What is 'momentum outflow'?

Question

Feynman, in his lecture, on Field momentum, used this term while relating the momentum of matter the field is interacting with & the field's momentum itself. Here is the excerpt:

Just as the field has energy, it will have a certain momentum per unit volume. Let us call that momentum density $\boldsymbol g.$ Of course, momentum has various possible directions, so that $\boldsymbol g$ must be a vector. Let’s talk about one component at a time; first, we take the $x$-component. Since each component of momentum is conserved we should be able to write down a law that looks something like this: \begin{equation*} -\frac{\partial}{\partial t} \begin{pmatrix} \text{momentum}\\ \text{of matter} \end{pmatrix} _x\!\!=\frac{\partial g_x}{\partial t}+ \begin{pmatrix} \text{momentum}\\ \text{outflow} \end{pmatrix} _x. \end{equation*}

The “momentum outflow” term, however, is strange. It cannot be the divergence of a vector because it is not a scalar; it is, rather, an $x$-component of some vector. Anyway, it should probably look something like $$\frac{∂a}{∂x}+\frac{∂b}{∂y}+\frac{∂c}{∂z},$$ because the $x$-momentum could be flowing in any one of the three directions. In any case, whatever $a$, $b$, and $c$ are, the combination is supposed to equal the outflow of the $x$-momentum. [...]

He didn't tell what 'momentum overflow' is. I'm not getting what this term actually means & how it entered the equation. Also, if it is a vector, why its $x$-component needs to get expressed in three directions? I really couldn't conceive that.

So, my questions are:

$\bullet$ What is 'momentum overflow'? How did it enter in the above equation?

$\bullet$ Why is the $x$-component vector expressed $\frac{∂a}{∂x}+\frac{∂b}{∂y}+\frac{∂c}{∂z}\;?$ $x$-momentum could be flowing in any one of the three directions- now, how can $x$-component of a vector can have components in other orthogonal directions?

Answer 1

Let's start with a simpler situation: conservation of mass rather than momentum, with just matter and no field. The differential form of the continuity equation for mass density can be written $$ -\frac{\partial\rho}{\partial t} = \nabla \cdot \vec{\jmath} = \frac{\partial j_x}{\partial x} + \frac{\partial j_y}{\partial y} + \frac{\partial j_z}{\partial z}. $$ (This is known as flux-conservative form.) $\vec{\jmath}$ is the vector flux of density -- basically it measures the current of mass per unit area going in some direction. For a well-behaved fluid in Newtonian mechanics, density flux is proportional to density and velocity: $\vec{\jmath} = \rho \vec{v}$. (You might recognize this as simply the momentum density: momentum per unit volume.) Here then we see the three derivatives Feynman alludes to, with the catch that in this case we really are taking the divergence of a vector.

Now let's consider conservation of momentum, a vector quantity. Since this coincidentally showed up in the mass conservation law, we'll keep calling the momentum density $\vec{\jmath}$. The relevant conservation law is $$ -\frac{\partial\vec{\jmath}}{\partial t} = \nabla \cdot \mathbf{T}. $$ Here $\mathbf{T}$ is a tensor, specifically the stress tensor of the matter. If you are not familiar with tensors (Feynman assumes you are not), think of it as a matrix. In component form, $\mathbf{T}$ has two indices: $T^{ij}$ is the $i$-flux of $j$-momentum. For example, $T^{xx}$ is the amount of $x$-momentum per unit area per unit time flowing in the $x$-direction, and $T^{yz}$ is the amount of $z$-momentum per unit area per unit time flowing in the $y$-direction. In our Newtonian fluid, $T^{ij} = \rho v^i v^j + p \delta^{ij}$, where $p$ is the fluid pressure.

The divergence of our tensor gives a vector. Explicitly, we could write momentum conservation as $$ -\frac{\partial\vec{\jmath}}{\partial t} = \left(\frac{\partial T^{xx}}{\partial x} + \frac{\partial T^{yx}}{\partial y} + \frac{\partial T^{zx}}{\partial z}\right) \hat{x} + \left(\frac{\partial T^{xy}}{\partial x} + \frac{\partial T^{yy}}{\partial y} + \frac{\partial T^{zy}}{\partial z}\right) \hat{y} + \left(\frac{\partial T^{xz}}{\partial x} + \frac{\partial T^{yz}}{\partial y} + \frac{\partial T^{zz}}{\partial z}\right) \hat{z}. $$ Taking just the $x$-component of this equation we recover Feynman's formula, with $a = T^{xx}$, $b = T^{yx}$, and $c = T^{zx}$. $T^{xx}$ measures the flow of $x$-momentum away from the point of interest in the $x$-direction, just as $T^{yx}$ and $T^{zx}$ measure the flow of $x$-momentum away in the $y$- and $z$-directions.

You need all three terms to account for where $x$-momentum is going, just as you need all three terms to account for where scalar mass is going. There are another three terms each in the corresponding equations for $y$-momentum and $z$-momentum.