What is the difference between the energy-momentum tensors $T^{\mu \nu}$, $T^\mu_{\ \nu}$ and $T_{\mu\nu}$? The energy-momentum tensor defined as
$$ T^{\mu \nu} = \frac{1}{\sqrt{-g}}\frac{\delta S}{\delta g_{\mu \nu}}.$$
If we are on a flat Minkowski spacetime with the metric $\eta_{\mu \nu} =\mathrm{diag}(1,-1,-1,\ldots)$, then we see that
$$T^{tt} = T^t_{\ t} = T_{tt},$$
where we raise and lower the indices with $\eta^{\mu\nu}$ and $\eta_{\mu \nu}$ respectively.
This is what we would identify as the energy density. However, if we were on a spacetime that had a non-trivial metric $g_{\mu \nu}$, such as the Schwarzschild metric, then this would no longer hold because we would find:
$$T^{tt} \neq T^t_{\ t} \neq T_{tt}. $$
Which quantity would I identify as the physically measured energy-density?
An example from literature
In Eq. (28) of this paper, the authors define the energy-momentum tensor of a Dirac field on a $(1+1)$D Schwarzschild background. They define the energy-momentum tensor with both indices down, $T_{\mu \nu}$, and idenfity $T_{tt}$ as their energy-density. The authors then proceed to calculate this quantity on the event horizon, which is the location for which the metric diverges, and they find a finite result for $T_{tt}$. However, if one tried to raise the indices with $g^{\mu \nu}$ to calculate $T^{tt}$, then one finds that this quantity is infinite due to the singularity of the metric. I am aware the singularity of the metric is a coordinate singularity, but we still find different results for $T^{tt}$ and $T_{tt}$ elsewhere. If I would discard the infinite $T^{tt}$ as unphysical due to the coordinate singularity, why would I trust $T_{tt}$ to be acceptable?
My question
My question is the following: what is the physical difference between the energy-momentum tensors $T^{\mu \nu}$, $T^\mu_{\ \nu}$ and $T_{\mu\nu}$, and which is the one I would measure in an experiment?
 A: A good rule of thumb is that in GR you never measure components of tensors - all physical measurements can be expressed in terms of scalars.
(I don't know if this works every time (though it should) but it hasn't failed me so far)
If you're measuring the energy density, the way to reframe the question is to realize that the energy density is the $tt$ component of the EM tensor in a given locally inertial frame. To put it mathematically, if you have an observer with four-velocity $U_o^\mu$, that observer will measure an energy density
$$\rho_o = T_{\mu\nu} U_o^\mu U_o^\nu. \tag{1}$$
You can move the indices up and down at will, since this is a scalar. In the simple case of special relativity, the observer's velocity is always $U_o^\mu = (1,0,0,0)$ in the observer's own coordinate system, so we recover the result $\rho_o = T_{tt} = T_t^t = T^{tt}$.
Another special case is that of a stationary observer, one whose spatial coordinates don't change and therefore whose four-velocity has only a time component. The normalization $g_{\mu\nu}U^\mu U^\nu = g^{\mu\nu}U_\mu U_\nu = 1$ shows that we must have $U^t = 1/\sqrt{g_{tt}}$ and $U_t = 1/\sqrt{g^{tt}}$, so we can write the energy density as measured by this observer in three ways:
$$\rho = \frac{1}{g_{tt}} T_{tt} = \frac{1}{\sqrt{g_{tt}g^{tt}}} T_t{}^t = \frac{1}{g^{tt}} T^{tt}.$$
In the extra special case that $g^{tt} = 1/g_{tt}$, which happens when the metric doesn't have mixed time-space components, we find $\rho = T_t{}^t$. And if instead of using general coordinates we use the observer's locally inertial frame as coordinates, we recover the special relativity formula. But in the general case, formula $(1)$ is always true.
A: In order to answer this question, we can simply look at the answer in Minkowski space, and generalize it in an invariant way. In Minkowski space, the energy density measured by an observer with four-velocity $u$ is $T^{\mu\nu} u_\mu u_\nu = T_{\mu\nu} u^\mu u^\nu$. This holds because it's the unique expression that generalizes $T^{00}$ (or $T_{00}$) in the case of a static observer, where $u^\mu = (1, 0, 0, 0)$ and $u_\mu = (1, 0, 0, 0)$.
This expression still holds in curved spacetime, but in this case the correct expressions for the components of the four-velocity, $u^\mu$ and $u_\mu$, can look very different. If your paper says $T_{00}$ is the energy density, that presumably means they've chosen coordinates where a static observer has $u^\mu = (1, 0, 0, 0)$. (This is the more natural choice, since the four-velocity is naturally a vector and not a covector, $u^\mu = dx^\mu/d\tau$.)
