Solve Physics Q1 from 2007 Paper61. Cambridge Maths Postgrad

[latentcorpse]

I'm trying Q1 in this past paper
http://www.maths.cam.ac.uk/postgrad/mathiii/pastpapers/2007/Paper61.pdf
and I am stuck on the very first part. Any hints as to how to get $n_{[a;b}n_{c]}=0$?

 

[dextercioby]

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

 

[latentcorpse]

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

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 $\beta(x)$.

 

[latentcorpse]

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

Thanks.

 

[dextercioby]

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

 

[jambaugh]

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

 

[dextercioby]

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.

 

[latentcorpse]

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.

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.

 

[dextercioby]

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.

 

[latentcorpse]

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.

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?

 

[dextercioby]

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

 

[latentcorpse]

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

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?

 

[dextercioby]

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.

 

[latentcorpse]

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.

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

Anyway, following your advice I have reduced it to 3 terms

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

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?

 

[dextercioby]

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.

 

[latentcorpse]

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.

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 $(\nabla n) n = n (\nabla n)$?

 

[dextercioby]

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.

 

[latentcorpse]

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.

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.

 

[dextercioby]

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.

 

[dextercioby]

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.

 

[latentcorpse]

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.

Hi again. I'm a little confused.

So $k_{a;bc}=-k_{b;ac}$ by the Killing property.

The Ricci identity (given at the end of the question) which holds for torsion free connections is that

$k_{a;bc}=k_{b;ac} + R_{a~bc}^{~d~~}k_d=-k_{a;bc} + R_{a~bc}^{~d~~}k_d$
$\Rightarrow 2k_{a;bc}=R_{a~bc}^{~d~~}k_d \Rightarrow k_{a;bc}= \frac{1}{2} R_{a~bc}^{~d~~}k_d$

Now my concern is the $\frac{1}{2}$ at the front and the fact that I have the $d$ index raised in the 2nd position instead of the 4th as it should be in the answer. What did you do here?

EDIT: Icopied the Ricci identity down wrong. Using the correct version I can't combine the two Killing vector terms into one term - how did you manage that?

Thanks.

 

[dextercioby]

Here's what I did

$$k_{a;bc} - k_{a;cb} = R_{a~bc}^{~d~~} k_d$$

which is the Ricci identity. And now I start playing around

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

$$-k_{b;ac} + R_{c~ab}^{~d~~} k_d + k_{c;ba} = R_{a~bc}^{~d~~} k_d$$

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

$$-k_{b;(ac)} - k_{b;[ac]} + k_{b;[ac]} = R_{a~bc}^{~d~~} k_d - R_{c~ab}^{~d~~} k_d$$

$$-2 k_{b;ac} = R_{a~bc}^{~d~~} k_d - R_{c~ab}^{~d~~} k_d - R_{b~ac}^{~d~~} k_d$$

$$- 2 k_{b;ac} = R_{a~bc}^{~d~~} k_d - R_{c~ab}^{~d~~} k_d + R_{b~ca}^{~d~~} k_d$$

Now I use the 1st Bianchi identity for the Riemann tensor $R_{[a~bc]}^{~~d~~~} =0$ to get

$$- 2 k_{b;ac} = - 2 R_{c~ab}^{~d~~} k_d$$

Dividing by -2 and making use of another property of the Riemann tensor I get

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

which is the result I wrote before and is confirmed by Ortin.

 

[latentcorpse]

Here's what I did

$$k_{a;bc} - k_{a;cb} = R_{a~bc}^{~d~~} k_d$$

which is the Ricci identity. And now I start playing around

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

$$-k_{b;ac} + R_{c~ab}^{~d~~} k_d + k_{c;ba} = R_{a~bc}^{~d~~} k_d$$

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

$$-k_{b;(ac)} - k_{b;[ac]} + k_{b;[ac]} = R_{a~bc}^{~d~~} k_d - R_{c~ab}^{~d~~} k_d$$

$$-2 k_{b;ac} = R_{a~bc}^{~d~~} k_d - R_{c~ab}^{~d~~} k_d - R_{b~ac}^{~d~~} k_d$$

$$- 2 k_{b;ac} = R_{a~bc}^{~d~~} k_d - R_{c~ab}^{~d~~} k_d + R_{b~ca}^{~d~~} k_d$$

Now I use the 1st Bianchi identity for the Riemann tensor $R_{[a~bc]}^{~~d~~~} =0$ to get

$$- 2 k_{b;ac} = - 2 R_{c~ab}^{~d~~} k_d$$

Dividing by -2 and making use of another property of the Riemann tensor I get

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

which is the result I wrote before and is confirmed by Ortin.

Ok. I can follow that. Just not sure I'd be able to reproduce it in an exam though!

As for the next bit, where do you recommend starting? I'm a bit confused by the $\left( k_{[a;b}k_{c]} \right)^{;c}$ since isn't that just $0^{;c}=0$ since the Killing vector $k_a$ is hypersurface orthogonal?

 

[dextercioby]

I refer to the screenshot above from Ortin. It's easy to make this deduction below

$$R_{ab|c}^{~~~~d} k_d = - k_{a;bc} \Rightarrow \Box k_a = k_b R^{b}_{~a}$$ (1)

We're asked to prove that

$$k_d R^{d}_{~a} k_{b} = k_d R^{d}_{~b} k_{a}$$ (2)

With (1) (2) becomes

$$\left(\Box k_a\right) k_b = \left(\Box k_b\right) k_a$$ (3)

So we should prove (3). Now follow the hint made the author (at least he's right with this hint) and obtain (3).

Post your calculations.

 

[latentcorpse]

I refer to the screenshot above from Ortin. It's easy to make this deduction below

$$R_{ab|c}^{~~~~d} k_d = - k_{a;bc} \Rightarrow \Box k_a = k_b R^{b}_{~a}$$ (1)

We're asked to prove that

$$k_d R^{d}_{~a} k_{b} = k_d R^{d}_{~b} k_{a}$$ (2)

With (1) (2) becomes

$$\left(\Box k_a\right) k_b = \left(\Box k_b\right) k_a$$ (3)

So we should prove (3). Now follow the hint made the author (at least he's right with this hint) and obtain (3).

Post your calculations.

First of all, can you explain what that vertical line in the Riemann tensor means?
What's the difference between $R_{ab|c}{}^d k_d$ and $R_{abc}{}^d k_d$?

Then, I verified (1) as follows:

$\Box k_a = k_{a;bc}g^{bc}=-g^{bc} R_{abc}{}^d k_d = g^{bc} R_{bac}{}^d k_d = R^c{}_{ac}{}^d k_d = R_a{}^d k_d = R^b{}_a k_b$
as required.

Now we consider $(k_{[a;b}k_{c]})^{;c}=0$. We know this is equal to zero because the Killing vector is hypersurface orthogonal and we use the result we showed at the start of the question.

So we expand to get:

$(k_{[a;b}k_{c]})^{;c}=0$
$\nabla^c ( \nabla_b k_a k_c + \nabla_a k_c k_b + \nabla_c k_b k_a - \nabla_a k_b k_c - \nabla_b k_c k_a - \nabla_c k_a k_b )=0$
$\nabla^c ( ( \nabla_b k_a) k_c) + \nabla^c ((\nabla_a k_c ) k_b ) + ( \Box k_b ) k_a - \nabla_c (( \nabla_a k_b ) k_c ) - \nabla_c (( \nabla_b k_c ) k_a ) - ( \Box k_a ) k_b =0$
So we're almost done - except for cancelling the terms we don't want. Now I assume this uses the Killing vector property $\nabla_a k_b = - \nabla_b k_a$ since we haven't used that yet but I can't get it to work...

 

[dextercioby]

The line stands as separation between the 2 groups of indices, but you can disregard it. It's used here http://arxiv.org/PS_cache/hep-th/pdf/0306/0306154v3.pdf and in other articles. Zinoviev, on the other hand, uses a different notation here http://arxiv.org/PS_cache/hep-th/pdf/0211/0211233v1.pdf, but I prefer the | one.

Coming back you your question,

$$\left(k_{[a;b} k_{c]}\right)^{;c}=0 \Rightarrow \left(k_{a;b} k_c + k_{c;a} k_b + k_{b;c} k_{a}\right)^{;c} = 0$$

Now expand to 6 terms. 4 of them will cancel each other out. The 2 you're left with are the ones you need, namely the 2 with the d'Alembertian.

 

[latentcorpse]

The line stands as separation between the 2 groups of indices, but you can disregard it. It's used here http://arxiv.org/PS_cache/hep-th/pdf/0306/0306154v3.pdf and in other articles. Zinoviev, on the other hand, uses a different notation here http://arxiv.org/PS_cache/hep-th/pdf/0211/0211233v1.pdf, but I prefer the | one.

Coming back you your question,

$$\left(k_{[a;b} k_{c]}\right)^{;c}=0 \Rightarrow \left(k_{a;b} k_c + k_{c;a} k_b + k_{b;c} k_{a}\right)^{;c} = 0$$

Now expand to 6 terms. 4 of them will cancel each other out. The 2 you're left with are the ones you need, namely the 2 with the d'Alembertian.

Yeah so

$\left(k_{[a;b} k_{c]}\right)^{;c}=0 \Rightarrow \left(k_{a;b} k_c + k_{c;a} k_b + k_{b;c} k_{a}\right)^{;c} = 0$ by expanding the square brackets.

Then we perform the derivative operator to get

$\nabla^c \nabla_b k_a k_c + \nabla_b k_a \nabla^c k_c + \nabla^c \nabla_a k_c k_b + \nabla_a k_c \nabla^c k_b + \nabla^c \nabla_c k_b k_a + \nabla_c k_b \nabla^c k_a=0$
$\nabla^c \nabla_b k_a k_c + \nabla_b k_a \nabla^c k_c - \nabla^c \nabla_c k_a k_b + \nabla_a k_c \nabla^c k_b + \nabla^c \nabla_c k_b k_a + \nabla_c k_b \nabla^c k_a=0$
where we used the Killing property on the third term. This then gives:
$\nabla^c \nabla_b k_a k_c + \nabla_b k_a \nabla^c k_c - \Box k_a k_b + \nabla_a k_c \nabla^c k_b + \Box k_b k_a + \nabla_c k_b \nabla^c k_a=0$

Now I was going to try and cancel the 4th and 6th terms by noting that $\nabla_c k_b \nabla^c k_a = \nabla^c k_b \nabla_a k_c = \nabla_a k_c \nabla^c k_b$
But this in fact would cause them to add rather than cancel each other? And as for getting rid of the first two terms, I have no idea how to do that because one of them has two covariant derivatives multiplied together and the other has a double covariant derivative?

 

[dextercioby]

Wow, what you wrote looks horrible. I'll use the semicolon notation.

$$\left(k_{a;b} k_c + k_{c;a} k_b + k_{b;c} k_{a}\right)^{;c} = 0$$

$$\Rightarrow k_{a;b}^{~~~;c} k_c + k_{a;b} k_{c}^{~;c} + k_{c;a}^{~~~;c} k_b + k_{c;a} k_{b}^{~;c} + \left(\Box k_b \right) k_a + k_{b;c} k_{a}^{~;c} = 0$$

 

[dextercioby]

The 6 terms I wrote above (and mentioned in post #26) can be simplified further. The second one is 0 (do you agree ?), while the 4th + 6th are one the opposite of the other, so when added, they cancel each other out (do you see why ?).

We're left with the first, the 3rd and the 5th. The first is 0, do you see why ?

The 3rd can be brought to a nice form involving the d'Alembert operator. The 5th is good the way it is.

Putting all together, you get the equality you need.

 

[latentcorpse]

The 6 terms I wrote above (and mentioned in post #26) can be simplified further. The second one is 0 (do you agree ?), while the 4th + 6th are one the opposite of the other, so when added, they cancel each other out (do you see why ?).

We're left with the first, the 3rd and the 5th. The first is 0, do you see why ?

The 3rd can be brought to a nice form involving the d'Alembert operator. The 5th is good the way it is.

Putting all together, you get the equality you need.

Yeah what I had is the same as what you had - just in a more ugly notation I suppose but I decided to work with $\nabla$'s just to make it easier for me to follow what was going on haha!

However - I don't see why the fourth and sixth cancel? This was what I was talking about trying to get to work in my last post.

Also, why does the second term vanish? I assume this is because $k_{;c}{}^{;c}=0$? Where does that come from?

And that leaves us with our 3rd and 5th terms which we can rewrite as our desired d'Alembertian terms but don't we also still have the first term left in there as well?

Thanks again.

 

[dextercioby]

A Killing vector is divergence-free, as it can be seen by contracting the 2 indices in their defining equation.

The 6th term is minus the 4th:

$$k_{b;c} k_{a}^{~;c} = k_{a}^{~;c} k_{b;c} = k_{a;c} k_{b}^{~;c} = - k_{c;a} k_{b}^{~;c}$$

Ok ?

I leave it to you to figure out why the first term is 0.

 

[latentcorpse]

A Killing vector is divergence-free, as it can be seen by contracting the 2 indices in their defining equation.

The 6th term is minus the 4th:

$$k_{b;c} k_{a}^{~;c} = k_{a}^{~;c} k_{b;c} = k_{a;c} k_{b}^{~;c} = - k_{c;a} k_{b}^{~;c}$$

Ok ?

I leave it to you to figure out why the first term is 0.

Yeah I made a stupid mistake when trying to get the 4th to cancel the 6th...

As for Killing vectors beign divergence free:
we contract the defining equation to get $\nabla^a k_a + \nabla_a k^a=0 \Rightarrow \nabla^a k_a + \nabla^a k_a = 0 \Rightarrow \nabla^a k_a =0$
i.e. $k_c{}^{;c}=0$
and hence the second term, $k_{a;b} k_c{}^{;c}$, will vanish.

The first term remains a mystery. Am I right in thinking that we want to show it's equal to negative itself and hence is zero?
The natural way to approach that would be to say
$k_{a;b}{}^{;c}k_c=-k_{b;a}{}^{;c}k_c$
Now the best I can come up with is arguing that whilst the two c indices are symmetric under exchange, the a and b indices are antisymmetric under exchange and therefore we have a symmetric term multiplying an antisymmetric term and so the whole thing will be zero. How's that? I have a feeling it's wrong because the c index terms aren't next to each other!

 

[dextercioby]

I will give you a hint, not post the solution this time. You need to use one of the 2 points already proven in this problem.

 

[latentcorpse]

I will give you a hint, not post the solution this time. You need to use one of the 2 points already proven in this problem.

Well, using the last thing we proved, we have that

$k_{a;b}{}^{;c}k_c=k_{a;bc}k^c=R_{abc}{}^dk^ck_d$

But the Riemann tensor is antisymmetric under exchange of indices $c \leftrightarrow d$ but $k^ck_d$ is symmetric and hence the whole thing vanishes.

Is that correct?

 

[dextercioby]

Yes. Finally :)