Talk:PlanetPhysics/Flux of Vector Field

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Let $$\vec{U} \;=\; U_x\vec{i}+U_y\vec{j}+U_z\vec{k}$$ be a \htmladdnormallink{vector field}{http://planetphysics.us/encyclopedia/NeutrinoRestMass.html} in $\mathbb{R}^3$\, and let $a$ be a portion of some surface in the vector field.\, Define one side of $a$ to be positive; if $a$ is a closed surface, then the positive side must be the outer surface of it.\, For any surface element $da$ of $a$, the corresponding {\em vectoral surface element} is $$d\vec{a} \;=\; \vec{n}\,da,$$ where $\vec{n}$ is the unit normal \htmladdnormallink{vector}{http://planetphysics.us/encyclopedia/Vectors.html} on the positive side of $da$.

The {\em \htmladdnormallink{flux}{http://planetphysics.us/encyclopedia/AbsoluteMagnitude.html}} of the vector $\vec{U}$ through the surface $a$ is the surface integral $$\int_a\vec{U} \cdot d\vec{a}.$$\\

\textbf{Remark.}\, One can imagine that $\vec{U}$ represents the \htmladdnormallink{velocity}{http://planetphysics.us/encyclopedia/Velocity.html} vector of a flowing liquid; suppose that the flow is stationary, i.e. the velocity $\vec{U}$ depends only on the location, not on the time.\, Then the \htmladdnormallink{scalar product}{http://planetphysics.us/encyclopedia/DotProduct.html} $\vec{U} \cdot d\vec{a}$ is the \htmladdnormallink{volume}{http://planetphysics.us/encyclopedia/Volume.html} of the liquid flown per time-unit through the surface element $da$; it is positive or negative depending on whether the flow is from the negative side to the positive side or contrarily.

\textbf{Example.}\, Let\, $\vec{U} = x\vec{i}+2y\vec{j}+3z\vec{k}$\, and $a$ be the portion of the plane \,$x+y+x = 1$\, in the first octant ($x \geqq 0,\; y \geqq 0,\, z \geqq 0$) with the positive normal away from the origin.

One has the constant unit normal vector: $$\vec{n} \;=\; \frac{1}{\sqrt{3}}\vec{i}+\frac{1}{\sqrt{3}}\vec{j}+\frac{1}{\sqrt{3}}\vec{k}.$$ The flux of $\vec{U}$ through $a$ is $$\varphi \;=\; \int_a\vec{U}\cdot d\vec{a} \;=\; \frac{1}{\sqrt{3}}\int_a(x+2y+3z)\,da.$$

However, this surface integral may be converted to one in which $a$ is replaced by its projection $A$ on the $xy$-plane, and $da$ is then similarly replaced by its projection $dA$; $$dA = \cos\alpha\, da$$ where $\alpha$ is the angle between the normals of both surface elements, i.e. the angle between $\vec{n}$ and $\vec{k}$: $$\cos\alpha \;=\; \vec{n}\cdot\vec{k} \;=\; \frac{1}{\sqrt{3}}.$$ Then we also express $z$ on $a$ with the coordinates $x$ and $y$: $$\varphi \;=\; \frac{1}{\sqrt{3}}\int_A(x+2y+3(1-x-y))\,\sqrt{3}\,dA \;=\; \int_0^1\left(\int_0^{1-x}(3-2x-y)\,dy\right)dx \;=\; 1$$

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