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Del dot E equals 0 and Del dot B equals 0, both in a vacuum

  1. Jul 28, 2009 #1
    So given the common explanation here for Del dot B equals 0 to be "There are no magnetic monopoles.", since the title indicates that the equations in a vacuum have the same form, would the meaning of Del dot E = 0 mean that "There are no ELECTRIC monopoles?"...which we assume to be false. Wherein lies the confilct?
     
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  3. Jul 28, 2009 #2

    LeonhardEuler

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    Both statements are true in a vacuum, because by definition there are no electric monopoles in a vacuum. If there were electric monopoles it wouldn't be a vacuum. In general
    [tex]\nabla\cdot \vec{E} = \frac{\rho}{\epsilon_{0}}[/tex]
    This applies even when electric monopoles are present.
     
    Last edited: Jul 28, 2009
  4. Jul 28, 2009 #3
    Do you mean you can't have electric monopoles in a vacuum because a vacuum can have no charges in them?
     
  5. Jul 28, 2009 #4

    LeonhardEuler

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    Yes, the definition of a vacuum is a place where there is no matter. Charges are made of matter, so a vacuum can contain no charge.
     
  6. Jul 28, 2009 #5
    You also cannot have magnetic monopoles (if they exist) in a vacuum since that would also imply the presence of matter, right?
     
  7. Jul 28, 2009 #6
    That's kind of a silly equation. Kind of like saying there is no laminar flow in a vacuum. There's no anything in a vacuum! lol I wonder why it was even stated?
     
  8. Jul 28, 2009 #7

    LeonhardEuler

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    It seems silly at first, but there are electric and magnetic fields in a vacuum even though there is no matter. This equation is important, for instance, in studying how an electromagnetic wave propagates in a vacuum.
     
  9. Jul 28, 2009 #8
    Oh, so you can have a point charge outside the defined vacuum and it permeates into the vacuum?
     
  10. Jul 28, 2009 #9

    LeonhardEuler

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    Yes, the electric field of a point charge will permeate all of space. If there is only 1 point charge, then in any region that does not include that charge,
    [tex]\nabla\cdot\vec{E} = 0 [/tex]
    but that is not true at the location of the charge.
     
  11. Jul 28, 2009 #10
    Not sure what you mean about "but that is not true at the location of the charge. Is it semantics to say there is not electric field at the exact charge point? I guess the electric field =0 there, right? But given that, then if there were another charge point, it's field woulcd be exactly at the first chage points position.
     
  12. Jul 28, 2009 #11
    Are you a famous physicist? You have a famous physicist type of name.
     
  13. Jul 28, 2009 #12

    LeonhardEuler

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    Sorry if that was unclear. I'll restate it like this: draw any box in space. If it doesn't contain the charge, then it is a vacuum and [tex]\nabla\cdot\vec{E} = 0[/tex] everywhere inside. If it does contain the charge, then it is not a vacuum and that equation does not apply everywhere inside.

    It's probably best not to think of the electric field as being 0 at the location of a point charge, just undefined. Certainly [tex]\nabla\cdot\vec{E}[/tex] is not defined there.

    No, I'm not actually Leonhard Euler. He lived in the 1700's and was one of the greatest mathematicians of all time. I just chose my name in honor of him.
     
  14. Jul 28, 2009 #13
    I wish to for sake of learning, argue that Del dot E is not zero inside the box, because in my limited knowledge I would say that the divergence of an electric field is a function of the interacting waves of many point charges and millions could be interacting inside a vacuum box. How would you respond to that? I though his name sounded familiar. lol
     
  15. Jul 28, 2009 #14

    LeonhardEuler

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    Yes, millions of charges outside the box could be influencing the field in the vacuum box, but the divergence would still be 0. The easiest way to see this is that if I have a lot of charges outside the box, call them q1, q2 ... q1000000, the total electric field at any point is the sum of the electric fields resulting from each charge alone (superposition principle):
    [tex]\vec{E}_{TOT} = \vec{E}_{1} + \vec{E}_{2} + ... + \vec{E}_{1000000}[/tex]
    The divergence is also a linear operator, so
    [tex]\nabla\cdot\vec{E}_{TOT} =\nabla\cdot \vec{E}_{1} + \nabla\cdot\vec{E}_{2} + ... + \nabla\cdot\vec{E}_{1000000}[/tex]

    You can fairly easily show that the divergence of the field of one charge is 0. That applies to all of the charges, so
    [tex]\nabla\cdot\vec{E}_{TOT} = 0 + 0 + ... + 0 = 0[/tex]
     
  16. Jul 28, 2009 #15
    Hmmm...sounds good but I'm still not getting divergence. I see it as the rate of change of an electric field and inside the box you've got flux lines (like in those 2D drawings of charge interactions) and at one point in the box more lines converge, more divergence and at other points in the box, less lines converge, less divergence. If what you say in true then you cannot have divergence in a vacuum. So outside of a vacuum, how do you demonstrate divergence? Sorry for being terminally thick on this but it's really difficult to conceptualize.
     
  17. Jul 28, 2009 #16

    LeonhardEuler

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    I'm sorry but I can't quite understand this. You don't have more divergence at some points in a vacuum and less at others. It is zero everywhere.

    A way to picture divergence is this: Draw a small region in space and calculate how much electric field enters and how much leaves. Let the size of this box shrink to 0. If more electric field is leaving, there is positive divergence. If more is entering, there is negative divergence.

    The divergence due to a single charge is 0, so the amount entering and leaving are in balance for each charge. Add them all together and they are still in balance.
     
  18. Jul 28, 2009 #17
    I think this is the most difficult concept to grasp. By your statement, you never have divergence NOT equal zero unless an electric field enters a closed 2D surface and vanishes magically. I don't think electric fields just vanish. And if they don't vanish, they proceed thru the surface, so you always have zero divergence. Correct me if I'm wrong here but maybe my example will clear up my inability to grasp this.
     
  19. Jul 29, 2009 #18

    LeonhardEuler

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    That's not quite right. If the electric field entered the box and disappeared, the divergence would NOT be zero, because it comes in but doesn't go out. Think of the vector field that describes the velocity of a liquid. If you draw any closed surface, the amount of liquid coming in to the surface has to equal the amount leaving. This means the divergence of the velocity field is 0. So too with the electric field.
     
  20. Jul 29, 2009 #19

    Born2bwire

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    The solution to this is that for a non-zero divergence, you must have a source or sink inside the volume. This is why they describe divergence as a measure of sources/sinks. The differential and integral forms of Maxwell's equations are equivalent. One can think of the differential form of Gauss' Law as being the limit of the integral form as the enclosed volume of integration goes to zero. Thus, the divergence of the electric field is zero everywhere except for the exact points in space where we have a charge, which acts as a source.
     
  21. Jul 29, 2009 #20
    Dude, seriously. Maxwell's equations are MATHEMATICALLY derived from VECTOR CALCULUS mapped to a physical situation. Saying Del dot B is a silly law is just retarded. Maxwell's equation are actually used, using MATH to SOLVE PROBLEMS. And each one is of extreme importance. One does not work without the other, in those 4 equations is EVERYTHING physical about classical EM. All the rest of classical EM is just a developed mathematical toolbox to help with certain types of problems (usually with high symmetry).

    P.S. as has been stated, Euler was one of the best mathematicians ever, not a physicist. There is a joke in math: "Why are there not many things named after Euler?", "Because in math things often get named after the SECOND person who discovered them and the first person was always Euler".
     
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