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linford86

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- Thread starter linford86
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linford86

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- #2

haushofer

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[tex]

T'(x) - T(x)

[/tex]

induced by the infinitesimal coordinate transformation

[tex]

x \rightarrow x + \xi

[/tex]

- #3

linford86

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Thanks for your reply! Can you give an example of either of these methods?

- #4

haushofer

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Thanks for your reply! Can you give an example of either of these methods?

I can give the rough idea :) Let's pick a tensor T

[tex]

T = T_{\mu\nu} dx^{\mu} \otimes dx^{\nu}

[/tex]

and perform the coordinate transformation

[tex]

x^{\lambda} \rightarrow x^{\lambda'} = x^{\lambda} - \xi^{\lambda}

[/tex]

We know that for our T we have

[tex]

T_{\mu' \nu '}(x') = \frac{\partial x^{\mu}}{\partial x^{\mu '}}\frac{\partial x^{\nu }}{\partial x^{\nu '}}T_{\mu\nu}(x) \ \ \ \ \ \ \ \ (1)

[/tex]

For our coordinate transformation we have that

[tex]

\frac{\partial x^{\mu}}{\partial x^{\mu '}} = \delta^{\mu}_{\mu '} + \frac{\partial }{\partial x^{\mu '}}\xi^{\mu}

[/tex]

You can plug this into (1). But the Lie derivative of our tensor T with respect to [itex]\xi[/itex] is given by (in components)

[tex]

T_{\mu ' \nu '}(x) - T_{\mu\nu}(x)

[/tex]

You can calculate this by performing a Taylor expansion,

[tex]

T(x) = T(x' + \xi) = T(x') + \xi^{\mu}\partial_{\mu}T(x')

[/tex]

and recognizing that, to first order in xi,

[tex]

\xi^{\mu}\partial_{\mu}T(x') = \xi^{\mu}\partial_{\mu}T(x)

[/tex]

It's been a while that I did these kind of calculations explicitly, but I think you should be able to do it. It maybe helps to first do it for a scalar field, then for a vector or covector, and so on. Notice that the signs depend on how you define your coordinate transformation with either a plus or minus sign! For a general tensor T there always will be a leading term

[tex]

\xi^{\mu}\partial_{\mu}T(x)

[/tex]

The other terms will be due to the "pulling back of the coordinate", but that will become clear if you read something about pushforwards and pullbacks; the GR text by Carroll or Nakahara's geometry book are very useful. Hope this helps :)

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haushofer

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haushofer

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Did you manage to do the calculation? :)

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linford86

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haushofer

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linford86

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Sorry it's taken me a while to respond. I've been pretty busy with grad school. At any rate, we want to define something like:

[tex]L_V T=[V,T][/tex]

or

[tex]L_V T=VT-TV[/tex]

for all tensors T and any vector V where L is the lie differential operator. Now, ordinarily, the lie derivative acting on a tensor is*defined* to be:

[tex]L_V T=V^\mu \partial_\mu T^{\mu_1 \mu_2 ... \mu_k}. _{\nu_1 \nu_2 ... \nu_l}-

(\partial_\lambda V^\mu_1)T^{\lambda \mu_2 ... \mu_k}. _{\nu_1 \nu_2 ... \nu_l}-

(\partial_\lambda V^\mu_2)T^{\mu_1 \lambda ... \mu_k}. _{\nu_1 \nu_2 ... \nu_l}-...

+(\partial_\nu_1 V^\lambda)T^{\mu_1 \mu_2 ... \mu_k}. _{\lambda \nu_2 ... \nu_l}

+(\partial_\nu_2 V^\lambda)T^{\mu_1 \mu_2 ... \mu_k}. _{\nu_1 \lambda ... \nu_l}+...

[/tex]

Obviously, [tex]V^\mu \partial_\mu T^{\mu_1 \mu_2 ... \mu_k}. _{\nu_1 \nu_2 ... \nu_l}[/tex] is just the action of V on T. Thus, our job is half done -- we have a natural way to define VT. What about the action of the tensor on the vector -- TV? Well, we can just define:

[tex]TV=

(\partial_\lambda V^\mu_1)T^{\lambda \mu_2 ... \mu_k}. _{\nu_1 \nu_2 ... \nu_l}+

(\partial_\lambda V^\mu_2)T^{\mu_1 \lambda ... \mu_k}. _{\nu_1 \nu_2 ... \nu_l}+...

-(\partial_\nu_1 V^\lambda)T^{\mu_1 \mu_2 ... \mu_k}. _{\lambda \nu_2 ... \nu_l}

-(\partial_\nu_2 V^\lambda)T^{\mu_1 \mu_2 ... \mu_k}. _{\nu_1 \lambda ... \nu_l}-...

[/tex]

And now we can write:

[tex]L_V T=VT-TV[/tex]

With a little computation, you can show that this satisfies all of the desirable properties.

[Note: the only way that I could make the indices appear in the right order was to include periods between the upstairs indices and the downstairs indices. If anyone knows how to make the indices go in the right order without doing that, let me know; it would be really helpful.]

[tex]L_V T=[V,T][/tex]

or

[tex]L_V T=VT-TV[/tex]

for all tensors T and any vector V where L is the lie differential operator. Now, ordinarily, the lie derivative acting on a tensor is

[tex]L_V T=V^\mu \partial_\mu T^{\mu_1 \mu_2 ... \mu_k}. _{\nu_1 \nu_2 ... \nu_l}-

(\partial_\lambda V^\mu_1)T^{\lambda \mu_2 ... \mu_k}. _{\nu_1 \nu_2 ... \nu_l}-

(\partial_\lambda V^\mu_2)T^{\mu_1 \lambda ... \mu_k}. _{\nu_1 \nu_2 ... \nu_l}-...

+(\partial_\nu_1 V^\lambda)T^{\mu_1 \mu_2 ... \mu_k}. _{\lambda \nu_2 ... \nu_l}

+(\partial_\nu_2 V^\lambda)T^{\mu_1 \mu_2 ... \mu_k}. _{\nu_1 \lambda ... \nu_l}+...

[/tex]

Obviously, [tex]V^\mu \partial_\mu T^{\mu_1 \mu_2 ... \mu_k}. _{\nu_1 \nu_2 ... \nu_l}[/tex] is just the action of V on T. Thus, our job is half done -- we have a natural way to define VT. What about the action of the tensor on the vector -- TV? Well, we can just define:

[tex]TV=

(\partial_\lambda V^\mu_1)T^{\lambda \mu_2 ... \mu_k}. _{\nu_1 \nu_2 ... \nu_l}+

(\partial_\lambda V^\mu_2)T^{\mu_1 \lambda ... \mu_k}. _{\nu_1 \nu_2 ... \nu_l}+...

-(\partial_\nu_1 V^\lambda)T^{\mu_1 \mu_2 ... \mu_k}. _{\lambda \nu_2 ... \nu_l}

-(\partial_\nu_2 V^\lambda)T^{\mu_1 \mu_2 ... \mu_k}. _{\nu_1 \lambda ... \nu_l}-...

[/tex]

And now we can write:

[tex]L_V T=VT-TV[/tex]

With a little computation, you can show that this satisfies all of the desirable properties.

[Note: the only way that I could make the indices appear in the right order was to include periods between the upstairs indices and the downstairs indices. If anyone knows how to make the indices go in the right order without doing that, let me know; it would be really helpful.]

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