- #1

TerryW

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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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- Thread starter TerryW
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- #1

TerryW

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

- #2

George Jones

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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.

Last edited:

- #3

TerryW

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For a=0, I reckon I'll get 1/2(g

I worked out that

d

I already know what the g

How am I doing?

Terry

- #4

George Jones

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For a=0, I reckon I'll get 1/2(g_{00,0}+ g_{11,0}+g_{22,0}+ g_{33,0}).

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.

I already know what the g_{αα}s 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

Altabeh

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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

TerryW

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Oh yes, I forgot about the (dx

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

TerryW

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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

- #8

George Jones

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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]

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

Altabeh

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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

TerryW

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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

- #11

TerryW

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- #12

Altabeh

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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

- #13

TerryW

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I thought I'd done that already

[tex]\frac{\partial K}{\partial \dot{x}^a} = g_{ab}\dot{x}^b}[/tex]

- #14

George Jones

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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?

[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.

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!

- #15

TerryW

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

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

Altabeh

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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?

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

TerryW

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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

Altabeh

- 660

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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

So all you need now is to start from the line element in spherical coordinates, ds

AB

- #19

George Jones

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We have been proceeding too generally.

Yes, it is! But the point is that if you wanted to derive the Christoffel symbolsfor 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

TerryW

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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.

Regards

Terry

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