# Energy density measured by an observer [duplicate]

"Because an observer with four-velocity $$v_{α}$$ measures the energy density to be $$T_{αβ}v^{α}v^{β}$$, we must have (...)"

We are talking about perfect fluid...

I am not understanding very well the mathematical expression above, why this $$T_{αβ}v^{α}v^{β}$$ is the energy density measured by the observer? Is this a definition?

I was expecting something like $$T_{00}v^{0}v^{0}$$, since only the 00 term says something about the density, what is wrong with my interpretation?

What you want is $${T'}^{00}$$ where the prime indicates that the component is being written down in the reference frame of the observer in question (another notation would be to use barred indices but this prime notation is also logical if you keep in mind that the prime refers to the components, not the tensor itself as a geometric object). Now the observer in question has four-velocity components $$v^a$$ in the frame you are working in, and all four of these components may be non-zero. But in their own rest frame the four-velocity will be simple: its components will be $$v'^a = (1,0,0,0)$$ (taking $$c=1$$). So in that frame we have $$T'_{00} \;\;\stackrel{^{(\rm in\; observer\; frame)}}{=} \;\; T'_{\alpha\beta} v'^{\alpha} v'^\beta$$ Note, this relation only holds in one frame. But the rhs is an invariant! So we can write $$T'_{00} = T_{\alpha\beta} v^{\alpha} v^\beta$$ This says "the covariant energy density component in the rest frame of an observer whose 4-velocity is $$\bf v$$ is equal to the invariant combination written on the right hand side". Another way to express this is to say that you are using the 4-velocity as one member of a tetrad, but if you do not know that concept then you can ignore it.
Finally, you may prefer to call $$T'^{00}$$ rather than $$T'_{00}$$ the energy density. They will be the same in special relativity but in GR they are not the same, but the term "energy density" is in practice used for either, I think, depending on circumstances.