Oz comes back the next day and knocks on the wizard's door.

"Go away! I'm busy!" shouts the wizard.

"But I think I'm getting somewhere!"

The door slowly opens, without any visible cause --- and the wizard, sitting at his desk, doesn't even turn around. "What is it?" he asks, still scribbling away at something. He waves away a small cloud of minus signs hovering over him. "Damn signs!"

"I think I'm getting somewhere on that third question! You know that one: 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? "

"Okay," says the wizard. He stops writing and turns around, smiling. "What have you got?"

"Well, first of all, it says in the course outline that if v is the velocity vector of a particle in the middle of a little ball of blah blah .... of volume V...

then d^2V/dt^2 = -R_{ab} v^a v^b. Hokay."

The wizard interrupts. "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 that's no big deal."

Oz continues. "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."

G. Wiz frowns. "Not `isotrophic' --- it's pronounced ISOTROPIC."

Oz nods. The wizard continues, "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". So 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."

Oz adds, "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."

The wizard gets a mischeivous glint in his eyes and smiles all the more. "Can you use the isotropy to say more about these off-diagonal terms T_{ij} (i,j = 1,2,3)?" In a barely audible voice he adds "Hint: yes, you can!" He continues in a normal voice, "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."

Oz says, "Hmm... Then we have T_{0n} and T_{n0} (n=1 to 3) like terms. What they represent physically is not too clear."

The wizard raises his eyebrows: "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?" Then he whispers, waggling his eyebrows, and with a parody of a mysterious look on his face, "Hint: yes!"

Oz, not wanting to get distracted, continues on with what he has prepared. "So assuming we can take a Minkowski metric in a small area, which seems unlikely..."

The wizard smiles. "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:

g_{ab} = -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, it's fine to pretend you're in Minkowski space, IF you do it only at a single (arbitrary) point."

Oz continues: "So, here's what we get." He draws with his index finger on the incredibly dusty floor of the wizard's study:

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, it's a bit untidy..."

The wizard "Yes, a bit untidy, so use isotropy to the hilt and boil it down to something a lot simpler!"

Oz decides he had better pay attention to these hints. "Okay, let's see. What about T_{0i} and T_{i0}? 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."

The wizard grins... Oz has seen through it! "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."

Oz continues, "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."

"You mean ISOTROPIC," interrupted the wizard, glaring.

Oz nods absently. "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."

He frowns. "How did I miss this before? Is it right? Sounds plausible anyway."

The wizard nods. "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," says Oz. "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} (i not equal to j) 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 wizard stands up. "The derivation is quite simple and I think I told you it before." He looks through some piles of paper, frowns, shrugs, and continues. "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!"

Oz gets up and says, "Anyway, throwing caution to the winds, and without a really proper explanation, lets set T_{ij} (i,j=1 to 3; i not equal to j) = 0 too. In for a penny in for a pounding!"

"Good!" says the wizard. "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."

Oz kneels down in the dust again, and says, "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...."

The wizard says, "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."

Oz smiles, thinks a bit, thinks a bit more, thinks a bit more, and slowly walks off, thinking.