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/* Example 2: Stress-energy tensor for an ideal dust */
====Example 2: Stress-energy tensor for an ideal dust====
Imagine a time-dependent distribution of identical, massive, non-interacting, electrically neutral particles. In general relativity, such a distribution is called a ''dust.'' Let's break down what this means.
;time-dependent
:The distribution of particles in our dust is not a constant; that is to say, the particles may be motion. The overall configuration you see when you look at the dust depends on the time at which you look at it, so the dust is said to be ''time-dependent.''
;identical
:The particles that make up our dust are all exactly the same; they don't differ from each other in any way.
;massive
:Each particle in our dust has some rest mass. Because the particles are all identical, their rest masses must also be identical. We'll call the rest mass of an individual particle <math>m_0</math>.
;non-interacting
:The particles don't interact with each other in any way: they don't collide, and they don't attract or repel each other. This is, of course, an idealization; since the particles are said to have mass <math>m_0</math>, they must ''at least'' interact with each other gravitationally, if not in other ways. But we're constructing our model in such a way that gravitational effects between the individual particles are so small as to be be negligible. Either the individual particles are very tiny, or the average distance between them is very large. This same assumption neatly cancels out any other possible interactions, as long as we assume that the particles are far enough apart.
;electrically neutral
:In addition to the obvious electrostatic effect of two charged particles either attracting or repelling each other — thus violating our "non-interacting" assumption — allowing the particles to be both charged and in motion would introduce electrodynamic effects that would have to be factored into the stress-energy tensor. We would ''greatly'' prefer to ignore these effects for the sake of simplicity, so by definition, the particles in our dust are all electrically neutral.
The easiest way to visualize an ideal dust is to imagine, well, dust. Dust particles sometimes catch the light of the sun and can be seen if you look closely enough. Each particle is moving in apparent ignorance of the rest, its velocity at any given moment dependent only on the motion of the air around it. If we take away the air, each particle of dust will continue moving in a straight line at a constant velocity, whatever its velocity happened to be at the time. This is a good visualization of an ideal dust.
We're now going to zoom out slightly from our model, such that we lose sight of the individual particles that make up our dust and can consider instead the dust as a whole. We can fully describe our dust at any ''event'' <math>P</math> — where ''event'' is defined as a point in space at an instant in time — by measuring the density <math>\rho</math> and the 4-velocity <math>u</math> at <math>P</math>. If we have those two pieces of information about the dust at every point within it at every moment in time, then there's literally nothing else to say about the dust: it's been fully described.
=====Density=====
Let's start by figuring out the density of dust at a the event <math>P</math>, as measured from the perspective of an observer moving along with the flow of dust at <math>P</math>. The density <math>\rho</math> is calculated very simply:
:<math>
\rho = m_0 n\,
</math>
where <math>m_0</math> is the mass of each particle and <math>n</math> is the number of particles in a cubical volume one unit of length on a side centered on <math>P</math>. This quantity is called ''proper density,'' meaning the density of the dust ''as measured within the dust's own reference frame.'' In other words, if we could somehow imagine the dust to measure ''its own density,'' the proper density is the number it would get.
Clearly proper density is a function of position, since it varies from point to point within the dust; the dust might be more "crowded" over here, less "crowded" over there. But it's also a function of time, because the configuration of the dust itself is time-dependent. If you measure the proper density at some point in space at one instant of time, then measure it at the same point in space at a different instant of time, you may get a different measurement. By convention, when dealing with a quantity that depends both on position in space and on time, physicists simply say that the quantity is a function of position, with the understanding that they're referring to a "position" in four-dimensional spacetime.
=====4-velocity=====
The other quantity we need is ''4-velocity.'' Four-velocity is an extension of three-dimensional velocity (or 3-velocity). In three dimensional space, 3-velocity is a vector with three components. Likewise, in four-dimensional spacetime, 4-velocity is a vector with four components.
Directly measuring 4-velocity is an inherently tricky business, since one of its components describes motion along a "direction" that we cannot see with our eyes: motion through time. The math of [[special theory of relativity|special relativity]] lets us calculate the 4-velocity of a moving particle given only its 3-velocity <math>v</math> (with components <math>v^i</math> where <math>i=1, 2, 3</math>) and the speed of light. The time component of 4-velocity is given by:
:<math>
u^0 = \gamma c\,
</math>
and the space components <math>u^1</math>, <math>u^2</math> and <math>u^3</math> by:
:<math>
u^i = \gamma v^i\,
</math>
where <math>\gamma</math> is the ''boost,'' or Lorentz factor:
:<math>
\gamma = \frac{1}{\sqrt{1-\frac{\|v\|^2}{c^2}}}\,
</math>
and where <math>\|v\|^2</math>, in turn, is the square of the Euclidean magnitude of the 3-velocity vector <math>v</math>:
:<math>
\|v\|^2 = (v^1)^2+(v^2)^2+(v^3)^2\,
</math>
Therefore, if we know the 3-velocity of the dust at event <math>P</math>, then we can calculate its 4-velocity. (For more details on the how and why of 4-velocity, refer to the article on [[special theory of relativity|special relativity]].)
Just as proper density is a function of position in spacetime, 4-velocity also depends on position. The 4-velocity of our dust at a given point in space won't necessarily be the same as the 4-velocity of the dust at another point in space. Likewise, the 4-velocity at a given point at a given time may not be the same as the 4-velocity of the dust at the same point at a different time. It helps to think of 4-velocity as the velocity of the dust ''through'' a point in both space and time.
=====Assembling the stress-energy tensor=====
Since the density and the 4-velocity fully describe our dust, we have everything we need to calculate the stress-energy tensor.
:<math>
T(x) = \rho (x) u(x)\otimes u(x)\,
</math>
where the symbol <math>\otimes</math> indicates a ''tensor product.'' The tensor product of two vectors is a tensor of rank two, so the stress-energy tensor must be a tensor of rank two. In an arbitrary coordinate frame <math>x^\mu</math>, the contravariant components of the stress-energy tensor for an ideal dust are given by:
:<math>
T^{\mu\nu} = \rho u^{\mu} u^{\nu}\,
</math>
From this equation, we can now calculate the contravariant components of the stress-energy tensor for an ideal dust.
======Time-time component======
We start with the contravariant time-time component <math>T^{00}</math>:
:<math>
T^{00} = \rho u^0 u^0 = \rho (\gamma c) (\gamma c) = \rho \gamma^2 c^2 \,
</math>
If we rearrange the terms in this equation slightly, something important becomes apparent:
:<math>
T^{00} = \gamma^2 (\rho c^2)\,
</math>
Recall that <math>\rho</math> is a ''density'' quantity, in mass per unit volume. By the [[mass-energy equivalence]] principle, we know that <math>E = m c^2</math>. So we can interpret this component of the stress-energy tensor, which is written here in terms of mass-energy, to be equivalent to an ''energy density.''<ref>Actually rewriting the equation for the time-time component in terms of energy density requires refining our proper density equation into a form that doesn't depend on counting particles in a unit volume. Such a refinement is beyond the scope of this discussion. In less abstract dust solutions, the mass density is usually either assumed to be constant over space (as in the [[Friedmann-Lemaître-Robertson-Walker solution|FLRW solution]] that models a homogenous, isotropic expanding or contracting universe) or is assumed to depend only on the radius of the distribution (as in the [[LTB solution]] that models gravitational collapse). At this point, it is sufficient merely to understand that matter density and energy density, and matter flux and energy flux, are equivalent concepts under general relativity.</ref>
======Off-diagonal components======
The off-diagonal components of the tensor — <math>T^{\mu\nu}</math> where <math>\mu</math> and <math>\nu</math> are not equal — are calculated this way:
:<math>
T^{10} = \rho u^1 u^0 = \rho (\gamma v^1) (\gamma c) = \gamma^2 c \rho v^1\,
</math>
Again, recall that <math>\rho</math> is a quantity of mass per unit volume. Multiplying a mass times a velocity gives ''momentum,'' so we can interpret <math>\rho v^1</math> as the ''density of momentum'' along the <math>x^1</math> direction, multiplied by constants <math>c</math> and <math>\gamma^2</math>. Momentum density is an extremely difficult quantity to visualize, but it's a quantity that comes up over and over in general relativity. If nothing else, one can take comfort in the fact that momentum density is mathematically equivalent to the product of mass density and velocity, both of which are much more intuitive quantities.
Note that the off-diagonal components of the tensor are equal to each other:
:<math>
T^{10} = \rho u^1 u^0 = \rho (\gamma v^1) (\gamma c) = \rho \gamma^2 c v^1\,
</math>
:<math>
T^{01} = \rho u^0 u^1 = \rho (\gamma c) (\gamma v^1) = \rho c \gamma^2 v^1\,
</math>
In other words, in the case of an ideal dust, the stress-energy tensor is said to be ''symmetric.'' A rank two symmetric tensor is said to be symmetric if <math>T^{ab}=T^{ba}</math>.
======Diagonal space components======
The diagonal space components of the stress-energy tensor are calculated this way:
:<math>
T^{11} = \rho u^1 u^1 = \rho (\gamma v^1) (\gamma v^1) = \gamma^2 \rho (v^1)^2\,
</math>
In this case, we're multiplying a four-dimensional mass density, <math>\rho</math>, by the square of a component of 4-velocity. By [[dimensional analysis]], we can see:
:<math>
\frac{\mbox{kg}}{\mbox{m}^4} \cdot \frac{\mbox{m}^2}{\mbox{s}^2} = \frac{\mbox{kg} \cdot \mbox{m}^2}{\mbox{m}^4 \cdot \mbox{s}^2} = \frac{\mbox{kg}}{\mbox{m}^2 \cdot \mbox{s}^2}
</math>
Recall that the [[force]] has units:
:<math>
\frac{\mbox{kg} \cdot \mbox{m}}{\mbox{s}^2}
</math>
If we divide the units of the diagonal space component by the units of force, we get:
:<math>
\frac{\mbox{kg}}{\mbox{m}^2 \cdot \mbox{s}^2} \cdot \frac{\mbox{s}^2}{\mbox{kg} \cdot \mbox{m}} = \frac{1}{\mbox{m}^3}
</math>
So the diagonal space components of the stress-energy tensor come are expressed in terms of force per unit volume. Force per unit ''area'' are, of course, the traditional units of ''pressure'' in three-dimensional mechanics. So we can interpret the diagonal space components of the stress-energy tensor as the components of "4-pressure"<ref>Not a standard term.</ref> in spacetime.
======The big picture======
We now know everything we know to assemble the entire stress-energy tensor, all sixteen components, and look at it as a whole.<ref>The stress-energy tensor is practically never written out in matrix form this way, even in textbooks. This is purely for illustration.</ref>
:<math>T_{\mu \nu }=\begin{pmatrix}
\gamma^2 \rho c^2 & \gamma^2 \rho v^1 c & \gamma^2 \rho v^2 c & \gamma^2 \rho v^3 c \\
\gamma^2 \rho c v^1 & \gamma^2 \rho (v^1)^2 & \gamma^2 \rho v^2 v^1 & \gamma^2 \rho v^3 v^1 \\
\gamma^2 \rho c v^2 & \gamma^2 \rho v^1 v^2 & \gamma^2 \rho (v^2)^2 & \gamma^2 \rho v^3 v^2 \\
\gamma^2 \rho c v^3 & \gamma^2 \rho v^1 v^3 & \gamma^2 \rho v^2 v^3 & \gamma^2 \rho (v^3)^2
\end{pmatrix}</math>
The large-scale structure of the tensor now becomes apparent. This is the stress-energy tensor of an ideal dust. The tensor is composed entirely out of the proper density and the components of 4-velocity. When velocities are low, the coefficient <math>\gamma^2</math>, even though it's a squared value, remains extremely close to one.
The time-time component includes a mass multiplied by the square of the speed of light, so it has to do with energy. The rest of the top row and left column all include the speed of light as a coefficient, as well as density and velocity; in the case of an ideal dust which is made up of non-interacting particles, the energy flux along any basis direction is the same as the momentum density along that direction. This is not the case in other, less simple models, but it's true here.
The diagonal space components of the tensor represent pressure. For example, the <math>T^{11}</math> component represents the pressure that would be exerted on a plane perpendicular to the <math>x^1</math> direction.
The off-diagonal space components represent shear stress. The <math>T^{12}</math> component, for instance, represents the pressure that would be exerted in the <math>x^2</math> direction on a plane perpendicular to the <math>x^1</math> axis.
The overall process for calculating the stress-energy tensor for any system is fairly similar to the example given here. It involves taking into account all the matter and energy in the system, describing how the system evolves over time, and breaking that evolution down into components which represent individual densities and fluxes along different directions relative to a chosen coordinate basis.
As can easily be imagined, the task of constructing a stress-energy tensor for a system of ''arbitrary'' complexity can be a very daunting one. Fortunately, gravity is an extremely weak interaction, as interactions go, so on the scales where gravity is interesting, much of the complexity of a system can be approximated. For instance, there is absolutely nothing in the entire universe that behaves ''exactly'' like the ideal dust described here; every massive particle interacts, in one way or another, with other massive particles. No matter what, a real system is going to be ''very'' much more complex than this approximation. Yet, the ideal dust solution remains a much-used approximation in theoretical physics specifically ''because'' gravity is such a weak interaction. On the scales where gravity is worth studying, many distributions of matter, including interstellar nebulae, clusters of galaxies, even the whole universe really do behave very much like an ideal dust.
===The left side of the equation: the Einstein curvature tensor===
