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.