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Here is the General mathematical definition of Flux on Wikipedia:

The frequent symbol is $j$, and a definition for scalar flux of physical quantity $q$ is the limit: $$j=\lim\limits_{A\to 0}\cfrac{I}{A}=\cfrac{\mathrm{d}I}{\mathrm{d}A}$$ where: $$I=\lim\limits_{\Delta t\to 0}\cfrac{\Delta q}{\Delta t}=\cfrac{\mathrm{d}q}{\mathrm{d}t}$$ is the flow of quantity $q$ per unit time $t$, and $A$ is the area through which theh quantity flows.For vector flux, the surface intergal of $j$ over surface S, followed by an integral over the time duration $t_1$ to $t_2$, gives the total amount of the property flowing through the surface in that time $(t_2 − t_1)$: $$q=\int_{t_1}^{t_2}\iint_S \mathbf{j}\cdot\hat{\mathbf{n}}\, \mathrm{d}A\mathrm{d}t$$

Although I have already know the relationship between $\mathrm{d}A$ and $\mathrm{d}\mathbf{A}$. However, what's the relationship between $j$ and $\mathbf{j}$? What's the definition of $\mathbf{j}$?

I ask this question because of this: problem.

I don't solve my problem from the two answers. I am not sure about the definition of heat flux. It is necessary for you to write rigorous description to this question.

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The relation between the scalar $j$ and the vector $\vec j$ is simply :

$$j = \vec j \cdot \vec n$$

where $\vec n$ is a unit vector normal to the surface $\mathrm{d}A$.

So, you could write:

$$j ~ \mathrm{d}A = \vec j \cdot \vec n ~ \mathrm{d}A = \vec j \cdot \vec {\mathrm{d}A}$$

with the notation $\vec {\mathrm{d}A} = \vec n ~ \mathrm{d}A$

So, choose the notation that you prefer, but it must be clear for you.

Suppose, for instance, that you are working with a conserved quantity $Q$. We may define a quantity density $\rho$, and a quantity flux $\vec j$, so that the local law of the conservation of the quantity is written:

$$\frac{\partial \rho}{\partial t} + \operatorname{div} \vec j = 0$$

Considering now a volume $V$, with boundary surface $A$. The variation of the quantity $Q$, in the volume V, between 2 times $t_1$ and $t_2$ is:

$$\Delta Q = \int_{t_1}^{t_2} \int_V \frac{\partial \rho(\vec x,t)}{\partial t} \mathrm{d}^3x = - \int_{t_1}^{t_2}\int_V \operatorname{div} \vec j \mathrm{d}^3x = - \int_{t_1}^{t_2} \int_A \vec j.\vec {\mathrm{d}A}$$

The last equality comes from Stokes Theorem.

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I'm not quite sure what you're asking here. The first couple of equations show that the scalar flux $j$ is the current density of the quantity $q$.

The last equation shows that the total amount of some, presumably different quantity, $q$ that has passed through the surface $S$ in some time $t_2-t_1$ is given by the time integral of the flux integral.

I'm not sure what you mean by the question "how j becomes j"

In the scalar case, $j$ is a scalar current density (one component) and in the vector case, j is vector current density (three components).

If this doesn't actually address your actual question, try editing your question to clarify.


What's the definition of j?

Imagine some "stuff" that fills space so that we can define a volume density $\rho$. In other words, the volume integral of $\rho$ is some quantity $Q$ of "stuff".

Imagine that this density $\rho$ changes with time and that, this "stuff" cannot be created or destroyed, it is conserved. That means that the "stuff" is flowing here and there, becoming more concentrated in some volumes and less concentrated in others.

How do we represent and characterize this flow? Think of each infinitesimal volume element of "stuff" as having a velocity so that we can form the product:

$$\vec J = \rho(\vec x) \, \vec v(\vec x)$$

Note this the product, the current density or flux is: (1) a vector and (2) has units of:

$$ \dfrac{\text{(stuff per second)}}{\text{(unit area)}}$$.

The flux integral over a surface $S$ then gives the flow of "stuff" through the surface.

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  • $\begingroup$ I modified my question. $\endgroup$ – Brooks Jun 6 '13 at 21:09
  • $\begingroup$ @Brooks, see my updated answer. $\endgroup$ – Alfred Centauri Jun 6 '13 at 22:59
  • $\begingroup$ In three dimension, it is $A\to 0$ or $V\to 0$? $\endgroup$ – Brooks Jun 7 '13 at 0:33
  • $\begingroup$ @Centauri, you take infinitesimal volume element then add them all together. But you also integral them over a surface $S$? How does it make sense? However, I got your idea. Thanks. $\endgroup$ – Brooks Jun 7 '13 at 0:37
  • $\begingroup$ @Brooks, the flux integral is a surface integral of a current density (flow per area). $\endgroup$ – Alfred Centauri Jun 7 '13 at 0:40
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Well, lets look at this: $$j=\lim\limits_{A\to 0}\cfrac{I}{A}=\cfrac{\mathrm{d}I}{\mathrm{d}A}$$ Now imagine that, while your area $A\to 0$, it simultaneously twists, so that its normal points in some other directions in the process. That wouldn't look like a flux at all, wouldn't it? That way you can see that some crucial detail is missing there.

Namely -- the fixed direction of the normal of the surface $A$: $$\mathbf{j}=\mathbf{n}\cdot\lim\limits_{A\to 0}\cfrac{I}{A}=\mathbf{n}\cdot\cfrac{\mathrm{d}I}{\mathrm{d}A}$$ This expression has all the information necessary. While in the one you quote it is supposed that the direction of the normal (and of the resulting flux) is clear from the context.

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  • $\begingroup$ While area $A\to 0$, it simultaneously twists. But why should I consider the normal points? $\endgroup$ – Brooks Jun 6 '13 at 20:18
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That is not a good definition of $\mathbf{j}$. It is better to think of $\mathbf{j}$ as an average velocity weighted by the phase space density. For simplicity, suppose all of the particles carry the same mass, $m$. If we define $\rho_\phi(\mathbf{x},\mathbf{p})\operatorname{d}^3 x \operatorname{d}^3 p$ as the average number of particles found in the real space volume $\operatorname{d}^3x$ and momentum space volume $\operatorname{d}^3p$ centered on the point $(\mathbf{x},\mathbf{p})$, then the current density, $\mathbf{j}$, is given by the integral: $$\mathbf{j}(\mathbf{x}) \equiv \int \frac{\mathbf{p}}{m} \rho_\phi(\mathbf{x},\mathbf{p}) \operatorname{d}^3p.$$ Notice that the average velocity at $\mathbf{x}$ is defined by: $$\langle \mathbf{v}(\mathbf{x})\rangle \equiv \frac{\int\frac{\mathbf{p}}{m} \rho_\phi(\mathbf{x},\mathbf{p}) \operatorname{d}^3p}{\int \rho_\phi(\mathbf{x},\mathbf{p}) \operatorname{d}^3p},$$ but the denominator of that expression is just the number density at $\mathbf{x}$, usually denoted $n(\mathbf{x})$. Thus: $\mathbf{j}(\mathbf{x}) = \langle \mathbf{v}(\mathbf{x})\rangle\, n(\mathbf{x}).$

There are a lot of complications that can be thrown in to this picture, depending on the type of particle(s) you're dealing with: heterogeneous masses, charges (for current densities), and the Pauli exclusion principle (for when only one particle can be at any given $\mathbf{x}$ at a time). In the first case, you just add sums over the different types of particles. In the second case, you transition to a probabilistic picture, but the idea that these quantities are some sort of averages remains the same.

Now, the flux, $I$, can be almost unambiguously defined as the net flow through an open surface, $A$, by the integral: $$I = \int_A \mathbf{j}(\mathbf{x}) \cdot \hat{n}(\mathbf{x}) \operatorname{d}^2 x,$$ where $\hat{n}(\mathbf{x})$ is a unit surface normal at the position $\mathbf{x}$. Note that the overall direction $\hat{n}$ points is ambiguous - it's usually defined by using the right hand rule and defining a path around the edge of $\mathbf{A}$ as clockwise. That makes $\hat{n}$ a pseudovector, and thus $I$ is a pseudoscalar. The "vectorness" attributed to $I$ is derived from the ambiguity in $\hat{n}$ and the need to keep in mind how $\hat{n}$ was defined to make physical sense of $I$.

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