Later that day G. Wiz was in his study, busily juggling indices. Greek and Roman superscripts and subscripts were flying through the air in neat, orderly paths... a miracle of precision. Then Oz burst into the study all of a sudden. Shocked, the wizard dropped all his indices, and they scurried away across the floor and under the curtain that led to that mysterious back room.
"Damn!" cried the wizard. "Now look what you've made me done. I have to start all over! Who said you could come in here, anyway?"
"Hi!" said Oz cheerfully. "I want to see how the Ricci tensor to the change in volume of a small ball of test particles, just like you said, using the geodesic deviation equation!"
The wizard scowled and asked, "Didn't I say something like you can derive it from the geodesic deviation equation, AT LEAST IF YOU ARE BETTER AT INDEX JUGGLING THAN I SUSPECT YOU ARE?" He pulled out a yellowing sheet of parchment from a towering stack on his desk and glanced at it. "Yes, I believe those were my exact words!" He stuck it back in the exact same spot.
Undaunted, Oz said, "Well, okay, but anyway... you said that
Riemann = Ricci + Weyl,
right?"
The wizard nodded. "In some vague sense, yeah."
Oz added "I thought you said that R^a_{bcd} = R^c_{bcd} + W^a_{bcd}."
The wizard hurled a fireball in Oz's general direction. "Wait a minute! You are violating the basic law of index juggling: all the indices appearing on the left hand side of the equation must appear on the right. For the purposes of this rule, repeated indices which are summed over --- like the c in R^c_{bcd} --- do not count!"
Oz said "You never told me that!"
The wizard hurled another, bigger fireball, and Oz stepped back. The wizard said "I know I never told you this rule, but I *warned* you not to stick your fingers into the machinery, so don't blame me if you do and they sliced right off! Think about it: this rule is obvious! Something like R^a_{bcd} is a tensor of rank (3,1), right? While something like R^c_{bcd}..."
"Well, R^c_{bcd} = R_{bd} is the Ricci tensor, right?" asked Oz.
"Yes, damn it, that's the point, it's a tensor of rank (0,2)! You can't go around adding tensors of different ranks; that's like adding vectors and numbers... which I bet you used to do as a kid, right?" The wizard scowled and hurled a still bigger fireball, this time singing Oz's left ear. "You're just the sort who would. I know your type.... So," the wizard continued, "the equation
R^a_{bcd} = R^c_{bcd} + W^a_{bcd}
makes no sense! By the way," he said, his voice dripping sarcasm as he hurled yet another fireball, singing Oz's right ear this time, "THIS IS EXACTLY WHY NOBODY EVER TOLD YOU THIS EQUATION."
"Gee whiz, G. Wiz! Give me a break!" cried Oz. "Students learn by making mistakes!"
"Is that what they think where you come from?" asked the wizard, shooting a few lightening bolts from the fingers of both hands in a casual gesture of impatience. "Around here, students learn by GETTING THINGS RIGHT! And if they don't learn fast, they..."
"Okay, okay! So what IS the right equation?"
"Hmm," said the wizard, thinking a minute, and seeming to lose interest in scolding Oz. "Hmm. I think it's something like
R^a_{bcd} = R_{bd} g^a_c + W^a_{bcd}.
Notice that the metric tensor g^a_c provides the indices which the Ricci tensor is missing. By the way, for any metric g we have g^a_c = delta^a_c, that is, it's 1 if a = c and 0 otherwise --- we call this delta gadget the "Kronecker delta". So we would usually write delta^a_c instead of g^a_c. But I don't think this formula for the Riemann tensor is exactly right. I'm pretty sure that
R^a_{bcd} = K R_{bd} delta^a_c + W^a_{bcd}
for *some* constant K, but we need to work out what the constant is. Just set a = c and sum over the resulting repeated index. Note that delta^c_c = 4 in 4 dimensions, so we get
R^c_{bcd} = 4K R_{bd} + W^c_{bcd}
Now W is defined to be the "trace free" part of the Riemann tensor, meaning that *by definition* W^c_{bcd} is zero, so we get
R^c_{bcd} = 4K R_{bd}.
But the left side is R_{bd} by definition so we must have K = 1/4. So I think
R^a_{bcd} = (1/4) R_{bd} delta^a_c + W^a_{bcd}.
Hmm. Let me look it up...." He unlocked a drawer of his desk, pulled out a large tome, and riffled through it. "My god!" he muttered "What a mess... this is more complicated than I thought" It said:
R^a_{bcd} = W^a_{bcd}
+ (1/2)(g_{ac}R_{db} - g_{ad}R_{cb} - g_{bc}R_{da} + g_{bd}R_{ca})
- R (g_{ac}g_{db} - g_{ad}g_{cb})
"Damn! This is more detailed than I want... they're splitting the
Riemann tensor up into a Weyl part, and a Ricci *scalar* part, and
another part (that big mess in the middle) that's probably some sort of
`traceless Ricci tensor' part. Hmm. I can't even tell, offhand, if
this is consistent or not with what I guessed!"
He let out a sigh. "I guess I should calculate it out." He reached for a sheet of parchment and a quill, and began scribbling.
When finally G. Wiz looked up, Oz was mysteriously nowhere to be found, so the wizard never got to finish explaining his teaching philosophy.