Covariant derivative - is this a typo ?

1. Oct 13, 2012

zn5252

hello,
please see the attached snapshot (taken from 'Problem book in relativity and gravitation'). In the last equation I think there would be no semicolon.
Here is why I believe (S is scalar by the way):
S;α[βγ] = 1/2 * ( S;αβγ - S;αγβ )
Now from the equation which precedes it, we have :
S;αβγ = S,αβγ - Sσ τ σ αβ,γ
and :
S;αγβ = S,αγβ - Sσ τ σ αγ,β
this means :
S;α[βγ] = 1/2 * ( Sσ Rσαβγ ) due to the first equation of the solution 9.8 on the top of the page.
We see that there is no semicolon. Where did I go wrong if at all ?
Thank you,

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2. Oct 13, 2012

bcrowell

Staff Emeritus
It's not possible to have S;α[βγ] = 1/2 ( Sσ Rσαβγ ) in general, because the right-hand side has no derivatives in it. If S is a constant, then the left-hand side vanishes but the right-hand side doesn't.

Since that line in your derivation is the first one that fails this test, I think the mistake must be at the final step of your derivation. Actually, looking at the result of 9.8 in the scan you posted, I don't understand how it would allow you to make that step. The result of 9.8 has a second derivative in it, but you're trying to apply it to something that's a third derivative.

As far as notation, are you writing $\tau$ when you mean $\Gamma$?

3. Oct 13, 2012

dextercioby

It can't be the torsion, it must be $\Gamma$.

4. Oct 13, 2012

zn5252

let us substract the second equation from the first and multiplying by a half= >
S;αβγ = S,αβγ - Sσ $\Gamma$ σ αβ,γ
MINUS
S;αγβ = S,αγβ - Sσ $\Gamma$ σ αγ,β

we get :

1/2 ( S,αβγ - Sσ $\Gamma$ σ αβ,γ - S,αγβ + Sσ $\Gamma$ σ αγ,β )

but since

S,αβγ = S,αγβ,

1/2 ( - Sσ $\Gamma$ σ αβ,γ + Sσ $\Gamma$ σ αγ,β )

which gives according to the equation at the top of the page for the Riemann tensor :

1/2 Sσ ( Rσ αβγ )

Yes that was the Christoffel symbol indeed...as I could not find the upper case Gamma, but now I found it.

5. Oct 13, 2012

bcrowell

Staff Emeritus
In post #4, you have no equals sign in many of your lines of math, which makes it hard to tell what you're asserting. Once you write everything out with equals signs, it should be easy for you to find at which step you've made a mistake. It's the first line that fails for a nonzero constant scalar S. If S is a constant, then its derivatives are all zero, so it can't be identically equal to an expression that doesn't involve any derivatives of S.

In some places you have $S_\sigma$, which can't be correct, since S is a scalar.

6. Oct 14, 2012

zn5252

Yes I'm also wondering whether the equation :

S;αβγ = S,αβγ - Sσ $\Gamma$ σ αβ,γ

is correct ?

In fact, wouldn't S;αβγ equal to (S,αβ) ? I'm a bit confused here...

What I know is that for a vector S:

[A] Sθ;α = Sθ,α - Sσ $\Gamma$ σ θα

But not sure whether :
Sθ;αλ = (Sθ;α)
or ?
[C] Sθ;αλ = (Sθ,α)
is correct regarding the position of the semicolon and the comma ?

In the attached scan, we can see the second formula in problem 9.8 which rather asserts that statement is true ?

But then if this is so then for third derivatives , we would have :
[D] Sθ;αλη = (Sθ;α),λη
or is it :
[E] Sθ;αλ = (Sθ;αλ)

Reminder : What is meant by Sθ;α is the covariant derivative of the θ component of the vector S in the direction of the vector u whose parameter is defined by α, i.e. u ≈ δ/δα.

Thanks for the clarification.

Last edited: Oct 14, 2012
7. Oct 14, 2012

bcrowell

Staff Emeritus
As I pointed out in #5, your S is a scalar, not a vector.

8. Oct 14, 2012

dextercioby

Oh, dear, to get things straight I wouldn't know where to start. This thread is the perfect example on why the comma-semicolon notation is confusing. You should use the partial-nabla notation.

You need to compute this object

$$\nabla_{\mu}\nabla_{\nu}\nabla_{\lambda} S$$

by yourself using the rules for covariant differentiation. Precisely,

$$\nabla_{\mu}\nabla_{\nu}\nabla_{\lambda} S = \nabla_{\mu}\left(\nabla_{\nu}\nabla_{\lambda} S\right) = \nabla_{\mu} T_{\nu\lambda}= \partial_{\mu}T_{\nu\lambda} \pm...$$

9. Oct 15, 2012

zn5252

thanks ! that helps indeed.