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. Hmm...."
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->-----QOz 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->-----QThe 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
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 to work this out." Oz knew his questions were mostly rubbish, but sometimes he did wish people would listen. HUMPF!!!"
The wizard smiled. "Believe it or not, I *had* listened to you; what I am doing now is making your remarks a wee bit more precise. Okay, so the relative acceleration is -R(w,v,v). But this is just the same as
since the Riemann tensor is defined so that it's skew-symmetric in the first two slots."
"Cool," says Oz, wondering what "skew-symmetric in the first two slots" meant.
"Oh, Oz," said the wizard. "You are always looking for deep mysteries where there are none. It just means that when you interchange what's in the first two arguments, the sign switches! My point is that this property is obvious from the definition of the Riemann tensor. Think about it!"
"Okay. This equation for the relative acceleration for geodesics is called the GEODESIC DEVIATION EQUATION. 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 spacetime. 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,
Aa = Rabcd vb wc vd
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. All of a sudden his 4d vision flickered and faded away.
"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 how 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!"