# What is 'momentum outflow'?

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?

## 1 Answer

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.

• It took me a day to read the 'Tensor' of Feynman's Lectures. Though problems like what 'energy ellipsoid' mean bothered me, it helped a quite to understand tensor just to conceive your answer. But one thing I want to ask is why we should need a tensor for the momentum flow? I can get why tensor is required for polarizability in asymmetric crystals; but am not getting why it's required here, in the momentum flow of the electromagnetic field. Can you explain that? – user36790 Dec 21 '15 at 13:49
• Take a look at Why is stress a tensor quantity? As for why an EM field has momentum flow even if it is stationary, it might help to remember that magnetic fields come from moving charges. – user10851 Dec 21 '15 at 19:41
• It also might help to think more relativistically, where energy is just the time component of momentum. Then the stress tensor goes from 3x3 to 4x4, with the added dimensions being energy and time. If you accept that fields have an energy density ($T^{00}$), it shouldn't be surprising they have momentum density ($T^{0i}$), energy flux ($T^{i0}$), and momentum flux ($T^{ij}$). – user10851 Dec 21 '15 at 19:42