Killing Vector Equations

TerryW

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?

George Jones

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

Hi George, and thanks for helping.

For a=0, I reckon I'll get 1/2(g_00,0 + g_11,0 +g_22,0 + g_33,0).

I worked out that
d_a (dx/ds^bdx/ds^c = 0 when I was verifying the workings on the previous page.

I already know what the g_ααs are.

How am I doing?


Terry

George Jones

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.
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

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

Thanks George,

Oh yes, I forgot about the (dx^a/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

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

Oops, I forgot the latex \dot. This should read

Now, what about

[tex]\frac{\partial K}{\partial \dot{x}^0}?[/tex]
I usually type the latex commands as I type my posts.

Altabeh

And was Schaum's Tensor Calculus useful and easygoing?

TerryW

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

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

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

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

Yes, this is correct. Did you arrive at this by using Kronecker deltas, as Altabeh suggested?
I'm not sure what this means.
Right!

TerryW

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 g_ab 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

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

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.