Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Vector Calculus: What do these terms mean?

  1. Dec 2, 2016 #1
    In our section on path independence, we were asked to find the potential function given a vector field. Our teacher says to use only line integrals to find the potential function, and not any other method. Like if we have ##F=\left< M,N,P \right> ## The first step is to determine if the domain is open/concave, then determine if M, N, P are "closed," and then determine if they are "exact." What do those terms mean?
  2. jcsd
  3. Dec 2, 2016 #2
    A vector field in 3-dimensional space is closed if its curl is 0. It is a theorem that a vector field over a simply connected domain is conservative (line integrals over that field are path independent) if and only if it is closed. (Simply connected just means there are no holes). If line integrals over the field are path independent, this implies the vector field is the derivative (gradient) of some scalar function, called the potential function.

    This implication follows because it implies the value of a line integral between two points depends only on the two points, not the path between them. Therefore, we may assign some value F(p) to a particular fixed point p of our choice, then assign the value F(q) to each other point q by using the value of any line integral between p and q, and equating it to F(q) - F(p), where we assign the appropriate value to F(q) to make this equation true. Note that the initial value we assign to F(p) is irrelevant: this arbitrary assignment takes the place of the "arbitrary constant of integration + C" we are used to in single-variable integrals of scalar functions.
    The fact that the gradient of any such created function F will give us back the original vector field is not obvious, but is a fundamental theorem demonstrated in each text on vector calculus. Try to prove it yourself, if you have the time.

    A vector field is exact if it is the derivative (gradient) of some scalar function. That is, if V is a vector field, and f is any scalar function that solves the equation [itex]V = \nabla f[/itex], then we say that V is exact. It is a theorem in each text on vector calculus that every closed vector field defined over a simply connected domain is exact.

    The function f is called a potential function because we can draw surfaces of constant value of f in space (surfaces defined by the equation f(x, y, z) = C for any constant C), and then ascertain that whatever the line integral of V represents is only non-zero when we move from one constant surface to a different one. Line integrals are usually used to determine work. If V is a gravitational field, for example, then these surfaces would represent surfaces of constant gravitational potential (movement within the surface yields a line integral of 0, so no work is being done). Changes in the value of f, associated with movement between different surfaces of constant potential, therefore correspond to changes in gravitational potential energy, and so forth. Electric and magnetic fields have the same setup. However, it is important not to equate the value of f with the potential energy of the particle: the values of f are arbitrary, as we determined above: but the changes in f are real changes in the value of the line integral.

    The terms "closed" and "exact" come from the study of differential equations and differential forms, which has a more general language to handle any amount of variables and dimensions (curl is only defined in 3 dimensions).
    Last edited: Dec 2, 2016
  4. Dec 3, 2016 #3
    Sory if I'm missing something, it seems like curl = 0 (closed) already mean that the there exists a potential function? Why do we need exactness? Also, what is simply-connected's connection with open/convex?
  5. Dec 3, 2016 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member
    2017 Award

    This is true locally in a contractible region (locally closed forms are exact - this is Poincaré's lemma). This need not be true if your region is not simply connected (essentially, if it has holes).

    Because if your domain has holes, exactness and closedness are not necessarily equivalent.
  6. Dec 3, 2016 #5
    I see, so according to the steps my teacher outlined, if the domain is open/convex, and the curl=0, then we have both closed and exact. I remember open means that there is no boundary, and convex mean connected in a straight line... what does that have to do with the potential problem?
  7. Dec 3, 2016 #6
    Unfortunately, it does not! A closed vector field only implies that it is an exact vector field if its domain is simply connected. A closed vector field defined over a domain with a hole, such as a vector field which is undefined at the origin, is not exact.
    In addition, there exist exact vector fields which are not closed. So closed and exact are not synonyms.

    As an example of the former case, consider the vector field that assigns the unit vector
    [tex]V(x, y, z) = \left\langle \frac{-y}{\sqrt{x^2 + y^2 + z^2}}, \frac{x}{\sqrt{x^2 + y^2 + z^2}}, \frac{z}{\sqrt{x^2 + y^2 + z^2}} \right\rangle[/tex]
    to each point (x, y, z) in 3-dimensional space, except the origin, where it is undefined. The domain of this vector field is therefore not simply connected, but its curl is 0 over its entire domain. So this is a closed vector field. However, it is not the gradient of any scalar function, so it is not exact. To see this, we note that being exact implies being conservative, as discussed in the previous reply. Therefore, line integrals over paths in an exact field must depend only on the endpoints, p and q, of the path, not the path itself. This implies that line integrals over a closed path, where the endpoints are the same point, must be 0 = F(p) - F(p).
    However, if we take a line integral of the above vector field over the unit circle, a closed path in the xy-plane, we get:
    [tex]\begin{align*}\int_0^{2\pi} \langle -\sin t, \cos t, 0\rangle \cdot \langle -\sin t, \cos t, 0\rangle \, dt &= \int_0^{2\pi} \sin^2 t + \cos^2 t \, dt\\
    &= \int_0^{2\pi} \, dt\\
    &= 2\pi
    Since this is not 0, the field is not conservative, and therefore cannot be exact (the contrapositive implication of the previous proposition).

    As an example of the latter case, consider the vector field

    [tex] V(x, y, z) = \begin{cases} \left\langle y\frac{x^2-y^2}{x^2+y^2}+\frac{4x^2y^3}{(x^2+y^2)^2}, x\frac{x^2-y^2}{x^2+y^2}-\frac{4x^3y^2}{(x^2+y^2)^2}, 2z \right\rangle, & x^2 + y^2 \neq 0 \\ \langle 0, 0, 2z\rangle, & x = y = 0\end{cases}[/tex]

    The curl of this vector field at the origin is [itex]\langle 0, 0, 2\rangle[/itex], so this vector field is not closed. However, this vector field is the gradient of the scalar function
    [tex] f(x, y, z) = \begin{cases} xy\frac{x^2 - y^2}{x^2 + y^2} + z^2, & x^2 + y^2 \neq 0\\ z^2, & x = y = 0\end{cases}[/tex], and is therefore an exact vector field that is not closed.
    Thus, we need to pay close attention to the directions of the implications in the theorems that we use: there are many pathological examples where the converse of the theorem's proposition is false!

    As for open/convex, these are conditions that guarantee the applicability of Poincaré's lemma, which states that over any contractible region, a closed vector field is synonymous with an exact vector field. So if you verify that the domain of the vector field is both open and convex, then verify that the vector field is closed by computing its curl, you have implied that said vector field is also exact, without having to do the work of trying to find the potential just yet. Imagine not knowing this lemma. Then we would have to do the hard work of trying to determine a potential function without ever knowing whether one even exists or not!
    Last edited: Dec 3, 2016
  8. Dec 3, 2016 #7
    Intuitively, what is the idea for all this? For example, having a domain be both simply connected and convex means that we can take any two points and take the path joining them right? Because having holes mean we can't connect some points, and convex means we have to go out of the region. if "open" means the domain doesn't have a boundary, what does this have to do with exactness?
  9. Dec 4, 2016 #8
    Not quite! We can also take any two points in 3-dimensional space without the origin and join them with a path that avoids the hole at the origin. Simply connected means more than just being able to join every pair of points with paths: it also means the absence of holes. Holes can be detected intuitively by considering any point in the space and drawing a collection of non-intersecting curves from that point to the boundary, making a type of star-shaped space. If there is no boundary, the curves may go to an arbitrary ball centered at the point. For every such ball, we must be able to continuously contract the ball to the center point by pulling the curves we drew inward, like fishing line. In the 3-dimensional space with the origin missing, once we make any ball that contains the hole at the origin, the ball gets caught on the hole when we try to shrink it, because the fishing lines we attached to the ball's surface are trying to rip the ball apart to either side of the hole, since none of them pass through the hole. Thus, the space is not contractible.
    These fishing line curves that we drew are the line integral paths that define the values of the potential function centered at our chosen point. Pulling the fishing lines inward means giving a continuous value of the potential function for every point along each curve. The hole introduces a discontinuity: intuitively, as we get nearer to the hole, our assigned values could diverge. There is no way to preserve the ball shape: a cone-shaped "hole" would have to appear in order to shrink the ball any further.
    The contractibility criterion is more properly referred to as the existence of a differentiable homotopy from each ball to the center point. By manipulating the properties of differentiability, Poincaré's lemma is able to associate the existence of such a homotopy with the ability to integrate any closed form in the space. The exact details are rather laborious, but can be found in any text on analysis/calculus on manifolds. (A short proof is here: http://math.stanford.edu/~dlitt/exposnotes/poincare_lemma.pdf )
    The original form of Poincaré's lemma is demonstrated for star-shaped regions, where the curves connecting the center of the star to the boundary are straight lines. This type of region is a more general form of convex: all convex regions are star-shaped, but there are star-shaped regions that are not convex. The lemma can be generalized to regions that are looser than star-shaped as well (simply connected regions), but in your text, they rely on the simpler criterion of convexity, which is easier for students to identify without going too deeply into matters of topology.
    The "open" criterion just means that each point in the space is contained in a ball of some non-zero radius that also only contains points from the space. This ball allows us to know that the derivative at the point is defined in the usual way, as the derivative is a limit of a particular quantity defined in the space as the radius of this ball approaches 0. Points on a boundary of a space contain points that are not elements of the space in which the vector field is defined, and as such it becomes very fiddly talking about how to define derivatives there when there are points arbitrarily close to our point for which the derivative's expression is undefined. So we ignore these fiddly boundary points and prefer to deal with points inside the space where the vector field, and therefore the derivative, is well defined in all directions.
    Last edited: Dec 4, 2016
  10. Dec 4, 2016 #9
    Oh, so open has to do with limits: the region must have no boundary, or else we can't apply the definition of a limit right? So first determine the region is open and concave, then make sure the partial derivatives are equal each other, then we have exactness, so we can use a line integral right? Let's say ##F(x,y)=\left< y{ e }^{ x }+siny,{ e }^{ x }+xcosy \right> ## We have open/concave and closed, that implies exactness. ##\alpha (s)=\left< s,0 \right> \quad \dot { \alpha } =\left< 1,0 \right> \quad 0\le s\le x\\ \beta (t)=\left< x,t \right> \quad \dot { \beta } =\left< 0,1 \right> \quad 0\le t\le y\\ f=\int _{ 0 }^{ x }{ F(\alpha ) } \cdot \dot { \alpha } ds+\int _{ 0 }^{ y }{ F(\beta ) } \cdot \dot { \beta } dt\\ f=\int _{ 0 }^{ x }{ 1 } ds+\int _{ 0 }^{ y }{ { e }^{ x }+xcos(t)dt } \\ f=x+y{ e }^{ x }+xsiny## But ##\nabla f\neq F##
    Last edited: Dec 4, 2016
  11. Dec 4, 2016 #10
    You've got it. However, one of your dot products was not evaluated properly. You should have:
    ##\alpha (s)=\left< s,0 \right> \quad \dot { \alpha } =\left< 1,0 \right> \quad 0\le s\le x\\ \beta (t)=\left< x,t \right> \quad \dot { \beta } =\left< 0,1 \right> \quad 0\le t\le y\\ f=\int _{ 0 }^{ x }{ F(\alpha ) } \cdot \dot { \alpha } ds+\int _{ 0 }^{ y }{ F(\beta ) } \cdot \dot { \beta } dt\\ f=\int _{ 0 }^{ x }\langle 0, e^s + s \rangle \cdot \langle 1, 0 \rangle ds+\int _{ 0 }^{ y }{ \langle t e^x + \sin t, e^x + x \cos t \rangle \cdot \langle 0, 1 \rangle dt } \\ f=\int _{ 0 }^{ x } 0 ds+\int _{ 0 }^{ y }{ e^x + x \cos t dt } \\ f= y{ e }^{ x }+xsiny##
  12. Dec 5, 2016 #11
    ha I don't know why I thought sin0 was 0, thanks for your help!
  13. Dec 5, 2016 #12
    No problem, but the value of sin(0) is indeed 0. :-)

    Edit: Oh, never mind. I see you meant to write "sin(0) was 1" now. No worries! It happens to all of us!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted