Calculating Lie Derivative for Metric Tensor with Given Coordinates and Vector

[trv]

Homework Statement

Calculate the lie derivative of the metric tensor, given the metric,

[itex]
g_{ab}=diag(-(1-\frac{2M}{r}),1-\frac{2M}{r},r^2,R^2sin^2\theta)
[/itex]

and coordinates (t,r,theta,phi)

given the vector

[itex]
E^i=\delta^t_0
[/itex]

Homework Equations

[itex]
(L_Eg)ab=E^cd_cg_{ab}+g_{cb}d_aE^c+g_{ac}d_bE^c
[/itex]

The Attempt at a Solution

[itex]
(L_Eg)ab=E^cd_cg_{ab}+g_{cb}d_aE^c+g_{ac}d_bE^c
[/itex]

all derivatives above being partial

Now the Last two terms go to zero, since E^i=Kronecker delta=constant and so its derivative is zero.

So,
[itex]
(L_Eg)ab=E^cd_cg_{ab}
[/itex]

[itex]
(L_Eg)ab=\delta^t_0 d_cg_{ab}
[/itex]

I'm unsure how to take it from here.

Firstly, I'm unsure what
[itex]
\delta^t_0
[/itex]

means. Does it means we get the result 1 at t=0 and zero for all other times?

How does it then affect the equation below.

[itex]
(L_Eg)ab=\delta^t_0 d_cg_{ab}
[/itex]

Please help.

 

[trv]

anyone?

 

[Dick]

I think what they meant to write is [itex]E^i=\delta^i_0[/itex] i.e. the vector field with a constant value 1 in the time component. So the Lie derivative is pretty trivial.

 

[trv]

Oh ok, so the 0 stands for the time component. That makes sense.