# Testing my understanding of flux

1. Feb 6, 2010

### Rasalhague

Here's what I understand by flux:

$$\int \textbf{F} \cdot \textup{d}\textbf{A},$$

the surface integral of a vector field. I gather the vector field that's integrated to give a flux is called the flux density (examples: the magnetic B field for magnetic flux, the Poynting vector for electromagnetic energy flux, mass "flow-rate" density for mass flux, and j charge density for current). I gather that the flux density through a surface is a scalar, the dot product of flux density (the vector) with a unit vector normal to the surface. For this, we need to specify an orientation of a surface at the relevant point, but not the size or shape of the surface. A related concept, when some surface is fully specified, is average flux density:

$$\frac{\int \textbf{F} \cdot \textup{d}\textbf{A}}{A}.$$

Exceptions: the flux density of electric flux is called the electric field, E, which differs from the related vector quantity called electric flux density, also called the electric displacement field, D. I don't know why, but then I don't know much about these concepts.

I also gather that the term "flux" is often loosely used in place of "flux density" (presumably in any of the senses of that latter). Are the 2nd, 3rd and 4th components of the top row of the stress-energy tensor [ http://en.wikipedia.org/wiki/Stress-energy_tensor ], here called "energy flux", actually energy flux density, the same kind of quantity as the Poynting vector? And does "energy density" here mean the limit of average energy per unit volume at a point as the volume goes to zero?

$$\frac{\mathrm{d} E}{\mathrm{d} V}$$

Given the analogy between mass flux and current, is current synonymous with "charge flux", if that term is used?

Last edited: Feb 6, 2010
2. Feb 6, 2010

### sai_2008

Hmm..while I wont repeat what you can find in a good physics text book regarding flux, just to point out, with current there is a "time aspect" to it - not just area: the amount or charge that traverses per second across any surface is the current... so the current is not a charge flux in that way, but a rate of charge movement.

3. Feb 6, 2010

### Rasalhague

This distinguishes current from "electric flux", but current still seems exactly analogous to "mass flux" as described by Davis & Snider in An Introduction to Vector Analysis, except that charge takes the place of mass, there being a time aspect to both quantities:

$$\nu = \text{particle density}, \frac{mol}{m^3}$$

$$\mu = m \nu = \text{mass density}, kg \cdot \frac{mol}{m^3}$$

$$\textbf{F} = \mu \textbf{v} = \text{mass flow-rate (i.e. flux?) density}, kg \cdot \frac{mol}{m^3} \cdot \frac{m}{s} = \frac{kg \cdot mol}{m^2 \cdot s}$$

$$\Phi = \text{the flux of } \textbf{F} \text{ through } \Delta S$$
$$\text{ } = \int \textbf{F} \cdot \hat{\textbf{n}} \; \Delta S$$

Versus

$$\nu = \text{particle density}, \frac{mol}{m^3}$$

$$\rho = q \nu = \text{charge density}, C \cdot \frac{mol}{m^3}$$

$$\textbf{j} = \rho \textbf{v} = \text{current density}, C \cdot \frac{mol}{m^3} \cdot \frac{m}{s} = \frac{C \cdot mol}{m^2 \cdot s}$$

$$\Phi = \text{the current through } \Delta S$$
$$\text{ } = \int \textbf{j} \cdot \hat{\textbf{n}} \; \Delta S$$

The physics textbooks that I've seen so far define various quantities such as "energy flux", "electric flux", "magnetic flux", but aren't so good on general definitions of flux per se. Are my definitions correct? Is this how the terminology is used?

4. Feb 7, 2010

### Rasalhague

I just came across this in Roger Penrose's The Road to Reality, describing the components of the charge-current 4-vector: "$J^0$ is the density of charge, and the three quantities $J^1, J^2, J^3$ provide the flux of the charge (i.e. the current)" (§ 19.5, Vintage 2005, p. 456). I think what he calls current is normally referred to as current density.

5. Feb 7, 2010

### Andy Resnick

Your final question is correct; electric current obeys a conservation law of the type:

$$\frac{\partial \rho}{\partial t} + \nabla \cdot j = 0$$.

As for the stress-energy tensor, since it a continuum mechanical tensor field, the components do involve 'density' (meaning a distributed energy/force/momentum as a concept; energy density, momentum density etc. The conservation laws for the various components of the tensor lead to assigning 'density' or current (i.e. flux) with certain components. Since the term 'density' appears in different contexts (i.e. a continuum concept and also a variable), you may be getting confused.

6. Feb 7, 2010

### Rasalhague

Could you elaborate on what you mean by "distrubuted energy/force/momentum". Does this sense of density mean that x-density is any quantity of the form x per n-volume?

7. Feb 7, 2010

### Rasalhague

I'd like to have general definitions of density, flux density and flux in this context. If it happens that multiple concepts are called by the same name, or that names are used inconsistently, I'd like at least to have distinct makeshift names that I can call the concepts by as I'm learning them, so that I don't get confused.

How many concepts are involved (when we talk about density, flux density and flux)? How are they defined? What names are normally given to each of them?

The two examples I found in Snider & Davis, mass flux and current, follow a similar pattern to each other which made me think they could be a clue towards a general definition. For some base quantity x, x density is x per volume. x flux density (which in the case of charge flux, if we can call it that, is given the special name, current) is x density times velocity. x flux density in some direction is x flux density dotted with a unit vector; an orientation is specified, but not the size or shape of a surface. Finally, x flux is x integrated over a specified surface.

But since electric flux, for example, doesn't involve a velocity, as sai pointed out, I suppose the above definitions must need modifying to make them more general. If flux density is defined in terms of flux as the vector field that's integrated to give flux, that would just leave density itself to be defined. Is there always a density for any flux? If so, what's the density for electric flux, say. If the electric field is the flux density of electric flux, what quantity is analogous to the mass flux density (for mass flux) and the charge density = (for current = charge flux)? Is it possible for two distinct fluxes to share the same density? Perhaps the density of electric flux is charge density too.

8. Feb 7, 2010

### Rasalhague

Trying to reason out what the generalised definitions might be, I'm experimenting with dimensional analysis by analogy with those two examples, mass flux and current.

Mass flux density: $\frac{kg}{m^2 \cdot s}$.

Current density: $\frac{C}{m^2 \cdot s}$.

Other fluxes I'll look at are those of momentum and energy. (I'll leave electric flux for now, as I asked about that in my previous post, but it seems like a useful example in searching for general definitions because of its lack of a velocity "aspect".)

By analogy with the name, if we substitute momentum for mass in the equation for mass flux density, momentum flux density would have units $\frac{kg}{m \cdot s^2} = \frac{N}{m^2}$. Ah, that looks promising! These are the units given for a quantity called "momentum flux" (i.e. momentum flux density?) in the Wikipedia article Flux. And likewise for energy flux (i.e. energy flux density?) they give units of $\frac{J}{m^2 \cdot s}$.

Looking at the intro to the Wikipedia article, maybe we could say that transport fluxes (per time) fluxes are a special and important subset of flux, defined more generally in an abstract way as surface integrals of a vector field. Which leaves the question of what the density (as opposed to the flux density) of the more abstract kind of flux is defined as, if such a quantity is defined at all for such a flux.

9. Feb 7, 2010

### Rasalhague

Is the reason that you used the partial symbol here that $\rho$ depends on position as well as time? I suppose, in what Blandford and Thorne call "slot-naming index notation", and others "abstract index notation", this would be

$$\frac{\partial J^\alpha}{\partial x^\alpha} = 0.$$

10. Feb 7, 2010

### diazona

Yes, that sounds right. Whenever a vector field is integrated over a surface to give a flux, the vector field is the associated flux density. Or more precisely, the component of the vector field normal to a given surface is the flux density across that surface.
Yes, there is. The flux per unit surface area is the flux density. The flux density corresponding to electric flux is the electric field.
Hmm... I suppose mass flux density would be mass density times velocity. And charge flux density would be charge density times velocity, which is indeed also known as current. These are specific cases in which there is a velocity field that factors into the definition of the flux density, but that doesn't always have to be the case. Any vector field can be a flux density, whether it is related to velocity or not.

Mathematically, for any vector field
$$\vec{V}(\vec{x})$$
the flux associated with this vector field and with a given surface S is
$$\Phi = \iint_S \vec{V}(\vec{x})\cdot\mathrm{d}\vec{A}$$
And the flux density associated with a given point in a given direction is
$$\hat{n}\cdot\frac{\mathrm{d}\Phi}{\mathrm{d}\vec{A}} = \hat{n}\cdot\vec{V}(\vec{x})$$
The density of electric flux is the electric field, and the density of charge is certainly not the same thing. I would say no, I don't believe two distinct fluxes can share the same flux density, because the flux is computed from the flux density, and if you have only one flux density you can compute only one flux.

Incidentally, you were right that density is a general term for the amount of something per unit n-volume. To be completely precise, we can talk about linear densities, area densities, volume densities, etc. For example, linear charge density is the amount of charge per unit length. Area charge density = charge per unit area, volume charge density = charge per unit volume. Similarly for linear mass density, area mass density, volume mass density, etc. Of course, there are a lot of potentially confusing conventions in place when you're talking about density, e.g. usually "charge density" or "mass density" refers to a volume density, but "current density" refers to an area density. Flux densities are always area densities (at least in 3D space; in an n-dimensional space, a flux density is the amount of flux per (n-1)-volume).

11. Feb 7, 2010

### diazona

I'd put it this way: a transport flux is one whose associated vector field is proportional to a velocity vector field. That is, any vector field which can be expressed in the form
$$\vec{V}(\vec{x}) = C(\vec{x})\vec{v}(\vec{x})$$
(lowercase v is velocity) has an associated flux that can be called a transport flux.

12. Feb 7, 2010

### Rasalhague

Excellent! Many thanks, diazona. I think it's starting to get clearer...

Typo on my part! I meant: which quantity is to electric flux what mass density (as opposed to mass flux density) is to mass flux?

I guess from this, if there is a quantity that stands in the same relation to electric flux as mass density stands to mass flux, it'd have to be a per 3-volume density. But I can't think what it would be if not charge density.

I'm not used to seeing a vector in the denominator of a derivative (or indeed any kind of denominator); should this be the magnitude

$$\mathrm{d}||\vec{A}||,$$

i.e. the area, rather than the vectorial surface element (if that's the right word)

$$\mathrm{d}\vec{A}$$

itself, given that the direction is already specified by $\hat{n}$?

13. Feb 8, 2010

### diazona

Well, mass flux would be calculated as
$$\Phi = \iint_S \rho \vec{v}\cdot\mathrm{d}\vec{A}$$
i.e. to get from mass density to mass flux, you multiply by velocity and integrate over a surface. So the question is, is there a quantity that, when multiplied by velocity and integrated over a surface, gives you electric flux? I'm pretty sure there is not. The reason there is one for mass is that the mass flux is a transport flux - in other words, its associated flux density $\rho\vec{v}$ is a product of a scalar field and a velocity vector field (specifically, the velocity of "elements" of the scalar field, roughly speaking). This is not true of the electric field. Electric fields are not created by some more fundamental entity moving around in space; the electric field itself is fundamental. (Unless you talk about the electric field being derivable from the vector potential, or the Faraday tensor in GR, but that's irrelevant for this discussion.)

Anyway, the point is that there is no quantity that bears the same relationship to electric flux that mass density bears to mass flux, because the mass flux is a transport flux but the electric flux is not.
You're right that if there were such a quantity, it would have to be a volume density. But as I said, I'm pretty sure there isn't one. The charge density would be associated with "current flux" (the surface integral of current density), not electric flux.
No, what I mean by that is
$$\frac{\mathrm{d}\Phi}{\mathrm{d}\vec{A}} = \frac{\mathrm{d}\Phi}{\mathrm{d}A_x}\hat{x} + \frac{\mathrm{d}\Phi}{\mathrm{d}A_y}\hat{y} + \frac{\mathrm{d}\Phi}{\mathrm{d}A_z}\hat{z}$$
where $\hat{x}$, $\hat{y}$, and $\hat{z}$ are the coordinate unit vectors. For example, one way to write what you probably know as the gradient is
$$\frac{\mathrm{d}\Phi}{\mathrm{d}\vec{x}} = \vec{\nabla}\Phi$$
The reason I wrote it that way is that flux density depends on which surface you use in the calculation. There are surfaces you could choose through which the flux would be zero, for example. If we had only one fixed surface to work on, then sure, I could get away with just writing $$\mathrm{d}\Phi/\mathrm{d}\lVert \vec{A}\rVert$$, but since we are working in 3D space where we can arbitrarily choose a surface, we need to write $$\mathrm{d}\Phi/\mathrm{d}\vec{A}$$ so that this derivative is sensitive to both the area and the orientation of the surface element.

14. Feb 8, 2010

### Rasalhague

Is there any special name for the quantity

$$\textbf{F}\cdot\hat{\textbf{n}} \; \mathrm{d}A$$

found by multiplying the (scalar) "flux in direction n" by a surface element?

(Coupla self-corrections: in #1, "j charge density" should read "j current density", and in #3, $\Delta S$ in the integrals should read $\mathrm{d}S$.)

15. Feb 8, 2010

### Rasalhague

Okay, thanks for clearing that up.

I see. I just wasn't familiar with the vector-symbol-in-the-denominator notation.

16. Feb 8, 2010

### Rasalhague

When you say "or more precisely", do you mean that it would be better not to call the vector field the "flux density"? If so, what precise term should I use for the vector field itself, that I've been calling "flux density", if I want to distinguish it from that scalar quantity that I've been calling "flux density in a (specified) direction" (i.e. $\textbf{v} \cdot \hat{\textbf{n}}$)?

Not if by flux density we mean "the vector field that's integrated to give the flux". I'd perhaps translate this (depending on how well I've understood it) into the terminology I've been using as "the flux density in a direction depends on which direction is chosen" (a bit laborious I know, but I really need unambiguous words to pin the ideas to, so that I can tell the concepts apart, at least until I'm familiar with them). What would be the conventional way of saying that the vector flux density doesn't depend on the direction, while the scalar flux density (in a specified direction) does depend on which direction is specified?

It seems that

$$\hat{n}\cdot\frac{\mathrm{d}\Phi}{\mathrm{d}\vec{A}} = \hat{n}\cdot\vec{V}(\vec{x}),$$

depends only on the direction of the unit normal vector, not on any other properties of the surface. Is that right?

Are these notations equivalent:

$$\hat{n}\cdot\frac{\mathrm{d}\Phi}{\mathrm{d}\vec{A}} = \hat{n}\cdot\vec{V}(\vec{x}) = \hat{n}\cdot\frac{\mathrm{d}\Phi}{\mathrm{d}\vec{x}} = \hat{n}\cdot \vec{\nabla} \Phi?$$

Does the surface element dA somehow encode or represent topological information about the surface or is it just a sort of dummy variable, the only real information about the surface in $\hat{\textbf{n}} \; dA$ being embodied in the unit normal vector that indicates its orientation?

Last edited: Feb 8, 2010
17. Feb 8, 2010

### diazona

Hmm... well, nobody's ever quizzed me on the precise meanings of these terms at this level of detail, so I'm working some of this out as I go. Thus your confusion is understandable (and good questions, by the way!) One thing I'm sure of is that in the conventional usage (at least as far as I'm familiar with it), "flux density" does refer to the vector field that is integrated to give the flux. I think an issue here is that "flux density across a given surface" means something slightly different from just "flux density." Mathematically I think I would express the former as
$$\frac{\mathrm{d}^2}{\mathrm{d}^2\lVert\vec{A}\rVert}\iint_S \vec{V}(\vec{x})\cdot\mathrm{d}^2\vec{A}$$
which is supposed to mean
$$\lim_{A_S\to 0}\frac{\iint_S \vec{V}(\vec{x})\cdot\mathrm{d}^2\vec{A}}{\iint_S \hat{n}\cdot\mathrm{d}^2\vec{A}}$$
and the latter (flux density) as
$$\frac{\mathrm{d}^2}{\mathrm{d}^2\vec{A}}\iint_S \vec{V}(\vec{x})\cdot\mathrm{d}^2\vec{A}$$
(by the way, I should have been writing $$\mathrm{d}^2\vec{A}$$ instead of $$\mathrm{d}\vec{A}$$ all along).
You're right, I guess in that particular instance I should have said that the flux density across a given surface (a scalar) depends on the surface used. The (vector) flux density certainly doesn't.
Yeah, that sounds right. It does, of course, also depend on the vector field $$\vec{V}(\vec{x})$$
Well... let's say this: your first equality
$$\hat{n}\cdot\frac{\mathrm{d}\Phi}{\mathrm{d}\vec{A}} = \hat{n}\cdot\vec{V}(\vec{x})$$
mathematically represents the fact that the component of the flux density in a particular direction is equal to the component of the associated vector field in the same direction. Sure it's true, but it's not like the two sides are just two different ways of writing the exact same thing (which is what I would consider "equivalent notation").

On the other hand, in your last equality
$$\hat{n}\cdot\frac{\mathrm{d}\Phi}{\mathrm{d}\vec{x}} = \hat{n}\cdot \vec{\nabla} \Phi$$
the two sides are nothing more than two different ways of writing the same thing.

And of course,
$$\hat{n}\cdot\vec{V}(\vec{x}) \neq \hat{n}\cdot\frac{\mathrm{d}\Phi}{\mathrm{d}\vec{x}}$$
Those two are completely different things. (Actually: it is possible for the gradient to be a flux density, since it is a vector field and so you can compute its flux through any given surface. So for selected choices of $$\vec{V}(\vec{x})$$, namely if you define it as $$\vec{V}(\vec{x}) = \tfrac{\mathrm{d}\Phi}{\mathrm{d}\vec{x}}$$, that would be true. But not all flux densities are gradients.)
It's a dummy variable in the same sense that any other differential is a dummy variable (like the $\mathrm{d}x$ in $\frac{\mathrm{d}V}{\mathrm{d}x}$).

18. Feb 8, 2010

### Rasalhague

I was just thinking of equations like this one for the line element in polar coordinates,

$$\left ( \textup{d}s \right )^2 = \left ( \textup{d}r \right)^2 + r^2 \; \textup{d}\theta,$$

where the variable $\textup{d}s$ seems to be given more meaning. If $\textup{d}s$ has this value, presumably not any differential could be equated with the right hand side. And I've come across a notation whereby $\textup{d}\Phi$ indicates the gradient of $\Phi$, but perhaps this is a different notational system not meant to be used alongside the notation in which $\mathrm{d}$ denotes a differential. I found a footnote in Roger Penrose's The Road to Reality that seems to relate to this issue, if I could but understand it... (Post #8 here https://www.physicsforums.com/showthread.php?p=2570815#post2570815 ).

Last edited: Feb 8, 2010
19. Feb 8, 2010

### diazona

Mathematicians seem to come up with all sorts of strange variations on derivatives... but this might be what you're thinking of: $\mathrm{d}\Phi$ can represent the gradient if the d is something called the exterior derivative. (I personally use a bold d to represent exterior derivatives.) It can be written as
$$\mathbf{d}\Phi = \frac{\partial\Phi}{\partial x}\mathbf{d}x + \frac{\partial\Phi}{\partial y}\mathbf{d}y + \frac{\partial\Phi}{\partial z}\mathbf{d}z$$
But here's the catch: in that expression, $\mathbf{d}x$ etc. are not necessarily differentials; they're basis one-forms, which are (roughly speaking) the equivalents in differential geometry of the basis vectors $$\hat{x}$$ etc. in coordinate geometry. I must admit I don't entirely understand it myself, but I think you'd be safe to forget about exterior derivatives and one-forms for the present discussion.

As for the line element, I'm not really sure what you're getting at with the idea of a dummy variable. If you mean "dummy variable" in the same sense that j is a dummy index in
$$\vec{p}\cdot\vec{x} = \sum_j p_j x^j$$
i.e. that the letter doesn't matter, then no, it's not a dummy variable at all. The differential $$\mathrm{d}^2\vec{A}$$ (again, I haven't been writing the superscript 2 but I should have been) is written with an A because it's a differential of area, as opposed to $$\mathrm{d} x$$ which is the differential of some length.

20. Feb 9, 2010

### Rasalhague

Hmm, I will come back to this, but I think I'll do some more reading first. It's an ongoing mystery for me. From Penrose's Road to Reality there does seem to be some connection between all these concepts: differentials, exterior derivatives, k-forms, integration variables, area and volume elements... But I'm a bit baffled at the moment as to how they all fit together. Still, thanks to your explanations, I feel like I've made some progress with flux.

Oh, right, so not because it's related to a second derivative in some way analogous to the relationship of an infinitesimal dx to a first derivative? (And presumably not an exterior derivative if the exterior derivative d obeys the equation d2 = 0...)

21. Feb 9, 2010

### diazona

Actually, it is, in a way. Here's my understanding of ordinary derivatives: basically each d represents one infinitesimal dimension. So for example, $\mathrm{d}^2A$ is the product of two infinitesimal dimensions, such as
$$\mathrm{d}^2A = \mathrm{d}x\mathrm{d}y$$
There is also a first differential of area, $\mathrm{d}A$ which would be the product of one finite dimension and one infinitesimal dimension. For example,
$$\mathrm{d}A = w\mathrm{d}y$$
With these definitions in mind, the derivative of flux (on a fixed 2D surface) would be more properly written
$$\frac{\mathrm{d}^2\Phi}{\mathrm{d}^2A}$$
because it's the limit of the flux per unit surface area as the area shrinks to a point. But nobody really writes that; it's just conventional to use
$$\frac{\mathrm{d}\Phi}{\mathrm{d}A}$$
The notation for derivatives is often confusing, or at least misleading... unfortunately I don't think there's any way around that because there doesn't seem to be any notation that accurately and intuitively represents all the properties of derivatives and differentials. (At least, nobody's come up with anything completely satisfying so far)
That's right, the exterior derivative is a separate thing from these differentials.

22. Feb 10, 2010

### Rasalhague

I suppose missing off the powers of two in this expression is similar to the shorthand convention whereby people write

$$\frac{\mathrm{d}^2 y}{\mathrm{d} x^2}$$

for the second derivative rather than

$$\frac{\mathrm{d}^2 y - \mathrm{d}^2x \frac{\mathrm{d} y}{\mathrm{d} x}}{\mathrm{d} x^2}.$$

23. Feb 10, 2010

### diazona

Well... I wouldn't really lump those two together, because
$$\frac{\mathrm{d}^2 y - \mathrm{d}^2x \frac{\mathrm{d} y}{\mathrm{d} x}}{\mathrm{d} x^2}$$
is just an unsimplified form. I don't think it really makes anything more clear than $\tfrac{\mathrm{d}^2 y}{\mathrm{d}x^2}$