How can Killing Vector Equations help find Christoffel symbols?

In summary, the conversation is about working through D'Inverno's Understanding GR and reaching Chapter 7 without any problems. However, the individual is stuck on p101 and problem 7.7, which involves two forms of the geodesic equations and a throwaway line about using these equations to read off the components of the Christoffel symbols. The individual has managed all the previous problems, but is struggling with this one. They ask for help and George provides assistance, discussing various mathematical calculations and techniques. They also briefly discuss the usefulness and difficulty of Schaum's Tensor Calculus.
  • #1
TerryW
Gold Member
191
13
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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  • #2
TerryW said:
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. :smile:
 
Last edited:
  • #3
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
 
  • #4
TerryW said:
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.
TerryW said:
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

https://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.
 
  • #5
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?
 
  • #6
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
 
  • #7
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.

RegardsTerry
 
  • #8
George Jones said:
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]
TerryW said:
Do you use Latex and then cut/paste the result into your posts?

I usually type the latex commands as I type my posts.
 
  • #9
TerryW said:
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?
 
  • #10
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
:smile:
 
  • #11
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!:smile:
 
  • #12
TerryW said:
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
:smile:

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
 
  • #13
Hi Altabeh,

I thought I'd done that already

[tex]\frac{\partial K}{\partial \dot{x}^a} = g_{ab}\dot{x}^b}[/tex]
 
  • #14
TerryW said:
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?
TerryW said:
[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.
TerryW said:
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
:smile:

Right!
 
  • #15
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?
 
  • #16
TerryW said:
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].)
 
Last edited:
  • #17
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.
 
  • #18
TerryW said:
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 divide 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
 
  • #19
Sorry, I have been busy with work all day.

We have been proceeding too generally.
Altabeh said:
Yes, it is! But the point is that if you wanted to derive the Christoffel symbols for the desired metric[/B


Exactly!

Or if you wanted to write down the equations of geodesic motion for a a given metric without first calculating the [itex]\Gamma^\alpha {}_{\mu \nu}[/itex] explicitly.

Let's start again. Write down the specific [itex]K[/itex] for the specific metric given in exercise 6.31.
 
  • #20
Hi George and Altabeh,

The penny has just dropped! I've just done [tex]\Gamma^0_{00}[/tex] for the Schwartzchild metric and can see my way to doing the rest.

Thanks very much both of you for helping with this.:smile:



Regards

Terry
 

Related to How can Killing Vector Equations help find Christoffel symbols?

What are Killing Vector Equations?

Killing Vector Equations are a set of equations in differential geometry that are used to determine symmetries of a given space or manifold. They are named after Wilhelm Killing, a German mathematician who first introduced them in the late 19th century.

What is the significance of Killing Vector Equations in physics?

Killing Vector Equations play a crucial role in the field of theoretical physics, particularly in general relativity. They are used to describe the symmetries of spacetime and help in solving Einstein's field equations, which describe the curvature of spacetime.

How are Killing Vector Equations related to Lie groups?

Killing Vector Equations are closely related to Lie groups, which are mathematical structures that represent continuous symmetries in a space. In fact, the solutions to Killing Vector Equations can be used to generate Lie groups, making them an important tool in the study of these groups.

What is the difference between a Killing vector and a Killing vector field?

A Killing vector is a vector field that satisfies the Killing Vector Equations, meaning it generates a symmetry in a space. On the other hand, a Killing vector field is a collection of Killing vectors that form a vector space. Essentially, a Killing vector field is a set of all possible Killing vectors in a given space.

How are Killing Vector Equations used in practical applications?

Killing Vector Equations have various applications in physics, including in the study of black holes, cosmology, and the behavior of spacetime in extreme conditions. They are also used in mathematics, specifically in the field of differential geometry, to study the symmetries of spaces and manifolds.

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