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,

<br /> g_{ab}=diag(-(1-\frac{2M}{r}),1-\frac{2M}{r},r^2,R^2sin^2\theta)<br />

and coordinates (t,r,theta,phi)

given the vector

<br /> E^i=\delta^t_0<br />

Homework Equations

<br /> (L_Eg)ab=E^cd_cg_{ab}+g_{cb}d_aE^c+g_{ac}d_bE^c<br />

The Attempt at a Solution

<br /> (L_Eg)ab=E^cd_cg_{ab}+g_{cb}d_aE^c+g_{ac}d_bE^c<br />

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,
<br /> (L_Eg)ab=E^cd_cg_{ab}<br />

<br /> (L_Eg)ab=\delta^t_0 d_cg_{ab}<br />

I'm unsure how to take it from here.

Firstly, I'm unsure what
<br /> \delta^t_0<br />

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.

<br /> (L_Eg)ab=\delta^t_0 d_cg_{ab}<br />

Please help.

 

[trv]

anyone?

 

[Dick]

I think what they meant to write is E^i=\delta^i_0 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.