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Divergence of curl

  1. Oct 20, 2013 #1
    Can anyone give me an intuitive/physical reason for why the divergence of the curl of a vector field is always zero? I know various methods to prove mathematically that it is so, but have not managed to find a physical reason. In other words, why is the curl of a vector field always incompressible.
  2. jcsd
  3. Oct 21, 2013 #2


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    There is no "physical" reason, in contrast to "maths".
    Rather, what you need is a strategy to VISUALIZE the content in concepts like the "curl" and the "divergence", so that's what I'll try to do here!

    Let's say the vector is the velocity field for a fluid!

    The curl tells you how, locally, the fluid rotates on a tiny disk element of it.

    The curl of the velocity field gives us, therefore, an image of that at each point, we have tiny disks of fluid spinning about their own axes, with their own angular velocities.

    But, the local DIVERGENCE of a vector tells us how a tiny element "exceeds" its bounds, for example the rate by which the element grows in size, effectively, summing up how the velocities directed outwards from the boundary of the element tells us of the element's flow out beyond its own borders.
    But, what about these tiny, rotating wheels in the curl field?

    They do NOT "exceed" their own bounds, their velocity is always TANGENTIAL to their own boundaries, and thus, the local divergence of these wheels must be..0
    Now, when the velocity field ITSELF is the curl of another vector (such as when the fluid is incompressible), it means that such a decomposition of the vector field into tiny, spinning wheels contains the FULL information about the behaviour of the velocity field, and that, globally, the divergence of it must be zero.
    Last edited: Oct 21, 2013
  4. Oct 21, 2013 #3


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  5. Oct 21, 2013 #4


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    The two most important facts about divergence and curl are:

    [itex]\int (\nabla \times \vec{F}) \cdot \vec{dS} = \int \vec{F} \cdot \vec{d\mathcal{l}}[/itex]

    where F is a vector field, and where the left side is an integral of the curl of F over a (2D) surface, and the right side is the line integral of F over the (1D) boundary of that surface.

    For divergence, we have:

    [itex]\int (\nabla \cdot \vec{G}) dV = \int \vec{G} \cdot \vec{d S}[/itex]

    where G is a vector field, and where the left side is an integral of the divergence of G over a (3D) volume, and the right side is a surface integral of G over the (2D) boundary of that volume.

    Putting these two together by letting [itex]\vec{G} = \nabla \times \vec{F}[/itex], gives:

    [itex]\int (\nabla \cdot (\nabla \times \vec{F})) dV = \int \vec{F} \cdot \vec{d\mathcal{l}}[/itex]

    where the left side is an integral over a volume, and the right side is an integral over the boundary of the boundary of that volume. What does "the boundary of the boundary" mean? Well, the boundary of a 3D volume is a 2D closed surface. And a closed surface has no boundary. So the right-hand integral is always zero. Since this has to be true for any volume whatsoever, that's only possible if [itex]\nabla \cdot (\nabla \times \vec{F}) = 0[/itex].

    To see intuitively what it means that the boundary of a boundary is zero, let's take the example of the surface of the Earth. If you consider the surface that is the northern hemisphere, then the boundary of that surface is the circle known as the equator. Now, imagine moving that circle to the south, to enclose more of the surface of the Earth. Then as the circle gets closer to the South pole, the surface consisting of all points north of the circle grows. When you actually reach the South pole, the boundary has shrunk to a single point, and the surface has grown to include the entire surface of the Earth. So the surface of the Earth has a boundary of size zero.

    The surface of the Earth is itself the boundary of the interior of the Earth. So if you start with a volume, and take its boundary, you get a closed surface. Then when you take its boundary, you get zero.
  6. Oct 22, 2013 #5
    Thanks a lot, arildno! Actually, I know what curl and divergence mean in terms of the physical operations that we do to the vector field to get them. I have the picture of rotating disks in my mind when I think of the curl of a vector field, and of the field strengthening or weakening locally when I think of the divergence. There is one problem, though.

    When I take the curl of the field, I associate a rotating disk with each point. Each disk will have its own angular velocity and direction (given by the right-hand rule for the direction of angular velocities). Now, when I take the divergence of the curl, I am essentially trying to find out how much the angular velocity changes in an infinitesimal volume. And this is where the confusion arises; the angular velocities of the disks (the curl at each point) varies, so, in general, the divergence should not be zero. For example, in going from A to C, the speeds of the disks at A, B and C could be omega, 2 times omega and 3 times omega. In this case, the divergence will not be zero because the velocities are increasing. In other words, the field obtained by the curl operation is strengthening in going from A to C. Hence the divergence, at least in the vicinity of A, B and C should not be zero.

    I hope my reasoning was not too complicated. I shall grateful if you could point out the flaw in my concept.
  7. Oct 22, 2013 #6


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    Suppose you have a small cube of volume [itex]\delta V[/itex], and each face has area [itex]\delta A[/itex]. Then

    [itex]\nabla \cdot (\nabla \times \vec{F}) \delta V \approx \sum_{faces} (\nabla \times \vec{F})_{normal} \delta A[/itex]

    where [itex](\nabla \times \vec{F})_{normal}[/itex] means the component of the curl that is normal to that face. For a single face, we have:

    [itex](\nabla \times \vec{F})_{normal} \delta A \approx \int \vec{F} \cdot \vec{d\mathcal{l}}[/itex]

    where [itex]\int \vec{F} \cdot \vec{d\mathcal{l}}[/itex] is the line integral of [itex]\vec{F}[/itex] around the border of the face, with the direction of the integration being counter-clockwise (when looking down on the face).

    As shown in the picture, the line integrals cancel exactly: The line integral for the top face (indicated by the red loop) overlaps one edge for the right face (indicated by the green loop), but the directions for the two integrals are opposite, so those two contributions cancel exactly. Similarly, the line integral for the top faces cancels the contribution for the front face (indicated by the blue loop) along one edge. Every one of the 12 edges of the cube belongs to two different faces, and so it gets integrated over twice in opposite directions. So all contributions cancel exactly.

  8. Oct 22, 2013 #7


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    You might either study closely the moreprecise arguments given by stevendaryl, but the rotating disk analogy works perfectly well also.
    Since the divergence on EACH such locally placed disk is zero, adding any number of them together adds up to zero as well, irrespective of what the disk's value of the local angular velocity is.
  9. Oct 22, 2013 #8


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    You might use [itex]\nabla\cdot(\nabla\times{\bf V})=(\nabla\times\nabla)\cdot{\bf V}=0.[/itex]
  10. Oct 23, 2013 #9
    The implication that [itex]\nabla \times \nabla =0 [/itex] is wrong! You can't treat differential operators like a vector.
    DON'T DO IT!

    The Laplacian of a vector is an example where this logic can lead one very astray.

    [itex]\nabla^2 \vec A =\nabla \left(\nabla \cdot \vec A \right) + \nabla \times \nabla \times \vec A [/itex]

    The last term is important and not zero!

    Now going back to the original question
    I think your problem is that you are associating all types of rotation with the curl. This is a common misconception of the curl, but it is not true. For example consider the flow [itex] V_{\theta} = \frac{C}{r} [/itex]. This flow is clearly rotation, but [itex] \nabla \times \vec V= 0 [/itex]. The rotational component of the flow that the curl picks out is going to be divergenceless.
    Last edited: Oct 23, 2013
  11. Oct 23, 2013 #10


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    Not that it's very important, but I would say that [itex]\nabla \times \nabla =0 [/itex] is correct (at least for flat spacetime). I would interpret [itex]\nabla \times \nabla =0 [/itex] to be a vector of operators [itex]O_i[/itex] defined by

    [itex]O_x = \nabla_y \nabla_z - \nabla_z \nabla_y[/itex]
    [itex]O_y = \nabla_z \nabla_x - \nabla_x \nabla_z[/itex]
    [itex]O_z = \nabla_x \nabla_y - \nabla_y \nabla_z[/itex]

    Since derivatives commute, all of these are zero.

    The expression [itex] \nabla \times \nabla \times \vec A [/itex] means [itex]\nabla \times (\nabla \times \vec{A})[/itex], not [itex](\nabla \times \nabla) \times \vec{A}[/itex]

    I think you're right, that you might get in trouble treating operators like vectors, but I don't think that's a good counterexample. A better example is the angular momentum operator from quantum mechanics:

    [itex]J_x = -i (y \partial_z - z \partial_y)[/itex]
    [itex]J_y = -i (z \partial_x - x \partial_z)[/itex]
    [itex]J_z = -i (x \partial_y - y \partial_x)[/itex]

    If you think of that as a vector [itex]\vec{J}[/itex], then you find

    [itex]\vec{J} \times \vec{J} = i \vec{J}[/itex], rather than zero.
  12. Oct 24, 2013 #11
    stevendaryl, thanks a lot! Your illustrated explanation is very clear. Now I have to see how it is equivalent to arildno's rotating disks analogy.

    the_wolfman, are you saying that the rotating disks analogy is over-simplistic and/or imperfect? If so, can you give another analogy? Thanks!
  13. Oct 24, 2013 #12


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    Not necessarily a misconception, but it does highlight a subtlety with rotation in fluid flow:
    There is no LOCAL rotation of fluid elements, but that doesn't mean the fluid cannot rotate as a whole.
    This is in contrast to, for example, the velocity field for a rotating rigid body, in which the local (and global!) angular velocity is half the curl.
    Last edited: Oct 24, 2013
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