious. Offhand, I can't see the derivation.
********************************************
Baez:
If we know the above quantity for all velocities v (even all
timelike velocities, which are the physically achievable ones), we can
reconstruct the Ricci tensor R_{ab}. But we might as well work in the
local rest frame of the particle in the middle of the little ball, and
use coordinates that make things look just like Minkowski spacetime
right near that point. Then
g_{ab} = -1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
and v^a = 1
0
0
0
So then --- here's a good little computation for you budding tensor
jocks --- we get
R_{ab} v^a v^b = -R_{00}
Oz*************************************************
No, I don't see this. v(0)=1, so the only non zero term should be
R_{00}.v(0).v(0) = R_{00} not -R_{00}
It would appear that g_{ab} ought to make an appearance here, but I
don't see where. Help!
**************************************************
Baez:
So in this coordinate system we can say the 2nd time derivative of the
volume of the little ball of test particles is just -R_{00}.
Oz
3. In the big bang model, what happens to the Ricci tensor as you go
back in past all the way to the moment of creation.
This has just GOTTA be a trick question.
Well, the stress-energy tensor has just got to go towards infinity,
surely. Both the mass and momentum density must zoom to infinity.
For minkowski space time Baez derived:
R_{00} = 1/2(T_{00} + T_{11} + T_{22} + T_{33})
The curvature of time in the time direction is R_{00}
This must surely tend to infinity because all the T's must surely head
to infinity. This would mean that the time direction is rotating every
which way. One begins to see why Hawking rather likes it to be
imaginary.
Well, lets take Baez's little derivation using Ted's given metric for a
Friedmann universe. A diagonal of (-1,q,q,q) where q=q(t)
Then with v^a=(1,0,0,0) we get taking a = b = 0
R_{ab} = R_{00} = T_{00} + [(-1/2) T^c_c g_{00}]
R_{00} = T_{00} + (1/2)T^c_c which is exactly the same as before!
BUT T^c_c must be calculated with the new metric.
With this metric T^c_c=g^(ca)T_{ac} which means we have to invert the
poxy metric. Oh bother. Ah, I think we can do it by guessing. It would
surely have to be diagonalised at (-1,1/q,1/q,1/q). <Nervous grin>
So T^c_c=-T_{00} + 1/q T_{11}+ 1/q T_{22}+ 1/q T_{33} and so
R_{00} = T_{00} - 1/2T_{00} + (1/2q)T_{11} +(1/2q)T_{22} +(1/2q)T_{33}
R_{00} = (1/2)T_{00} +(1/2q)T_{11} +(1/2q)T_{22} +(1/2q)T_{33}
Now if q=pt, which might not sound unreasonable (ho, ho) then lo and
behold, everything blows up into infinities when t=0. On the other hand
possibly a better description would be q = pt + m, when it doesn't.
Well, at least not in this straightforward and unambiguous way. By golly
Oz will be turned into a dung-beetle egg yet!
Anyway, it's probably all wrong in any case. <Sigh, I tried, honest.>
I have a niggling hesitation in ignoring x,y,z, but Baez is boss.
Oh well, I suppose one ought pretend to be able to do it in the
x-direction. So setting a = b = 1 <Who always burnt his fingers?>
R_{ab} = R_{11} = T_{11} + [(-1/2) T^c_c g_{11}]
R_{11} = T_{11} + [(-1/2) T^c_c q] since g_{11} = q
Now T^c_c=-T_{00} + 1/q T_{11}+ 1/q T_{22}+ 1/q T_{33} and so
R_{11} = T_{11} + q/2T_{00} - (q/2q)T_{11} -(q/2q)T_{22} -(q/2q)T_{33}
R_{11} = -(q/2)T_{00} +(1/2)T_{11} -(1/2)T_{22} -(1/2)T_{33}
Which is probably both wrong and meaningless. Anyway, press on
regardless. It must be some sort of practice anyway. Lets say that q=mt
for simplicity. Then at t=0 (and totally ignoring the fact that 1/q is
infinite) we get q=0 to give us
R_{11} = (1/2)T_{11} -(1/2)T_{22} -(1/2)T_{33}
In a uniform and homogenous universe surely T_{11} = T_{22} = T_{33}
so R_{11} = -(1/2)T_{11}
Now this is odd. Well, this also seems to have a distinct tendency to
want to zoom off to minus infinity as all of the T_{1 to 3} head that
way. Indeed if the universe is indeed uniform and homogenous R_{11}
would appear to be heading towards infinite negative curvature. Er, um,
whatever that means in detail. Sounds sort of more like time.
Wow! Isn't this stuff wierd?
Oh well, enough of making a fool of myself. Let's post it off to be
buried under derision. Lets let the pro's, and even the tyro's, have a
good laugh. I can take it!
Oz crawled into this hole in the rock and prepared for the Wiz's wrath.
By gosh he had really done it now.
Oz
vital questions. The second was:
2. Explain how, in the standard big bang model, where the universe is
homogeneous and isotropic --- let us assume it is filled with some fluid
(e.g. a gas) --- the curvature of spacetime at any point may be(?)
determined at each point.
Well, obviously the Riemann tensor is required. So we have to go in our
little loop and find out the rotation of our iddy-biddy vector. Now this
sounds absolutely fine except that it's not very easy to go on a little
rectangular loop whilst simultaneously zooming along the t-axis at c.
Let's try and figure out how it might be done. Well, the first thing to
realise is that we can choose our basis vectors and luckily they need
not be orthogonal, although one could imagine that 'various corrections'
could be made to make them so. Hmmmm ....
Ok, how about this way. You have two identical stable 'clock and laser'
flashers at your origin. You take one a suitable distance away from
your origin and set it down in the same frame as your origin. You know
you have done this because a laser beaming from the origin has exactly
the same frequency as when you had the equipment side by side at the
origin. Ok, now you have a point in spacetime. You know how far you went
because you measured it locally with your 1mm ruler. You do this very
slowly. Heck, it's a 1mm ruler and you are going light seconds at least,
in any case you should be able to count the flashes from your laser at
the origin, both out and back, to 'synchronise' time somehow or other.
Anyway, whatever, you send off a laser pulse to say all is 'at rest'.
Now you have the equivalent of a distance in, well, let's call it the
'out' direction. Now you trundle back to your origin and see what
transpired. Well, you (in curved spacetime) will see two things. Firstly
the distant laser started off with the same frequency as the origin
laser (once you set it up), but slowly it will drift as spacetime
curvature moves it away (or towards). You will also see a change in the
time pulses (much the same thing really) as the distance gets further
away. So at the very least you have a little parallelogram in the time-
out plane.
*
|\ The '*' are my little parallelogram
| \ for my Riemann loop.
| \ The light pulses are at 45 deg as req'd.
* *
.\ |
. \ |
. \|
. * All set up point.
. .
. .
. . Trundle out leg.
..
..
.
.
Now exactly how you interpret this as a rotation of your tangent vector,
I am not rightly sure. My guess is that you deduce that the distant leg
is 'really' at a small angle determined by it's relevant velocity
accumulated over the time it travelled (vertically) , which being a
dx/dt thingy, looks like a slope to me.
Oh well, maybe worth 25%?? ..... 20%......
Right ball park? .. Not even wrong <gloom>.
Notes:
a) The universe has to be uniform and isotropic to avoid you interacting
with a bendy, lumpy bit of spacetime that throws everying off.
b) This give you a bit of the Riemann tensor, obviously it should be the
same for vectors in the other space/time directions by symmetry. How you
do it in the space-space planes I am less certain. Well, I don't have a
clue, actually. Probably so simple you can't see it.
c) I suppose somewhere I am assuming the underlying metric is a
ds^2 = -a dt^2 + dx^2 + dy^2 + dz^2
sort of thingy with appropriate choice of co-ordinates etc.
Mostly 'cos that's what Ted sed.
Waiting in trepidation,
Brian J Flanagan
> Oz has returned from his weekend labours. Now he has to answer the three
> vital questions. The second was:
>
> 2. Explain how, in the standard big bang model, where the universe is
> homogeneous and isotropic --- let us assume it is filled with some fluid
> (e.g. a gas) --- the curvature of spacetime at any point may be(?)
> determined at each point.
>
> Well, obviously the Riemann tensor is required. So we have to go in our
> little loop and find out the rotation of our iddy-biddy vector. Now this
> sounds absolutely fine except that it's not very easy to go on a little
> rectangular loop whilst simultaneously zooming along the t-axis at c.
> Let's try and figure out how it might be done. Well, the first thing to
> realise is that we can choose our basis vectors and luckily they need
> not be orthogonal, although one could imagine that 'various corrections'
> could be made to make them so. Hmmmm ....
>
> Ok, how about this way. You have two identical stable 'clock and laser'
> flashers at your origin. You take one a suitable distance away from
> your origin and set it down in the same frame as your origin. You know
> you have done this because a laser beaming from the origin has exactly
> the same frequency ...
(& therefore the same color/vector in M space)
(Voice from left field: Batter up!)
>
> Notes:
> a) The universe has to be uniform and isotropic to avoid you interacting
> with a bendy, lumpy bit of spacetime that throws everying off.
>
> b) This give you a bit of the Riemann tensor, obviously it should be the
> same for vectors in the other space/time directions by symmetry. How you
> do it in the space-space planes I am less certain. Well, I don't have a
> clue, actually. Probably so simple you can't see it.
>
> c) I suppose somewhere I am assuming the underlying metric is a
> ds^2 = -a dt^2 + dx^2 + dy^2 + dz^2
> sort of thingy with appropriate choice of co-ordinates etc.
> Mostly 'cos that's what Ted sed.
>
> Waiting in trepidation,
>
Consider your worst fears reduced to broad farce:)
Chow,
Brian
john baez
From within he heard a mumbled "I'm busy. Go away."
He stood there for a second, and then said "Umm, but I had some problems
with the course notes."
The door popped open and a very grumpy-looking face leaned out.
"Problems?" asked the wizard. "You have PROBLEMS with them???"
"Ah, err, yes sir. First of all, you say that
R_{00} = T_{00} + (1/2) T^c_c
and later:
T^c_c = -T_{00} + T_{11} + T_{22} + T_{33}
and conclude that
R_{00} = (1/2) T_{00} + T_{11} + T_{22} + T_{33}.
Shouldn't that be
R_{00} = (1/2) [T_{00} + T_{11} + T_{22} + T_{33}] ?"
The wizard didn't reply, and only stared at Oz. St. Elmo's fire
began flickering around his head. Oz continued, "Also, umm, sir, you said
that R_{ab} v^a v^b when v = (1,0,0,0), but
I don't see this. v(0)=1, so the only non zero term should be
R_{00}.v(0).v(0) = R_{00} not -R_{00}
It would appear that g_{ab} ought to make an appearance here, but I
don't see where. Help!"
The last word, "Help!", was uttered by Oz when the wizard, in a rage,
began to hurl fireballs hither and thither, cursing up a storm.
"Factors of two and sign errors! Factors of two and sign errors! Here
I am trying to quantize gravity using n-categories and you bother me
with FACTORS OF TWO AND SIGN ERRORS??? When I get through with you,
you'll wish you were a planarium, you ninny! You'll even wish you were
a dung beetle! There are things worse than that, you know!" He
advanced towards Oz, swinging his staff, and Oz retreated, desperately
pleading for mercy. At the last minute, just as Oz felt the whishing of
the staff right before his nose, a brightly lit sphere appeared in the
air. The staff bounced off it and fell to the ground.
"Hi," said the sphere. Oz recalled it now... it was the sphere of radius
r! "Say, G. Wiz, aren't you being a trifle petulant? Don't you
remember when you were an apprentice like Oz is now, and your master
hided you for making sign errors? Are you saying it's okay to make sign
errors now?"
Oz glared at the sphere. "Ah yes, my professor. He always was a real
nitpicker." He looked down thoughtfully for a minute. "Still, he did
teach me well." He sighed. "All right, I won't turn Oz into an
intestinal nematode just yet. Yes, Oz, the factor of 1/2 goes in front of
the whole thing:
R_{00} = (1/2) [T_{00} + T_{11} + T_{22} + T_{33}]
As for that other thing, yes,
R_{00} v^0 v^0 = R_{00},
not -R_{00}. I just got it wrong: the second time derivative of the
volume of the ball of coffee grounds is MINUS R_{ab} v^a v^b times the
volume of the ball, where v is the velocity 4-vector of the ball." A
swarm of minus signs appeared around the wizard's head, buzzing like
midges. He glared at them and swatted them with his staff. "Minus
signs! I *hate* them!" The sphere, hoping to keep the wizard from
becoming angry again, chased after the minus signs, and they all flew away
down the tunnels of the wizard's keep.
"By the way," the wizard said, and here he picked up his staff and waved
it threateningly at Oz once more, "don't write v(0) when you mean v^0,
since then it's ambiguous whether you mean v^0 or v_0, and these are
different... in the example I gave, we'd have v_0 = -v^0."
Oz asked, "Umm, by the way, how do you derive that bit about the second time
derivative of the volume of the ball being, umm, MINUS R_{ab} v^a v^b?
It seems like it should follow from something or other about curvature,
but..."
"I didn't explain that part yet!" interrupted the wizard. "Just take my
word for it for now! I'll explain it sometime when I'm less busy, if
you pass that test. Now get out of here... and hand in the answers to
those questions SOON." He went back into his room and slammed the door.
Oz thought, "Hmm, so he is human after all. In fact, he makes just as
many mistakes as I do, I bet! He sure gets upset about it, though."
The wizard's door creaked open, though the wizard remained hunched over
his desk scribbling. "What was that you were thinking?" asked the
wizard, not turning around.
"Oh, nothing," said Oz, and scurried away.
Oz
but not speak. This explains the uncharacteristic silence from Riverside. He
has asked me to post this reply.
Since my newsreader will not forge headers note that below is from John Baez.
=============================================================================
(John Baez)
Oz knocked hesitantly at the wizard's door. "Ahem?"
From within he heard a mumbled "I'm busy. Go away."
He stood there for a second, and then said "Umm, but I had some problems
with the course notes."
The door popped open and a very grumpy-looking face leaned out.
"Problems?" asked the wizard. "You have PROBLEMS with them???"
"Ah, err, yes sir. First of all, you say that
R_{00} = T_{00} + (1/2) T^c_c
and later:
T^c_c = -T_{00} + T_{11} + T_{22} + T_{33}
and conclude that
Shouldn't that be
Wiz glared at the sphere. "Ah yes, my professor. He always was a real
nitpicker." He looked down thoughtfully for a minute. "Still, he did
teach me well." He sighed. "All right, I won't turn Oz into an
intestinal nematode just yet. Yes, Oz, the factor of 1/2 goes in front of
the whole thing:
R_{00} = (1/2) [T_{00} + T_{11} + T_{22} + T_{33}]
As for that other thing, yes,
R_{00} v^0 v^0 = R_{00},
not -R_{00}. I just got it wrong: the second time derivative of the
volume of the ball of coffee grounds is MINUS R_{ab} v^a v^b times the
volume of the ball, where v is the velocity 4-vector of the ball." A
swarm of minus signs appeared around the wizard's head, buzzing like
midges. He glared at them and swatted them with his staff. "Minus
signs! I *hate* them!" The sphere, hoping to keep the wizard from
becoming angry again, chased after the minus signs, and they all flew away
down the tunnels of the wizard's keep.
"By the way," the wizard said, and here he picked up his staff and waved
it threateningly at Oz once more, "don't write v(0) when you mean v^0,
since then it's ambiguous whether you mean v^0 or v_0, and these are
different... in the example I gave, we'd have v_0 = -v^0."
Oz asked, "Umm, by the way, how do you derive that bit about the second time
derivative of the volume of the ball being, umm, MINUS R_{ab} v^a v^b times
its original volume? It seems like it should follow from something or
other about curvature, but..."
"I didn't explain that part yet!" interrupted the wizard. "Just take my
word for it for now! I'll explain it sometime when I'm less busy, if
you pass that test. Now get out of here... and hand in the answers to
those questions SOON." He went back into his room and slammed the door.
Oz thought, "Hmm, so he is human after all. In fact, he makes just as
many mistakes as I do, I bet! He sure gets upset about it, though."
The wizard's door creaked open, though the wizard remained hunched over
his desk scribbling. "What was that you were thinking?" asked the
wizard, not turning around.
"Oh, nothing," said Oz, and scurried away.
(Ex John Baez)
john baez
away as he works on his test:
"Well, now the third thaumaturlogical test.
3. In the big bang model, what happens to the Ricci tensor as you go
back in past all the way to the moment of creation.
This has just GOTTA be a trick question."
The wizard rolls his eyes and shakes his head.
"Well, the stress-energy tensor has just got to go towards infinity,
surely. Both the mass and momentum density must zoom to infinity."
For minkowski space time Baez derived:
R_{00} = 1/2(T_{00} + T_{11} + T_{22} + T_{33})
The curvature of time in the time direction is R_{00}
This must surely tend to infinity because all the T's must surely head
to infinity. This would mean that the time direction is rotating every
which way. One begins to see why Hawking rather likes it to be
imaginary."
The wizard frowns. Where the hell is he coming up with that stuff from?
Since when is R_{00} the "curvature of time in the time direction"? He
explained so many times the physical interpretation of R_{00} in terms
of the the change of volume of a ball of initially comoving geodesics...
has it all been in vain? Time direction rotating every which way?!?
Imaginary time?!? Why did he ever mention imaginary time to the
impressionable youth. What lunacy! Sighing, he listens some more:
"Well, lets take Baez's little derivation using Ted's given metric for a
Friedmann universe. A diagonal of (-1,q,q,q) where q=q(t)
Then with v^a=(1,0,0,0) we get taking a = b = 0
R_{ab} = R_{00} = T_{00} + [(-1/2) T^c_c g_{00}]
R_{00} = T_{00} + (1/2)T^c_c which is exactly the same as before!
BUT T^c_c must be calculated with the new metric.
With this metric T^c_c=g^(ca)T_{ac} which means we have to invert the
poxy metric. Oh bother. Ah, I think we can do it by guessing. It would
surely have to be diagonalised at (-1,1/q,1/q,1/q). <Nervous grin>"
The wizard looks relieved. At least the boy knows how to invert
diagonal matrices. There may be some use he can be put to.
"So T^c_c=-T_{00} + 1/q T_{11}+ 1/q T_{22}+ 1/q T_{33} and so
R_{00} = T_{00} - 1/2T_{00} + (1/2q)T_{11} +(1/2q)T_{22} +(1/2q)T_{33}
R_{00} = (1/2)T_{00} +(1/2q)T_{11} +(1/2q)T_{22} +(1/2q)T_{33}
Now if q=pt, which might not sound unreasonable (ho, ho) then lo and
behold, everything blows up into infinities when t=0. On the other hand
possibly a better description would be q = pt + m, when it doesn't.
Well, at least not in this straightforward and unambiguous way. By golly
Oz will be turned into a dung-beetle egg yet!"
"Anyway, it's probably all wrong in any case. <Sigh, I tried, honest.>"
Oz crawled into this hole in the rock and prepared for the Wiz's wrath.
By gosh he had really done it now.
The wizard, however, was in a more merciful mood than usual. Startling
Oz mightily, he simply leaned his head in and said: "You seem to be
making rather heavy weather of this, but what is your basic qualitative
conclusion about the Ricci tensor as we approach the moment of the big
bang? You are correct that the energy density zooms to infinity. That
means that T_{00} goes to infinity. How about the other components of
T_{ab}? Can you guess what they do? And what would you guess happens
to R_{ab}, then? A formula for R_{ab} in terms of T_{ab} would be nice
at this point, no? I told you one, no? Do you remember it?"
Sometimes of course the wizard could be rather nerve-racking even when
in a merciful mood, what with all these questions!
Edward Green
Whoops! I should read the whole post before hitting "send"...
>
>Baez: ......
>
>geodesics. Remember, if v is the velocity vector of the particle in the
>middle of a little ball of initially comoving test particles in free
>derivative of the volume of the ball is
>R_{ab} v^a v^b times V.
>
>Oz******************************************
>I expect the above is obvious. Offhand, I can't see the derivation.
>********************************************
>
>Baez:
>If we know the above quantity for all velocities v (even all
>timelike velocities, which are the physically achievable ones), we can
>reconstruct the Ricci tensor R_{ab}. But we might as well work in the
>local rest frame of the particle in the middle of the little ball, and
>use coordinates that make things look just like Minkowski spacetime
>right near that point. Then
>
>g_{ab} = -1 0 0 0
>0 1 0 0
>0 0 1 0
>0 0 0 1
>
>and v^a = 1
>0
>0
>0
>
>So then --- here's a good little computation for you budding tensor
>jocks --- we get
>
>R_{ab} v^a v^b = -R_{00}
>
>Oz*************************************************
>
>R_{00}.v(0).v(0) = R_{00} not -R_{00}
>
>It would appear that g_{ab} ought to make an appearance here, but I
>don't see where. Help!
Are you really worried about a sign? :-) We can always find one
somewhere... Well, I suppose they may occasionally be rather important...
Yes, it looks like it should be +R_{00} to me too. But as far as the
metric, maybe I can straighten that one out. We already cast the Ricci
with two lower indices, and the vectors with upper indices (I see why
people talk like this now, saves from embarrassment when you can't
remember terminology) , and so, in the timeless Baezian description,
neither need have a sex-change operation before they can mate. So maybe we
did include the metric, but shouldn't have?
-erg
Edward Green
>
>More precisely, the second time derivative of the volume of this little
>ball is approximately
>
>
>better and better in the limit as the ball gets smaller and smaller.
>
>Oz******************************************
>So are we take it that 'smaller and smaller' means that contributions
>from the Weyl Tensor becomes even more smaller and smaller. Or/and are
>we to take it that local spacetime becomes flatter and flatter in the
>absence of T so that we only need consider curvature due to T. Hmm, I
>suppose that doesn't make sense. Perhaps I mean that spacetime can be
>considered flat at a reasonable distance away from our coffee grain of
>momentum density. Well, what are you neglecting by taking a smaller and
>smaller volume?
>********************************************
becomes a better approximation as "dx gets smaller and smaller"... it's
one of those differential relationship thingies...
Pray carry on sir.
john baez
"Dear Oz --
At this moment, the wizard was in the back room reading a parchment
that his slaves had fetched him... a recently written disquisition on
n-categories, by a wizard far across the seas. He frowned. It seemed
his competitor was catching up... he had been spending too much time on
that apprentice of his, and neglecting his serious work. He decided to
put in a good several hours trying to see just what the other fellow had
done. Just as he was reaching for his quill, to take some notes, a
large bell off in the corner began ringing. "Damn! What is it now?"
the wizard cried. "Someone is making a conceptual error somewhere...
Bell! Tell me, who is screwing up?"
"It's Oz, sir," said the bell, in a melodious high-pitched voice. "He's
trying to solve a general relativity by treating gravity as a force!"
"Gravity as a FORCE??? What? Are you sure it's Oz? I told him
millions of times that in GR, freely falling objects simply follow
geodesics! The whole point of GR is that gravity is not described as a
force! I'm sure he knows that. It can't possibly be Oz! Tell me it's
not!"
"Yes, it's Oz sir, I can sense his aura very clearly. He's in his cave."
"FORCE? What the hell is he trying to do, anyway?"
"He's trying to solve that problem you gave him, on why sufficiently
dense matter collapses to form a black hole."
"God's wounds, why doesn't he just use the course notes! I told him the
exact formula for how the stress energy tensor curves spacetime and
makes initially parallel geodesics converge! Doesn't he see what that
implies?"
"Apparently not, sir."
"Why, I'll go out and hide that fellow, teach him a thing or two about
force..." The wizard stormed towards the door, cape swirling behind him
and static electricity building for an enormous discharge, when all of a
sudden he halted... he had a thought. Why not get that other fellow, Ed
Green, to help Oz out? Why do all the work himself? True, Ed never
seemed to enter into the wizard fantasy realm, but if he could be
reached in his own realm, he could probably be coaxed into helping Oz on
these test problems. It would probably be good for Ed, too.
"Courier!" The wizard clapped his hands, and a phoenix appeared in a
puff of smoke. "Tell Ed Green to help Oz with those test questions.
You will have to leave this realm and take on a form suitable for the
realm of Late-Twentieth Century America. I think he lives in
Manhattan. Perhaps you should be bicycle courier."
"Is he wired?"
"Yes, certainly. I've communicated with him that way before."
"Well then," said the Courier, "With your permission, I will simply take
a little snippet of the spacetime continuum, say the last five minutes
in Oz's cave and this room here. Don't worry, I won't actually cut it
out; I'll just copy it, edit and compress it a bit, translate it into
ASCII form, and email it to him. That's more efficient than sending me
off to Manhattan, no?"
The wizard smiled. "You're so clever. I can never get used to these
labor-saving conveniences. Go right ahead. Post it to sci.physics
while you're at it. Ask him to help Oz solve this GR problem without
referring to that archaic and misleading notion of the "force" of gravity."
Oz
writes
>
>The wizard, however, was in a more merciful mood than usual. Startling
>Oz mightily, he simply leaned his head in and said: "You seem to be
>making rather heavy weather of this, but what is your basic qualitative
>conclusion about the Ricci tensor as we approach the moment of the big
>bang? You are correct that the energy density zooms to infinity. That
>means that T_{00} goes to infinity. How about the other components of
>T_{ab}? Can you guess what they do? And what would you guess happens
>to R_{ab}, then? A formula for R_{ab} in terms of T_{ab} would be nice
>at this point, no? I told you one, no? Do you remember it?"
>
>Sometimes of course the wizard could be rather nerve-racking even when
>in a merciful mood, what with all these questions!
>
>
Oz has decided to keep it simple. Flights of fancy are fun, but just end
up with a staff stuffed somewhere unpleasant with a mighty mean Wiz on
the end of it. And that's if he is in a GOOD mood.
OK, start from the beginning.
1) If v is the velocity vector of a particle in the middle of a little
ball of blah blah .... of volume V.
so d^2V/dt^2 = -R_{ab} v^a v^b. Hokay.
2) Now to find R_{ab}
R_{ab} = T_{ab} - (1/2)T^c_c g_{ab}
and T^c_c = -T_{00} + T_{11} + T_{22} + T_{33}
Now since it's uniform and isotrophic we should be able to say that
Tc = T_{11} = T_{22} = T_{33} which also head towards infinity as V->0
and probably the cross terms are equal, at least some of them.
Well, surely
T_{12} = T_{21} = T_{13} = T_{31} = T_{23} = T_{32} = Td
must be equal too in this situation. They presumably give the density of
momentum flow in one direction travelling in another. Maybe something to
do with things that spin or suchlike.
Then we have T_{0n} and T_{n0} (n=1 to 3) like terms. What they
represent physically is not too clear. Let's see. A photon travels at an
angle of pi/4 to the t-axis. It has momentum. So presumably this
momentum can be resolved into a bit parallel to the t-axis and a bit
parallel to a space axis. No, don't follow this. Anyway they may not be
zero.
So assuming we can take a minkowski metric in a small area, which seems
unlikely, but still, we get
R_{ab} =
[(1/2)(T_{00}+T_{11}+T_{22}+T_{33}) , T_{01} , T_{02} , T_{03}]
[T_{10} , (1/2)(T_{00}+T_{11}-T_{22}-T_{33}), T_{12}, T_{13}]
[T_{20} , T_{21}, (1/2)(T_{00}-T_{11}+T_{22}-T_{33}), T_{23}]
[T_{30} , T_{31}, T_{32} , (1/2)(T_{00}-T_{11}-T_{22}+T_{33})]
=
[(1/2)(T_{00} + 3Tc) , T_{01} , T_{02} , T_{03}]
[T_{10} , (1/2)(T_{00} - Tc), T_{12}, T_{13}]
[T_{20} , T_{21}, (1/2)(T_{00} - Tc), T_{23}]
[T_{30} , T_{31}, T_{32} , (1/2)(T_{00} - Tc)]
Well, a bit untidy, however we can see that
R_{11} = (1/2)(T_{00} - Tc) [Nb Tc=T_{11} etc]
Now, as the universe gets hotter and smaller, one could imagine that Tc
gets bigger. In fact one could imagine that Tc could even get to be as
big as T_{00}. Lots of interactions producing massless particles and
all. In this case R_{11} *might* become zero. With T_{00} going
infinite, this is attractive. <groan>
3) Now let's see.
d^2V/dt^2 = R_{ab} v^a v^b.
Now I can pick out each term by my choice of a & b. However it's not
very clear, without explanation, what for example R_{12} means. So I
will skip this for now. One optimistically hopes they turn out to be
zero.
Anyway, I don't believe that if I choose a=b=1, that I will get a
d^2V/dt^2 term. Surely this is a slice out of a hypervolume? I would
have expected something along the lines of:
d^2V/dt^2 = -R_{00} with V a volume of type (0,1,1,1)
d^2V/dx^2 = R_{11} with V a volume of type (1,0,1,1) and it may be +ve.
2R_{11} = T_{00} - T_{11}
This is the curvature, not actually the size, although they must surely
be related. One is tempted to imagine a cone filled with tangential
spheres from an itsy bitsy one at t=epsilon to a big one centered at
t=10^10 years. However this can't be quite right because the curvature
of a sphere is constant. In the 4D thingy the curvature is forever
decreasing from t=0 to t=10^10y. Well, not quite. Throughout the entire
period c remains constant. In this case a line of length c drawn on each
sphere would show different curvature, from wrapping round several times
(t=epsilon) to being very nearly uncurved at t=10^10y.
So our sphere of type (1,0,1,1) centered epsilon from t=0 *could* have a
small d^2V/dx^2 so it's rate of decrease in size may be rather modest,
but on the other hand dV/dx could be almost anything.
I need more 'obvious' information to proceed further.
Edward Green
Rudely seized from his mundane life in a late twentieth century metropolis
by the wizard's messenger, Ed finds himself hurled through some
mind-bending special effects thingies involving swirling colors and the end
of the popular song by the dung-beetles, 'I am the Large Marine Mammal'.
["Hmm..." thinks Ed, "that doesn't sound quite right... must be some sort
of reality dislocation"] His trip ends in a thunk as he falls into a snow
drift, and rolls over to see the messenger receeding, laughing to
himself... "ASCII... Applied Strategem to Corral Innocent Inebriates....
ha ha ha ha... "
[The wizard employed many creatures of dubious antecendents bound by debts
of service, but he was not always able to fully control their impish
aspects]
"Ow!" said Ed, "What have I been drinking?"
He finds himself wearing the scratched and dented armor of an apprentice,
one who has served many masters but never completed his apprenticeship. A
bit graying around the temples for an apprentice... On the breast plate is
the curious device "erg", held on by a few rivets.
Before erg has time to realize just how cold this feebly insulated suit of
armor is in a snow drift, he is roused from his post translational reverie
by an approaching troll, one who has been eating stone tablets, and has
a touch of indigestion.
"ARGGHHHESTH" comments the troll, drolly. Ed leaps up and attempts to
dislodge his sword from the scabbard, but apparently his fantastic
alter-ego has not kept his equipment in shape, and it is rusted solidly
into place. "Ohmygod" he thinks, giving one last desperate pull at the
weapon. The impulse is sufficient to overcome the coefficient of static
friction between steel boots and snow, an he falls in a heap before the
charging troll, who stubs his toes on the armor, leaving fresh toe-dents,
trips over the mighty warrior and rolls into a ball, gaining momentum on
the icy slope and finally flying out into space like a boulder, smashing
into smithereens on the rocks below. Seeing that the fearsome warrior
knows JUDO, the other trolls beat a tasty retreat (munching the remaining
tablets).
"Whoa!" thinks the hero, nursing his bruised arm and slowly pulling
himself up. "This place is dangerous!" Just then he spies a feeble bit
of firelight in the mouth of a nearby cavern, and painfully heads in that
direction. In the cavern, of course, is his old friend, Oz.
"What, you too?" says Oz, once a wizard in his own right, in a distant
story. "Has your geodesic entered this place of no return?"
"Yes" says the erg, pulling a stone up to the fire, and drawing his sword
to dry it on some straw, finding that the fall in the heroic trollomachy
was sufficient to dislodge it from the scabbard, "and the odd thing is, I
am not even sure I remember crossing the horizon.
And Oz said to him....
[Cheap b*stard... ain't I... :-) ]
john baez
>>R_{ab} v^a v^b
That's right. Whenever I speak of something being true for "small" this
or that I really mean it's true for infinitesimal this or that, which is
the sort of thing one makes precise using limits. As for the Weyl
tensor, recall that it describes the change in *shape* of the little
ball of initially comoving coffee grounds, *apart* from its change in
volume. In other words, as the initially round little ball changes into
an ellipsoid, the Ricci tensor describes its change in volume, while the
Weyl tensor describes the rate at which the *shape* of the ellipsoid
changes, completely neglecting the change in volume. (Technical note:
this is the sort of thing one can easily do with tensors: the Ricci is
the "trace" of a certain matrix, while the Weyl is the "trace-free
part".) So the Weyl tensor has nothing to do with change in volume.
When we take a smaller and smaller volume, we are just neglecting the
complexities of what a finite-sized ball of coffee grounds can really do
under an arbitrary coordinate transformation --- it can turn into the
shape of a banana! Instead, we are considering the linear approximation
applicable to infinitesimally small ball. Under a linear
transformation, a ball merely turns into an ellipsoid.
john baez
vital questions. The second was:
2. Explain how, in the standard big bang model, where the universe is
homogeneous and isotropic --- let us assume it is filled with some fluid
(e.g. a gas) --- the curvature of spacetime at any point may be(?)
determined at each point.
Oz ponders this for an hour or two, and walks over to the wizard's door
and knocks on it. "Come on in!" cries the wizard. As Oz enters, G. Wiz
slips out from behind a curtain, shaking some dust off his hands.
He then wipes his brow and asks, "So, how are you doing on those
questions? I presume you're starting with question 2, right?"
That mind-reading ability always takes Oz aback. He nods and says,
"Well, obviously the Riemann tensor is required. So we have to go in our
little loop and find out the rotation of our iddy-biddy vector. Now this
sounds absolutely fine except that it's not very easy to go on a little
rectangular loop whilst simultaneously zooming along the t-axis at c.
Let's try and figure out how it might be done. Well, the first thing to
realise is that we can choose our basis vectors and luckily they need
not be orthogonal, although one could imagine that 'various corrections'
could be made to make them so. Hmmmm ...."
The wizard also says "Hmmm ...."
Oz continues. "Ok, how about this way. You have two identical stable
'clock and laser' flashers at your origin. You take one a suitable
distance away from your origin and set it down in the same frame as your
origin. You know you have done this because a laser beaming from the
origin has exactly the same frequency as when you had the equipment side
by side at the origin. Ok, now you have a point in spacetime. You know
how far you went because you measured it locally with your 1mm ruler.
You do this very slowly. Heck, it's a 1mm ruler and you are going light
seconds at least, in any case you should be able to count the flashes
from your laser at the origin, both out and back, to 'synchronise' time
somehow or other. Anyway, whatever, you send off a laser pulse to say
all is 'at rest'. Now you have the equivalent of a distance in, well,
let's call it the 'out' direction. Now you trundle back to your origin
and see what transpired. Well, you (in curved spacetime) will see two
things. Firstly the distant laser started off with the same frequency as
the origin laser (once you set it up), but slowly it will drift as
spacetime curvature moves it away (or towards). You will also see a
change in the time pulses (much the same thing really) as the distance
gets further away. So at the very least you have a little parallelogram
in the time- out plane."
He scratches the following picture in the dust on the floor:
*
|\ The '*' are my little parallelogram
| \ for my Riemann loop.
| \ The light pulses are at 45 deg as req'd.
* *
.\ |
. \ |
. \|
. * All set up point.
. .
. .
. . Trundle out leg.
..
..
.
.
"Now exactly how you interpret this as a rotation of your tangent vector,
I am not rightly sure. My guess is that you deduce that the distant leg
is 'really' at a small angle determined by it's relevant velocity
accumulated over the time it travelled (vertically) , which being a
dx/dt thingy, looks like a slope to me."
The wizard again says "Hmm ...." and stares off abstractedly in a rather
unnerving way.
Oz asks: "Oh well, maybe worth 25%?? ..... 20%...... Right ball
park?... Not even wrong?"
The wizard turns back to Oz, smiles and says "Well, it has a certain
charm to it." Oz breathes a sigh of relief, but then notices a curious
gleam in the wizard's eye... and begins to fear he will not be let off
quite so easy.
"Maybe you should think of it this way," says the wizard. "Start with
two clocks next to each other, out in the wilderness of empty space!"
As he speaks, the room dims and Oz seeks two clocks next to each other,
floating in starry emptiness. "Now drag one a foot away from the other
and do your best to leave it at rest relative to the first." One clock
moves over a foot, starting at rest, but then the clocks begin to drift
away from each other at an accelerating rate. "You you let them float
out there in the wilderness of space. They are in free fall, so they
trace out geodesics. These geodesics may converge or diverge, and the
rate of this "geodesic deviation" may be measured as you suggest, by
shining a laser from one to the other and measuring the redshift."
"By the way: note that since the clocks are relatively close to each
other the notion of the distance between them, as measured between two
clocks that start out at rest with respect to each other --- and indeed
the notion of starting out at rest with respect to each other! --- are
unproblematic. Really we should do this with clocks that are
infinitesimally close to each other, but a distance of say, a
lightsecond is darn close to infinitesimal on cosmic scales."
"Now, what does geodesic deviation have to do with curvature, exactly?
Well, let's look at a spacetime diagram of the situation. Hang on while
I equip you with 4-dimensional vision." He passes his hands over Oz's
head and mutters an almost inaudible syllable, and all of sudden Oz is
shocked to find that he can see the clocks, not just in space, but in
spacetime. They trace out curves in spacetime, and he sees these curves
in their entirety as a static whole! Yet he is simultaneously able to
see how at any given "moment" --- i.e., any given slice of spacetime ---
the clocks occupy positions in "space" just as they did in the previous
scene! Furthermore, as before, he can see, as in a movie, how as time
passes the clocks begin to accelerate away from each other. Somehow he
is seeing not just space but spacetime!
"Hey, how are you doing that?" he asks. "I bet if I could do this
myself these problems would be a whole lot easier!"
"I'm just using my limited telepathic ability to show you how I think of
these things," the wizard says. "For you to do it too, all you need is
practice."
Oz frowns, unsure as to how he'd practice this. He decides to keep
quiet and pay careful attention; maybe he'll get the habit of 4d vision.
"Wise move," says the wizard, smiling.
"Now, let's use "v" to denote the velocity vector of the first clock at
time zero, and let "w" denote the vector from the first clock to the
second." Oz sees something like the following, but of course everything
he sees is in 4 dimensions:
| |
| |
v^ ^
| w |
P->-----Q
"I've labelled the initial positions of the two clocks at rest by P and
Q. Note that the velocity vector of the clock at Q is just the result
of PARALLEL TRANSPORTING v in the direction w."
Oz asked, "Wait? Is w the path from P to Q, or just a tangent vector at
P?"
The wizard smiled. "Good question! Roughly speaking, we can pretend
the path from P to Q is a vector because the path is so short... it
should really be infinitesimally short, of course."
"Now, suppose we let each clock wait a second. They now have new
positions (in spacetime) P' and Q'." Oz now sees something like this:
P'->----Q'
| |
| |
v^ ^
| w |
P->-----Q
Oz asks, "Say, shouldn't v be infinitesimal too?"
"Right, if we want to think of the path from P to P' as a vector we
should really be waiting just an *infinitesimal* amount of time to get
from P to P', not a second; but a second is close enough for practical
purposes."
They stare at the infinitesimal rectangle a while and then the wizard
continues: "Now, what's the velocity vector of the clock at Q'? Well,
think how we got it: first we parallel transported v over to Q along w.
Then we parallel tranported the result over to Q', since the curve from
Q to Q' is a geodesic, which means its velocity vector is parallel
translated along itself." Oz sees a helpful green dot go across from P
to Q and then forwards in time from Q to Q', carrying a green copy of
the tangent vector v with it.
"Now for the big question! We want to know if the clock at Q' is moving
away from the clock at P'. To answer this, we compare its velocity
vector to the following vector: what we get by first parallel
translating v along itself over to P', and then over to Q'. That's the
velocity the clock at Q' would have it it were at rest relative to the
clock at P'." Oz now sees a red dot go forwards in time from P to P'
and then across from P' to Q', carrying a red copy of the tangent vector
v with it.
P'->----Q'
| |
| |
v^ ^
| w |
P->-----Q
The resulting red tangent vector at Q' is a bit different from
the green one representing the actual velocity of the clock at Q'.
"Curvature!" cries Oz in a moment of revelation.
"Right! We are taking the vector v and parallel translating it two
different ways from P to Q' and getting two slightly different answers.
If the answers were the same, the second clock would remain at rest
relative to the first. But in fact they are not, and the difference
tells us how the second one begins accelerating away from the first."
"Now remember how curvature works: the result of
dragging v from P to Q to Q'
minus the result of
dragging v from P to P' to Q'
is going to be
-R(w,v,v)
where R is the Riemann tensor. Right?"
Oz nods unconvincingly, and promises himself that he'll reread the
course notes for the definition of the Riemann tensor.
"Good," says G. Wiz. "But this is just the same as
R(v,w,v)
since the Riemann tensor is defined so that it's skew-symmetric in the
first two slots."
"Cool," says Oz.
"This is called the GEODESIC DEVIATION EQUATION, by the way. I've
cheated in 2 or 3 places in deriving it here --- for example, I hid some
epsilons under the table --- but the result is correct. Let me
summarize." The wizard had a tendency towards pedagogical pedantry
which Oz forgave him only because of his tendency to hurl thunderbolts
when interrupted. "Two initially comoving particles in free fall will
accelerate relative to one another in a manner determined by the
curvature of space. Suppose the velocity of one particle is v, and the
initial displacement from it to the second is small, so that it may be
represented as a vector w. Then the acceleration A of the second
relative to the first is given by R(v,w)v. Or if you like indices,
A^a = R^a_{bcd} v^b w^c v^d
So..." the wizard paused. "So, we can really determine the Riemann
curvature using experiments as you proposed, and using this formula."
"Now," asked the wizard, "remember how there are 20 components to the
Riemann tensor, of which 10 are determined by the Ricci curvature and 10
by the Weyl? If we are doing the case of a homogeneous isotropic big
bang model, most of those darn components should be redundant, thanks to
the symmetry. Can you figure out how many components we really need to
worry about in this big bang model?"
Oz gulped.
"By the way, you may enjoy remembering the definition of the Ricci and
Weyl tensor in terms of how a bunch of initially comoving test particles
begins to change volume and shape. We have a nice formula for the Ricci
tensor along these lines. A while back you asked we derived it. I said
"wait and see". Well, now you can derive it from the geodesic deviation
equation, at least if you are better at index juggling than I suspect
you are. After all the geodesic deviation equation says exactly what
those coffee grounds are going to do."
Oz suddenly got the feeling G. Wiz was going to assign him more
homework, so he casually looked at the grandfather clock in the corner
and said "Gee whiz! It's late! I have to start polishing the
doorknobs! You told me to finish them all by tonight, remember?"
G. Wiz smiled and nodded as Oz beat a hasty retreat. "Anyway," he
called after Oz, "try to see what you can say about the Ricci and Weyl
curvature in the big bang cosmology, using the coffee grounds business
and all the symmetry. That would really improve your grade on this
question!"
Oz
<egr...@nyc.pipeline.com> writes
>"Whoa!" thinks the hero, nursing his bruised arm and slowly pulling
>himself up. "This place is dangerous!" Just then he spies a feeble bit
>of firelight in the mouth of a nearby cavern, and painfully heads in that
>direction. In the cavern, of course, is his old friend, Oz.
>
>"What, you too?" says Oz, once a wizard in his own right, in a distant
>story. "Has your geodesic entered this place of no return?"
>
>"Yes" says the erg, pulling a stone up to the fire, and drawing his sword
>to dry it on some straw, finding that the fall in the heroic trollomachy
>was sufficient to dislodge it from the scabbard, "and the odd thing is, I
>am not even sure I remember crossing the horizon.
>
"Well," said Oz
"it's quite nice here really. In the summer you can walk over to the
village and have a drink at the village pub. Of course at the moment
it's a mite chilly."
"You gotta Wizard here?" said Ed
"Yup," said Oz "not a bad one really. Usual bad temper on occasion and
he has this tendency to drift off when you are talking to him. He starts
thinking about some higher thing and you gotta hold him down from fair
floating away".
"How powerful is he, then." asked Ed "Hmm, a fair sized keep, but not
one of them with a whole faculty to terrorise."
"Oh, he's pretty good." said Oz "Not that old either. He has to use a
wizening spell to look the part. Really enjoys his subject too, which
makes a change. You should see the pleasure he got from turning me into
a turd^H^H^H^H toad. He is trying to teach me about General Relativity,
but I'm making heavy weather of it. He *says* it's really powerful, I
hope it's powerful enough to hex the wench at the pub, must be ..
surely. Anyway he does it in a funny way, we have been at it for weeks
and we still haven't got to any sargents, let alone any generals.
There's all this stuff he gives me that I have to carve onto these stone
tablets."
Oz gestures casually at a mighty pile of inscribed stone tablets
balanced precariously over the chasm.
"It's a full time job keeping the Trolls away from them. These local
trolls claim to have a sophisticated palate. Humpf" sighed Oz.
"Anyway, Wiz is having a thaumaturlogical battle with some smart-ass
upstart from over the water at the moment. No work for me as he is glued
to his desk producing magic spells to win the battle. You want to have a
look at some of this stuff? There's a few questions that he doesn't like
my answers to, no sir, not at all." Groaned Oz
Ed walked over to the heap of tablets and pulled one out. On it, in
flickering glowing letters, were the Wizard's questions.
"Hey, no sweat." said Ed "Lets go through it slowly."
And then Ed said .......
Edward Green
befell that a stray piece of spacetime dilation drifted in from one of the
wizard's experiments in his fell castle, high up on the troll infested
mountain on a bleak outcropping. As this disturbance dropped into the
cave the part where Oz stood kind of slowed down, then seemed to black out
altogether! Only a few faint sparks jumping across the seams in erg's rusty
armor told him that the odd microwave and radio signal were still coming
from that part of the room.
"Oh great", erg thought, "Just what we need; a usenet timelag"
(Uniform Spacelike Entrapment; a wizardly technology using the coarse
woven fabric of space as a net to catch the unwary. Stray blobs of
redshift were either an industrial waste products or failed prototypes,
it's not clear which.)
"Well", mused the knight erring, "normally it's bad form to hold a
conversation with one's self, but there is no telling how long this could
last, or in whose proper time." So, looking around the visible portion of
the cave's floor, and careful not to put his hand in the menacing looking
red aura surrounding the missing region, he came upon the fell paper:
>"Dear Oz --
>If you wish to learn more general relativity I am afraid you will need
>to pass a test of your valor. So: answer me the following questions:
wedge shaped runes in the margin mean? Maybe if I held it sideways..."
A image of the wizard suddenly appeared on the sheet, laughing in a rather
unpleasant manner. "And stop saying 'fell', nitwit!"
"Yow!" he remarked, dropping the paper. "Maybe I should just try to
answer the questions. Let's see...
"1. Explain why, when the energy density within a region of space is
sufficiently large, a black hole must form, no matter how much the
pressure of whatever substance lying within that region attempts to
resist the collapse."
General Relativity pressure is just a form of energy, and a very high
pressure is just the same as a very high energy density; so "pressure"
sort of gets hoisted on its own petard: An extremely high pressure
*provides* its own containment. Or something.
Somehow I feel the answer should involve geodesics... It's a most
peculiar thing. A blackhole is a region in space where geodesics can get
*into* but cannot get *out*. But soft... isn't a geodesic a geodesic in
either direction? Of course it is. But one direction is "physical" and
the other is not. In particular, one direction of a timelike geodesic
describes a particle moving "forward in time" and the other, well, you
know... So GR evidently has an arrow of time build into it. Fair
statement? And what about "spacelike" geodesics. I suppose they would
represent a superluminal trajectory in all reference frames... or should I
say, all "proper" reference frames, the way timelike geodesics represent
sub-luminal trajectories in all proper reference frames...?
Is there an anti-black hole? A region of space(time) where stuff can only
*leave* from, but never get back in? Sure! The complement of a blackhole
is such a region... but is it strange that all blackholes are roughly
spherica in space, and all "anti-blackholes" are the complement of such
regions, or in fact the *one* anti-blackhole is the complement of all such
regions...
What happens to geodesics after they get *into* a blackhole? I suppose
they all wind up at a singularity. Is a black hole a mini big bang in
reverse? What if we embedded another big bang into *this* universe... would
that look like a spherical anti-blackhole...?
But all this makes blackholes sound like complicated global results about
tracing geodesics... I don't think it was supposed to be this
complicated...
"2. Explain how, in the standard big bang model, where the universe is
homogeneous and isotropic --- let us assume it is filled with some fluid
each point."
Oh damn. This is *just* what got me started on all that analyticity stuff.
*Part* of the curvature of spacetime at each point is dertermined by local
conditions. I think it was 10 of the 20 independent components of the
Riemann tensor, the 10 in the Ricci, the Weyl being free. So... maybe
in the standard big bang model, we just set the Weyl equal to zero? Why?
It seems too simple. Maybe it turns out that the partial differential
equations governing these components are the non-elliptic ones? That they
therefore represent something propagating, rather than something that is
locally evolving? After all, the "standard big bang model" must just be a
name for a simplified pet solution to the general equations... but how do
we enforce this simplification. Boundry conditions? Arbitrary
assignment?
"3. In the big bang model, what happens to the Ricci tensor as you go
back in past all the way to the moment of creation."
for. No doubt the density of stuff becomes infinite, so the Ricci becomes
infinite, and so forth. I know! It's a trick question. "Happens"
implies evolution forward in time... Nothing "happens" going back in the
past. :-)
Suddenly it becomes clear our apprentice hasn't a clue how to answer these
questions quantitatively, or even sharply qualitatively. Perhaps one of
the spectators will throw a rock with a clue wrapped around it into the
fantasy world of G. Wiz.
But finally the time shift seems to be clearing up, so let's hear what Oz
has to say...
--
E. R. (Ed) Green / egr...@nyc.pipeline.com
"All coordinate systems are equal,
but some are more equal than others".
Oz
<egr...@nyc.pipeline.com> writes
>
>Suddenly it becomes clear our apprentice hasn't a clue how to answer these
>questions quantitatively, or even sharply qualitatively. Perhaps one of
>the spectators will throw a rock with a clue wrapped around it into the
>fantasy world of G. Wiz.
>
>But finally the time shift seems to be clearing up, so let's hear what Oz
>has to say...
>
Oz re-appears in a kaleidoscope of naked women and strange goings on.
"Hey!" said Oz "There's a serious transdimensional time warping going on
in the link to Wiz-time over in the United Plates. A whole lotta demons
are using the lode-lines over there for their own nefarious purposes,
and the lode-masters have gone on short time working. "
"Gosh" said Ed "you had better be careful. You definitely wouldn't want
to cross-link with some of those. You couldn't IMAGINE what you might
drop into."
"It's really nice to talk to someone who doesn't think a quick ZAP! will
make you think better." sighed Oz. "I guess we can chat over this in
peace and quiet. What we gotta show Wiz is that d^2V/dt^2 heads negative
as V-> something small no matter what the momentum flow is."
Now d^2V/dt^2 = -(1/2)[T_{00} + T_{11} + T_{22} + T_{33}]
lets say T_{11} + T_{22} + T_{33} = T_{ss} for simplicity.
and as you say there must be a relationship between T_{00} and T_{ss}
and for expansion T_{ss} has gotta be negative. Now it seems to me that
T_{00} is a volume term so if we keep the same amount of energy in the
volume V the energy density will vary as 1/V or perhaps better 1/r^3 if
r is the characteristic radius of the volume V. If T_{ss} is really a
pressure term then you would expect it to vary as the surface area or as
1/r^2. The trouble is that I don't know of this is right. Dimensionally
it doesn't look right, but there are so many c's that have dissapeared
off as 1, that it's hard to be sure.
Anyway if this is about right then as r gets very small the T_{00} term
will always win out in the end over the T_{ss} term. Then d^2V/dt^2 will
become negative and the volume will start to decrease (as long as
dV/dt=0 at some point) and as it decreases so it will decrease even
faster, and eventually become (horror) zero.
Then there's this E^2 = p^2 + m^2 thingy I came across years ago. I'm
sure we ought to be able to work this in somewhere. What do you think?
Ed wiped his rusty sword, jumped up and swiped it around impressively.
Oz dived at the floor as it swung round over his thinning hair.
"Well," interjected Ed "what I think is this ......"
Oz
writes
>
>"Now for the big question! We want to know if the clock at Q' is moving
>away from the clock at P'. To answer this, we compare its velocity
>vector to the following vector: what we get by first parallel
>translating v along itself over to P', and then over to Q'. That's the
>velocity the clock at Q' would have it it were at rest relative to the
>clock at P'." Oz now sees a red dot go forwards in time from P to P'
>and then across from P' to Q', carrying a red copy of the tangent vector
>v with it.
>
> P'->----Q'
> | |
> | |
> v^ ^
> | w |
> P->-----Q
>
>The resulting red tangent vector at Q' is a bit different from
>the green one representing the actual velocity of the clock at Q'.
Well, of course in his heart of hearts Oz rather felt that the Wiz was
having it both ways. He rather felt that if he had drawn little square
rightangled thingies then G. Wiz would have said,
"Aha, using tachions again I see. Yes, well it *does* make it all much
easier doesn't it? Now go away and do it properly. zzzZZZAP!!!"
On the other hand, Wiz being very wise, always knows which bit of
approximation he can get away with. Hmmmm something about "experience"
Oz thought. Oz rather hoped he could catch a dose of experience, but it
did sound unpleasant.
>"Curvature!" cries Oz in a moment of revelation.
[Well, it paid to humor the old boy occasionally]
>"Right! We are taking the vector v and parallel translating it two
>different ways from P to Q' and getting two slightly different answers.
>If the answers were the same, the second clock would remain at rest
>relative to the first. But in fact they are not, and the difference
>tells us how the second one begins accelerating away from the first."
>
>"Now remember how curvature works: the result of
>
> dragging v from P to Q to Q'
>
>minus the result of
>
> dragging v from P to P' to Q'
>
>is going to be
>
> -R(w,v,v)
>
>where R is the Riemann tensor. Right?"
>
>Oz nods unconvincingly, and promises himself that he'll reread the
>course notes for the definition of the Riemann tensor.
Oz went red. Very, very red. He looked like he was going to explode.
Steam started coming out of his ears and his eyes started slowly and
deliberately popping out of his head. A small, very tiny, blue corona
started flickering faintly around his head. It looked like he was going
to explode. With a monumental effort of will he managed to slam down the
dampers before he went critical. It would probably only have resulted on
a pzzzittt instead of a zap, or probably only a ..... phit.
"Ahem, Wiz, er Wi-hiz, woo-hoo, Wizziness. That's exactly what I have
been going on about when I said that going in a little RECTANGLE will
leave you not back where you started. The vector joining where you end
up to where you started from must surely relate to the Riemann tensor.
In this case the little vector is directly related to the velocity
picked up by Q as it went to Q'. I will leave you (Wiz) to work this
out. Oz knew his questions were mostly rubbish, but sometimes he did
wish people would listen. HUMPF!!!"
>
>"Good," says G. Wiz. "But this is just the same as
>
> R(v,w,v)
>
>since the Riemann tensor is defined so that it's skew-symmetric in the
>first two slots."
>
>"Cool," says Oz.
Wondering what *exactly* "skew-symmetric in the first two slots" really
meant!
>"This is called the GEODESIC DEVIATION EQUATION, by the way. I've
>cheated in 2 or 3 places in deriving it here --- for example, I hid some
>epsilons under the table --- but the result is correct. Let me
>summarize." The wizard had a tendency towards pedagogical pedantry
>which Oz forgave him only because of his tendency to hurl thunderbolts
>when interrupted. "Two initially comoving particles in free fall will
>accelerate relative to one another in a manner determined by the
>curvature of space. Suppose the velocity of one particle is v, and the
>initial displacement from it to the second is small, so that it may be
>represented as a vector w. Then the acceleration A of the second
>relative to the first is given by R(v,w)v. Or if you like indices,
>
> A^a = R^a_{bcd} v^b w^c v^d
>
>So..." the wizard paused. "So, we can really determine the Riemann
>curvature using experiments as you proposed, and using this formula."
Was Oz pleased about this. You betcha. As soon as he had done the
washing up, he would have to have a look at R^a_{bcd} v^b w^c v^d. Well,
maybe after a beer or two with Ed, first. Well, maybe several. Oz noted
that the Wiz dropped into acceleration, rather assuming that curvature
only really existed in the t-direction. He also noted that all these
little deltas wandering about had differentiated themselves into
acceleration rather nicely. He kinda felt that these tensor jobbies
tended to do this without you really noticing. He would have to watch
out for that. It was the sort of thing that Wiz tended to assume was
obvious. He hoped they hid a lot of integration too. Preferably as much
as possible.
>"Now," asked the wizard, "remember how there are 20 components to the
>Riemann tensor, of which 10 are determined by the Ricci curvature and 10
>by the Weyl? If we are doing the case of a homogeneous isotropic big
>bang model, most of those darn components should be redundant, thanks to
>the symmetry. Can you figure out how many components we really need to
>worry about in this big bang model?"
>
>Oz gulped.
Well the BB couldn't almost by definition have any Weyl bits since it
must surely be described entirely by itself, so to speak. Of course it
would help if we knew which components in the Weyl were zero. Since the
Ricci tensor can be derived from the Stress-energy tensor we only (ha)
need to sort that out. Well, obviously T_{00} is important but you can
see that T_{11} = T_{22} = T_{33} and that these won't be zero, in fact
these should be rather large and possibly even comparable to T_{00}.
Then we have the spacial cross terms. Now I don't really know what these
represent. These are momentum in one direction being transported in
another. Now the only thing I can think of that might do that would be a
spinning particle. I suppose if we had a lot of spinning particles then
they might well cancel each other out if they were nice and random, but
I'm not totally convinced. I wonder what Wiz thinks.
Then we have the space-time cross terms T_{0X} and T_{X0}. I expect it's
obvious once you know, and probably also to do with spinning thingies
(actually I might have all this mixed up with the spacial ones now I
think about it) with luck they will all cancel out too.
I hope Wiz can discuss these cross thingies without getting, ahem,
cross.
Oz remembered all the chores he had to do. Sometimes they looked
positively attractive. He picked up his broom and ......
john baez
>OK, start from the beginning.
>1) If v is the velocity vector of a particle in the middle of a little
>ball of blah blah .... of volume V.
>
>so d^2V/dt^2 = -R_{ab} v^a v^b. Hokay.
It's actually
d^2V/dt^2 = -R_{ab} v^a v^b V,
since the rate of change of volume is proportional to the initial volume
as well as everything else. But this is no big deal here.
>2) Now to find R_{ab}
>
>R_{ab} = T_{ab} - (1/2)T^c_c g_{ab}
>
>and T^c_c = -T_{00} + T_{11} + T_{22} + T_{33}
>
>Now since it's uniform and isotrophic we should be able to say that
>Tc = T_{11} = T_{22} = T_{33} which also head towards infinity as V->0
As we discussed a few times, the diagonal entries T_{ii} (i = 1,2,3) are
just the pressure in the x, y, and z directions. Certainly isotropy
implies that they are equal, so we can just think of them all as
"pressure". Ted pointed out that your intuition is right; the pressure
approaches infinity as we go back in time towards the big bang,
particularly in the "radiation-dominated era" when there was a lot more
energy in the form of light. For light, pressure is comparable to
energy density (in sensible units where G = c = hbar = 1). In the
current "matter-dominated epoch", where the stars don't bump into each
other all that much, the pressure is negligible compared to the energy
density.
>and probably the cross terms are equal, at least some of them.
>Well, surely
>T_{12} = T_{21} = T_{13} = T_{31} = T_{23} = T_{32} = Td
>must be equal too in this situation. They presumably give the density of
>momentum flow in one direction travelling in another. Maybe something to
>do with things that spin or suchlike.
Can you use the isotropy to say more about these off-diagonal terms
T_{ij} (i,j = 1,2,3)? Hint: yes, you can! Certainly what you have said
so far is true, but it doesn't nearly exhaust the consequences of
rotational symmetry. You have only used rotations that switch the x, y,
and z axes. (Please, folks, don't post the answer to this question; let
my victims Oz and Ed tackle this one first.)
>Then we have T_{0n} and T_{n0} (n=1 to 3) like terms. What they
>represent physically is not too clear.
Remember, T_{ab} is the flow in the a direction of b-momentum. So the
component T_{0i} represents the density of i-momentum (i.e., x-momentum,
y-momentum, or z-momentum), while T_{i0} represents the flow of energy
in the i direction. Can you use isotropy to say something about these?
Hint: yes!
>So assuming we can take a minkowski metric in a small area, which seems
>unlikely...
You can't do it in a small finite region unless spacetime is flat in
that region --- since the Minkowski metric is flat. But you *can* do it
at a *single point*. Indeed, we may as well take the definition of a
Lorentzian metric --- the sort of metric we're interested in --- as one
which takes the form
-1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
in some coordinates at any given point. This is the essential
mathematical content of the "equivalence principle": at every point,
there are coordinate systems in which the metric looks just like that of
good old Minkowski space.
So, what you do below is fine, since you're doing it at a single
(arbitrary) point:
>but still, we get
R_{ab} =
[(1/2)(T_{00}+T_{11}+T_{22}+T_{33}) , T_{01} , T_{02} , T_{03}]
[T_{10} , (1/2)(T_{00}+T_{11}-T_{22}-T_{33}), T_{12}, T_{13}]
[T_{20} , T_{21}, (1/2)(T_{00}-T_{11}+T_{22}-T_{33}), T_{23}]
[T_{30} , T_{31}, T_{32} , (1/2)(T_{00}-T_{11}-T_{22}+T_{33})]
=
[(1/2)(T_{00} + 3Tc) , T_{01} , T_{02} , T_{03}]
[T_{10} , (1/2)(T_{00} - Tc), T_{12}, T_{13}]
[T_{20} , T_{21}, (1/2)(T_{00} - Tc), T_{23}]
[T_{30} , T_{31}, T_{32} , (1/2)(T_{00} - Tc)]
>Well, a bit untidy...
Yes, it's a bit untidy, so use isotropy to the hilt and boil it down to
something a lot simpler!
>however we can see that
>
>R_{11} = (1/2)(T_{00} - Tc) [Nb Tc=T_{11} etc]
I don't like calling that "Tc"; it reminds me too much of the critical
temperature of a superconductor or something. Since you know that
T_{11} = T_{22} = T_{33} is equal to the pressure in this isotropic
situation, why don't you just call it 3P?
>Now, as the universe gets hotter and smaller, one could imagine that Tc
>gets bigger. In fact one could imagine that Tc could even get to be as
>big as T_{00}. Lots of interactions producing massless particles and
>all. In this case R_{11} *might* become zero. With T_{00} going
>infinite, this is attractive. <groan>
Yes indeed, the pressure can dominate the energy.
>3) Now let's see.
>
>d^2V/dt^2 = R_{ab} v^a v^b.
>
>Now I can pick out each term by my choice of a & b. However it's not
>very clear, without explanation, what for example R_{12} means. So I
>will skip this for now. One optimistically hopes they turn out to be
>zero.
As for a simple conceptual understanding, it's easiest to understand
R_{00}, as we've done. As for your optimism, you will have to follow my
hints above to see if it's well-founded.
By the way, you can't "pick" a and b, exactly. Remember, we sum over
all a and b in the formula above! You can pick v, if you like.
>One is tempted to imagine a cone filled with tangential
>spheres from an itsy bitsy one at t=epsilon to a big one centered at
>t=10^10 years. However this can't be quite right because the curvature
>of a sphere is constant. In the 4D thingy the curvature is forever
>decreasing from t=0 to t=10^10y.
It's certainly true that we can picture a closed big-bang universe as
a sphere (a 3-sphere) whose radius increases with time. The radius is
not exactly a linear function of time so it's not exactly a cone. But
you are right that as the radius increases, the curvature decreases.
Note how that fits with the fact that pressure and energy density are
decreasing.
>I need more 'obvious' information to proceed further.
I hope to have provided some. But what you really need to proceed
further is to ponder isotropy a bit more.
You're doing fine, by the way.
Oz
writes
>
>>and probably the cross terms are equal, at least some of them.
>>Well, surely
>>T_{12} = T_{21} = T_{13} = T_{31} = T_{23} = T_{32} = Td
>>must be equal too in this situation. They presumably give the density of
>>momentum flow in one direction travelling in another. Maybe something to
>>do with things that spin or suchlike.
>
>Can you use the isotropy to say more about these off-diagonal terms
>T_{ij} (i,j = 1,2,3)? Hint: yes, you can! Certainly what you have said
>so far is true, but it doesn't nearly exhaust the consequences of
>rotational symmetry. You have only used rotations that switch the x, y,
>and z axes. (Please, folks, don't post the answer to this question; let
>my victims Oz and Ed tackle this one first.)
Well, I took 'isotrophic' to mean isotrophic in space. It's evolving in
time so I don't feel that I ought to start making it isotrophic there
without a good argument.
>>Then we have T_{0n} and T_{n0} (n=1 to 3) like terms. What they
>>represent physically is not too clear.
>
>Remember, T_{ab} is the flow in the a direction of b-momentum. So the
>component T_{0i} represents the density of i-momentum (i.e., x-momentum,
>y-momentum, or z-momentum), while T_{i0} represents the flow of energy
>in the i direction. Can you use isotropy to say something about these?
Oops. I thought I had set T_{0i} (i=1..3) equal and T_{i0} (i=1..3)
equal, but I hadn't. Clearly they are equal. It would be worth
investigating if we could plausibly make these zero as this would
simplify things quite a bit. Well T_{i0} representing the flow of energy
in the i direction looks a good candidate. There should be as much going
in the -i direction as the +i direction so I vote for T_{i0} = 0.
Now this makes R_{ab} look terribly asymmetric with those T_{0i} terms
representing the density of i-momentum. Not good for an isotrophic
uniform universe. Oh, well the argument is the same, momentum is a
vector. So in any small volume if we add all the little vectors up, that
are travelling in the i-direction, +ve and -ve, we will get zero net
density. Er, I think.
How did I miss this before? Is it right? Sounds plausible anyway.
Oh dear. Now all those T_{ij} (i,j = 1 to 3) look terribly out of place.
Wouldn't it be nice to have them zero as well? Let's see. T_{ij} must be
the flow in the i direction of momentum in the j direction. Now we gotta
be careful here or we will convince ourselves that T_{ii} equals zero
too. I have the distinct feeling that this will result in a period of
time spent as a colonic parasite, which is even worse than an intestinal
one. Well, T_{ii} is a pressure. It's a little hard to see how T_{ij}
(j=/=i) could be a pressure as long as we keep our hands waving at all
times since the flow is perpendicular to the action, so to speak. Of
course if we knew the derivation of T_{ii} being a pressure then it would
almost certainly be perfectly clear. Anyway, throwing caution to the
winds, and without a really proper explanation, lets set T_{ij} (i,j=1 to
3; j=/=i) = 0 too. In for a penny in for a pounding.
so R_{ab} =
[(1/2)(T_{00}+T_{11}+T_{22}+T_{33}) , T_{01} , T_{02} , T_{03}]
[T_{10} , (1/2)(T_{00}+T_{11}-T_{22}-T_{33}), T_{12}, T_{13}]
[T_{20} , T_{21}, (1/2)(T_{00}-T_{11}+T_{22}-T_{33}), T_{23}]
[T_{30} , T_{31}, T_{32} , (1/2)(T_{00}-T_{11}-T_{22}+T_{33})]
or 2R_{ab} =
[(T_{00} + 3P),0 ,0 ,0 ]
[0 ,(T_{00} - 3P),0 ,0 ]
[0 ,0 ,(T_{00} - 3P),0 ]
[0 ,0 ,0 ,(T_{00} - 3p)]
Which is noticeably simpler. Also possibly wrong, anyway
d^2V/dt^2 = -2V R_{ab} v^a v^b
so now we want to know what this ball is doing.
No, I'm going to need some hints as to where to go next. Obviously I can
now pick various vectors v=(1,0,0,0) and v=(0,1,0,0) or even if one felt
masochistic v=(1,1,1,1). Well, let's do two to show willing.
1) v=(1,0,0,0) Or how the volume varies in the t-direction?
d^2V/dt^2 = -2V R_{ab} v^a v^b = -2V(T_{00} + 3P)
>>Now, as the universe gets hotter and smaller, one could imagine that 3P
>>gets bigger. In fact one could imagine that 3P could even get to be as
>>big as T_{00}. Lots of interactions producing massless particles and
>>all.
I assume that 3P is in fact negative. In other words it resists the
contraction of the ball. Now of course we haven't considered dV/dt. One
has the intuitive feel that 3P and dV/dt ought to be related rather
strongly, more so if much of the energy is composed of mass. If all the
energy were photons then one might not be astonished if somebody who had
considered this carefully were to suggest that d^2V/dV^2 = 0 or even
became positive.
Then there is the problem of blackholiness. Now this (one above)
derivation (ho-ho-ho) is odd in that we are enforcing a minkowski metric
at every small point. We are also (at least I am at the moment) setting
the Weyl tensor at zero. So we are 'sort of' saying that curvature due to
distant thingies cancels out due to the uniform isotrophy of everything.
This doesn't seem to be the same case I would expect from a blackhole
derivation where 'distant' space is minkowskian and the Weyl tensor
presumably dominates due to the 'point' source. In other words the
blackhole situation is very far from uniform and isotrophic.
2) v=(0,1,0,0) Or how the volume varies in the x-direction???????
d^2V/dt^2 = -2V R_{ab} v^a v^b = -2V(T_{00} - 3P)
Well, I have a conceptual problem here. I have a gut feeling that the
volume here ought to be a (0,1,1,1) type volume and we should be
considering d^2V/dt^2 or possibly dV/dt.dv/dx. Anyway because so many
derivations are skipped (probably thankfully) it is not very clear what
this represents. So someone will have to tell me. One can guess all over
the place, but it's better to be told. Or perhaps hinted at heavily, very
heavily.
PS I hope I am not toooooooo far out.
Oz
writes
>
>Oz gulped.
>
>"By the way, you may enjoy remembering the definition of the Ricci and
>Weyl tensor in terms of how a bunch of initially comoving test particles
>begins to change volume and shape. We have a nice formula for the Ricci
>tensor along these lines. A while back you asked we derived it. I said
>"wait and see". Well, now you can derive it from the geodesic deviation
>equation, at least if you are better at index juggling than I suspect
>you are. After all the geodesic deviation equation says exactly what
>those coffee grounds are going to do."
Well, Oz was well known for not having enough common sense to keep his
fingers out of the fire. So he came back with a few questions.
1) Riemann = Ricci + Weyl
R^a_{bcd} = R^c_{bcd} + W^a_{bcd} was given somewhere or other.
and R^c_{bcd} = R_{bd)
So lets look at the Ricci tensor. This says that if we have three
vectors (lets make them basis for simplicity) then the Ricci tensor will
output a vector giving the curvature along one of the *input* vectors.
Anyway something related to curvature. Now this is a subset of the
Riemann tensor. It's some sort of subset of diagonals of the Riemann
tensor but being 4D it's a little hard to pick it out. What we can say
is that it has 16 elements. Now as you can imagine I am slowly sinking,
but in for a penny.
Now you all say there are only ten independent items in the 16 elements.
This makes one want to set the cross terms equal (ie R_ij = R_ji). Well,
one would anyway since this is nothing more than switching your input
vectors about so it looks reasonable.
So one imagines that something *analogous* to the following is going on:
[00,01,02,03] = [00,01,02,03] + [- ,- ,- , -] in a wierd but clear
[10,11,12,13] [10,- ,- ,- ] [- ,11,12,13] notation.
[20,21,22,23] [20,- ,- ,- ] [- ,21,22,23]
[30,31,32,33] [30,- ,- ,- ] [- ,31,32,33]
= [00,01,02,03] +
[10,- ,- ,- ] [11,12,13]
[20,- ,- ,- ] [21,22,23]
[30,- ,- ,- ] [31,32,33]
Now lets look a what Wiz says about the Riemann tensor R^a_{bcd}
If x^a,x^b,x^c are basis vectors then it spits out a
vector a *whose component in the x^a direction* is R^a_{bcd}
so presumably the component in the x^c direction is R^c_{bcd} which is
the Ricci tensor. Whew! Took me a bl**dy long time to work that out!!!
Now what does this mean. Well, it means that if I go in a little loop
along v^b and w^d carrying a vector u^c then u^c, ooops. Thats what
skew-symmetric in the first two slots means. It means R(wvv)=-R(vwv).
Anyway, we can begin to see why it's a rate of change of volume term. If
we define a small volume as being of radius (or side of a cube if you
like) epsilon. Then the volume is epsilon^3. Basis vectors x^a,x^b,x^c
spacially, and x^d temporally.
Now let's evolve this little system dt along the time axis. Compared to
the center of our little sphere, in a distance dt x^a has evolved into a
new vector whose incremental change in the x^a direction is epsilon dt
R(x^a,x^d,x^d) and similarly for x^b and x^c. So ignoring a few higher
order terms the new volume is (e=epsilon)
[e+ e dt R(x^a,x^d,x^d)][e+ e dt R(x^b,x^d,x^d)][e+ e dt R(x^c,x^d,x^d)]
or dV=[e dt R(x^a,x^d,x^d)][e dt R(x^b,x^d,x^d)][e dt R(x^c,x^d,x^d)]
or dV=e^3 dt^3 R(x^a,x^d,x^d) R(x^b,x^d,x^d) R(x^c,x^d,x^d)
Oh shit! I'm nearly there but I have this darned dt^3 that should be dt
and it's not a rate of change. Somewhere I should have had a dx/dt type
term. I am not quite grasping this yet. It's a velocity that I get out
of the Riemann tensor really and not a displacement. Anyway, I haven't
any more time I can spend on this now.
I humbly request some wizardly help.
And how you do it using index juggling I haven't a clue. I am pretty
certain I know why. This index juggling is differentiating I think, but
without a course or help, I don't quite see just yet.
Heeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeelp!
Oz
Anyway, we can begin to see why it's a rate of change of volume term. If
we define a small volume as being of radius (or side of a cube if you
like) epsilon. Then the volume is epsilon^3. Basis vectors u,v,w
spacially, and t temporally.
Now let's evolve this little system dt along the time axis. Compared to
the center of our little sphere, in a distance dt then v has evolved
into a new vector whose incremental change in the v direction is epsilon
dt R(v,t,t) and similarly for u and w. So ignoring a few higher order
terms the new volume after time dt is (e=epsilon) is V' =
[e+ e dt R(w,t,t)][e+ e dt R(v,t,t)][e+ e dt R(w,t,t)]
so dV = V - V'= e^3 - (that lot above) in time dt
or dV/dt = e^3 dt[ R(u,t,t)+ R(v,t,t) + R(w,t,t)]
so d^2V/dt^2 = V [ R(u,t,t)+ R(v,t,t) + R(w,t,t)]
Well, I think I've fiddled this a bit but it kinda looks plausible.
Er, doesn't it?? Ahem, cough cough.
So now all I have to do is show that
R(u,t,t)+ R(v,t,t) + R(w,t,t) is equivalent to R^a_{bcd} v^a w^c v^d
At which point I roll over and play dead. Oh, I don't know, if I squint
my eyes enough .........
This isn't kosher. I think it sort of signposts the way to go though.
Heeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeelp!!
john baez
>Well, I took 'isotrophic' to mean isotrophic in space. It's evolving in
>time so I don't feel that I ought to start making it isotrophic there
>without a good argument.
By the way, it's ISOTROPIC, not "isotrophic"... my gentle hints (like
spelling the word correctly) seem not to be getting through...
Yes, the big bang universe is only isotropic in space, and using this
you should be able to conclude a lot about the space components
T_{ij} (i,j = 1,2,3) of the stress-energy tensor... and similarly for
the Ricci tensor!
>>>Then we have T_{0n} and T_{n0} (n=1 to 3) like terms. What they
>>>represent physically is not too clear.
>>Remember, T_{ab} is the flow in the a direction of b-momentum. So the
>>component T_{0i} represents the density of i-momentum (i.e., x-momentum,
>>y-momentum, or z-momentum), while T_{i0} represents the flow of energy
>>in the i direction. Can you use isotropy to say something about these?
>Oops. I thought I had set T_{0i} (i=1..3) equal and T_{i0} (i=1..3)
>equal, but I hadn't. Clearly they are equal. It would be worth
>investigating if we could plausibly make these zero as this would
>simplify things quite a bit. Well T_{i0} representing the flow of energy
>in the i direction looks a good candidate. There should be as much going
>in the -i direction as the +i direction so I vote for T_{i0} = 0.
Yes, indeed, isotropy makes these zero! The net flow of energy in any
given direction through a point is zero, or there'd be a preferred
direction there. So T_{i0} = 0.
>Now this makes R_{ab} look terribly asymmetric with those T_{0i} terms
>representing the density of i-momentum. Not good for an isotrophic
>uniform universe. Oh, well the argument is the same, momentum is a
>vector. So in any small volume if we add all the little vectors up, that
>are travelling in the i-direction, +ve and -ve, we will get zero net
>density. Er, I think.
>How did I miss this before? Is it right? Sounds plausible anyway.
It's right. The density of momentum in the i-direction must be zero, or
there'd be a preferred direction.
Okay, so now you've got rid of those irksome T_{i0} and T_{0i} guys.
>Oh dear. Now all those T_{ij} (i,j = 1 to 3) look terribly out of
>place.
Indeed.
>Wouldn't it be nice to have them zero as well?
Indeed.
>Let's see. T_{ij} must be
>the flow in the i direction of momentum in the j direction. Now we gotta
>be careful here or we will convince ourselves that T_{ii} equals zero
>too. I have the distinct feeling that this will result in a period of
>time spent as a colonic parasite, which is even worse than an intestinal
>one. Well, T_{ii} is a pressure. It's a little hard to see how T_{ij}
>(j=/=i) could be a pressure as long as we keep our hands waving at all
>times since the flow is perpendicular to the action, so to speak. Of
>course if we knew the derivation of T_{ii} being a pressure then it would
>almost certainly be perfectly clear.
The derivation is quite simple and I think I said it before. For
example, T_{xx} is the flow in the x direction of x-momentum. Suppose
for visual vividness that this flow is carried by little ball-shaped
atoms. Then if you put a wall in their way they push on the wall in
the x direction, with a certain pressure. This pressure is a force per
unit area, which is just the same as "momentum per time per unit area".
This is given by the flow of x-momentum in the x-direction!
>Anyway, throwing caution to the
>winds, and without a really proper explanation, lets set T_{ij} (i,j=1 to
>3; j=/=i) = 0 too. In for a penny in for a pounding.
Good! Your intuitions are leading you right here... those off-diagonal
terms T_{ij} (i,j=1,2,3) are also zero.
By the way, a tensor like T_{ij} --- a symmetric (0,2) tensor on space
--- can only be invariant under rotations if it is diagonal and all the
diagonal entries are equal. That is how the math experts out there
would have figured this out. But you triumphed using just the physics
of this example! And I knew you could... believe it or not, I am not
setting you off on an impossible quest... just testing your valor,
that's all.
>so R_{ab} =
>[(1/2)(T_{00}+T_{11}+T_{22}+T_{33}) , T_{01} , T_{02} , T_{03}]
>[T_{10} , (1/2)(T_{00}+T_{11}-T_{22}-T_{33}), T_{12}, T_{13}]
>[T_{20} , T_{21}, (1/2)(T_{00}-T_{11}+T_{22}-T_{33}), T_{23}]
>[T_{30} , T_{31}, T_{32} , (1/2)(T_{00}-T_{11}-T_{22}+T_{33})]
>or 2R_{ab} =
>[(T_{00} + 3P),0 ,0 ,0 ]
>[0 ,(T_{00} - 3P),0 ,0 ]
>[0 ,0 ,(T_{00} - 3P),0 ]
>[0 ,0 ,0 ,(T_{00} - 3P)]
>Which is noticeably simpler. Also possibly wrong...
It looks right to me. Why don't we call T_{00} something like E, the
energy density. So, you've seen what the Ricci tensor is in terms of
E and P. Now you are pretty close to answering the question. What was
the question, anway? Oh yes...
3. In the big bang model, what happens to the Ricci tensor as you go back in
past all the way to the moment of creation?
So, all you need is to tell me what happens to E and P.
>d^2V/dt^2 = -2V R_{ab} v^a v^b
Let's see, I don't think there should be a 2 there.
>so now we want to know what this ball is doing.
>No, I'm going to need some hints as to where to go next. Obviously I can
>now pick various vectors v=(1,0,0,0) and v=(0,1,0,0) or even if one felt
>masochistic v=(1,1,1,1). Well, let's do two to show willing.
>1) v=(1,0,0,0) Or how the volume varies in the t-direction?
>d^2V/dt^2 = -2V R_{ab} v^a v^b = -2V(T_{00} + 3P)
Yes, this is the most interesting case. If you keep with it, by the
way, you can do a lot more than I hoped for... you can figure out pretty
much everything about the big bang solution! But let's see, I'm not
sure this stuff about the ball of particles will help you figure out
what happens to R_{ab} as you approach the initial singularity. (It
will help you understand what it *means* though.)
>I assume that 3P is in fact negative.
Why should the pressure be negative? Pressure is usually positive.
>In other words it resists the contraction of the ball.
Hmm. Sounds fishy to me.
>Now of course we haven't considered dV/dt.
>Then there is the problem of blackholiness.
Aren't we talking about the big bang in this problem? Have you shifted
over to considering a black hole? I'm confused.
>We are also (at least I am at the moment) setting the Weyl tensor at
>zero.
Oh yeah? Why are you doing that? I'm not saying it's bad, but explain
why you're doing that? (Hint: isotropy, isotropy, isotropy.)
john baez
O...@upthorpe.demon.co.uk writes:
John Baez wrote:
P'->----Q'
| |
| |
v^ ^
| w |
P->-----Q
>>"We are taking the vector v and parallel translating it two
>>different ways from P to Q' and getting two slightly different answers.
>>If the answers were the same, the second clock would remain at rest
>>relative to the first. But in fact they are not, and the difference
>>tells us how the second one begins accelerating away from the first."
>>"Now remember how curvature works: the result of
>> dragging v from P to Q to Q'
>>minus the result of
>> dragging v from P to P' to Q'
>>is going to be
>> -R(w,v,v)
>>where R is the Riemann tensor. Right?"
>Oz went red. Very, very red. He looked like he was going to explode.
>Steam started coming out of his ears and his eyes started slowly and
>deliberately popping out of his head. A small, very tiny, blue corona
>started flickering faintly around his head. It looked like he was going
>to explode. With a monumental effort of will he managed to slam down the
>dampers before he went critical. It would probably only have resulted on
>a pzzzittt instead of a zap, or probably only a ..... phit.
>"Ahem, Wiz, er Wi-hiz, woo-hoo, Wizziness. That's exactly what I have
>been going on about when I said that going in a little RECTANGLE will
>leave you not back where you started. The vector joining where you end
>up to where you started from must surely relate to the Riemann tensor.
>In this case the little vector is directly related to the velocity
>picked up by Q as it went to Q'. I will leave you (Wiz) to work this
>out. Oz knew his questions were mostly rubbish, but sometimes he did
>wish people would listen. HUMPF!!!"
The wizard smiled and said, "Yes. Though it might not have occured to
you, I was indeed listening. When you drew this picture:
*
|\ The '*' are my little parallelogram
| \ for my Riemann loop.
| \ The light pulses are at 45 deg as req'd.
* *
.\ |
. \ |
. \|
. * All set up point.
. .
. .
. . Trundle out leg.
..
..
.
.
it was clear you were on the right track; you knew that the Riemann
curvature could be measured experimentally using a setup along these
lines. But you were, shall we say, less than fully forthcoming about
EXACTLY how the components of the Riemann curvature were to be computed.
I recall you saying:
"Now exactly how you interpret this as a rotation of your tangent vector,
I am not rightly sure. My guess is that you deduce that the distant leg
is 'really' at a small angle determined by it's relevant velocity
accumulated over the time it travelled (vertically) , which being a
dx/dt thingy, looks like a slope to me."
So at this point I stepped in and kindly gave you the precise formula
you were after, namely the geodesic deviation equation. The relative
acceleration of the two clocks is exactly -R(w,v,v) where w is the
initial displacement between the two clocks and v is the velocity of the
first clock. By the way, it doesn't matter much whether w is chosen to
be timelike, as in my picture, or lightlike, as in your experimentally
more practical setup. The formula remains the same!
So, it's -R(w,v,v).
>>"Good," says G. Wiz. "But this is just the same as
>>
>> R(v,w,v)
>>since the Riemann tensor is defined so that it's skew-symmetric in the
>>first two slots."
>>"Cool," says Oz.
>Wondering what *exactly* "skew-symmetric in the first two slots" really
>meant!
"Oh Oz," said the Wizard, shaking his head. You are always looking for
deep mysteries where there are none. "Skew-symmetric in the first two
slots simply means that -R(w,v,v) = R(v,w,v)! When you switch the first
two guys, the sign changes. The point is that if you recall the
definition of the Riemann tensor, it's blitheringly obvious that it has
this property."
>Was Oz pleased about this. You betcha. As soon as he had done the
>washing up, he would have to have a look at R^a_{bcd} v^b w^c v^d. Well,
>maybe after a beer or two with Ed, first. Well, maybe several. Oz noted
>that the Wiz dropped into acceleration, rather assuming that curvature
>only really existed in the t-direction.
[Of course, the Wizard was doing no such thing. He was simply
concentrating on curvature in the time direction because that was what
affected the behavior of nearby *timelike* geodesics.]
>>"Now," asked the wizard, "remember how there are 20 components to the
>>Riemann tensor, of which 10 are determined by the Ricci curvature and 10
>>by the Weyl? If we are doing the case of a homogeneous isotropic big
>>bang model, most of those darn components should be redundant, thanks to
>>the symmetry. Can you figure out how many components we really need to
>>worry about in this big bang model?"
>>
>>Oz gulped.
>
>Well the BB couldn't almost by definition have any Weyl bits since it
>must surely be described entirely by itself, so to speak.
Good intuition, but it might help to recall the rough definition of the
Weyl bits we are currently working with.
>Well, obviously T_{00} is important but you can
>see that T_{11} = T_{22} = T_{33} and that these won't be zero, in fact
>these should be rather large and possibly even comparable to T_{00}.
>Then we have the spacial cross terms. Now I don't really know what these
>represent. These are momentum in one direction being transported in
>another.
Isotropy, isotropy, isotropy.
Oz
writes
>By the way, it's ISOTROPIC, not "isotrophic"... my gentle hints (like
>spelling the word correctly) seem not to be getting through...
Ah, one of those deeply ingrained spelling errors. If years of spelling
it incorrectly have failed to make an impression, your gentle chiding is
unlikely to have much effect. I will make an effort to spot it to avoid
you in too much agitation. I can provide plenty of that simply by
struggling to answer your tests!
R_{ab} =
[(E + 3P),0 ,0 ,0 ] . (1/2)
[0 ,(E - 3P),0 ,0 ]
[0 ,0 ,(E - 3P),0 ]
[0 ,0 ,0 ,(E - 3P ]
>
>3. In the big bang model, what happens to the Ricci tensor as you go back in
>past all the way to the moment of creation?
>
>So, all you need is to tell me what happens to E and P.
>
>>d^2V/dt^2 = -V R_{ab} v^a v^b
>
>>1) v=(1,0,0,0) Or how the volume varies in the t-direction?
>
>>d^2V/dt^2 = -V R_{ab} v^a v^b = -(1/2)V(E + 3P)
>
>Yes, this is the most interesting case. If you keep with it, by the
>way, you can do a lot more than I hoped for... you can figure out pretty
>much everything about the big bang solution! But let's see, I'm not
>sure this stuff about the ball of particles will help you figure out
>what happens to R_{ab} as you approach the initial singularity. (It
>will help you understand what it *means* though.)
>
>>I assume that 3P is in fact negative.
>
>Why should the pressure be negative? Pressure is usually positive.
>
>>In other words it resists the contraction of the ball.
>
>Hmm. Sounds fishy to me.
Eh? Here you are looking at a ball of hot gas. It's a mini-universe. If
3P is positive then R_{00} is always positive and the ball will always
contract. Oh-ho well, perhaps ...., after all our universe is expanding
but at a decreasing rate. Well, I'm still not convinced. Lets take an
almost massless ball of hot gas stuck all alone in minkowski space. Hey
looky here, it expands, presto! Oz gets a large cannon and blows his
head clean off. OK, you are right. The CURVATURE is positive. That's
what I had in mind way back when I first partially answered way up the
thread. I said that R_{00} was always positive so would go to infinity.
Forget the coffee grains. Let's stick to curvature.
R_{00} = (1/2)(E + 3P) Both E & P must -> infinity as size->0
So the curvature sure tends to infinity in the time direction.
>
>>Now of course we haven't considered dV/dt.
Yeah. The coffee grounds are not an appropriate analogy.
>
>>Then there is the problem of blackholiness.
>
>Aren't we talking about the big bang in this problem? Have you shifted
>over to considering a black hole? I'm confused.
I was trying to pick out how the BB differs from the BH.
This way I avoid making too many stupidities.
>>We are also (at least I am at the moment) setting the Weyl tensor at
>>zero.
>
>Oh yeah? Why are you doing that? I'm not saying it's bad, but explain
>why you're doing that? (Hint: isotropy, isotropy, isotropy.)
Eh? I thought I gave a quasi-plausible reason way way back at the
beginning of the thread. My argument went as follows:
a) There ain't no boundary conditions since we include the entire
universe. From what has been said about the Weyl tensor, some of it's
terms relate to boundary conditions. Maybe tidal effects 'n stuff. These
gotta be zero.
b) The universe is uniformly isotroph^Hic so there ain't no waves or
other transient stuff wandering about. This should take out another
whole load of terms from the Weyl.
c) So far we have only considered a teeny point-like bit of this
universe. I have a vague memory that we have, at some point, made use of
the approximation that in a teeny volume, spacetime is minkowskian.
However I can't recall exactly where. On the other hand the R_{ab} we
have derived is far from flat so hopefully this is a valid approximation
in the limit.
I am a mite concerned that the Weyl does include static curvature from
distant bits. Hopefully in this situation these end up zero too.
<nervous grin>
2) In the spacial directions.
R_{11} = (1/2)(E - 3P) Now hang on, what IS *R*_{11}?
R_{ab} = R^c_{acb} aaaah, this is the key.
We have calculated a subset of the Riemann tensor, and if all the other
elements (if the Weyl elements =0) are zero then we have calculated the
Riemann tensor itself. I think I'll go and lie down.
The problem now is to get a proper handle on what curvature 'feels'
like. I mean actual results. Now a curvature of 1 ought to mean that we
end up with ordinary flat pythagorean space stuff, I guess. No that's
too general a statement, all those cross terms. Let's call on Riemann
himself, perhaps.
So if we input a trio of vectors into the Riemann we have derived.
Well, let's try v = (1,0,0,0) = a = b = c.
This should output a vector in the c-direction giving the curvature in
that direction. Oooooooo dear.
Perhaps I'll post later on this.
john baez
>In article <h8i5lNAW...@upthorpe.demon.co.uk>, Oz
><O...@upthorpe.demon.co.uk> writes
>Poor old Oz is depressed. He fears that the Wiz is going to be terribly,
>terribly cross with him. Her needs that most exquisitly boring of
>subjects, an elementary Tensor revision course. <abject embarrasement>
Boring, boring, boring. Next we'll be reviewing partial differential
equations and Fourier transforms.
>Q1) Example T^c_c = g^(ca)T_{ac} = scalar
>
>Oz feels that there should also be a product that is another tensor.
A product of g and T that's a tensor? Sure, like g_{ab}T_{cd}.
That's rank (0,4): eats four vectors for breakfast and spits out a
scalar at noon.
By the way, why are you using a mix of {}'s and ()'s? There's no point
in doing that here, and it could even be confusing if I didn't know that
you couldn't possibly know what () meant in this context. We just use
^{}'s and _{}'s as a substitute for writing batches of superscripts or
subscripts, respectively.
>He
>also wonders if the reversal of the sub and superscript order is
>important. He also wonders if g_{ca}T^(ac) is different, and why?
Well, the order is very important in general, but the metric and the
stress energy just HAPPEN to be symmetric: g_{ab} = g_{ba}, and T_{ab} =
T_{ba}. Figure out why.
Show that g_{ac} T^{ac} = g^{ac} T_{ac}, using the fact that g^{ab} is
the inverse matrix of g_{ab}, and maybe some other facts.
>He
>suspects that g_{ab}T_{cd} would be a tensor of form U_{abcd}.
Yup.
>Q2) R^a_{bcd} This takes in three vectors and outputs one.
>He thinks this could be evaluated as something that takes in three
>vectors u^b,v^c,w^d to output a vector q^a like this:
>q^n=sum(i=0 to 3){sum(j=0 to 3)[sum(k=0 to 3)<R(n,i,j,k).u^i.v^j.w^k>]}
>Is this right (if you can work out what I actually said!)?
Almost; you mean R^n_{ijk} where you wrote R(n,i,j,k). Also, the Einstein
summation convention is designed to prevent such an obscene proliferation
of summation signs. The quick way to write what you wrote is:
q^n = R^n_{ijk} u^i v^j w^k
>In other words R^a_{bcd} is a 4x4x4x4 matrix of 256 elements.
Yes, but thankfully, it possesses enough symmetries to reduce down to
only 20 independent elements. Exercise: show from the the definition
that R^a_{bcd} = -R^a_{cbd}. You know, that "skew-symmetric in the
first two slots" businesss.
>Q3) Now when we raise an index how does that alter the terms in the
>array? I suspect it's not trivial.
Well, index raising is done using the metric and the Einstein summation
convention, so e.g. if we want to raise the first index on X_{abcd} we
do this
X^a_{bcd} = g^{an} X_{nbcd}
where we sum over n. So yes, this can alter the terms in a severe way.
>Q4) Now I suspect that to get from R_{ab} to a form like R^a_{bcd) I
>would need to blend a g_{cd} with R_{ab} to get a R_{abcd} then raise an
>index to get R^a_{bcd}.
That's right.
>I think this will be very, very messy.
Nah, it's easy. Maybe my message didn't get across the Atlantic yet,
but the formula for the Riemann in terms of the Weyl and the Ricci is
R^a_{bcd} = (1/4) R_{bd} delta^a_c + W^a_{bcd}
You can also think of this the way you suggest. Starting with
R_{abcd} = (1/4) R_{bd} g_{ac} + W_{abcd}
we can raise the index a and get
R^a_{bcd} = (1/4) R_{bd} delta^a_c + W^a_{bcd}
using the fact that g^a_c = delta^a_c.
>Also g_{ab} is which metric.
It's whatever the hell metric is the actual metric on spacetime that you
happen to be studying!!!!!!!!
Crucial, crucial point. Remember, the metric describes the geometry of
spacetime. All sorts of things depend on the metric. You gotta use the
metric you actually are studying, you can't just pull one out of the hat.
>Can it be simply the minkowskian one, despite
>the evidence that the metric is likely to be non-minkowskian?
No way. That'd be like randomly picking an electric field and
stuffing it into the problem you're solving, instead of using the
physically correct one.
>Q5) I also suspect that this brute force way of doing it is silly. All
>we need to know is in R_{ab}, and a diagonalised form at that. How
>should I proceed trying to see what this means?
It depends on what you're doing.
john baez
>In article <4hg378$b...@guitar.ucr.edu>, john baez <ba...@guitar.ucr.edu>
>writes
>>But this is merely speculation, and your argument is merely
>>quasi-plausible. Why don't you concoct a 100% rigorous argument [for
>>why the Weyl curvature vanishes in the big bang cosmology] based
>>on what we know about the Weyl curvature? Remember: the Ricci
>>curvature tells how your little ball of freely falling coffee grounds
>>changes volume, while the Weyl curvature tells how it changes shape into
>>an ellipsoid.
>Aw, you gave it away. Ok, ok it's a better way of saying it than
>'boundary conditions' and stuff. The universe couldn't be or stay
>uniformly isotroph^hic if Weyl was anything but zero. Almost by
>definition, well probably actually by definition.
Sorry, I should have given you a sneakier hint and let you have more of
the fun of figuring it out. Yes, if a little sphere of initially
comoving particles in free fall turned into an ellipsoid, there would
necessarily be some preferred directions (the axes of the ellipsoid),
hence anisotropy.
So you have worked out the Weyl curvature of the big bang universe and
also the Ricci curvature --- given the energy density E and the pressure
P. No heavy-duty calculations, just symmetry! Cool, huh?
Now, if you read that nasty lecture from G. Wiz on the formula for the
Riemann tensor in terms of the Ricci and the Weyl, you will know how to
calculate the Riemann tensor for the big bang universe as a
function of E and P. These in turn depend on time, but not on position
in space, since things are homogeneous. Figuring out exactly *what*
functions of time they are requires one to make some assumptions as to
what the universe is made of... freely falling dust (no pressure),
radiation, or whatever.
I would say you're almost done with questions 2 and 3:
2. Explain how, in the standard big bang model, where the universe is
homogeneous and isotropic --- let us assume it is filled with some fluid
(e.g. a gas) --- the curvature of spacetime at any point may is
determined at each point.
3. In the big bang model, what happens to the Ricci tensor as you go
back in past all the way to the moment of creation?
I still would like to hear a bit more on what happens to the components
R_{ii} of the Ricci as we approach the moment of creation. (Note: it's
standard to use letters like i,j,k to range from 1 to 3. You told me
R_{00} went to infinity, so I'm asking about the other diagonal terms.)
But that's mainly it. You have come so close to completely
working out the big bang model, that when you answer question number 1,
I will reward you by finishing it up. By the way, I'm tremendously
enjoying this. I never really understood the big bang model as well as
I do now... I'll have to tell you what I've realized, after you answer
question number 1.
john baez
>Now, I am perfectly sure T^c_c = T_c^c , but right now I can only offer
>my "proof by it would just be too weird otherwise".
See, erg has good mathematical intuition; he knows when something would be
too weird to be true. But he's wrong about it not being able to prove
it's true. Earlier in the post I'm quoting he wrote both:
> T_c^c = g^{ca}T_{ac}
and
> T^c_c = g^{ca}T_{ac}
Since both of these are true, we definitely have T^c_c = T_c^c!
Of course T_{ab} is symmetric, i.e. T_{ab} = T_{ba}. This allows us to
get away with certain sins that would otherwise be mortal. But let's
show that for any old tensor X_{ab} we have X^c_c = X_c^c. Here we will
be very careful about orders of things. We will only use the symmetry
of the metric: g_{ab} = g_{ba}.
Okay. Time for index gymnastics! Stand upright with arms loosely at
your sides... do a few neck rolls, and then, note:
X_c^c = g^{cd} X_{cd}
while on the other hand
X^c_c = g^{cd} X_{dc}
But on the right side of the second equation we can switch the "dummy
indices" c and d --- dummies because they are summed over and are purely
arbitrary names --- and get
X^c_c = g^{dc} X_{cd}
So then we get
X^c_c = g^{cd} X_{cd}
using the symmetry of the metric.
>I think this is a rule of thumb: Almost all tensors in GR are symmetric
>or antisymmetric, so we can always sloppily sway indices, and at worst
>introduce a stray minus sign, which can be fudged out later. Well,
>maybe.
That rule has a certain truth to it, though of course if you don't
happen to already know the right answer, it can be important to be able
to get the right sign. Someday, if you folks get really interested in
the inner workings of curvature, I'll go over the symmetries of the
Riemann tensor. If we define
R_{abcd} = g_{ax} R^x_{bcd}
(note, this is a controversial convention and perhaps a bad one, but
it's the one I used in my book), we have
R_{abcd} = -R_{dbca}
and
R_{abcd} = -R_{acbd},
and
R_{abcd} = R_{cdab}.
I don't think these are all the symmetries (I know others which I don't
think are consequences of these), but they suffice to prove "Green's
Theorem", namely that if you switch any two indices of the Riemann
tensor when all indices are down, at most you get a sign error.
john baez
>>He suspects that g_{ab}T_{cd} would be a tensor of form U_{abcd}.
>I know nothing about outer products, however...
The tensor U with U_{abcd} = g_{ab}T_{cd} used to be called the outer
product of g and T; nowadays folks would call it the tensor product.
Oz
writes
>
>You've got the basic idea, there are just a few mistakes here and there.
>You want to start with an orthonormal basis of spacelike vectors u, v, w.
>Consider one of them, say u. Then you want to look at this picture
>
> P'->----Q'
> | |
> | |
> t^ ^
> | u |
> P->-----Q
>
>where following your notation I'm using t for a unit timelike vector,
>the velocity of the particle at P. NOTA BENE: this is what I had been
>calling *v*, since it's a *velocity*!
^^^^^^^^^^^^^^^^^^^^^^
> This terminological confusion can
>get us confused if we let it.
Too true. However it has confused me mightily, and this has offered up a
philosophical question that bears discussion.
I am seeing what is effect a static volume of spacetime by dropping one
spacial dimension and fitting in a time dimension in it's place.
So what I have marked vector t in the time dimension is a *distance*.
How do I know how long it is? Well, it's ct long in P's restframe, or I
suppose P's geodesic. Of course if you choose units where c=1, then it's
t long. I don't really have velocities as such in this universe, I have
dx/dt's. Really there isn't a flow of time, just a path. I am seeing it
exactly the same way as I see an x-y graph on paper. I can read off the
x-position and get back a y-position. I can get the slope, it's dy/dx,
and move about at will. So whilst velocity has a meaning, it's not got
the uniquness that time has in calculus. It's just an ordinary axis
(well fairly ordinary) and I can read off distances with my ruler.
Of course it lives in a rather odd space where transformation of co-
ordinates are not linear in the sense that if I transform to a new
origin that is moving wrt another the transformation is non-linear. On
top of that the co-ordinates may not be equally spaced and may curve,
but no bother, I just get out my bent ruler and measure stuff off.
Now what could be easier, so where am I going wrong?
Oz
<O...@upthorpe.demon.co.uk> writes
>Wiz:
>1. Explain why, when the energy density within a region of space is
>sufficiently large, a black hole must form, no matter how much the
>pressure of whatever substance lying within that region attempts to
>resist the collapse.
>
Oz has been looking back to the beginning of this thread.
Suddenly he is puzzled. A post seems to be missing. In fact his second
post. It was his first hesitant stab at the above question. He remembers
that it was never replied to. Perhaps it has expired? Perhaps he never
sent it? Perhaps the Demon system that provides his link to the
WizardWorld screwed up. Well, it has servers that seem to spend most of
their time in some relativistic place called 'down time'. I think this
has something to do with 'imaginary time'.
Now if he remembers rightly his original answer was not (to him) very
satisfactory because it didn't offer a finite radius where 'interesting
things' happened. It went something like this:
We have a little ball of coffee grains. Naturally, being massive, they
come together. However, being pointlike, they can get as close together
as they like. Now
d^2V/dt^2 = - (1/2)(E+3P) V
So as V gets smaller, d^2V/dt^2 gets ever larger since both E and P are
positive, so the volume should tend to zero. The density will thus tend
to infinity. I guess this might (just) be called a singularity.
Certainly the curvature gets to be infinite which is good enough for me.
Personally I find this too glib and too simplistic. However Ed and co
might like to take it another step to make it a better answer to the
question. At least it's a starting place.
Whaddyathink then Ed and Doug? The ball's in your court.
john baez
impossible to think when they're around."
"Let's just wait around," said Oz, "until the smoke returns. After
all, if it really wants to know the answer, it'll come back. Why waste
our time fighting these damn signs?"
"Gee whiz!" said G. Wiz. "A great idea."
So they hung out by the campfire and talked about other subjects until
the signs, seeing there would be no more food, buzzed off. Eventually,
hours later, the smoke returned!
"I hate those signs too", said the smoke. "I've found that raising
and lowering indices serves as a breeding ground for them, which is
why I suggest this process be avoided whenever possible!!
That's why I recommend using T^a_b, G^a_b and R^a_b"
"Well," said the wizard, "that's fine as long as you don't ever need to
use R_{ab}, but typically you do. One needs to battle signs occaisionally in
this business... perhaps your method would help mimimize it, but not
eliminate it."
"I am now getting:
T^a_b =
-E 0 0 0
0 P 0 0
0 0 P 0
0 0 0 P
i.e. T^0_0 = -E, T^i_i = P (pressure) for i=1,2,3."
The wizard said "Good, that's what I get."
"Did you really work it out," asked the smoke, "or are you just taking
my word for it? I have a feeling you're not all that eager to do this
sort of calculation."
The wizard smiled. "You're right, but if you look in the course notes
you'll see I have that in there somewhere. Check out
http://math.ucr.edu/home/baez/outline2.html "
The smoked approximated a nod. "Furthermore, I find that R = R^c_c = -Tc_c
i.e. R= -(T^0_0 + T^1_1 + T^2_2 + T^3_3) = E-3P"
"Yes, that's what I got too," said G. Wiz.
Since R^a_b = T^a_b + .5 R g^a_b . (g^a_b = 1 iff a=b, 0 otherwise)
R^0_0 = .5^(T^0_0 - T^1_1 - T^2_2 - T^3_3) = .5(E+3P)
"This indicates that P makes things contract! (Or at least go in the
same direction as E.) This seems weird, but" the cloud turns slightly
pinkish, "we are talking about d^2 V/dt^2, not dV/dt. It's obvous that
dV/dt is positive when a high pressure area is in empty space, but that
says nothing about d^2V/dt^2"
"Wait a minute!" said G. Wiz. "Your formula for R^0_0 agrees with the
stuff in the course outline, and you are right that it means the
gravitational effect of pressure is to cause things to contract. In
layman's lingo: like energy, pressure causes an attractive gravitational
force."
Oz added "I thought you said gravity wasn't considered a `force' in
general relativity."
"It's not," admitted the wizard, "that's why I said, `in layman's
lingo'."
"BUT," added G. Wiz, "the formula relating R^0_0 to d^2V/dt^2 only applies
to a ball of particle following geodesics! That is, particles in free
fall, feeling no force other than gravity (which, ahem, is not really a
force either). In reality, when we have a gas with different pressures
at different regions, the atoms of the gas *do* feel forces other than
gravity. So pressure does have the usual effect of making stuff want to
expand, in addition to the special GR effect of making it want to
contract. Just don't mix them up; they are very different."
"Something else that is interesting. If we take a simple particle with
no pressure term, the stress energy tensor is"
T^0_0 = -E
all other terms are zero."
"If we take this and boost half of it in the -x direction at some velocity
v, (energy E/2) and boost half of it in the +x direction at the same
velocity v so that the average momentum/velocity is zero and add, we get"
T^0_0 = -gamma^2E
T^1_1 = v^2gamma^2E
where gamma = 1/sqrt(1-v^2) [c assumed to be 1]
"The other terms all cancel out." The cloud swirls as it attempts to cross
it's nonexistent fingers as it makes this remark.
"This gives us some insight into the pressure term, the above anisotropic
case has a positive pressure in the x direction."
"It seems odd at first that T^0_0 scales as gamma^2, but it's
energy / volume, not energy, and volume is defintely not Lorentz
invariant!"
"Hmm," said the wizard. "Interesting. I'll have to ponder that."
john baez
>In article <4i08t9$g...@guitar.ucr.edu>, john baez <ba...@guitar.ucr.edu>
>writes
>>In article <MDyEWBAj...@upthorpe.demon.co.uk> O...@upthorpe.demon.co.uk
>>writes:
>>>We have a little ball of coffee grains. Naturally, being massive, they
>>>come together. However, being pointlike, they can get as close together
>>>as they like. Now
>>>d^2V/dt^2 = - (1/2)(E+3P) V
>>What can you say about the more realistic case where, after they
>>get close enough together, they "bump into each other". Doesn't this keep
>>the black hole from forming?
>Well, after a while our little ball of coffee grounds (iron filings
>would have been better) is so dense that the grounds start bumping into
>each other.
Yes. That's how things normally resist collapse. But...
>This is nothing I couldn't have said before this course started (except
>for swapping curvature for 'gravitational field').
Well, there is something you know now that you might not have known then.
I claim if you think about how the coffee grounds "try to resist
collapse", you'll see that under certain circumstances it's futile.
All I want is a rough argument, but there should be something
distinctively GR-ish about it, which explains the "futility" mentioned
above.
I am NOT asking you to prove the Penrose-Hawking singularity theorems.
These guys got famous for showing *rigorously* that singularities must
form under certain circumstances, but they spent a bit more time
learning GR than you have before doing this. A rough argument will do,
and you have done half of it.
Doug Merritt
Oz <O...@upthorpe.demon.co.uk> wrote:
>Suddenly he is puzzled. A post seems to be missing. In fact his second
>post. It was his first hesitant stab at the above question. He remembers
>that it was never replied to. Perhaps it has expired? Perhaps he never
>sent it?
I was confused by your reference to it, since I hadn't seen it, but
assumed that perhaps I just was too hasty in reading one of your posts.
Give your comment here, however, perhaps it did in fact disappear into
the ozone.
>Now if he remembers rightly his original answer was not (to him) very
>satisfactory because it didn't offer a finite radius where 'interesting
>things' happened. It went something like this:
I believe it's ok not to offer a *specific*, because G. Wiz didn't provide
tools for calculating a threshold; that would appear to come later. (I
cheated and drew on a tiny little factoid about Gaussian curvature, myself :-)
>d^2V/dt^2 = - (1/2)(E+3P) V
>
>So as V gets smaller, d^2V/dt^2 gets ever larger since both E and P are
>positive, so the volume should tend to zero. The density will thus tend
>to infinity. I guess this might (just) be called a singularity.
>Certainly the curvature gets to be infinite which is good enough for me.
>
>Personally I find this too glib and too simplistic. However Ed and co
IMHO this is in fact the sort of thing you were supposed to come up with,
thus the Wiz commenting about you "making heavy weather of it" :-)
I doubt that you gave quite this answer before, however, because
previously you were subtracting P from E. That and several other
issues (such as the abbreviations P and E) would have arisen quite
some time after the point of your mysteriously missing post.
However, in my next post watch the creative use to which I put the
incandescent thunderbolt the wizard is flinging your way... :-)
Doug
--
Doug Merritt do...@netcom.com
Professional Wild-eyed Visionary Member, Crusaders for a Better Tomorrow
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Oz
<do...@remarque.berkeley.edu> writes
>In article <MDyEWBAj...@upthorpe.demon.co.uk>,
>Oz <O...@upthorpe.demon.co.uk> wrote:
>>Suddenly he is puzzled. A post seems to be missing. In fact his second
>>post. It was his first hesitant stab at the above question. He remembers
>>that it was never replied to. Perhaps it has expired? Perhaps he never
>>sent it?
>
>I was confused by your reference to it, since I hadn't seen it, but
>assumed that perhaps I just was too hasty in reading one of your posts.
>Give your comment here, however, perhaps it did in fact disappear into
>the ozone.
I quickly checked some of my outfile. What a lot of garbage I post!
There seems to have been at least three that didn't end up on the group,
but had been posted according to my system. Two relate to Q1. All look
amazingly inept so I guess I have learned a small amount in the process.
>
>>Now if he remembers rightly his original answer was not (to him) very
>>satisfactory because it didn't offer a finite radius where 'interesting
>>things' happened. It went something like this:
>
>I believe it's ok not to offer a *specific*, because G. Wiz didn't provide
>tools for calculating a threshold; that would appear to come later. (I
>cheated and drew on a tiny little factoid about Gaussian curvature, myself :-)
>
>>d^2V/dt^2 = - (1/2)(E+3P) V
>>
>>So as V gets smaller, d^2V/dt^2 gets ever larger since both E and P are
>>positive, so the volume should tend to zero. The density will thus tend
>>to infinity. I guess this might (just) be called a singularity.
>>Certainly the curvature gets to be infinite which is good enough for me.
>>
>>Personally I find this too glib and too simplistic. However Ed and co
>
>IMHO this is in fact the sort of thing you were supposed to come up with,
>thus the Wiz commenting about you "making heavy weather of it" :-)
>
>I doubt that you gave quite this answer before, however, because
>previously you were subtracting P from E. That and several other
>issues (such as the abbreviations P and E) would have arisen quite
>some time after the point of your mysteriously missing post.
Correct. I have since found it in my files. I originally did it with all
those T_{ab} thingies. Hey, this was from memory, and about half or a
quarter the length. It said exactly the same thing, but untidily.
>However, in my next post watch the creative use to which I put the
>incandescent thunderbolt the wizard is flinging your way... :-)
Does this mean you will field it for me? No? Oh dear, I thought not.
Ed, Ed can I borrow you armour? Please Ed, hey Ed ..... Ed ...
Oz
writes
>In article <MDyEWBAj...@upthorpe.demon.co.uk> O...@upthorpe.demon.co.uk
>writes:
>>We have a little ball of coffee grains. Naturally, being massive, they
>>come together. However, being pointlike, they can get as close together
>>as they like. Now
>>
>>
>>So as V gets smaller, d^2V/dt^2 gets ever larger since both E and P are
>>positive, so the volume should tend to zero. The density will thus tend
>>to infinity. I guess this might (just) be called a singularity.
>>Certainly the curvature gets to be infinite which is good enough for me.
>
>>Personally I find this too glib and too simplistic.
>
>glib, simplistic, yet correct.
>
Well, it was not on my newsreader and I'm pretty sure I have set all
this thread to 'keep', yet I have a record of posting a similar one. I
think it never got past the Demons. The problem I have now is
remembering to do it from scratch, ignoring any posts I may have posted.
So this time we will start from the beginning.
>So, keeping in mind this stuff, think a bit about the wizard's comment
>about a ball of dust... and tell me what you can about *curvature
>singularities*... places where the Riemann curvature blows up. Also,
>how are your coffee grounds moving... are they perchance following
>geodesics? (I guess so, since you're using the formula that applies
>then.) What can you say about the more realistic case where, after they
>get close enough together, they "bump into each other". Doesn't this keep
>the black hole from forming?
See! I was right! It *was* too glib and simplistic. At least I got
something right, even if it's nothing to do with the question.
OOooo dear. If we aren't careful we is going to get into trouble
defining energy again. These poor coffee grains that started at rest
with each other are, after a period, flying towards each other at
significant speed. (Oz hears the dratted minus signs revving their wings
in their nest high in the wizards cave). Yes, yes, I know, they are
still following geodesics but this is what is bringing them together
ever faster.
Hmmmm. I think a good wheeze is to do it in little steps and see if we
are heading in the right direction each time. This is called fireball
delaying tactics.
Well we have:
R_{00} = (1/2)(E + 3P)
Well, the momentum flow in our little ball of particles would seem to be
increasing but I would guess that the ball is isotropic and uniform in
the spacial dimensions so the above looks a reasonable guide to the
curvature. We note that the density (E & P) are getting bigger and
bigger as the ball contracts, so the curvature is getting bigger and
bigger too.
Well, after a while our little ball of coffee grounds (iron filings
would have been better) is so dense that the grounds start bumping into
each other. Some of them at the surface get ejected and start to head
away from the ball. Unfortunately for them they are now in very curved
space. Since the ejected grounds now follow geodesics away from the ball
we can use our little formula:
d^2V/dt^2 = -(1/2)(E + 3P)
except this time as the particle speeds outwards the volume of space
enclosing the mass is increasing so the effective density is decreasing
and so d^2V/dt^2 is decreasing as it heads outwards. (Well, nobody has
actually said this, indeed they implied that it was a naughtly thing to
say, however this way we should get a definitive fireball reply one way
or the other).
Now some lucky speedy little grounds will escape all the way to
infinity, but other slower doomed grounds will find that their velocity
wrt the main ball eventually becomes negative and the curvature bends
them round to head inexorably back to the still-contracting main ball.
In this way the velocity between grounds will be reduced. Essentially by
evaporation of high velocity coffee grounds, to below the ever
increasing escape velocity of the ball. We note in passing that these
coffee grounds feel no acceleration as they travel along their curved
path.
So the relative velocities of the coffee grounds (note that they are
infinitely small coffee grounds) within the ball can increase as the
curvature increases. As the ball gets smaller it takes faster and faster
coffee grounds to escape from the increasingly curved spacetime that
surrounds the ball. Now lets give these relative velocities a more
useful name, say 'temperature'.
Now in passing it's clear that any given curvature will have some
characteristic temperature where it will be in equilibrium. In effect
all the coffee grounds are orbiting the centre. It will inevitably lose
some energy continually by evaporation. If you had an energy source that
put heat into the ball at the rate it is lost, it could sit at that
curvature until the power source ran out. It would then continue to
contract. These are called 'suns'. Lo and behold we note that low
density suns have a low temperature, and high density suns have a high
temperature. This is why a red giant is red, it has very low density.
Massive stars have to have a high temperature because of their high
density. We ignore neutron stars and dwarves because they have a
different mechanism to prevent collapse.
However our coffee grounds have no such source. They continue to
contract and the temperature continues to rise as the coffee grounds
have to get faster and faster to escape to infinity due to the ever
increasing curvature of the ever increasing density of the ever
decreasing ball. I feel that there should be a mechanism that allows the
ball to completely evaporate, but starting where we started I am not too
sure there is one.
One might wonder if there is a maximum velocity. If there were then at
some point the curvature gets so curved that nothing can escape to
infinity. This would be the event horizon, and I *still* think that is
where the curvature is one (or something like that). The worldline of a
photon then goes from 45 degrees to being parallel to the time axis.
Anyway, our hot little ball of incinerated coffee grounds would still
continue to contract, and contract, and contract, until it was
infinitely small and the space around it infinitely curved. I guess this
counts as a Baez curvature 'blow up'.
This is nothing I couldn't have said before this course started (except
for swapping curvature for 'gravitational field'). It's a lot easier, of
course, if I don't have to pull this stuff out of equations I only
partly understand (well, you gotta be optimistic). So where have I used
my (ho, ho, ho, ROTFL) newfound knowledge?
Nowhere. <sob, sob, booo-hoo.>
I hope this will do for starters. It's not, however, an adequate answer
for this course. IMHO.
Bronis Vidugiris
)"Ack!" says the wizard, swatting madly. "It's these damn signs. It's
)impossible to think when they're around."
)
)[To be continued...
"I hate those signs too", said the smoke. "I've found that raising
and lowering indices serves as a breeding ground for them, which is
why I suggest this process be avoided whenever possible!!.
That's why I recommend using T^a_b, G^a_b and R^a_b"
I am now getting:
T^a_b =
-E 0 0 0
0 P 0 0
0 0 P 0
0 0 0 P
i.e. T^0_0 = -E, T^i_i = P (pressure) for i=1,2,3
Furthermore, I find that R = R^c_c = -Tc_c
i.e. R= -(T^0_0 + T^1_1 + T^2_2 + T^3_3 = E-3P
Since R^a_b = T^a_b + .5 R g^a_b . (g^a_b = 1 iff a=b, 0 otherwise)
R^0_0 = .5^(T^0_0 - T^1_1 - T^2_2 - T^3_3) = .5(E+3P)
"This indicates that P makes things contract! (Or at least go in the
same direction as E.) This seems weird, but" the cloud turns slightly
pinkish, "we are talking about d^2 V/dt^2, not dV/dt. It's obvous that
dV/dt is positive when a high pressure area is in empty space, but that
says nothing about d^2V/dt^2"
"Something else that is interesting. If we take a simple particle with
john baez
>john baez (ba...@guitar.ucr.edu) wrote:
>: Sorry, I should have given you a sneakier hint and let you have more of
>: the fun of figuring it out. Yes, if a little sphere of initially
>: comoving particles in free fall turned into an ellipsoid, there would
>: necessarily be some preferred directions (the axes of the ellipsoid),
>: hence anisotropy.
>True, but we are only talking about spacelike directions here.
>What about the timelike components?
Yes... I overlooked this at first but then realized this fly in the
ointment. In one of my G. Wiz posts I hinted darkly at this problem...
but I never yet got around to figuring out its resolution.
>No doubt this is an easy
>consequence of the symmetries of the Weyl tensor.
I wish it were. I'm a bit nervous about that, though. We can recover
the Weyl tensor if we know how balls of dust moving along at arbitrary
velocity change shape into ellipsoids (in their own rest frame). But we
can only apply the isotropy argument to balls of dust which are at rest
in the "cosmic rest frame" of the big bang. I think we only get 5
components of the Weyl tensor to vanish by this argument. (There are
5 linearly independent 3x3 symmetric traceless matrices.) 5 out of 10
ain't bad... but it doesn't look like we'll be able to get all 10
components of the Weyl to vanish without some extra reasoning.
Can anyone think of a way to fix my argument?
>This is an interesting way of looking at the FRW metric! If indeed
>one can get conformal flatness directly from local isotropy [....]
If!
Oz
<do...@remarque.berkeley.edu> writes
>
>You are much closer than you seem to think, though. The first 2/3rds
>of John's questions above are simply probing you to state slightly
>more explicitly what you've already said. The final question is
>merely pointing out that you haven't said explicitly why pressure
>doesn't prevent the black hole from forming...but you did say why
>*implicitly* (very implicitly :-),
Yes. Well before I wrote my "Oz trudges off into the night" piece, I DID
shove an email 'under his door'. It had a selection of answers including
the trivial answer John required. So now it's up to the Wizard to catch
Oz before he reaches the village to tell him he's passed the test.
>I think you can answer all of John's remaining questions in a
>single paragraph that is just as glib as the other one. :-)
>
>Go for the obvious.
It wasn't so much obvious as trivial. Indeed I was absolutely certain
that I had already mentioned it several times. In a considerable huff I
went back over the postings to prove same. You are right. I never
explicitly stated it anywhere, I had simply assumed it as obvious. I
mean, when you say the curvature/volumetric acceleration goes as (E+3P)
several times, and say that most of the energy will go into P but that
doesn't matter .... Oh, well. The Wizard likes his little tricks :-)
Brett McInnes
: In article <4hj79h$3...@nuscc.nus.sg> matm...@leonis.nus.sg (Brett McInnes) writes:
: >john baez (ba...@guitar.ucr.edu) wrote:
: >: Sorry, I should have given you a sneakier hint and let you have more of
: >: the fun of figuring it out. Yes, if a little sphere of initially
: >: comoving particles in free fall turned into an ellipsoid, there would
: >: necessarily be some preferred directions (the axes of the ellipsoid),
: >: hence anisotropy.
: >True, but we are only talking about spacelike directions here.
: >What about the timelike components?
: Yes... I overlooked this at first but then realized this fly in the
: ointment. In one of my G. Wiz posts I hinted darkly at this problem...
: but I never yet got around to figuring out its resolution.
: >No doubt this is an easy
: >consequence of the symmetries of the Weyl tensor.
: I wish it were. I'm a bit nervous about that, though.
What I was thinking was this. Your argument seems to show that
Wijkl { where ijkl are always subscripts} is zero whenever both
k and l are in the range 123.{This is probably the part I am getting
wrong}.
But then because W is traceless,
W0j0l = W1j1l + W2j2l + W3j3l = 0+0+0 = 0 and so on. So the whole
thing has to be zero. Seems too simple to be right,though, and
I don't think well when rushing to a class, as at present....
john baez
>'do...@remarque.berkeley.edu (Doug Merritt)' wrote:
>Whichever, it sounds like T is actually a source for the Weyl....
>
>Of bloody course!!!!
>
>It must be this way. T is *not* a source for the Einstein, which somehow
>seems to locally mimic it, but for the Weyl, or part of it. Somehow this
>fact must be hidden in Einstein's equation, and some minor assumption
>about the Riemann... like differentiability. The Weyl components must be
>twisted around by the Einstein components in order for the Riemann to
>exist... in some way I don't understand.... and when the Einstein goes
>away, the Weyl can relax... but only so fast... relax to its value at
>infinity...
If I had anything interesting to say about this interesting notion, I
would, but I don't. I'm saying *that* just to let you know that my
silence on this sort of issue is not one of smug superiority or bland
indifference. The thing is, as I've said:
1) Most textbooks scarcely breathe a word about the Weyl tensor. I have
created a monster by telling you folks about the Weyl/Ricci split!
Now you are trying to understand everything in these terms, which seems
reasonable to try... but I can't actually help you much more than I've
done, not without working my butt off, anyway.
The idea you mention about the Weyl tensor being twisted around by the
Einstein tensor, and being able to relax "only so fast" when Einstein is
zero, seems to have some truth to it. As I explained earlier,
both Weyl and Einstein are computed in a complicated nonlinear
way (soon to be disclosed) from the metric. So the relation is subtle.
More precisely:
Using Einstein's equation, the Einstein tensor is determined by the
stress-energy tensor, and from this one can try to solve ---
non-uniquely --- for the metric. The Weyl tensor is then determined by this
metric. Since there isn't a unique metric with a given Einstein tensor,
the Weyl tensor is not uniquely determined by the stress-energy tensor,
but it isn't completely arbitrary, either. Since the equations are
nonlinear, it would take serious work to sort out in general exactly how
much the Weyl tensor is determined, and how much it's arbitrary. (I bet
folks have considered this... at least in a linearized analysis about a
fixed solution.)
2) But: as I said before, in most realistic physics problems one is not
handed a stress-energy tensor and left to compute the metric!!!! Nobody
says: "here is how the binary pulsar orbits, you figure out the
gravitational field". (Indeed, in addition to the nonuniqueness
mentioned above, there may not *exist* a metric having a given Einstein
tensor.) That's why I don't feel more guilty than I already do, about
not being able to completely sort out all this stuff. If we ever get
around to actually solving Einstein's equation in this tutorial, you'll
see how it's actually done.