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 John Armstrong | Differential Geometry, Geometry

6 Comments »

[…] 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
[…] 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
[…] 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
[…] 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
[…] 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
[…] 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

About this weblog

This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.

RSS Feeds

RSS - Posts
RSS - Comments
Feedback

Got something to say? Anonymous questions, comments, and suggestions at Formspring.me!
Subjects

Subjects
Archives

December 2011

M T W T F S S

1 2 3 4

5 6 7 8 9 10 11

12 13 14 15 16 17 18

19 20 21 22 23 24 25

26 27 28 29 30 31

« Nov Jan »
Search for:

The Unapologetic Mathematician

Mathematics for the interested outsider