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Curl(V) = 0

  1. Sep 14, 2013 #1
    Hi, upon studying vector calculus and more precisely about the curl I stumbled upon a question : why is it that there is always a potential function of a vector field when the curl of this vector field is equal to 0?
     
  2. jcsd
  3. Sep 14, 2013 #2

    WannabeNewton

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    Is this a homework question? If so, what have you already tried? You have to show your work. If you need a hint, then consider Stokes' theorem ##\oint F\cdot dr = \int (\nabla \times V)\cdot dA##. Using this you can very easily show that there exists a ##\varphi## such that ##V = \nabla \varphi##; try to show it.

    Just for the sake of being complete, I should warn you that this is not always true; ##\nabla \times V = 0## only implies ##V = \nabla \varphi## if ##V## is defined on an open subset of ##\mathbb{R}^{3}## that is simply-connected i.e. every path in the open subset can be continuously deformed to a point. If this doesn't hold then ##\nabla \times V = 0## will not imply that ##V = \nabla \varphi##.
     
  4. Sep 14, 2013 #3
    There are a number of ways to answer this question. Mathematically, you can [pretty much] always take a vector field and separate it into a part with zero curl and a part with zero divergence--this is called Helmholtz decomposition. In other words, given a vector field [itex]\mathbf{F}[/itex], we can write it as: [tex]\mathbf{F}=-\nabla\phi+\nabla\times\mathbf{A}[/tex]
    Where the first term [itex]-\nabla\phi[/itex] has zero curl by to the identity Curl(Grad(f))=0 for any scalar field f, and the second term has zero divergence by the identity Div(Curl(v))=0 for any vector field v. The second term can be chosen to represent only the curl of the field, so if we set it to zero, then we have [itex]\mathbf{F}=-\nabla\phi[/itex]. This is a very mathematical way to answer your question and it is not that intuitive.

    A more intuitive way might be to think about what a curl means physically. Let's assume there is a nonzero curl in a vector field, let's call it an electric field. This implies that if you sit a charge down at rest in a region with nonzero curl, the field could force it to start going in a big loop and return to its original spot, arriving back at the spot with a nonzero velocity (and it could continue looping, picking up more and more velocity each loop.). In other words, the field returned the particle to its original spot but added some kinetic energy. In this sense the field does not conserve kinetic energy, so we call it a "non-conservative field".

    On the other hand, if we assume a particle's gain in kinetic energy between any two points is independent of the path it were to take between those points, then a particle could never traverse a closed loop and pick up kinetic energy (since that loop is equivalent to having stood still the whole time, since the particular path doesn't matter--just the endpoints.) This is the case that corresponds to zero curl. Because the kinetic energy gained between any two points is independent of path, we could totally describe this kind of field by picking a reference point called "0 energy" and then labeling all other points with the value of the kinetic energy that the particle would have gained or lost going from the reference point to each other point. Thus we can think of such a path-independent field as one which does conserve energy, or a "conservative field". This labeling of all points by a scalar energy is basically exactly what the potential field [itex]\phi(\mathbf{x})[/itex] is, so we have constructed our representation of a conservative field with a scalar field.

    Wikipedia has a nice discussion about your question: http://en.wikipedia.org/wiki/Conservative_field
     
    Last edited: Sep 14, 2013
  5. Sep 14, 2013 #4
    (Actually, it's not strictly true: Every conservative field is irrotational (curl vanishes), since the curl of the gradient of an arbitrary vector field is zero, but every irrotational field is conservative only in a simply connected domain. It does, however, hold in most physical circumstances, for example.)

    In simply connected region, it can be proved like this: A potential function exists iff the line integral of the field over an arbitrary closed curve vanishes (use fundamental theorem of line integrals), ie, the field is path-independent (also used as the definition of a conservative field). You can then write the said integral, consider a surface that has the curve as its boundary and use Stokes' theorem to show that the integral vanishes if the curl is zero to prove this.

    I seriously should stop leaving tabs open for like 30 minutes, coming back to them and posting a reply without refreshing the page.
     
    Last edited: Sep 14, 2013
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