Testing my understanding of flux

In summary, flux can refer to the surface integral of a vector field or the vector field itself. The vector field is called the flux density and examples include the electric field and mass "flow-rate" density. The flux density is a scalar obtained by taking the dot product of the vector field with a unit vector normal to the surface. Average flux density refers to the limit of the surface integral divided by the surface area. However, there are exceptions such as the electric field and electric flux density being different quantities. The term "flux" is often used loosely instead of "flux density". Current is an example of a flux with a time aspect to it, specifically the rate of charge movement. The stress-energy tensor involves density and flux of energy
  • #1
Rasalhague
1,387
2
Here's what I understand by flux:

[tex]\int \textbf{F} \cdot \textup{d}\textbf{A},[/tex]

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:

[tex]\frac{\int \textbf{F} \cdot \textup{d}\textbf{A}}{A}.[/tex]

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?

[tex]\frac{\mathrm{d} E}{\mathrm{d} V}[/tex]

Given the analogy between mass flux and current, is current synonymous with "charge flux", if that term is used?
 
Last edited:
Physics news on Phys.org
  • #2
Hmm..while I won't repeat what you can find in a good physics textbook 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
sai_2008 said:
Hmm..while I won't repeat what you can find in a good physics textbook 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.

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:

[tex]\nu = \text{particle density}, \frac{mol}{m^3}[/tex]

[tex]\mu = m \nu = \text{mass density}, kg \cdot \frac{mol}{m^3}[/tex]

[tex]\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}[/tex]

[tex]\Phi = \text{the flux of } \textbf{F} \text{ through } \Delta S[/tex]
[tex]\text{ } = \int \textbf{F} \cdot \hat{\textbf{n}} \; \Delta S[/tex]

Versus

[tex]\nu = \text{particle density}, \frac{mol}{m^3}[/tex]

[tex]\rho = q \nu = \text{charge density}, C \cdot \frac{mol}{m^3}[/tex]

[tex]\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}[/tex]

[tex]\Phi = \text{the current through } \Delta S[/tex]
[tex]\text{ } = \int \textbf{j} \cdot \hat{\textbf{n}} \; \Delta S[/tex]

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
I just came across this in Roger Penrose's The Road to Reality, describing the components of the charge-current 4-vector: "[itex]J^0[/itex] is the density of charge, and the three quantities [itex]J^1, J^2, J^3[/itex] 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
Rasalhague said:
<snip>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?

[tex]\frac{\mathrm{d} E}{\mathrm{d} V}[/tex]

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

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

[tex]\frac{\partial \rho}{\partial t} + \nabla \cdot j = 0[/tex].

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
Andy Resnick said:
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.

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
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
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: [itex]\frac{kg}{m^2 \cdot s}[/itex].

Current density: [itex]\frac{C}{m^2 \cdot s}[/itex].

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 [itex]\frac{kg}{m \cdot s^2} = \frac{N}{m^2}[/itex]. 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 [itex]\frac{J}{m^2 \cdot s}[/itex].

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

[tex]\frac{\partial \rho}{\partial t} + \nabla \cdot j = 0[/tex].

Is the reason that you used the partial symbol here that [itex]\rho[/itex] 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

[tex]\frac{\partial J^\alpha}{\partial x^\alpha} = 0.[/tex]
 
  • #10
Rasalhague said:
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.
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.
Rasalhague said:
Is there always a density for any flux? If so, what's the density for electric flux, say.
Yes, there is. The flux per unit surface area is the flux density. The flux density corresponding to electric flux is the electric field.
Rasalhague said:
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)?
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
[tex]\vec{V}(\vec{x})[/tex]
the flux associated with this vector field and with a given surface S is
[tex]\Phi = \iint_S \vec{V}(\vec{x})\cdot\mathrm{d}\vec{A}[/tex]
And the flux density associated with a given point in a given direction is
[tex]\hat{n}\cdot\frac{\mathrm{d}\Phi}{\mathrm{d}\vec{A}} = \hat{n}\cdot\vec{V}(\vec{x})[/tex]
Rasalhague said:
Is it possible for two distinct fluxes to share the same density? Perhaps the density of electric flux is charge density too.
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
Rasalhague said:
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.
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
[tex]\vec{V}(\vec{x}) = C(\vec{x})\vec{v}(\vec{x})[/tex]
(lowercase v is velocity) has an associated flux that can be called a transport flux.
 
  • #12
Excellent! Many thanks, diazona. I think it's starting to get clearer...

diazona said:
Hmm... I suppose mass flux density would be mass density times velocity.

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?

diazona said:
[...] 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).

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.

diazona said:
Mathematically, for any vector field
[tex]\vec{V}(\vec{x})[/tex]
the flux associated with this vector field and with a given surface S is
[tex]\Phi = \iint_S \vec{V}(\vec{x})\cdot\mathrm{d}\vec{A}[/tex]
And the flux density associated with a given point in a given direction is
[tex]\hat{n}\cdot\frac{\mathrm{d}\Phi}{\mathrm{d}\vec{A}} = \hat{n}\cdot\vec{V}(\vec{x})[/tex]

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

[tex]\mathrm{d}||\vec{A}||,[/tex]

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

[tex]\mathrm{d}\vec{A}[/tex]

itself, given that the direction is already specified by [itex]\hat{n}[/itex]?
 
  • #13
Rasalhague said:
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?
Well, mass flux would be calculated as
[tex]\Phi = \iint_S \rho \vec{v}\cdot\mathrm{d}\vec{A}[/tex]
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 [itex]\rho\vec{v}[/itex] 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.
Rasalhague said:
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.
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.
Rasalhague said:
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

[tex]\mathrm{d}||\vec{A}||,[/tex]

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

[tex]\mathrm{d}\vec{A}[/tex]

itself, given that the direction is already specified by [itex]\hat{n}[/itex]?
No, what I mean by that is
[tex]\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}[/tex]
where [itex]\hat{x}[/itex], [itex]\hat{y}[/itex], and [itex]\hat{z}[/itex] are the coordinate unit vectors. For example, one way to write what you probably know as the gradient is
[tex]\frac{\mathrm{d}\Phi}{\mathrm{d}\vec{x}} = \vec{\nabla}\Phi[/tex]
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 [tex]\mathrm{d}\Phi/\mathrm{d}\lVert \vec{A}\rVert[/tex], but since we are working in 3D space where we can arbitrarily choose a surface, we need to write [tex]\mathrm{d}\Phi/\mathrm{d}\vec{A}[/tex] so that this derivative is sensitive to both the area and the orientation of the surface element.
 
  • #14
Is there any special name for the quantity

[tex]\textbf{F}\cdot\hat{\textbf{n}} \; \mathrm{d}A[/tex]

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, [itex]\Delta S[/itex] in the integrals should read [itex]\mathrm{d}S[/itex].)
 
  • #15
diazona said:
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.

Okay, thanks for clearing that up.

diazona said:
No, what I mean by that is
[tex]\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}[/tex]
where [itex]\hat{x}[/itex], [itex]\hat{y}[/itex], and [itex]\hat{z}[/itex] are the coordinate unit vectors. For example, one way to write what you probably know as the gradient is
[tex]\frac{\mathrm{d}\Phi}{\mathrm{d}\vec{x}} = \vec{\nabla}\Phi[/tex]

I see. I just wasn't familiar with the vector-symbol-in-the-denominator notation.
 
  • #16
diazona said:
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.

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. [itex]\textbf{v} \cdot \hat{\textbf{n}}[/itex])?

diazona said:
The reason I wrote it that way is that flux density depends on which surface you use in the calculation.

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

[tex]\hat{n}\cdot\frac{\mathrm{d}\Phi}{\mathrm{d}\vec{A}} = \hat{n}\cdot\vec{V}(\vec{x}),[/tex]

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:

[tex]\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?[/tex]

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 [itex]\hat{\textbf{n}} \; dA[/itex] being embodied in the unit normal vector that indicates its orientation?
 
Last edited:
  • #17
Rasalhague said:
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. [itex]\textbf{v} \cdot \hat{\textbf{n}}[/itex])?
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 :wink: (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
[tex]\frac{\mathrm{d}^2}{\mathrm{d}^2\lVert\vec{A}\rVert}\iint_S \vec{V}(\vec{x})\cdot\mathrm{d}^2\vec{A}[/tex]
which is supposed to mean
[tex]\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}}[/tex]
and the latter (flux density) as
[tex]\frac{\mathrm{d}^2}{\mathrm{d}^2\vec{A}}\iint_S \vec{V}(\vec{x})\cdot\mathrm{d}^2\vec{A}[/tex]
(by the way, I should have been writing [tex]\mathrm{d}^2\vec{A}[/tex] instead of [tex]\mathrm{d}\vec{A}[/tex] all along).
Rasalhague said:
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?
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.
Rasalhague said:
It seems that

[tex]\hat{n}\cdot\frac{\mathrm{d}\Phi}{\mathrm{d}\vec{A}} = \hat{n}\cdot\vec{V}(\vec{x}),[/tex]

depends only on the direction of the unit normal vector, not on any other properties of the surface. Is that right?
Yeah, that sounds right. It does, of course, also depend on the vector field [tex]\vec{V}(\vec{x})[/tex]
Rasalhague said:
Are these notations equivalent:

[tex]\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?[/tex]
Well... let's say this: your first equality
[tex]\hat{n}\cdot\frac{\mathrm{d}\Phi}{\mathrm{d}\vec{A}} = \hat{n}\cdot\vec{V}(\vec{x})[/tex]
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
[tex]\hat{n}\cdot\frac{\mathrm{d}\Phi}{\mathrm{d}\vec{x}} = \hat{n}\cdot \vec{\nabla} \Phi[/tex]
the two sides are nothing more than two different ways of writing the same thing.

And of course,
[tex]\hat{n}\cdot\vec{V}(\vec{x}) \neq \hat{n}\cdot\frac{\mathrm{d}\Phi}{\mathrm{d}\vec{x}}[/tex]
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 [tex]\vec{V}(\vec{x})[/tex], namely if you define it as [tex]\vec{V}(\vec{x}) = \tfrac{\mathrm{d}\Phi}{\mathrm{d}\vec{x}}[/tex], that would be true. But not all flux densities are gradients.)
Rasalhague said:
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 [itex]\hat{\textbf{n}} \; dA[/itex] being embodied in the unit normal vector that indicates its orientation?
It's a dummy variable in the same sense that any other differential is a dummy variable (like the [itex]\mathrm{d}x[/itex] in [itex]\frac{\mathrm{d}V}{\mathrm{d}x}[/itex]).
 
  • #18
diazona said:
It's a dummy variable in the same sense that any other differential is a dummy variable (like the [itex]\mathrm{d}x[/itex] in [itex]\frac{\mathrm{d}V}{\mathrm{d}x}[/itex]).

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

[tex]\left ( \textup{d}s \right )^2 = \left ( \textup{d}r \right)^2 + r^2 \; \textup{d}\theta,[/tex]

where the variable [itex]\textup{d}s[/itex] seems to be given more meaning. If [itex]\textup{d}s[/itex] has this value, presumably not any differential could be equated with the right hand side. And I've come across a notation whereby [itex]\textup{d}\Phi[/itex] indicates the gradient of [itex]\Phi[/itex], but perhaps this is a different notational system not meant to be used alongside the notation in which [itex]\mathrm{d}[/itex] 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:
  • #19
Mathematicians seem to come up with all sorts of strange variations on derivatives... but this might be what you're thinking of: [itex]\mathrm{d}\Phi[/itex] 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
[tex]\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[/tex]
But here's the catch: in that expression, [itex]\mathbf{d}x[/itex] etc. are not necessarily differentials; they're basis one-forms, which are (roughly speaking) the equivalents in differential geometry of the basis vectors [tex]\hat{x}[/tex] 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
[tex]\vec{p}\cdot\vec{x} = \sum_j p_j x^j[/tex]
i.e. that the letter doesn't matter, then no, it's not a dummy variable at all. The differential [tex]\mathrm{d}^2\vec{A}[/tex] (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 [tex]\mathrm{d} x[/tex] which is the differential of some length.
 
  • #20
diazona said:
Mathematicians seem to come up with all sorts of strange variations on derivatives... but this might be what you're thinking of: [itex]\mathrm{d}\Phi[/itex] 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
[tex]\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[/tex]
But here's the catch: in that expression, [itex]\mathbf{d}x[/itex] etc. are not necessarily differentials; they're basis one-forms, which are (roughly speaking) the equivalents in differential geometry of the basis vectors [tex]\hat{x}[/tex] 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.

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.

diazona said:
The differential [tex]\mathrm{d}^2\vec{A}[/tex] (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 [tex]\mathrm{d} x[/tex] which is the differential of some length.

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
Rasalhague said:
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?
Actually, it is, in a way. Here's my understanding of ordinary derivatives: basically each d represents one infinitesimal dimension. So for example, [itex]\mathrm{d}^2A[/itex] is the product of two infinitesimal dimensions, such as
[tex]\mathrm{d}^2A = \mathrm{d}x\mathrm{d}y[/tex]
There is also a first differential of area, [itex]\mathrm{d}A[/itex] which would be the product of one finite dimension and one infinitesimal dimension. For example,
[tex]\mathrm{d}A = w\mathrm{d}y[/tex]
With these definitions in mind, the derivative of flux (on a fixed 2D surface) would be more properly written
[tex]\frac{\mathrm{d}^2\Phi}{\mathrm{d}^2A}[/tex]
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
[tex]\frac{\mathrm{d}\Phi}{\mathrm{d}A}[/tex]
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)
Rasalhague said:
(And presumably not an exterior derivative if the exterior derivative d obeys the equation d2 = 0...)
That's right, the exterior derivative is a separate thing from these differentials.
 
  • #22
I suppose missing off the powers of two in this expression is similar to the shorthand convention whereby people write

[tex]\frac{\mathrm{d}^2 y}{\mathrm{d} x^2}[/tex]

for the second derivative rather than

[tex]\frac{\mathrm{d}^2 y - \mathrm{d}^2x \frac{\mathrm{d} y}{\mathrm{d} x}}{\mathrm{d} x^2}.[/tex]
 
  • #23
Rasalhague said:
I suppose missing off the powers of two in this expression is similar to the shorthand convention whereby people write

[tex]\frac{\mathrm{d}^2 y}{\mathrm{d} x^2}[/tex]

for the second derivative rather than

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

1. What is flux and why is it important in science?

Flux is a measure of the flow of a physical quantity through a given area. It is important in science because it helps us understand and quantify the movement of energy, mass, or other physical quantities through a system.

2. How is flux calculated and what are the units of measurement?

Flux is calculated by multiplying the magnitude of the quantity by the area through which it is flowing and then taking the dot product with the normal vector to the area. The units of flux depend on the physical quantity being measured, but can include joules per second for energy flux, mass per second for mass flux, or watts per square meter for radiant flux.

3. What is the difference between flux and flow rate?

Flux and flow rate are related concepts, but they are not the same. Flux is a measure of the flow of a quantity through a given area, while flow rate is a measure of the amount of a quantity that flows through a specific point in a given amount of time. Flux is a rate of flow per unit area, while flow rate is a total amount of flow per unit time.

4. How is the concept of flux applied in different fields of science?

Flux is a fundamental concept that is applied in a wide range of scientific fields, including physics, chemistry, biology, and engineering. It is used to describe the flow of energy, mass, and other physical quantities in systems such as fluid dynamics, electromagnetism, and heat transfer.

5. How can I use flux to test my understanding of a scientific concept?

Flux can be used as a tool to test your understanding of a scientific concept by applying it to real-world situations and problems. By using flux to analyze and quantify the flow of a physical quantity in a system, you can gain a deeper understanding of the underlying principles and relationships at play.

Similar threads

Replies
5
Views
895
Replies
2
Views
610
  • Electromagnetism
Replies
2
Views
938
Replies
2
Views
271
  • Electromagnetism
Replies
30
Views
2K
  • Electromagnetism
Replies
1
Views
1K
Replies
1
Views
2K
Replies
1
Views
1K
Replies
4
Views
1K
Replies
25
Views
4K
Back
Top