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Divergence-free field has closed field lines?

  1. Oct 15, 2008 #1
    Is the restriction that a field is divergence-free enough to ensure the field lines will be closed? How can one (dis-) prove such a statement?
  2. jcsd
  3. Oct 15, 2008 #2


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    Just invoke the definition of divergence and I don't mean the general formula for calculating divergence. It involves the limit of a shrinking volume and the flux through the closed surface. What does that say about where the field lines terminate?
  4. Oct 16, 2008 #3
    to me divergence free means:

    at a point (still a very small, macroscopic however, region) the net flux is zero, unless there is a source or sink contained in that small volume.

    how about the example of the point charge? any point is divergence-free but the point that contains the charge (delta density).
    The E field lines are not closed...........
  5. Oct 17, 2008 #4
    What is a `field line'?
  6. Oct 17, 2008 #5


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    I meant flux lines. This comes from electromagnetism and the concept of electric flux lines. Didn't you use the term "field line" in the title of the thread?
  7. Oct 17, 2008 #6
    I'm not the OP, but I am asking a follow up question. Given an electric field, how do you determine the electric field lines?
  8. Oct 17, 2008 #7


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    Sorry I didn't notice you were a different poster. The electric field lines are just the vector field lines. Suppose you have a vector function and this function assigns every point in 3D space with a direction. If you were to start at any one point and draw a continuous line always following the directional arrows assigned by the vector function you'll get field lines.
  9. Oct 17, 2008 #8
    Field lines are always explained in qualitative terms, but looking at them closer, how are they actually defined?

    Clearly, they must be paths where at each point, the tangent of the line at that point is the same as the field's value at that point. But I don't think that uniquely defines them, since it does not pay respect to the magnitude of force at each point.

    So, basically I'm curious. Given a path in a field, how do you determine if that path is part of a field line? Or equivalently, given a field and a point, how do you generate the set of points comprising the field line that contains that point?
  10. Oct 17, 2008 #9


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    Field lines are a qualitative and intuitive way of visualising vector fields and functions. If you want something more precise you'll have to invoke the vector function itself. It's true that field lines don't tell you anything about the magnitude of the vector function evaluated at the point, but that's true only if you draw all the arrows of equal size. No one says you can't draw larger arrows when the field at the point has greater magnitude.

    As for your question of "paths", it depends on how that path is derived. Is it simply an arbitrary path drawn without any regards to the vector field? This is reminiscent of a 2nd year E&M course where you'll get questions about a given path and you're told to find the work done by moving a charge along that path in the vector field.

    But as you've said above, if that path is part of a field line, then the vector arrows evaluated at any point along that path is always tangent to it. You've already answered your question as above.

    I believe your 2nd question is more interesting. Are you asking if for vector fields in general (not just the E&M case) field lines would never cross each other? That's something I can't give a definitive answer to. But I'll look at it this way: If they ever crossed each other at any point, that would imply that at that point the vector function assigns two different vectors for the same point in space.
  11. Oct 18, 2008 #10
    I found some useful things in some books I have.

    Suppose a field-line is a curve that is represented parametrically as [tex]\mathbf{r} = \mathbf{r}(\theta)[/tex]. Given a vector field [tex]\mathbf{v}[/tex], then [tex]\mathbf{r}[/tex] must satisfy the differential equation [tex]\frac{d\mathbf{r}}{d\theta} \times \mathbf{v} = \mathbf{0}[/tex].

    In response to the OP's question about closed field lines in divergenceless fields, it is easy to cook up an example. Consider the vector field [tex]\mathbf{v} = \alpha \mathbf{i}[/tex], where [tex]\alpha[/tex] is a real constant. This field is divergenceless and with a little work you can convince yourself that the field lines are straight lines that do not 'close'.
    Last edited: Oct 18, 2008
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