# What are the units of every vector in the codomain of the vector field $\textbf{F}$?

I don't understand how the rate at which a fluid flows along the bottom edge of a rectangular region A in the direction of i is approximately $\textbf{F}(x,y)\dot{i} \Delta x$. In physics, the rate as which fluid flows is known as the speed measured in terms of meters per second. The rate as which a fluid flows in some direction is its velocity which is the magnitude of the speed times its direction. What are the units of each vector in the codomain of the vector field $\textbf{F}$? I think the units of each vector in the codomain of the vector field would have to have units of $\frac{1}{s}$. Recall that $\frac{1}{s}$ is the SI units of hertz, so the units of each element in the range of the vector field $\textbf{F}$ is hertz? I assume $\Delta x$ can be measured in meters. So the units of each element in the vector vector field multiplied by its direction multiplied by the units of meters gives us meters per second which is the units of the rate at which a fluid flows along a bottom edge of a rectangular region $A$?

But I think F is supposed to be a velocity vector field whose codomain consists of velocity vectors. When we dot F with i and then $\Delta x$ we should have meters squared per second Which doesn't make sense as a rate at which fluid flows along an edge. Meters squared per second is viscosity which isn't rate at which fluid flows.

This is a very good question, and I can see where the confusion is coming from. The source should be much more specific.

What are the units of each vector in the codomain of the vector field $\mathbf{ F}$?

That all depends on how the rate at which the fluid flows is measured. Usually, when some “thing” flows, this flow is measured as flux. Flux is a measurement of a quantity / (area ⋅ time), much like how density is a measurement of a quantity per measure of space (i.e., length, area, or volume). If you’re curious about how many kilograms of fluid pass per second, you’ll measure mass flux; if you’re curious about how many liters of fluid pass per second, you’ll measure volumetric flux.

You’re probably wondering where the area quantity came from, since this is what distinguishes flux from speed. The idea is that the bigger the passage, the more fluid can flow through. Vectors are also important, because the angle of the passage relative to the direction in which the fluid flows matters as well.

So, to determine the dimensions of $\mathbf{ F}$, we have to know the dimensions of $\mathbf{F}\cdot \mathbf{ i}\,\Delta x$, which could be volumetric flux, mass flux, or any other kind of flux.

Since we don’t know what kind of flux $\mathbf{F}\cdot \mathbf{ i}\,\Delta x$ is, let’s call it $\mathsf{R}$-flux, so that

$$\dim(\mathbf{F}\cdot \mathbf{ i}\,\Delta x) = \mathsf{R}^1\,\mathsf{L}^{-2}\,\mathsf{T}^{-1}$$

(which is some quantity of dimension $\mathsf{R}$ per area-time).

Just by the properties of how dimensions work,

\begin{align} \dim(\mathbf{F}\cdot \mathbf{ i}\,\Delta x) &= \dim(\mathbf{F})\dim(\mathbf{ i})\dim(\Delta x) \\ \mathsf{R}^1\,\mathsf{L}^{-2}\,\mathsf{T}^{-1} &= \dim(\mathbf{ F}) \, 1 \, \mathsf{L} \\ \mathsf{R}^1\,\mathsf{L}^{-3}\,\mathsf{T}^{-1} &= \dim(\mathbf{ F}) \\ \end{align}

This implies that that unit of $\mathbf{ F}$ should be $$\frac{\operatorname{unit}(\mathsf{R})}{\mathrm{m^3}\cdot\mathrm{s}}$$

Again, the book should have been more specific by telling us what is traveling per second.

. . . so the units of each element in the range of the vector field $\mathbf F$ is hertz?

No, not even if the flow rate is measured in $\left.\mathrm{ m }\middle/\mathrm{ s }\right.$.

These are the exact words from the BIPM’s handbook for the SI:

The hertz is used only for periodic phenomena, and the becquerel is used only for stochastic processes in activity referred to a radionuclide.

So, the unit you would have to use would be the reciprocal second or inverse second, $\mathrm{s}^{-1}$.

• @Cake You can up-vote answers in addition to giving the $\color{green}{\checkmark}$ as well – gen-z ready to perish Nov 19 '17 at 1:40
• Shouldn't the actual rate at which fluid flows along the bottom of a rectangular region R be in units of $R/(m^3*s)$? If the actual rate at which fluid flows along the bottom of a rectangular Region R were in units of $R/(m^3*s)$?, then the units of the vectors in the codomain of F would be $R/(m^4*s)$ which wouldn't be any where surprising.Realistically. fluids are three dimensional. This problem claims that the rate at which fluid flows along the bottom is in units of $R/(m^2*s)$ which is contrary to my intuition that fluid flows through a 3-space not a 2-space. @ChaseRyanTaylor – Cake Nov 19 '17 at 1:51
• @Cake Not exactly. You can have $\mathsf{R}$ be measured in $\mathrm{m}^3$, but only area matters for the denominator, not volume. This is because in a single instant, the fluid is passing through a cross-sectional area. – gen-z ready to perish Nov 19 '17 at 1:54
• @Cake Good question. Grams, kilograms, moles, liters, cubic meters, etc. Really I just $\mathsf{R}$ being mass, volume, or amount of substance. In other fields, flux could involve coulombs, newtons, teslas (as in electric, gravitational, and magnetic flux), of anything you could think of imagining! – gen-z ready to perish Nov 19 '17 at 2:00
• R could be in units of teddy bears? – Cake Nov 19 '17 at 2:01