Oz and the Wizard -

Tensors

Oz and John Baez

The Wiz, rightly, tends to gloss over details. Oz also has a poor memory at times and it takes a little while for things to sink in.

Basically Oz has been looking at ways to find which elements of R^a_bcd correspond to his 16 elements of R_ab. He realises that he is not as sure as he should be about some basic tensor stuff. He might have come a fair way, but his knowledge is still appallingly elementary. Basically he does it mechanically, and not very well at that.

He hopes that Wiz will find the time to answer a few basic questions, or at least some other wandering Wiz can deign to... he walks up to the Wiz's door and knocks.

"Come in," says the Wiz.

"Could review a bit of, umm, tensor stuff with me?" asks Oz, looking down at the floor and blushing.

"Boring, boring, boring. Next we'll be reviewing partial differential equations and Fourier transforms. But okay...."

"Okay." Oz scratches on the floor:

T^c_c = g^(ca)T_ac = scalar.

"I feel that there should also be a product that is another tensor. I suspect that g_abT_cd would be a tensor of form U_abcd...."

The wizard nods. "A product of g and T that's a tensor? Sure, like g_abT_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."

"I also wonder if the reversal of the sub and superscript order is important. I also wonders if g_caT^(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."

"Also: 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."

Oz scratches another tensor in the dust on the floor:

R^a_bcd

"This takes in three vectors and outputs one. I think 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:

qⁿ = sum(i=0 to 3){sum(j=0 to 3)[sum(k=0 to 3) R(n,i,j,k) uⁱ v^j w^k ]}

Is this right (if you can work out what I actually said!)?"

"Almost; you mean Rⁿ_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:

qⁿ = Rⁿ_ijk uⁱ v^j w^k"

Oz nods. "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' business."

Oz asks, "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."

"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."

"Well, to get from R_ab to something of the same general form as the Riemann tensor we can just form something like R_bd delta^a_c. Note this has the same number of indices up and down as the Riemann tensor, which is R^a_bcd. In fact, you are right that there is some R_bd delta^a_c lurking in the Riemann tensor, waiting to burst free, and also some Weyl tensor stuff... but you are also right that the exact formula is a mess."

"What's that delta^a_c thing? " asks Oz.

"Well, it's the Kronecker delta, which is 1 if a = c and 0 otherwise, but it's closely related to the metric, since we get it by raising one index on the metric: g^a_c = delta^a_c."

"Also... g_ab is which metric?" asks Oz.

The wizard becomes a bit angry. "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."

"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." The wizard pauses, subliminally aware of some disturbance outside in the hall. Continued...