Strictly speaking, a matrix is not a tensor, it is a representation of a tensor in a particular basis. You can't tell whether a given matrix is a tensor using only its components. You would have to know how it transforms to different reference frames.
For the electromagnetic field tensor, for example, you could write the equations for some physical configuration of electromagnetic fields, and then write the equations that describe the same physical configuration in a different reference frame, and show that applying the corresponding Lorentz transformation twice converts from one to the other.
$$F^{\mu\nu}_\text{frame 2} = [\Lambda(1,2)]_\alpha^\mu [\Lambda(1,2)]_\beta^\nu F^{\alpha\beta}_\text{frame 1}$$
where I've used $\Lambda(1,2)$ to denote the Lorentz transformation that transforms from frame 1 to frame 2. Doing this calculation out in full gets somewhat tedious, which is why many textbooks don't go through it in full detail, they just make it seem plausible. (But at least Einstein had to do it to show that the theory worked this way.)
Anyway, the point is that the tensorial nature of a quantity is really a consequence of the transformation law, not the representation (the matrix).
The answer you're looking for seems to be contained in
Rezzolla & Zanotti: Relativistic Hydrodynamics (Oxford U.P. 2013)
https://books.google.com/books/?id=KU2oAAAAQBAJ
but it is not a trivial generalization. Quoting Disconzi's On the well-posedness of relativistic viscous fluids (Nonlinearity 27 (2014) 1915, arXiv:1310.1954):
we still lack a satisfactory formulation of viscous phenomena within
Einstein's theory of general relativity. [... T]here have been different proposals for what the correct $T_{\alpha\beta}$ should be. [... A]ttempts to formulate a viscous relativistic theory based on a simple covariant generalization of the classical (i.e., non-relativistic) stress-energy tensor for the Navier-Stokes equations have also failed to produce a causal theory.
Similar problems exist for elastic materials.
You may also take a look at Chapter 15 ("Relativistic continuum mechanics") of Maugin's Continuum Mechanics Through the Twentieth Century (Springer 2013), and its references:
https://books.google.com/books?id=-QhAAAAAQBAJ
and at Bressan's Relativistic Theories of Materials (Springer 1978):
https://books.google.com/books?id=kMTuCAAAQBAJ
General-relativistic continuum mechanics unfortunately has not been given a clear mathematical and conceptual framework yet. Newtonian continuum mechanics is easy to summarize:
- We choose a reference frame (preferably but not necessarily inertial).
- We have a set of 11 spacetime-dependent fields with clear physical meaning: mass, momentum or deformation, stress, body force, internal energy, heating flux, body heating, temperature, entropy, entropy flux, body entropy supply. Of these, the "body" ones represent external interventions.
- We have 5 balance equations: mass, force-momentum, torque-rotational momentum, energy, entropy. They are clearly written in terms of the fields above and are valid for any material.
- We choose a set of independent fields (usually mass, momentum or deformation, temperature).
- We choose constitutive equations (compatibly with the balance ones) that relate the remaining fields to the independent ones. These equations express the peculiar properties (fluid, solid, elastic, plastic, with/without memory...) of the material under study.
And at this point we have a well-defined set of partial differential equations in a number of unknown fields, for which we can set up well-defined initial- & boundary-value problems to be solved analytically or numerically. (An expanded but analogous framework accommodates electromagnetism and continua with internal structure.)
This framework and steps are very neat – we clearly know what the fields are, which of them are dependent and which independent; what are the equations valid for all materials, and what are the equations constitutive to each material. I've never seen a clearly defined procedure like the one above for general relativity, although I believe it could be extracted from Rezzolla & Zanotti's or Bressan's books. Moreover, the core of general-relativistic community uses a different jargon and way of thinking.
Most general-relativity books tell you that the Einstein equations determine everything, but they are not so clear about which fields in them are independent and which dependent; even Misner et al.'s Gravitation (ch. 21) has a long discussion and explanation about this point. It was only with 3+1 formulations and the work of Arnowitt, Deser, Misner, York, and others around the 1970s that this point got clarified. Then they tell you that we need "special" additional equations for the stress tensor – that is, constitutive equations. Sometimes other conservation equations, like baryonic number (basically rest-mass), are added with no real explanation. This is a sample of books where "constitutive equations" are mentioned explicitly (only once or twice in most of them):
- Rezzolla & Zanotti above (and they explain what a "constitutive equation" is as though it was an exotic concept)
- Choquet-Bruhat: General Relativity and Einstein's Equations
- Anile & Choquet-Bruhat: Relativistic Fluid Dynamics
- Bertotti et al.: General Relativity and Gravitation
- Puetzfeld et al.: Equations of Motion in Relativistic Gravity
- Tonti: The Mathematical Structure of Classical and Relativistic Physics
- Bini & Ferrarese: Introduction to Relativistic Continuum Mechanics
- Tolman (obviously): Relativity, Thermodynamics, and Cosmology
but they constitute a very small minority in the huge relativistic literature.
Yet, the general-relativistic community cannot be criticized for the confused conceptual state and somehow confused language of the subject. Newtonian continuum mechanics can be neatly formulated today because it has been refined over several centuries. General relativity is still very young instead, and its conceptual refinement still in progress. Some of the steps in the Newtonian framework become extremely complicated in general relativity. For example: step 1. (choose an inertial frame) cannot be done so simply. The Einstein equations, evolved from initial conditions, construct a reference frame "along the way", while they determine the dynamics. This gives rise to peculiar fields like "lapse" and "shift", which aren't really physical, and all sorts of redundancy (gauge freedom) in the equations.
Another example: the metric becomes a dynamical field variable, and you suddenly realize that it is hidden almost everywhere in the Newtonian framework – divergences, curls, vectors/covectors... So its evolution can't be easily divided among some new balance and constitutive equations (like we can do with electromagnetism instead). Are all of its appearances in the Newtonian framework dynamically significant? or can the metric be eliminated from some places? There's some research today on this "de-metrization" of Newton's equations; see for example Segev's Metric-independent analysis of the stress-energy tensor, J. Math. Phys. 43 (2002) 3220. This line of research has shown that some Newtonian physical objects actually don't need a metric: they be expressed via differential forms and other metric-free differential-geometrical objects (e.g., van Dantzig's On the geometrical representation of elementary physical objects and the relations between geometry and physics, Nieuw Archief voor Wiskunde II (1954) 73; there is a vast literature on this, let me know if you want more references). This is still work in progress – which means that it's obviously not completely clear yet how mass-energy-momentum-stress and metric are coupled.
To conclude, I think another good starting point to understand how things work in general-relativistic continuum mechanics is to look in books on numerical formulations of general relativity and matter dynamics. The conceptual framework in them is a bit confused, but you can see how they actually do it. If from the practice of these books you manage to reverse-engineer a framework like the Newtonian one above, please write a pedagogical paper about it!
Here are some books and reviews on numerical relativity with continua:
- Rezzolla & Zanotti above
- Gourgoulhon: 3+1 Formalism in General Relativity (Springer 2012, arXiv:gr-qc/0703035)
- Baumgarte & Shapiro: Numerical Relativity (Cambridge U.P. 2010)
- Alcubierre: Introduction to 3+1 Numerical Relativity (Oxford U.P. 2008)
- Palenzuela-Luque & Bona-Casas: Elements of Numerical Relativity and Relativistic Hydrodynamics (Springer 2009)
- Lehner: Numerical relativity: a review, Class. Quant. Grav. 18 (2001) R25, arXiv:gr-qc/0106072
- Guzmán: Introduction to numerical relativity through examples, Rev. Mex. Fis. S 53 (2007) 78
I'm happy to provide or look for additional references.
Best Answer
This is a quote by Tommaso Toffoli, in "Entropy? Honest!". Entropy 18, 247 (2016). doi: 10.3390/e18070247. Reading your question reminded me of it because it is pretty much the point: we don't need to write our equations in tensorial form, we could indeed just write them in terms of each individual component (important note: those would not be scalars, for they wouldn't transform as scalars). However, this often will hide interesting properties of what we are dealing with that might make our life considerably easier (see, e.g., this brilliant answer by Terence Tao on a similar question in Math Overflow).
In the case of tensors, our main interest is that their components have well defined transformation properties under changes of coordinates and are, deep down, geometrically invariant, and that allows you to see and comprehend much more than you otherwise would.
For example, take the expression $\mathbf{F} = m \mathbf{a}$ in Classical Mechanics. This is written in terms of rank-1 tensors (vectors), but we could write it as three equations in components. It would read $$ \left\lbrace \begin{aligned} F_x &= m a_x, \\ F_y &= m a_y, \\ F_z &= m a_z, \end{aligned} \right. $$ where I employed Cartesian coordinates. Let us suppose you already measured $F_x$, $F_y$, and $F_z$ experimentally and now wants to determine the acceleration in each direction. However, you notice you messed up your setup and actually you wanted the components of the acceleration in a slightly different angle, rotated in $45º$ around the $z$ axis with respect to the system you chose. Then you compute the transformation and finds that the previous equations read now, after we transform them, $$ \left\lbrace \begin{aligned} \frac{\sqrt{2}}{2} F_x - \frac{\sqrt{2}}{2} F_y &= m a_{x'}, \\ \frac{\sqrt{2}}{2} F_x + \frac{\sqrt{2}}{2} F_y &= m a_{y'}, \\ F_z &= m a_{z'}, \end{aligned} \right. $$ assuming I didn't get anything wrong. Had you chosen to work with vectors, the equation would read $$\mathbf{F} = m \mathbf{a},$$ because while vector components transform under a change of basis, vectors themselves are geometrical objects which do not depend on the basis.
Of course, this is just an example, you'd have to compute the components to get to your final computation either way. However, notice how writing in vector notation employs automatically the invariance properties of vectors. You can choose to write component-wise, but that won't destroy the vectors that lie beneath your computations, it just hides them. The components still transform as vectors and the vectors are still there, you are just not looking at them, and, as a consequence, you are missing some information that could make things way simpler. The point is not that they are a way of organizing calculations (you'll often have to work with the components sooner or later), but that there are "hidden" symmetries and properties that become explicit once you cast the formulae into an appropriate notation.
At this point, it is worth pointing out the comment made by Dvij D.C. (which I'm adding here in case it is deleted in the future)
Notice how this is similar to us writing Newton's second law in components and in vector form. Vectors are simply rank-1 tensors, and when employing them our equations automatically are expressed in an invariant way under rotations.
As a second example, consider the expression $\mathbf{p \cdot q}$ written in two other different notations. The first one, in Einstein notation, which would read $$\mathbf{p \cdot q} = p_i q^i,$$ and the second being the expression in Cartesian coordinates, $$\mathbf{p \cdot q} = p_x q_x + p_y q_y + p_z q_z.$$
Both expressions are tautological. Neither of them adds any intrinsic knowledge to us. However, the expression in Einstein notation makes it clear that the object is invariant under rotations, while the second one has a lot of objects that transform in a sort of complicated way, and you are not sure whether the expression would change if you chose to work with different coordinates.
In principle, one can component-wise everything we do with tensors. It is similar to searching for a needle in a haystack: you can do it by looking carefully and eventually you'll find it, but you can also use the extra knowledge that the needle is made of iron and use a magnet.
Extra Examples
Quantum Mechanics
One other example of usage of tensors in Physics, although in a particularly specific notation, is within Quantum Mechanics. For simplicity, I'll stick to the rotation example. Suppose you made a measurement, noticed you should have rotated your coordinates before, and so on. If you denote the states in terms of wavefunctions, you'll figure out that the probability a particle prepared in the state $\psi(x)$ to be measured in the state $\phi(x)$ is $$P(\phi|\psi) = \left\vert \int \phi^*(x) \psi(x) \mathrm{d}^3{x} \right\vert^2$$ before the rotation and, after the rotation, $$P'(\phi|\psi) = \left\vert \int \phi'^*(x) \psi'(x) \mathrm{d}^3{x} \right\vert^2,$$ where $\psi'$ is the wavefunction $\psi$ after the rotation of the system of coordinates. Opening up the appropriate expression for $\psi'$ and $\phi'$, one will eventually find out that the probability is, of course, the same. In Dirac notation, however, which employs the fact that the objects we're dealing with are vectors, one would write $$P(\phi|\psi) = \vert\langle \phi \vert \psi \rangle\vert^2$$ before the rotation and $$P'(\phi|\psi) = \vert\langle \phi' \vert \psi' \rangle\vert^2.$$ Since rotations are implemented in Quantum Mechanics by means of unitary operators, one has then that $\vert\psi'\rangle = U \vert\psi\rangle$ for some unitary operator $U$. Hence, the rotated version is $$P'(\phi|\psi) = \vert\langle \phi' \vert \psi' \rangle\vert^2 = \vert\langle \phi \vert U^\dagger U \vert \psi \rangle\vert^2 = \vert\langle \phi \vert \psi \rangle\vert^2 = P(\phi|\psi),$$ and the result is immediate using properties of linear algebra, without the need to manipulate anything with calculus.
Another example is how one can solve the quantum harmonic oscillator in terms of ladder operators instead of solving differential equations. It exploits the linear structure hidden in the wavefunctions and allows one to get a solution with much less labor.
Relativity
Expressions written in terms of tensors in Relativity are assured to work in all reference frames, since tensors are geometrical objects. This is similar to my previous example of how working in vector notation made taking the rotations into account much easier.
Take, for example, the famous expression $$E^2 = p^2 c^2 + m^2 c^4.$$ In this form, it is not obvious that this formula holds in any inertial frame (unless, of course, you are already well acquainted with Relativity and know the fact by heart). However, the same formula can be written as $$p^\mu p_\mu = m^2 c^2,$$ where $p^\mu = (\frac{E}{c}, \mathbf{p})^\intercal$, which explicitly shows the expression is invariant.
Another example is the fact that the relativistic Doppler effect for a source in uniform linear motion boils down to noticing that $k^\mu u_\mu$ is an invariant, $k^\mu = (\frac{\omega}{c}, \mathbf{k})^\intercal$ being the wave's four-momentum and $u^\mu$ being the source's four-velocity. By computing the invariant in two different frames, one of which has the source at rest, one arrives at the expression for the frequency shift in an incredibly straightforward way. Indeed, suppose that in the frame of rest of the source the wave has $k^\mu = (\frac{\omega_0}{c}, \mathbf{k}_0)^\intercal$, while in some other frame the source has $u^\mu = (\gamma c, \gamma \mathbf{v}$ and the wave has $k^\mu = (\frac{\omega}{c}, \mathbf{k})^\intercal$. Then $$ \begin{align} k^\mu u_\mu\vert_{\text{rest}} &= k^\mu u_\mu\vert_{\mathbf{v}}, \\ - \omega_0 &= \gamma(-\omega + \mathbf{k \cdot v}). \end{align} $$
Let's write $\mathbf{k} = \frac{\omega}{c} \mathbf{\hat{n}}$, where $\mathbf{\hat{n}}$ is a unit vector (the fact that this is possible comes from usual wave mechanics). Then $$ \omega_0 = \omega\gamma\left(1 - \frac{\mathbf{n \cdot v}}{c}\right), $$ and we conclude that $$\frac{\omega}{\omega_0} = \frac{\sqrt{1 - \frac{v^2}{c^2}}}{1 - \frac{\mathbf{n \cdot v}}{c}}.$$
This derivation just comes in such a direct way because we noticed from the start that $k^\mu u_\mu$ is a relativistic invariant, something that is favored by tensor notation: the absence of "free" indices tells us that the quantity is an invariant.