Home

Blog

Projects

Notes

Links

Publications

Twitter

Github

Relativistic Matter Current

29 Mar 2019

What is the expression for the fully relativistic, conserved spacetime matter current of a fluid or continuous body? Let’s figure it out.

Context

I have been swimming in continuum mechanics lately. It is tremendously enjoyable. Not only are the situations studied very relatable, the mathematics are extremely satisfying.

Physical situations get more symmetrical as they become relativistic. This is because at relativistic velocities, time and space start to get mixed up. I have learned that in analyzing a physical situation, starting relativistically and then making non-relativistic approximations yields a lot of insight into the physics of what’s happening.

I have been following the books “Classical Field Theory” by Davison Soper and “Physics of Continuous Matter” by Benny Lautrup. These two books are easily two of the best textbooks I’ve ever read.

Introduction

Let’s consider a fluid or continuum body. We will parameterize our system with three fields, $$R_1(x), R_2(x), R_3(x)$$. These fields tell us the material coordinates of whatever material particle is currently at spacetime point x. Additionally, suppose we know the function $$n(R)$$ that tells us the number of atoms or whatever in a small volume of material around the material coordinates R.

The essence of figuring out currents is computing volumes of various 3-dimensional subspaces of 4-dimensional spacetime. Think about it: the amount of a substance in a given volume of space is just the density times the volume of that space.

Intuitively, if something is conserved, this means that the change in the amount of that something in a volume of space is solely due to the material leaving or entering the volume through its sides. This flow can also be expressed as a spacetime volume.

The flow is given by an amount per unit area per unit time. But in spacetime, time is just another axis, so the “per unit area per unit time” is the same as a volume!

Our relativistic matter current is going to be a spacetime vector, so it has four components. The components are going to be the amount of material in the 3-D volume “perpendicular” to the coordinate axis. For example, the 0-th component will be the amount of substance in a volume spanned by the 3 space axes. This is just the density expressed in an inertial frame.

Note however, that due to relativistic effects like length contraction, this density by itself is NOT Lorentz invariant… we need the other components. The components of the matter current give the density and flow observed by an observer at rest in that inertial frame.

Constructing the Matter Current

Ok, so how do we construct these amounts? Well formally, amount = density * volume.
We have the density in material coordinates, so we need to compute the volume in material coordinates.

To do this, we need to take the spacetime vectors that span our volume and map them to material coordinate vectors. The map to do this is the Jacobean matrix:

Once we have our spacetime vectors mapped to material vectors, we compute the volume using the completely antisymmetric symbol $$\epsilon^{abc}$$

It’s clear that this tensor maps three spacetime vectors to a real number.

Which volumes do we want to compute? Well, for the i-component of our matter current, we want the volume spanned by the axes perpendicular to i. So, again, the t-component will be spanned by the x, y, and z axes. The x-component will be spanned by the y, z, and t axes, and so on. Since the coordinate axes have particularly simple coordinate representations as vectors, the volumes turn out to be specific components of this completely antisymmetric tensor.

Thus, $$J^0$$ will be

$$J^1$$ will be

And so on. Because the map is completely antisymmetric, we can tidy things up by rewriting the current as:

This is the expression for the relativistic matter current.

You can go through the motions and check that it satisfies the Lorentz transformations, but that should be clear from the form of the expression.

Density and Fluid Four-Velocity

So now what? One quantity of considerable interest is the local density of matter. Why the local density? Well, one of the most important principles of physics is that the laws of physics look the same in all inertial frames. So, the only density that can affect the physics of a fluid at a given space time point is the density measured in a local inertial frame at rest with respect to the fluid at that point.

This quantity must be a spacetime scalar - the amount of material in a volume of spacetime can’t change depending on who observes it. However, $$J^0$$ is the density as seen in whatever inertial frame we are working in - it is not a scalar.

Well, the simplest scalar we can construct out of $$J^\mu$$ is its magnitude. So let’s define the density as:

In a local inertial frame, $$J^0 = \rho$$ and $$J^k =0$$. If we divide by $$\rho$$, we get the four velocity of the fluid in the local rest frame of the fluid at that point. But since $$\rho$$ is a Lorentz scalar and $$J^\mu$$ is a Lorentz vector, the quantity

is always a Lorentz vector, and so in all reference frames it gives the four-velocity of the fluid at that point.

Now here’s something nice. Intuitively, you would expect the flow of matter to be equal to the density times the local velocity. Let’s prove this to be the case.

Since

and

then $J^k = \rho u^0 v^k = J^0 v^k$

So the matter flow in a given inertial frame is equal to the fluid density measured in that frame times the fluid velocity in that frame, as expected.

An Approximation

Let’s suppose that the deformations are such that we can write the material coordinates of a fluid particle, as a function of spacetime coordinates, in the following form:

We will compute the matter current in this parameterization. Let’s assume further that the deformation gradients are small. This means that as we compute the current, we will ignore terms that contain two or more factors of the deformation gradient.

Clearly, the matrices $\partial_\mu R_a$ have the form

Let’s compute $J^0$, or the matter density measured in the given inertial frame

The order 0 term consists of all the delta functions multiplied together. When we sum over the indices of the Levi-Civita symbol, we get $\epsilon^{0 1 2 3}$ which is simply 1. The first order terms consist of two delta function factors and one deformation gradient factor. So the sum of the order 0 and order 1 terms becomes

Summing the delta functions gives

Now, due to the antisymmetry of the Levi-Civita symbol, there is only one non-zero term in each sum! We finally get something very familiar:

This is incredible. What this equation is saying is that to first order, the density of the fluid is equal to its density in the reference state minus a term proportional to the divergence of the deformation field. This makes too much sense: if the deformation field is diverging (positive divergence), then we should expect the density to drop! Conversely, if the deformation field is converging (negative divergence), then the density increases. We also see that, as expected, computing the density to first order yields a linear expression for the dependency of density on the deformation.

Starting from something highly symmetric but not intuitive, and then breaking it down by applying physical assumptions to give a readily interpretable expression that makes sense with our every day experiences is incredibly satsifying to me.

But what about the other components of the matter current? We can get them pretty quickly in this approximation:

Since if any two indices of the delta function are the same, it gives zero, the zeroth-order term of this expression is 0. We get

And likewise for the other spatial componets. So this gives our full expression for the matter current, to first order:

Killer!

Remember we said that the rest-frame density was given by $\sqrt{J^\mu J_\mu}$ ? Well, this becomes

And, as it should be, if the fluid has low velocity $% $, we get

Thanks for reading!