## Sunday, August 19, 2012

### Conservative Fields: Theory

Next stop: Vector Calculus!

To continue in the present direction--a discussion of Energy methods in Particle Dynamics--it’s necessary to take a quick stop over in mathematician territory. An unfriendly place, it is. Look alive.

The basic idea that will recur in this discussion is that, in certain types of vector fields, the net work done on an object moving from one point to another is independent of the path taken. This is called path-independence, and is a very important property of conservative fields

Conceptually, I like to think of a river:

 Wisconsin is great, but the morning commute can get tiresome.
Yeah, there's one. The path you take in swimming across this river has no effect on how much work you do: Maybe I swim straight across, fighting the current the whole way. Maybe you swim far upstream, float downstream, and step out at the same point I did. The river has done the same amount of net work on both of us, regardless of our paths. In this context, it's easy to see what's going on. Things can get trickier, though.

Conservative fields are both physically and mathematically important, and there are a lot of ways to answer the question, "What is a conservative field?" These competing definitions are all equivalent, but can get confusing. I'll start with my personal favorite: A conservative field is irrotational, or contains no vortices. I like this because it makes physical sense. The whole idea of conservative fields and forces is that the direction you travel through them does not matter. Imagine a whirlpool: Swimming in circles with the current will give you a free ride. Swimming in circles against the current will cost you. A whirlpool cannot be conservative.

Mathematically, this is equivalent to saying "If F is conservative, the curl of F is zero:"

$\overrightarrow{\bigtriangledown} \times \overrightarrow{F} = \overrightarrow{0}$

Carrying out the cross product leads to a 3-part component test for a vector (Mi + Nj + Pk):

For 2-dimensional vectors in the x-y plane, only the third equality need be tested.

So what now? If the vector field passes this component test, we know it is conservative, and we know that the work through that field is path-independent. Specifically, the Fundamental Theorem of Line Integrals, (much like the Fundamental Theorem of Calculus), states that:

For conservative F. What about little f, though? Well, it's obviously a scalar function. In fact, the vector field F is actually the gradient of scalar f, and so we call little f the scalar potential of a vector field F.

I don't know who was the genius who decided to relate capital and lowercase F like this. It's difficult enough to read, it's even worse trying to explain verbally. I often refer to f as the "integral" of F; even though that's not precise, it's the same idea.

Anyway, finding a vector field's potential function is fairly straightforward, I'm not going into it here. Probably later. The important thing is that, much like standard one-dimensional differentials from elementary calculus, the value of the conservative work function depends only on two endpoints--this leads to the loop property, which state: "Travelling on a closed loop through a conservative field yields no net work."

$\oint\overrightarrow{F} \cdot d\overrightarrow{r}= 0$

So, to review:

- Work in a conservative field is path-independent.

- We can check for the conservancy of a field with a 3-part component test. (Single-component in 2 dimensions).

- If a field is conservative, then it must also have a potential function, which is a scalar function. The field in question will be that function's gradient.

- The net work can then be evaluated by taking values in that potential function between the two endpoints we're interested in.