Derivative of the mixed metric tensor

In summary, the conversation discusses the derivative of the mixed metric tensor $$\delta^\mu_\nu$$ and how it can be calculated using the identity matrix. The relationship between the two is shown to be the same, and the derivative is proven to be equal to zero. The conversation also mentions the use of chain derivatives and the properties of the delta function.
  • #1
Brage
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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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  • #2
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]
 
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  • #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?
 
  • #4
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]
 
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  • #5
But that derivation is really stupid once you have shown it with the [itex]g_{ab} g^{bc} = \delta_a^c[/itex]
 
  • #6
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.
 

FAQ: Derivative of the mixed metric tensor

What is the mixed metric tensor?

The mixed metric tensor is a mathematical object used in the field of differential geometry to describe the curvature of a space. It is a combination of the metric tensor, which measures distances and angles in a space, and the inverse metric tensor, which relates the coordinates of a space to the corresponding coordinates of the dual space.

What is the derivative of the mixed metric tensor?

The derivative of the mixed metric tensor is a mathematical operation that describes how the mixed metric tensor changes as the coordinates of a space are varied. It can be calculated using the Christoffel symbols, which are related to the curvature of the space.

Why is the derivative of the mixed metric tensor important?

The derivative of the mixed metric tensor is important because it allows us to calculate the curvature and other geometric properties of a space. It is also an essential component in the Einstein field equations, which describe the relationship between the curvature of spacetime and the distribution of matter and energy.

How is the derivative of the mixed metric tensor used in general relativity?

In general relativity, the derivative of the mixed metric tensor is used to calculate the curvature of spacetime and the geodesic equation, which describes the motion of particles in a curved space. It is also used in the Einstein field equations to relate the curvature of spacetime to the distribution of matter and energy.

Are there any practical applications of the derivative of the mixed metric tensor?

Yes, the derivative of the mixed metric tensor has practical applications in various fields such as physics, engineering, and computer graphics. It is used in the study of black holes, gravitational waves, and cosmology. It is also used in the development of algorithms for computer simulations of physical phenomena in curved spaces.

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