Covectors physics help

1. Jan 4, 2011

latentcorpse

I'm trying Q1 in this past paper
and I am stuck on the very first part. Any hints as to how to get $n_{[a;b}n_{c]}=0$?

Last edited by a moderator: May 5, 2017
2. Jan 5, 2011

dextercioby

Re: Covectors

What's the definition of <normal> ? What does the first sentence look like in mathematical notation ?

3. Jan 5, 2011

latentcorpse

Re: Covectors

it would be $n_aT^a=0$ if $T^a$ is tangent to the surface $\beta(x)=0$ but I don't know how to relate $T^a$ to $\beta(x)$.

4. Jan 13, 2011

latentcorpse

Re: Covectors

Do you have any more hints for this because I'm still unable to get anywhere really?

Thanks.

5. Jan 13, 2011

dextercioby

Re: Covectors

I didn't get anywehere, else I would have written to you on this thread.

6. Jan 13, 2011

jambaugh

Re: Covectors

Found the following related homework exercise on a google search (see problem 2) http://www.physics.umd.edu/grt/taj/776b/hw3.pdf" [Broken]

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7. Jan 13, 2011

dextercioby

Re: Covectors

I've used the file kindly provided by jambaugh to make the proof requested in the exercise.

Can you prove number 2 in the file provided ? Then jumping to point 3 is almost trivial.

8. Jan 14, 2011

latentcorpse

Re: Covectors

Thanks for providing this file guys.

First things first, he says just above (2) that $V_a$ and $\nabla_a S$ are both orthogonal to the hypersurface. Well $V_a$ is (essentially by definition) but why is $\nabla_a S$?

Anyway, I couldn't prove (2):

So far I have

$\nabla_{[a}V_{b]}=\frac{1}{2} \nabla_a V_b - \frac{1}{2} \nabla_b V_a= \frac{1}{2} \nabla_a(f \nabla_b S) - \frac{1}{2} \nabla_b ( f \nabla_a S) = \frac{1}{2} ( \nabla_a f)( \nabla_b S) + \frac{1}{2} f \nabla_a \nabla_b S - \frac{1}{2}( \nabla_b f)( \nabla_a S) - \frac{1}{2} f \nabla_b \nabla_a S$

But in order to get something of the form $V_{[a}W_{b]}$, I have to start regrouping terms into $V$'s but everything is tangled up and I don't see how I can do this?

Thanks.

9. Jan 14, 2011

dextercioby

Re: Covectors

S is a scalar, so the commutator of the covariant derivatives (under the assumption of zero torsion) applied to it is 0, so 2 of the 4 terms in the RHS vanish. The other 2 can be grouped as needed by recalling the relationship between V and S.

The gradient to a surface in a point is orthogonal to that surface, it's a thing you learn in 1st year of colleage.

10. Jan 14, 2011

latentcorpse

Re: Covectors

Are we justified in assuming the connection is torsion free? Is that quite a common thing to do? I wasn't sure if I was allowed to do that or not which is why I left those terms in.

So using the torsion free property we get

$\frac{1}{2} (\nabla_a f)(\nabla_b S) - \frac{1}{2}(\nabla_b f)(\nabla_a S) = -\frac{1}{2}(\nabla_b f)(f^{-1} V_a) + \frac{1}{2} ( \nabla_b f)(f^{-1} V_b) = \frac{1}{2} V_a W_b - \frac{1}{2} V_b W_a = V_{[a}W_{b]}$

where $W_b=-(\nabla_b f) f^{-1}$

Although I am not 100% convinced by this because I am not sure if I am allowed to just have that $f^{-1}$ hanging off the end like that???

11. Jan 14, 2011

dextercioby

Re: Covectors

Yes, that's exactly the result I've got. So proving (3) shouldn't be an issue anymore, right ?

12. Jan 14, 2011

latentcorpse

Re: Covectors

I'm afraid not.

So to prove the "if" case. I am to expand out and contract with $X^c$

$V_{[a} \nabla_b V_{c]} X^c = \frac{1}{6} \left( V_a \nabla_b V_c + V_b \nabla_c V_a + V_c \nabla_a V_b - V_b \nabla_a V_c - V_a \nabla_c V_b - V_c \nabla_b V_a \right) X^c$

But I can't move the $X^c$ past the covariant derivatives can I so even though they contract with the "c" term I don't see how this is going to become any simpler?

Should I write that first term as $V_a \nabla_b ( V_c X^c)$?

And would the second one become $V_a \nabla_X V_c$? I'm not sure this is allowed because $\nabla_X = X^c \nabla_c$ and that requires moving the $X^c$ from the right of the covariant derivative tot he left of the covariant derivative which, as I said above, I'm not sure that I'm able to do?

13. Jan 15, 2011

dextercioby

Re: Covectors

Why would you need the $X^c$ ? You need to show that

$$V_{[a}\nabla_b V_{c]} = 0$$.

You've already written down the expansion in the 6 terms of the total antisymmetrization. Now use point (2), either the smart way, or the hard way.

I mean the 6 terms you wrote can be restrained first to 3 including W, then expanded back to 6 without the antisymmetrization brackets on the indices.

14. Jan 15, 2011

latentcorpse

Re: Covectors

But doesn't it say to use $X^c$ in the question??

$\frac{1}{3} \left( V_a V_{[b} W_{c]} + V_b V_{[c}W_{a]} + V_c V_{[a} W_{b]} \right)$

But then you say to expand it back to six terms? So

$=\frac{1}{6} \left( V_aV_bW_c - V_aV_cW_b + V_bV_cW_a - V_bV_aW_c + V_cV_aW_b - V_cV_bW_a \right)=0$
which is obviously zero since the $V$'s can commute

ok. so we have shown that in the "if" direction. Now what about the "only if"

Also, here we have shown that $V_{[a} \nabla_b V_{c]}=0$
But in the original question we were asked to show that $n_{[a;b}n_{c]}=0$ (*)
Are these the same thing?
Isn't (*) the same as $\nabla_{[b}n_an_{c]}=0$? which would be different from what we just showed, no?

Last edited: Jan 15, 2011
15. Jan 15, 2011

dextercioby

Re: Covectors

What you've shown is the same think you wanted to prove in the original exercise, because it doesn't matter what the indices are called, they can be renamed without any problem.

16. Jan 15, 2011

latentcorpse

Re: Covectors

I'm not concerned about the indices but rather the position of the $\nabla$. It would be at the front in the original exercise (since it acts on the first n) but in what we showed it's in the middle i.e. acting on the second n.

Or is this irrelevant seeing as $(\nabla n) n = n (\nabla n)$?

17. Jan 15, 2011

dextercioby

Re: Covectors

Yes, it's irrelevant. The <n> with an index or another can stay in front of the differentiated <n>, or after it, depending on your preference, since it's a commutative variable.

18. Jan 15, 2011

latentcorpse

Re: Covectors

Thanks. Can you recommend anything for the next part of the question. So far I have done:

$k_{[a;bc]}=\frac{1}{6} \left( \nabla_c \nabla_b k_a + \nabla_b \nabla_a k_c + \nabla_a \nabla_c k_b - \nabla_b \nabla_c k_a - \nabla_c \nabla_a k_b - \nabla_a \nabla_b k_c \right)$

But Killing vectors satisfy $\nabla_a k_b + \nabla_b k_a=0 \Rightarrow -\nabla_a k_b = \nabla_b k_a$

And hence

$k_{[a;bc]}=\frac{1}{3} \left( \nabla_c \nabla_b k_a + \nabla_b \nabla_a k_c + \nabla_a \nabla_c k_b \right)$

But I don't know how to get this to relate to the Riemann tensor as required?

Thanks.

19. Jan 15, 2011

dextercioby

Re: Covectors

Well, what you did it's wrong, because

$$k_{[a;bc]} = - R_{[cb|a]}^{~~~~~d} k_d = 0$$

by the symmetry properties of the Riemann tensor. You need to show something else, namely that

$$k_{a;bc} = R_{ab|c}^{~~~~d} k_d$$

which is a different animal. I'll try to the calculations and see what i get.

Last edited: Jan 15, 2011
20. Jan 15, 2011

dextercioby

Re: Covectors

Using the author's conventions and the Killing property

$$k_{a;bc} = -k_{b;ac}$$

I get the author's formula with a minus in front of the RHS, namely

$$k_{a;bc} = - R_{ab|c}^{~~~~d} k_d$$

I double-checked my work and it looks ok.

Hmm, Wolfram on this page mentions the same result as your paper http://mathworld.wolfram.com/KillingVectors.html (and here he's probably quoting Weinberg, but I don't have Weinberg's book before me, though).

Taking a look at Ortin (screenhot attached), I see that we agree, though I didn't check his conventions.

I just took that

$$\nabla_a \nabla_b k_c = k_{c;b;a} = k_{c;ba}$$

as a connection between the 2 distinct notations for the covariant derivative.

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Last edited: Jan 15, 2011