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Killing Vector Equations

by TerryW
Tags: equations, killing, vector
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TerryW
#1
Jan14-10, 12:35 PM
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I'm working my way through D'Inverno's Understanding GR and have reached Chapter 7 with no real problems. I'm stuck though on p101 (and the corresponding problem 7.7). D'inverno sets out two forms of the geodesic equations, one of which is the Lagrangian with the Killing Vectors and then has a throwaway line (paraphrased as) '..it is possible ...using these two equations to read off the components of the Christoffel symbols..'

I have managed all the problems so far but I just cannot see this one. Am I missing something blindingly obvious?
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George Jones
#2
Jan14-10, 12:58 PM
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Quote Quote by TerryW View Post
I'm working my way through D'Inverno's Understanding GR and have reached Chapter 7 with no real problems. I'm stuck though on p101 (and the corresponding problem 7.7). D'inverno sets out two forms of the geodesic equations, one of which is the Lagrangian with the Killing Vectors and then has a throwaway line (paraphrased as) '..it is possible ...using these two equations to read off the components of the Christoffel symbols..'

I have managed all the problems so far but I just cannot see this one. Am I missing something blindingly obvious?
Welcome to Physics Forums!

Let's work through this. Can you find

[tex]\frac{\partial K}{\partial x^a}?[/tex]

for [itex]x^a[/itex] equal to [itex]t[/itex], [itex]r[/itex], [itex]\theta[/itex], and [itex]\phi[/itex]

If you can, post what you get; if you can't, ask more questions.
TerryW
#3
Jan14-10, 02:06 PM
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P: 65
Hi George, and thanks for helping.

For a=0, I reckon I'll get 1/2(g00,0 + g11,0 +g22,0 + g33,0).

I worked out that
da (dx/dsbdx/dsc = 0 when I was verifying the workings on the previous page.

I already know what the gααs are.

How am I doing?


Terry

George Jones
#4
Jan14-10, 02:55 PM
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Killing Vector Equations

Quote Quote by TerryW View Post
For a=0, I reckon I'll get 1/2(g00,0 + g11,0 +g22,0 + g33,0).
For a diagonal metric, this is almost, but not quite, correct.

Since [itex]K = g_{ab} \dot{x}^a \dot{x}^b /2[/itex],

[tex]\frac{\partial K}{\partial x^0} = \frac{1}{2} \left( \frac{\partial g_{ab}}{\partial x^0} \right) \dot{x}^a \dot{x}^b = \frac{1}{2} \left[ \left( \frac{\partial g_{00}}{\partial x^0} \right) \left( \dot{x}^0 \right)^2 + \left( \frac{\partial g_{11}}{\partial x^0} \right) \left( \dot{x}^1 \right)^2 + \left( \frac{\partial g_{22}}{\partial x^0} \right) \left( \dot{x}^2 \right)^2 + \left( \frac{\partial g_{33}}{\partial x^0} \right) \left( \dot{x}^3 \right)^2 \right][/tex]

for a diagonal metric.
Quote Quote by TerryW View Post
I already know what the gααs are.

How am I doing?
Very well.

Now, what about

[tex]\frac{\partial K}{\partial x^0}?[/tex]

You can learn about entering mathematical expression in the thread

http://www.physicsforums.com/showthread.php?t=8997.

If you want to see the code for a mathematical expression, click on the expression, and the code will be displayed in a code window. As you click on different expressions (in any post), new code windows don't appear, the contents of a single code window change.

This is a time consuming process that is sometimes (and sometimes not) worth the effort. For now, don't worry too much about learning how to do this.
Altabeh
#5
Jan14-10, 03:20 PM
P: 665
I'm working my way through D'Inverno's Understanding GR and have reached Chapter 7 with no real problems.

Sorry for asking an irrelevant question: Is this book really great for a self-study, George, or are there any other books which can be more helpful and interesting than D'Inverno's book?
TerryW
#6
Jan14-10, 04:51 PM
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P: 65
Thanks George,

Oh yes, I forgot about the (dxa/ds)2 terms. It's the next part of the Lagrangian which is the problem though and it seems to be a bit of a circular argument which brings you back to the first equation for the geodesics with d2x0/ds2 and the Christoffel symbols.

Do you use Latex and then cut/paste the result into your posts?

I'm off to bed now. Hope to hear from you tomorrow.

Regards


Terry
TerryW
#7
Jan14-10, 04:54 PM
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P: 65
Hi Altabeh,

I'm finding it OK but I did work my way all the way through Schaum's Tensor Calculus (doing all the problems!) before I started. I think I might have struggled with D'Inverno's chapters on Tensor Calculus if I hadn't done this.

Regards


Terry
George Jones
#8
Jan14-10, 06:55 PM
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Quote Quote by George Jones View Post
Now, what about

[tex]\frac{\partial K}{\partial x^0}?[/tex]
Oops, I forgot the latex \dot. This should read

Now, what about

[tex]\frac{\partial K}{\partial \dot{x}^0}?[/tex]
Quote Quote by TerryW View Post
Do you use Latex and then cut/paste the result into your posts?
I usually type the latex commands as I type my posts.
Altabeh
#9
Jan15-10, 04:17 AM
P: 665
Quote Quote by TerryW View Post
Hi Altabeh,

I'm finding it OK but I did work my way all the way through Schaum's Tensor Calculus (doing all the problems!) before I started. I think I might have struggled with D'Inverno's chapters on Tensor Calculus if I hadn't done this.

Regards


Terry
And was Schaum's Tensor Calculus useful and easygoing?
TerryW
#10
Jan15-10, 05:13 AM
PF Gold
P: 65
Hi George,

My first venture into Latex:

[tex]
\frac{\partial K}{\partial \dot{x}^a} = g_{ab}\dot{x}^b \ \ \ \ (2K = g_{ab}\dot{x}^a\dot{x}^b\ \ \ \frac{\partial g_{ab}}{\partial \dot{x}^a}\ =\ \frac{\partial g_{ab}}{\partial {x}^a}\frac{\partial{x}}{\partial \dot{x}^a} \rightarrow \0\ as\ \ x\rightarrow0)
[/tex]

Now you are going to ask me what happens when I take the derivative of

[tex]
\frac{\partial K}{\partial \dot{x}^a}
[/tex]

with respect to the affine parameter represented by the dot over the x
TerryW
#11
Jan15-10, 05:24 AM
PF Gold
P: 65
Schaum's Tensor Calculus was very useful, but you have to do the problems. There are quite a few printing errors however which leave you pondering for a while. Whether or not it is easy going depends on you!
Altabeh
#12
Jan15-10, 05:33 AM
P: 665
Quote Quote by TerryW View Post
Hi George,

My first venture into Latex:

[tex]
\frac{\partial K}{\partial \dot{x}^a} = g_{ab}\dot{x}^b \ \ \ \ (2K = g_{ab}\dot{x}^a\dot{x}^b\ \ \ \frac{\partial g_{ab}}{\partial \dot{x}^a}\ =\ \frac{\partial g_{ab}}{\partial {x}^a}\frac{\partial{x}}{\partial \dot{x}^a} \rightarrow \0\ as\ \ x\rightarrow0)
[/tex]

Now you are going to ask me what happens when I take the derivative of

[tex]
\frac{\partial K}{\partial \dot{x}^a}
[/tex]

with respect to the affine parameter represented by the dot over the x
See that you are getting professional in using Latex...

[tex]
\frac{\partial K}{\partial \dot{x}^a} = \frac{\partial}{\partial \dot{x}^a}(\frac{1}{2}g_{bc}\dot{x}^b\dot{x^c}) = ... [/tex]

Now you continue this calculation and use the Kronecker's delta to manage combining two terms appearing in the operation. (Remember that g_ab is symmetric).

AB
TerryW
#13
Jan15-10, 06:44 AM
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P: 65
Hi Altabeh,

I thought I'd done that already

[tex]\frac{\partial K}{\partial \dot{x}^a} = g_{ab}\dot{x}^b}[/tex]
George Jones
#14
Jan15-10, 07:06 AM
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Quote Quote by TerryW View Post
Hi George,

My first venture into Latex:

[tex]
\frac{\partial K}{\partial \dot{x}^a} = g_{ab}\dot{x}^b
[/tex]
Yes, this is correct. Did you arrive at this by using Kronecker deltas, as Altabeh suggested?
Quote Quote by TerryW View Post
[tex](2K = g_{ab}\dot{x}^a\dot{x}^b\ \ \ \frac{\partial g_{ab}}{\partial \dot{x}^a}\ =\ \frac{\partial g_{ab}}{\partial {x}^a}\frac{\partial{x}}{\partial \dot{x}^a} \rightarrow \0\ as\ \ x\rightarrow0)[/tex]
I'm not sure what this means.
Quote Quote by TerryW View Post
Now you are going to ask me what happens when I take the derivative of

[tex]
\frac{\partial K}{\partial \dot{x}^a}
[/tex]

with respect to the affine parameter represented by the dot over the x
Right!
TerryW
#15
Jan15-10, 08:11 AM
PF Gold
P: 65
Hi George,

Yes I did arrive at my answer by Kronecker deltas. The other bit was just an explanation of why the partial derivative of gab disappears.

So for the next bit;

[tex]\frac{d}{du}(g_{ab}\dot{x}^b) = \partial_{c}g_{ab}\dot{x}^b\dot{x}^c+g_{ab}\ddot{x}^b[/tex]

Where now?
Altabeh
#16
Jan15-10, 11:30 AM
P: 665
Quote Quote by TerryW View Post
Hi George,

Yes I did arrive at my answer by Kronecker deltas. The other bit was just an explanation of why the partial derivative of gab disappears.

So for the next bit;

[tex]\frac{d}{du}(g_{ab}\dot{x}^b) = \partial_{c}g_{ab}\dot{x}^b\dot{x}^c+g_{ab}\ddot{x}^b[/tex]

Where now?
Nice. Now you are required to calculate [tex]\frac{\partial}{\partial x^{a}}(\frac{1}{2}g_{bc}\dot{x}^b\dot{x}^c)[/tex]. If you've already calculated these, then arrange all the terms obtained so far in the order given by the geodesic equation (7.46). After that just try to get rid of [tex]g_{ab}[/tex] in [tex]g_{ab}\ddot{x }^b[/tex]. What should you do to lead to pure terms like [tex]\ddot{x }^b[/tex]? (Note: Don't take the index b of [tex]\ddot{x}^b[/tex] seriously here. It must be something else if one still uses terms including [tex]\dot{x}^b\dot{x}^c[/tex].)
TerryW
#17
Jan15-10, 01:50 PM
PF Gold
P: 65
Hi Altabeh,

Is all this leading towards:

[tex]\ddot{x}^a +\Gamma^a_{bc}\dot{x}^b\dot{x}^c = 0[/tex] ? (7.42 in D'Inverno)

If it is, it isn't the point of my original question which was to find out what D'Inverno means by "It is possible, by (7.42) (ie the equation above) to read off directly from

[tex]\frac{\partial{K}}{\partial{x}^a} - \frac{d}{du}\left(\frac{\partial{K}}{\partial\dot{x}^a}\right)=0[/tex] (7.46 in D'Inverno)
the components of the connection [tex]\Gamma^a_{bc}[/tex], and this proves to be a very efficient way of calculating [tex]\Gamma^a_{bc}[/tex]."

I know various techniques for working out the [tex]\Gamma^a_{bc}[/tex] values for a given metric, I thought that this was alluding to an even smarter way of doing it but I can't see it.
Altabeh
#18
Jan15-10, 02:24 PM
P: 665
Quote Quote by TerryW View Post
Hi Altabeh,

Is all this leading towards:

[tex]\ddot{x}^a +\Gamma^a_{bc}\dot{x}^b\dot{x}^c = 0[/tex] ? (7.42 in D'Inverno)

If it is, it isn't the point of my original question which was to find out what D'Inverno means by "It is possible, by (7.42) (ie the equation above) to read off directly from

[tex]\frac{\partial{K}}{\partial{x}^a} - \frac{d}{du}\left(\frac{\partial{K}}{\partial\dot{x}^a}\right)=0[/tex] (7.46 in D'Inverno)
the components of the connection [tex]\Gamma^a_{bc}[/tex], and this proves to be a very efficient way of calculating [tex]\Gamma^a_{bc}[/tex]."

I know various techniques for working out the [tex]\Gamma^a_{bc}[/tex] values for a given metric, I thought that this was alluding to an even smarter way of doing it but I can't see it.

Yes, it is! But the point is that if you wanted to derive the Christoffel symbols for the desired metric here directly from the Euler-Lagrange equation, you would end up leading to what D'Inverno asked for in reality. We just gave a proof of how one would extract (7.42) from (7.46) which is the Euler-Lagrange equation.

So all you need now is to start from the line element in spherical coordinates, ds2, and devide both sides of it by du2, so this would be your [tex]\frac{1}{2}g_{bc}\dot{x}^b\dot{x}^c[/tex]. Remember that here [tex](ds/du)^2=2K[/tex] and for the sake of convenience, simply ignore that factor 2 which won't make any trouble ahead.

AB


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