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Different authors seem to attach different levels of importance to keeping track of the exact tensor valences of various physical quantities. In the strict-Catholic-school-nun camp, we have Burke 1980, which emphasizes that you don't always have a metric available, so it may not always be possible to raise and lower indices at will. Burke makes firm pronouncements, e.g., that force is a covector (I recapped his argument here). At the permissive end of the spectrum, Rindler 1997 has a disclaimer early on in the book that he doesn't want to worry about distinguishing upper and lower indices, and won't do so until some later point in the book. Sometimes it feels a little strained to try to maintain such distinctions, especially in relativity, which we don't even know how to formulate without a nondegenerate metric. E.g., Burke argues that momentum is really a covector, because you can get it by differentiation of the Lagrangian with respect to $\dot{x}$. But then a perfectly natural index-gymnastics expression like $p^i=m v^i$ becomes something wrong and naughty.

I find this particularly confusing when it comes to higher-rank tensors and questions about which form of a tensor is the one that corresponds to actual measurements. Measurements with rulers measure $\Delta x^i$, not $\Delta x_i$, which is essentially a definition that breaks the otherwise perfect symmetry between vectors and covectors. But for me, at least, it gets a lot more muddled when we're talking about something like the stress-energy tensor. For example, in this question, I was working through a calculation in Brown 2012 in which he essentially writes down $T^\mu_\nu=\operatorname{diag}(\rho,P,0,0)$ for the stress-energy tensor of a rope hanging in a Schwarzschild spacetime. It's not obvious to me that this corresponds better with measurements than writing down the same r.h.s. but with $T^{\mu\nu}$ or $T_{\mu\nu}$ on the left. Misner 1973 has a nice little summary of this sort of thing on p. 131, with, e.g., a rule stating that $T^\mu_\nu v_\mu v^\nu$ is to be interpreted as the density of mass-energy seen by an observer with four-velocity $v$. Most, but not all, of their rules are, like this one, expressed as scalars. This is very attractive, because we have identities such as $a^ib_i=a_ib^i$, which means that it makes absolutely no difference whether we discuss an object like $a^i$ or its dual $a_i$, and we never have to discuss which form of a tensor matches up with measurements, because our measurements are scalars.

Related: Type/Valence of the stress tensor

Different authors seem to attach different levels of importance to keeping track of the exact tensor valences of various physical quantities. In the strict-Catholic-school-nun camp, we have Burke 1980, which emphasizes that you don't always have a metric available, so it may not always be possible to raise and lower indices at will. Burke makes firm pronouncements, e.g., that force is a covector (I recapped his argument here). At the permissive end of the spectrum, Rindler 1997 has a disclaimer early on in the book that he doesn't want to worry about distinguishing upper and lower indices, and won't do so until some later point in the book. Sometimes it feels a little strained to try to maintain such distinctions, especially in relativity, which we don't even know how to formulate without a nondegenerate metric. E.g., Burke argues that momentum is really a covector, because you can get it by differentiation of the Lagrangian with respect to $\dot{x}$. But then a perfectly natural index-gymnastics expression like $p^i=m v^i$ becomes something wrong and naughty.

I find this particularly confusing when it comes to higher-rank tensors and questions about which form of a tensor is the one that corresponds to actual measurements. Measurements with rulers measure $\Delta x^i$, not $\Delta x_i$, which is essentially a definition that breaks the otherwise perfect symmetry between vectors and covectors. But for me, at least, it gets a lot more muddled when we're talking about something like the stress-energy tensor. For example, in this question, I was working through a calculation in Brown 2012 in which he essentially writes down $T^\mu_\nu=\operatorname{diag}(\rho,P,0,0)$ for the stress-energy tensor of a rope hanging in a Schwarzschild spacetime. It's not obvious to me that this corresponds better with measurements than writing down the same r.h.s. but with $T^{\mu\nu}$ or $T_{\mu\nu}$ on the left. Misner 1973 has a nice little summary of this sort of thing on p. 131, with, e.g., a rule stating that $T^\mu_\nu v_\mu v^\nu$ is to be interpreted as the density of mass-energy seen by an observer with four-velocity $v$. Most, but not all, of their rules are, like this one, expressed as scalars. This is very attractive, because we have identities such as $a^ib_i=a_ib^i$, which means that it makes absolutely no difference whether we discuss an object like $a^i$ or its dual $a_i$, and we never have to discuss which form of a tensor matches up with measurements, because our measurements are scalars.

Related: Type/Valence of the stress tensor

Finally, in the example above, suppose that we have an asymptotically flat spacetime, and suppose that at large distances we have $A^2\rightarrow 1$ and $B^2\rightarrow 1$. Then $|T^0_0/T_{00}|=A^{-2}>1$ corresponds to a gravitational redshift factor seen by an observer at infinity. Matter falling toward a black hole has its rest mass $m$ boosted to $mA^{-2}$ in terms of its contribution to the mass-energy detected at infinity.

Finally, in the example above, suppose that we have an asymptotically flat spacetime, and suppose that at large distances we have $A^2\rightarrow 1$ and $B^2\rightarrow 1$. Then $|T^0_0/T_{00}|=A^{-2}>1$ corresponds to a gravitational redshift factor seen by an observer at infinity.

Related: Type/Valence of the stress tensor

Update

After chewing this over with Cristi Stoica and Trimok, I think I've understood the issue about $T^\mu_\nu$ versus $T_{\mu\nu}$ better. Contrary to what I said above, the expression $T_{\mu\nu}=(\rho+P)v_\mu v_\nu+P g_{\mu\nu}$ (with $-+++$ signature) for a perfect fluid is really scalar-ized, in the sense that $\rho$ and $P$ have no tensor indices, so they're notated as scalars. This makes sense, because $\rho$ and $P$ are defined by reference to a particular frame of reference, the rest frame of the fluid. This is exactly analogous to the way in which we define the scalar proper time by reference to the rest frame of a clock.

Now suppose we have coordinates in which the metric is diagonal, e.g., the Schwarzschild metric written in Schwarzschild coordinates. Let $g_{\mu\nu}=\operatorname{diag}(-A^2,B^2,\ldots)$. Furthermore, let's assume that we have some perfect fluid whose rest frame corresponds to zero coordinate velocity in these coordinates. The velocity vector of this rest frame is $v^\mu=(A^{-1},0,0,0)$, or, lowering an index, $v_\mu=(-A,0,0,0)$. Simply plugging in to the expression for $T$, we have $T_{00}=A^2\rho$, $T_{11}=B^2P$, $T^0_0=-\rho$, $T^1_1=P$. So this is the justification for Brown's use of the mixed-index form of $T$ -- it simply happens to be the one in which the factors of $A$ and $B$ don't appear. But that doesn't mean that the mixed-index form is the "real" one. What a static observer actually measures is the scalars $\rho$ and $P$. Similarly, it's not really $\Delta x^0$ or $\Delta x_0$ that an observer measures on a clock, it's the scalar $\Delta s$. When we say that coordinate differences correspond to upper-index vectors $\Delta x^i$, we're actually making a much more complicated statement that refers not to a simple measurement with a single device but rather to some much more extensive setup involving surveying, gyroscopes, and synchronization of clocks.

I think one of the pitfalls here is that I wasn't keeping in mind the fact that in an equation like $T_{\mu\nu}=\operatorname{diag}(\ldots)$, the right-hand side circumvents the rules of index gymnastics. This makes it preferable to work with equations like $T_{\mu\nu}=(\rho+P)v_\mu v_\nu+P g_{\mu\nu}$, where both sides are valid index-gymnastics notation, and we can raise and lower indices at will without worrying about invalidating the equation.

Finally, in the example above, suppose that we have an asymptotically flat spacetime, and suppose that at large distances we have $A^2\rightarrow 1$ and $B^2\rightarrow 1$. Then $|T^0_0/T_{00}|=A^{-2}>1$ corresponds to a gravitational redshift factor seen by an observer at infinity. Matter falling toward a black hole has its rest mass $m$ boosted to $mA^{-2}$ in terms of its contribution to the mass-energy detected at infinity.

Related: Type/Valence of the stress tensor