# Derivative of the mixed metric tensor

1. Jun 16, 2015

### Brage

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. Jun 16, 2015

### ChrisVer

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:

$g_{\mu \nu} \partial_\beta g^{\mu \nu} = g_{\mu \nu} \partial_\beta ( g^{\rho \mu} g^{\sigma \nu} g_{\rho \sigma})$

Applying a chain derivative:

$g_{\mu \nu} \partial_\beta g^{\mu \nu} = g_{\mu \nu}g^{\rho \mu} g^{\sigma \nu} \partial_\beta g_{\rho \sigma} + g_{\mu \nu}g^{\rho \mu} g_{\rho \sigma} \partial_\beta g^{\sigma \nu} + g_{\mu \nu}g^{\sigma \mu} g_{\rho \sigma} \partial_\beta g^{\rho \nu}$

$g_{\mu \nu} \partial_\beta g^{\mu \nu} = g^{\rho \sigma} \partial_\beta g_{\rho \sigma} + \delta_{\nu}^\rho g_{\rho \sigma} \partial_\beta g^{\sigma \nu} + \delta_{\nu}^\sigma g_{\rho \sigma} \partial_\beta g^{\rho \nu}$

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

$g_{\mu \nu} \partial_\beta g^{\mu \nu}-2g_{\mu \nu} \partial_\beta g^{\mu \nu} =g^{\rho \sigma} \partial_\beta g_{\rho \sigma}$

$-g_{\mu \nu} \partial_\beta g^{\mu \nu} =g^{\rho \sigma} \partial_\beta g_{\rho \sigma}$

Last edited: Jun 16, 2015
3. Jun 16, 2015

### Brage

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?

4. Jun 16, 2015

### ChrisVer

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.

$g_{ab}\partial_c g^{ak} = g_{ab} \partial_c (g^{ar} g^{kw} g_{rw})= g_{ab}g^{ar} g^{kw} \partial_c g_{rw} +g_{ab}g_{rw} g^{kw} \partial_c g^{ar}+ g_{ab}g_{rw} g^{ar} \partial_c g^{kw}$

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

$g_{ab}\partial_c g^{ak} =g^{kw} \partial_c g_{bw}+ g_{ab}\partial_c g^{ak}+g_{ab} \partial_c g^{ka}$

$- g_{ab} \partial_c g^{ak} = g^{ka} \partial_c g_{ba}$

5. Jun 16, 2015

### ChrisVer

But that derivation is really stupid once you have shown it with the $g_{ab} g^{bc} = \delta_a^c$

6. Jun 16, 2015

### ChrisVer

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

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

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