# The Unapologetic Mathematician

## Conservative Vector Fields

For a moment, let’s return to the case of Riemannian manifolds; the vector field analogue of an exact $1$-form $\omega=df$ is called a “conservative” vector field $X=\nabla f$, which is the gradient of some function $f$.

Now, “conservative” is not meant in any political sense. To the contrary: integration is easy with conservative vector fields. Indeed, if we have a curve $c$ that starts at a point $p$ and ends at a point $q$ then fundamental theorem of line integrals makes it easy to calculate:

$\displaystyle\int\limits_c\nabla f\cdot ds=f(q)-f(p)$

I didn’t go into this before, but the really interesting thing here is that this means that line integrals of conservative vector fields are independent of the path we integrate along. As a special case, the integral around any closed curve — where $q=p$ — is automatically zero. The application of such line integrals to calculating the change of energy of a point moving through a field of force explains the term “conservative”; the line integral gives the change of energy, and whenever we return to our starting point energy is unchanged — “conserved” — by a conservative force field.

This suggests that it might actually be more appropriate to say that a vector field is conservative if it satisfies this condition on closed loops; I say that this is actually the same thing as our original definition. That is, a vector field is conservative — the gradient of some function — if and only if its line integral around any closed curve is automatically zero.

As a first step back the other way, it’s easy to see that this condition implies path-independence: if $c_1$ and $c_2$ go between the same two points — if $c_1(0)=c_2(0)$ and $c_1(1)=c_2(1)$ — then

$\displaystyle\int\limits_{c_1}X\cdot ds=\int\limits_{c_2}X\cdot ds$

Indeed, the formal sum $c_1-c_2$ is a closed curve, since $\partial(c_1-c_2)=\partial c_1-\partial c_2=0$, and so

$\displaystyle\int\limits_{c_1}X\cdot ds-\int\limits_{c_2}X\cdot ds=\int\limits_{c_1-c_2}X\cdot ds=0$

Of course, this also gives rise to a parallel — and equivalent — assertion about $1$-forms: if the integral of $\omega$ around any closed $1$-chain is always zero, then $\omega=df$ for some function $f$. Since we can state this even in the general, non-Riemannian case, we will prove this one instead.

December 15, 2011 - Posted by | Differential Geometry, Geometry

1. […] we prove the assertion from last time: if is a -form on a manifold such that for every closed curve we […]

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2. […] any closed curve . But this means that every closed -form is path-independent, and path-independent -forms are exact. And so we conclude that , as […]

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3. […] Last time we spent a while noting that the fraction here is secretly a closed -form in disguise, so its divergence is zero. This time, I say it’s actually a conservative vector field: […]

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4. […] is conservative, this amounts to the difference in “potential energy” between the start and end of the […]

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5. […] this electric field is conservative, and so its integral around the closed circuit is automatically zero. Thus there is no […]

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6. […] we can write it down in a formula: . And for most forces we’re interested in the force is a conservative vector field, meaning that it’s the (negative) gradient (fancy word for […]

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