2
$\begingroup$

Green's Theorem gives that the flux on a vector field $\mathbf{\vec F}$ over a closed curve C is equal to the double integral over the enclosed region of C of the divergence of $\mathbf{\vec F}$ (provided the region is continuously differentiable), namely, $\oint_C\mathbf{\vec F} \cdot \mathbf{\hat n}\,ds = \iint_R \left(\frac{\partial M}{\partial x} + \frac{\partial N}{\partial y}\right)\,dA$, where $\mathbf{\vec F} = M(x,y) \mathbf{\hat i} + N(x,y) \mathbf{\hat j}$ and represents a velocity field (fluid flow field).

The intuition proceeds to explain the integrand $\left(\frac{\partial M}{\partial x} + \frac{\partial N}{\partial y}\right)\,dA$ as follows. Take a very small rectangle in the region R, where the bottom left of the rectangle has coordinates $(x,y)$, and which has a width and height of $\Delta x$ and $\Delta y$, respectively. Calculating the flux across the boundary of this rectangle (going in the counterclockwise direction) gives approximately the integrand in question.

To see this, consider the top of the rectangle. The flux across this piece will be approximately $\mathbf{\vec F}(x,y+\Delta y) \cdot \mathbf{\vec j}\,\Delta x = N(x,y+\Delta y)\Delta x$. We've assumed that the rectangle is small enough to where $\mathbf{\vec F}$ is constant across the top of the rectangle, when in reality it would change, so instead of taking the integral at each point to calculate net flux, we just use the dot product to calculate the area of the parallelogram.

Similarly, the flux across the bottom of the rectangle is approximately $\mathbf{\vec F}(x,y) \cdot -\mathbf{\vec j}\,\Delta x = -N(x,y) \Delta x$, again assuming that $\mathbf{\vec F}(x+\Delta x, y)$, and everywhere in between, is the same as $\mathbf{\vec F}(x,y)$.

Adding the two fluxes, we get that the flux is approximately $(N(x,y+\Delta y) - N(x,y))\Delta x$. Since $\frac{N(x,y+\Delta y) - N(x,y)}{\Delta y} \approx \frac{\partial N}{\partial y}$, we can rewrite the flux as approximately $(\frac{\partial N}{\partial y}\Delta y)\Delta x$.

Similar reasoning shows that the flux across the left and right sides of the rectangle is approximately $(\frac{\partial M}{\partial x}\Delta x)\Delta y$. Therefore, the total flux is approximately the sum of these two effects, namely, $(\frac{\partial M}{\partial x} + \frac{\partial N}{\partial y})\Delta x \Delta y$, which gives the integrand.

Here is the part that I don't understand:

Why is it that the flux across the top is given by

$\mathbf{\vec F}(x,y+\Delta y) \cdot \mathbf{\vec j}\,\Delta x = N(x,y+\Delta y)\Delta x$?

I don't understand the dot product fully. I understand that $\mathbf{\vec F}(x,y+\Delta y) \cdot \mathbf{\vec j}$ gives us the Component of the Vector field F in the J direction.

I'm wondering why we have to multiple this dot product of $\mathbf{\vec F}(x,y+\Delta y) \cdot \mathbf{\vec j}$ another time by $\Delta x$??

What does it mean to multiply the component of the vector field in a specific direction by some really small number $\Delta x$?

Are we saying that the longer the edge of the rectangle, the greater the flux along that edge? Because when $\Delta x \rightarrow \infty$, then the flux across that edge must be $\infty $? Why is the flux proportional to the length of the rectangle's sides? Why does flux have anything to do with the size of the rectangle? Assume we're talking about the rate of fluids.

What are the units of $\mathbf{\vec F}(x,y+\Delta y)$ in the physics sense?

What are the units of $\mathbf{\vec F}(x,y+\Delta y) \cdot \mathbf{\vec j}$ ?

What are the units of $\mathbf{\vec F}(x,y+\Delta y) \cdot \mathbf{\vec j}\,\Delta x$?

When we talk about $\mathbf{\vec F}(x,y+\Delta y)$, is $(x,y)$ some arbitrary point on the boundary of the rectangle either on the bottom, the top, the right, or left?

$\endgroup$
2
  • $\begingroup$ and I should also replace every instance of flux with flow. $\endgroup$
    – Cake
    Nov 18, 2017 at 23:11
  • $\begingroup$ I do intend to talk about green's theorem, so the flux and incorrect things are mistakes. $\endgroup$
    – Cake
    Nov 18, 2017 at 23:12

1 Answer 1

0
$\begingroup$

The statement you copied is wrong when it says that $\mathbf F$ is a velocity.

The units of $\mathbf F$ are "whatever"/distance, where "whatever" is the substance that flows. Note that $\mathbf F$ is a vector.

When you have $\mathbf F\cdot \mathbf j$, now you get a scalar, still of the form substance/distance.

When you multiply by $\Delta x$, you get the amount of substance that is flowing over such length.

The rectangle with vertices $(x,y)$, $(x+\Delta x, y)$, $(x,y+\Delta y)$, $(x+\Delta x,y+\Delta y)$ is anywhere in $R$. The idea is that to measure the flux you measure over all these small rectangles that cover $R$, and the fluxes on adjacent sides of touching rectangles will cancel each other when you sum.

Finally, of course that the longer the side, the more substance is flowing through that border. Think of a river; if you fix a net across the river, the bigger the net, the more water flows across it.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .