# Component of vector parallel to boundary while calculating divergence

1. Nov 5, 2013

### Urmi Roy

So when we calculate divergence (especially referring to the gauss divergence theorem), why aren't the components of the vector field parallel to the boundary considered?

I mean even of, say fluid, is travelling parallel to the boundary when it comes out, fluid is exiting, or diverging out, right?

2. Nov 5, 2013

### UltrafastPED

The parallel portion of a flow has no divergence ... the same amount flows in as flows out.

Hence the name: divergence - a change in density or intensity with distance.
The details are covered here: http://mathworld.wolfram.com/Divergence.html

3. Nov 5, 2013

### Urmi Roy

Well from an intuitive point of view, if I imagine fluid scattering out of a boundary, some of it might scatter out tangentially from the boundary and just travel out tangentially...then what goes out, stays out.

4. Nov 5, 2013

### UltrafastPED

The pipe would be a source of water to the local space ... so the divergence at the end of a pipe would be non-zero. You can map out the flow fields with a high speed camera.

5. Nov 6, 2013

### Staff: Mentor

Suppose that you have rain falling vertically downward, and you have a vertical window. How much rain goes through the window? Now suppose you again you have rain falling vertically downward, and you have a horizontal window. The rate of rain flow through the window is equal to the downward velocity of the rain times the volumetric concentration (i.e., number concentration) of raindrops, times the area of the window. Now suppose you again have rain falling vertically downward and the window is oriented at an angle θ to the vertical (i.e., the normal to the window makes and angle θ with the vertical). The rate of rain flow through the window is equal to the downward velocity of the rain times the volumetric concentration of raindrops, times the area of the window, times the cosine of the angle θ. But, the downward velocity of the rain times the cosine of the angle θ is equal to the component of the rain velocity normal to the window. Just as in the case of a vertical window, the component of the velocity parallel to the window makes no contribution to the rate of flow of rain through the window.

Chet

6. Nov 15, 2013

### Urmi Roy

Chestermiller, what you said makes sense to me except in the instance I've shown in the diagram...maybe I'm missing something, but it seems that in this case, even the tangential component goes out, never to return inside the region again!

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7. Nov 15, 2013

### Staff: Mentor

Let N represent the number of particles per unit volume, and let $\vec{v}$ represent the velocity vector of the particles. Then the so-called flux of particles is $N\vec{v}$, and the number of particles per unit time passing through an element of area dS oriented normal to the velocity vector is NvdS. But, if the surface is not oriented perpendicular to the velocity vector, the number of particles per unit time passing through the element of area is $N(\vec{v}\centerdot \vec{n})dS$, where $\vec{n}$ is a unit vector normal to the surface element dS. In your diagram, if there is a circular arc rΔθ with a normal vector $\vec{i_r}$ and a velocity vector $\vec{v}=v_r\vec{i_r}+v_θ\vec{i_θ}$, the velocity vector dotted with the normal vector is $\vec{v}\centerdot \vec{n}=(v_r\vec{i_r}+v_θ\vec{i_θ})\centerdot \vec{i_r}=v_r$. The number of particles per unit time passing across the arc (per unit depth into the page) is $N(\vec{v}\centerdot \vec{n})rΔθ=v_rrΔθ$. Notice that the tangential component makes negligible contribution to the flow through the arc. This is because, if you have a differential element of area, even if the surface is curved, on the scale of the differential element, it is virtually flat. The effect of the curvature along a finite length of arc is picked up by the normal component.

8. Nov 16, 2013

### HallsofIvy

Staff Emeritus
That is your error. If the fluid is moving "tangential to the boundary" it cannot cross the boundary so cannot come out or go in.