The Unapologetic Mathematician

Mathematics for the interested outsider

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.

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

6 Comments »

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

    Pingback by Path-Independent 1-Forms are Exact « The Unapologetic Mathematician | December 15, 2011 | Reply

  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 [...]

    Pingback by Simply-Connected Spaces and Cohomology « The Unapologetic Mathematician | December 17, 2011 | Reply

  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: [...]

    Pingback by Gauss’ Law for Magnetism « The Unapologetic Mathematician | January 12, 2012 | Reply

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

    Pingback by Electromotive Force « The Unapologetic Mathematician | January 13, 2012 | Reply

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

    Pingback by Faraday’s Law « The Unapologetic Mathematician | January 14, 2012 | Reply

  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 [...]

    Pingback by The Higgs Mechanism part 1: Lagrangians « The Unapologetic Mathematician | July 16, 2012 | Reply


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