# Homework Help: Flux and Divergence

1. Sep 14, 2013

### thatguy14

1. The problem statement, all variables and given/known data

This is a coursework problem. I am having issues understanding the concepts on this one topic - divergence and how it relates to flux. I have attached screenshots that honestly give the best representation of my issue but I will set up the issue I am having none-the-less:

We are trying to find an equation for the divergence. He gives that div v = lim τ->0 $\frac{\ointvdot da}{Δτ}$. So then he sets up a rectangular prism. He didn't go through this with us as he just posted in notes scanned. So v is some vector function da is the normal area to the right plane. In the image he also puts a bunch of points. I guess he choose an x,y,z point contained in the rectangular prism (FIRST QUESTION: Does the x,y,z coordinate location matter? Could I have done it in one of the quarters of the shape rather than in the middle?). He then talks about how x,y,z vary in that right plane.

SECOND QUESTION: why does he say "Since we are going to be multiplying by Δx and Δz, any variation in x and z in these ranges will give rise to (Δx)^2 or (Δz)^2 terms, which we can ignore.

He then sets up v dot da as the y component of it (parallel to the normal of that plane.

I think I should leave it there actually because I feel once I understand the second question, that will give me clues to what is going on. I have been trying to figure this out for quite sometime now and I am lost on how he went from the v-da to the stuff on page 22. If you look later on page 22 at the rectanglular prism that is sliced on a diagonal that is a homework problem we have to do (show that it doesn't matter what shape we choose). I want to try that on my own but I need to know how he set his up first. Any help would be greatly appreciated

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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2. Sep 14, 2013

### vela

Staff Emeritus
Yes, you could have located (x,y,z) at one of the corners. If you go through the derivation with this assumption, you'll get the same result.

Concentrating on just the x-direction for now, we have to first order that the quantity $v_y$ will be some value plus a deviation proportional to $\Delta x$.
$$v_y(x+\Delta x/2, y+\Delta y/2, z) = v_y(x, y+\Delta y/2, z) + \alpha \Delta x$$ This is just a Taylor expansion. When you multiply by $\Delta x$, you get
$$v_y(x+\Delta x/2, y+\Delta y/2, z) \Delta x = v_y(x, y+\Delta y/2, z)\Delta x + \alpha (\Delta x)^2.$$ In the limit $\Delta x \to 0$, the second-order term become negligible compared to the first-order term. That's why you can just drop it right from the start. The same thing happens if you consider what happens in the z-direction.

Last edited: Sep 14, 2013
3. Sep 14, 2013

### thatguy14

Unfortunatly I am still confused. I guess I am incredibly rusty with my calculus. In that first step you did there, could you explain to me how those are equal or reference me to something that tells me what you did there? Maybe I am over thinking it.

I should ask, why does he even say that instead of saying that since we are dotting v with j (the normal) the other terms fall out so to speak? Is that incorrect to say?

Last edited: Sep 14, 2013
4. Sep 14, 2013

### vela

Staff Emeritus
http://en.wikipedia.org/wiki/Taylor_series

Note that above the two sides aren't strictly equal because I didn't bother to write the higher-order terms. You could tack 'em on, but they'll disappear as well in the limit.

5. Sep 14, 2013

### thatguy14

Sorry not the taylor series thing, what rule allows you to bring out the deltax/2 from the vector?

6. Sep 14, 2013

### TSny

If you want to be meticulous about the approximations, you can approach it as follows.

The flux through the right face is given exactly by

$$\Phi = \int_{z-\Delta z/2}^{z+\Delta z/2} \int_{x-\Delta x/2}^{x+\Delta x/2} V_y(x',y+\Delta y/2, z')\,dx' \,dz'$$

Introduce new variables of integration $u = x' - x$ and $v = z'-z$. So,

$$\Phi = \int_{-\Delta z/2}^{\Delta z/2} \int_{-\Delta x/2}^{\Delta x/2} V_y(x+u,y+\Delta y/2, z+v)\,du \,dv$$

Now, follow vela and Taylor expand the integrand about the center point $(x,y+\Delta y/2, z)$

$$V_y(x+u,y+\Delta y/2, z+v) = V_y(x,y+\Delta y/2, z) + \frac{\partial V_y}{\partial x}(x,y+\Delta y/2, z)\,u + \frac{\partial V_y}{\partial z}(x,y+\Delta y/2, z)\,v + \cdots$$

The first term of the expansion will lead to the desired result. The second and third terms of the expansion actually yield zero because

$$\int_{-\Delta x/2}^{\Delta x/2}u \,du = 0 \;\;\text{and}\; \int_{-\Delta z/2}^{\Delta z/2} v \,dv = 0$$

The next higher order terms in the expansion will be quadratic in $u$ and $v$. The $u^2$ term will lead to a contribution to the flux proportional to

$$\int_{-\Delta x/2}^{\Delta x/2}u^2 \,du = \frac{(\Delta x)^3}{12}$$.

And so forth. These higher order terms will go to zero when taking the final limit.

7. Sep 14, 2013

### vela

Staff Emeritus
That is the Taylor expansion.
$$f(w) = f(w_0) + f'(w_0)(w-w_0) + \cdots$$ In the derivation you have $w = x + \frac{\Delta x}{2}$ and $w_0 = x$ so that
$$f(x + \frac{\Delta x}{2}) = f(x) + f'(x)\frac{\Delta x}{2}+\cdots$$

8. Sep 14, 2013

### thatguy14

Is there a resource you can provide where I can read more about taylor series expansions of vectors? I see now that I am quite behind in my understanding.

Edit: You guys have been great. I can't believe so much knowledge was assumed when my professor was writing the notes. This is a 3rd year E+M course so I didn't think it would be that involved. I am going to take a break for a bit at this point but I will be coming back to my assignment and I am sure I will need help with it you guys don't mind.

Thanks again

9. Sep 14, 2013

### thatguy14

Hey guys, quick question, going back to TSny's response, in his taylor expansion he has 2 first derivatives and this is because he skipped a step and just did a taylor expansion with respect to x and z at the same time correct? Sorry if that's a silly question.

10. Sep 14, 2013

### vela

Staff Emeritus
11. Sep 16, 2013

### thatguy14

Another question vela (you both have been great help but I wanted to focus on the one dimensional analysis for now):

in post 7 you wrote f(x + delta x) and then (I think I am going to say this right), you evaluated around the center point x. Why did you start off with f(x + delta x) (which represents the bottom left hand corner of the face) and not somewhere else? What am I missing here?