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Derivative of the mixed metric tensor

  1. Jun 16, 2015 #1
    So i am studying GR at the moment, and I've been trying to figure out what the derivative (not covarient) of the mixed metric tensor $$\delta^\mu_\nu$$ would be, since this tensor is just the identity matrix surely its derivative should be zero. Yet at the same time $$\delta^\mu_\nu = g_{\alpha\nu}g^{\alpha\mu}$$ which means $$\partial_\beta \delta^\mu_\nu = g_{\alpha\nu}\partial_\beta g^{\alpha\mu} + g^{\alpha\mu}\partial_\beta g_{\alpha\nu}$$ so I cannot see why it would be equal to zero. Can anybody help me out?
     
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  3. Jun 16, 2015 #2

    ChrisVer

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    you proved that ##g_{\mu \nu} \partial_\beta g^{\mu \nu} = - g^{\mu \nu} \partial_\beta g_{\mu \nu}##

    another way (the "stupid" way) is:

    [itex]g_{\mu \nu} \partial_\beta g^{\mu \nu} = g_{\mu \nu} \partial_\beta ( g^{\rho \mu} g^{\sigma \nu} g_{\rho \sigma})[/itex]

    Applying a chain derivative:

    [itex] g_{\mu \nu} \partial_\beta g^{\mu \nu} = g_{\mu \nu}g^{\rho \mu} g^{\sigma \nu} \partial_\beta g_{\rho \sigma} + g_{\mu \nu}g^{\rho \mu} g_{\rho \sigma} \partial_\beta g^{\sigma \nu} + g_{\mu \nu}g^{\sigma \mu} g_{\rho \sigma} \partial_\beta g^{\rho \nu}[/itex]

    [itex]g_{\mu \nu} \partial_\beta g^{\mu \nu} = g^{\rho \sigma} \partial_\beta g_{\rho \sigma} + \delta_{\nu}^\rho g_{\rho \sigma} \partial_\beta g^{\sigma \nu} + \delta_{\nu}^\sigma g_{\rho \sigma} \partial_\beta g^{\rho \nu} [/itex]

    The last two is just [itex]g_{\rho \sigma} \partial_{\beta}g^{\rho \nu}[/itex] , so they give 2 times the LHS, bring it there and you get:

    [itex]g_{\mu \nu} \partial_\beta g^{\mu \nu}-2g_{\mu \nu} \partial_\beta g^{\mu \nu} =g^{\rho \sigma} \partial_\beta g_{\rho \sigma} [/itex]

    [itex]-g_{\mu \nu} \partial_\beta g^{\mu \nu} =g^{\rho \sigma} \partial_\beta g_{\rho \sigma} [/itex]
     
    Last edited: Jun 16, 2015
  4. Jun 16, 2015 #3
    Hi Chris, yes but I have ##g_{\alpha\nu}g^{\alpha\mu}##, not ##g_{\mu\nu}g^{\nu\mu}=4##, is the relationship the same?
     
  5. Jun 16, 2015 #4

    ChrisVer

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    Ah sorry I mixed the indices...

    The two things are not the same since the one is the trace of the other... but their derivatives still apply.

    [itex]g_{ab}\partial_c g^{ak} = g_{ab} \partial_c (g^{ar} g^{kw} g_{rw})= g_{ab}g^{ar} g^{kw} \partial_c g_{rw} +g_{ab}g_{rw} g^{kw} \partial_c g^{ar}+ g_{ab}g_{rw} g^{ar} \partial_c g^{kw}[/itex]

    [itex]g_{ab}\partial_c g^{ak} = \delta_{b}^r g^{kw} \partial_c g_{rw} +g_{ab}\delta_r^k \partial_c g^{ar}+ g_{ab}\delta^a_w \partial_c g^{kw}[/itex]

    [itex] g_{ab}\partial_c g^{ak} =g^{kw} \partial_c g_{bw}+ g_{ab}\partial_c g^{ak}+g_{ab} \partial_c g^{ka}[/itex]

    [itex] - g_{ab} \partial_c g^{ak} = g^{ka} \partial_c g_{ba}[/itex]
     
  6. Jun 16, 2015 #5

    ChrisVer

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    But that derivation is really stupid once you have shown it with the [itex]g_{ab} g^{bc} = \delta_a^c[/itex]
     
  7. Jun 16, 2015 #6

    ChrisVer

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    If you ask why the [itex]\partial \delta=0[/itex] then you should remember that [itex]\delta_a^b= \begin{cases} 1 & ,a=b \\ 0 & ,\text{else} \end{cases}[/itex].

    That's because the [itex]\delta_a^b = g_{ac}g^{cb}[/itex] are the unit matrix elements since the [itex]g^{cb}[/itex] are the inverse metric's elements to the metric [itex]g_{ac}[/itex]. In matrix form you would write [itex] \textbf{g} \cdot \textbf{g}^{-1} = I[/itex].

    Then the derivative of a constant number (1 or 0) is zero.
     
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