What does this notation mean?

  • #1

Summary:

The killing vector equation reads: ##\nabla_{(\mu K_{\nu}) = 0## What do the parenthesis mean explicitly?

Main Question or Discussion Point

Hi all,

The killing vector equation reads: ##\nabla_{(\mu K_{\nu})} = 0## What do the parenthesis mean explicitly?

Moreover, I know that ##\nabla_\mu x^\nu = \partial_\mu x^\nu+ \Gamma_{\rho \mu}^\nu x^\rho##

So if the parentheses mean symmetric the Killing equation will read:

##\frac{1}{2} ( \partial_\mu k_\nu + \partial_\nu k_\mu) - \Gamma_{\nu \mu}^\rho k^\rho##

Is this correct?
 
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Answers and Replies

  • #2
Nugatory
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You have mismatched curly braces that are breaking the Latex.... you should be able to to edit the post to fix that, but of you are having trouble let me or any of the other mentors know and we can help.
 
  • #3
vanhees71
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I guess, you wanted to write ##\partial_{(\mu} K_{\nu)}## then it usually means "symmetrization", i.e., for 2nd-rank tensor components it means
$$A_{(\mu \nu)}=\frac{1}{2}(A_{\mu \nu}+A_{\nu \mu}).$$
 
  • #4
Orodruin
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Fixed LaTeX:
Summary:: The killing vector equation reads: ##\nabla_{(\mu} K_{\nu)} = 0## What do the parenthesis mean explicitly?

Hi all,

The killing vector equation reads: ##\nabla_{(\mu} K_{\nu)} = 0## What do the parenthesis mean explicitly?

Moreover, I know that ##\nabla_\mu x^\nu = \partial_\mu x^\nu+ \Gamma_{\rho \mu}^\nu x^\rho##
So if the parentheses mean symmetric the Killing equation will read:

##\frac{1}{2} ( \partial_\mu k_\nu + \partial_\nu k_\mu) - \Gamma_{\nu \mu}^\rho k^\rho##

Is this correct?
No, it is not correct, you have an index mismatch since your ##\rho## appears twice as a covariant index. Because of this it cannot be the correct expression. The ##\rho## on the vector component should be lowered. In other words:
$$
\nabla_{(\mu} K_{\nu)} = \frac{1}{2} ( \partial_\mu k_\nu + \partial_\nu k_\mu) - \Gamma_{\nu \mu}^\rho k^\rho = 0.
$$
 
  • #5
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In other words:
$$
\nabla_{(\mu} K_{\nu)} = \frac{1}{2} ( \partial_\mu k_\nu + \partial_\nu k_\mu) - \Gamma_{\nu \mu}^\rho k^\rho = 0.
$$
This doesn't seem correct either.
 
  • #6
vanhees71
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Well, isn't
$$\nabla_{\mu} K_{\nu}=\partial_{\mu} K_{\nu} - \Gamma^{\rho}_{\mu \nu} K_{\rho}$$
and then
$$\nabla_{(\mu} K_{\nu)} = \frac{1}{2} (\partial_{\mu} K_{\nu} + \partial_{\nu} K_{\mu}) -\Gamma^{\rho}_{\mu \nu} K_{\rho}?$$
So @Orodruin is right, except that he forgot to lower the index in the last term too :-).
 
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  • #7
Orodruin
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except that he forgot to lower the index in the last term too :-).
The connection between brain and fingers sometimes does not work ... I blame copy-paste.
 
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