Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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?
  2. jcsd
  3. Jun 16, 2015 #2


    User Avatar
    Gold Member

    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


    User Avatar
    Gold Member

    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


    User Avatar
    Gold Member

    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


    User Avatar
    Gold Member

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook