Energy-momentum tensor perfect fluid raise index

[binbagsss]

Homework Statement

This should be pretty simple and I guess I am doing something stupid?

$T_{bv}=(p+\rho)U_bU_v-\rho g_{bv}$
compute $T^u_v$:
$T^0_0=\rho, T^i_i=-p$

Homework Equations

$U^u=\delta^t_u$
$g_{uv}$ is the FRW metric,in particular $g_{tt}=1$
$g^{bu}T_{bv}=T^u_v$
$g^{ab}g_{ca}=\delta^{(4)b} _c=4$ if b=c(where I haven't paid attention to order of the indices when lowering and raising since both metric and energy-momentum tensor are symmetric objects)

The Attempt at a Solution

[/B]
$T^u_v=g^{bu}T_{bv}=g^{bu}(p+\rho)U_bU_v-\rho g_{bv}g^{bu}$
$= (p+\rho)U^bU_v-p\delta^u_v$

similarly $U^u=\delta^u_t$

Now I get the correct components if $\delta^u_v=1$ but $\delta^u_v=4$ if u=v because we are in 4d-d space-time I thought?

Many thanks

 

[Dick]

$g^{ab}g_{ca}=\delta^{(4)b} _c=4$ if b=c

This is wrong. $\delta^{b} _c$ is $1$ if $b=c$ and $0$ otherwise. You may be confusing it with $\delta^b_b$ which is $4$ because the repeated index b is summed over.

 

[binbagsss]

This is wrong. $\delta^{b} _c$ is $1$ if $b=c$ and $0$ otherwise. You may be confusing it with $\delta^b_b$ which is $4$ because the repeated index b is summed over.

Omg ok thanks

But I got this from contracting over the metric, I had it in my head this was $4$ for some reason but . Ahh but I guess it's the same idea so I had:
$g_{ab}g^{ac}=\delta^b_c$
as a pose to $g_{ab}g^{ab}=4=\delta^b_b$?

 

[Dick]

Omg ok thanks

But I got this from contracting over the metric, I had it in my head this was $4$ for some reason but . Ahh but I guess it's the same idea so I had:
$g_{ab}g^{ac}=\delta^b_c$
as a pose to $g_{ab}g^{ab}=4=\delta^b_b$?

Right. The difference comes from the Einstein summation convention.