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The second Maxwell equation

  1. Dec 11, 2007 #1
    I need to prove Maxwell's second equation [tex]\nabla[/tex][tex]\bullet[/tex]B=0
    I tried few things, but never came with result... and all i could find on internet was finished formula with explanation that magnetic monopols doesnt exist. so is there any proof of this equation?
    Last edited: Dec 11, 2007
  2. jcsd
  3. Dec 11, 2007 #2

    Claude Bile

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    The absence of magnetic monopoles is sufficient justification for the law, simply because if magnetic monopoles did exist, the law would not hold true.

    I should remark that Maxwell's laws are empirical laws and thus cannot be "proven" since that would imply the existence of more fundamental rules. As far as we know, Maxwell's equations are the most fundamental set of laws we know of concerning electromagnetism.

  4. Dec 11, 2007 #3
    I always thought that the absence of magnetic monopoles was just the assumption that they don't exist, based on the fact no1 has ever observed one, and so the 'law' that magnetic monopoles don't exist (or at least aren't thought to exist) was just the result of physical observation.
  5. Dec 11, 2007 #4


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    What is your starting point?
    and what do you mean by "prove"?
    As others have mentioned, it is a statement of the absence of isolated magnetic monopoles.
    However, if you assume something else [for example, the relativistic form of the Maxwell Equations], you could "derive" from it [tex]\nabla \cdot \vec B=0[/tex].
  6. Dec 12, 2007 #5
    Reiterating what people said:[tex] \nabla \cdot B = 0[/tex], just like all of Maxwell's equations, is an empirical fact and cannot be proven from first principles. It does, however, follow from some more fundamental expressions. Namely, [tex]B = \nabla \times A[/tex]. Since [tex]\nabla \cdot \nabla \times {\rm anything} = 0[/tex], [tex] \nabla \cdot B = 0[/tex] follows.

    Now, [tex]B = \nabla \times A[/tex] is in fact where the classical EM theory meets empirical observations. On the theoretical side, we derive the equations of motion for a charged particle using the least action principle. If we were armed with this principle alone, without knowing anything about electric and magnetic fields, we would eventually conclude that for a reasonable theory we need 4 quantities that will fully characterize the field. These quantities we will call the scalar potential [tex]\phi[/tex] and the vector potential [tex]\vec{A}[/tex]. There exists a general prescription of how to use the least action principle to obtain equations of motion for a particle. Next, we look at the equations, and realize that the quantity [tex]\nabla \times A[/tex] plays the role of the magnetic field in determining the force on the moving particle, for instance. Hence, we have [tex]B = \nabla \times A[/tex]. [tex] \nabla \cdot B = 0[/tex] is, in a sense, mathematically equivalent.

    For the equation to not be true, we would have to go back all the way to the original expression for the action and meddle with that. But then, all the kinematic equations of how charged matter interacts with the field would change.
  7. Dec 12, 2007 #6
    Thx for your responses.
    I read several texts and watched several videos about this subject yesterday and came with same conclusion...
    Our professor at the college told us to prove this equation... so I thought it to be possible...
    Obviously it's not:rofl:
  8. Dec 12, 2007 #7


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    From introduction to electrodynamics (Griffiths, page 223), he starts with the Biot-Savart law, takes the divergence of both sides, and uses [tex]\nabla\cdot(\vec{J} \times \frac{\vec{r}}{r^2}}) = \frac{\vec{r}}{r^2}\cdot(\nabla\times\vec{J}) - \vec{J}\cdot(\nabla\times\frac{\vec{r}}{r^2})[/tex], and concludes that the curl of J is 0 because "J doesn't depend on the unprimed variables (x,y,z)" ("The integration is over the primed coordinates, the divergence and curl are to be taken with respect to the unprimed coordinates), and the curl of r/r^2 is 0, so the divergence of B is 0.
  9. Dec 12, 2007 #8
    if you wanna "prove" (rather derive) the equations there is an article by D. Kobe

    Derivation of Maxwell's equations from the local gauge invariance of quantum mechanics

    where he ,as the article title says, derive it from the principle of local gauge invariance, it is actually a good article, that gives a good idea about how physesists thinks when they use the least action principle.
  10. Dec 12, 2007 #9
    If you have access to a good library, you can find this article in Dyson's collected works which presents Feynman's derivation of the Maxwell equations.

    Here's the abstract:

    http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=AJPIAS000072000003000345000001&idtype=cvips&gifs=yes [Broken]
    Last edited by a moderator: May 3, 2017
  11. Dec 13, 2007 #10
    I just had this on a physics final. Thanks to the forum I knew exactly what the "total flux through a closed cylindrical surface" was. Thanks!

    -searcher of magnetic monopoles
  12. Dec 20, 2007 #11
    I think you should prove that this equation can be derived from the Faraday-Neumann law, i.e. from the first Maxwell equation:
    [tex]\nabla \times E = - \frac{\partial B}{\partial t}[/tex].
    If you take the divergence of both sides you obtain
    [tex]0 = - \frac{\partial }{\partial t} \nabla \cdot B[/tex].
    So [tex] \nabla \cdot B[/tex] is constant with respect to the time. If the fields are zero for t<t0, where t0 is the time when the sources are turned on, then [tex]B(t)=0[/tex] for t<t0 and [tex] \nabla \cdot B(t)=0[/tex] for t<t0.
    Since[tex] \nabla \cdot B(t)=const[/tex], [tex]\forall t[/tex], you obtain that [tex] \nabla \cdot B=0[/tex].
  13. Dec 20, 2007 #12


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    Yes, but that law itself assumes no magnetic monopoles (since it does not include a magnetic current). Hence the result of your derivation.

    The essential reason for the form of that law is the observation of the general absence of magnetic monopoles. There may well be a deeper physical truth tied up in the QM unification of EM phenomena, but I can't comment on that.
  14. Dec 21, 2007 #13
    nope, each of maxwell's equations stands for itself. As my professor said to any1 who tried that: "it's a virus... and another one... and another one..."

    you start from Biot-Savart law, transform it and u get -[tex]\nabla \times {\rm ...}[/tex]
    and take divergence of that and u get [tex] \nabla \cdot B = \nabla \cdot \nabla \times {\rm ...} [/tex]

    and yeah, thx for help... i was the only one who did this right, from Biot-Savart law:smile:
    p.s i haven't figured out how to write this equations completely, so i couldn't post whole method
    Last edited: Dec 21, 2007
  15. Dec 28, 2007 #14
    It doesn't come up too often in physics, but you've inadvertently stated a classical logical fallacy:

    If monopoles existed, then there would be a divergence of B. THEREFORE:

    If B never diverges, then monopoles do not exist. HOWEVER:

    That fails to prove the converse, namely that without monopoles
    there can be no divergence. For example, there might be some way of constructing a field purely out of various currents, which turns out to have a divergence somewhere. That must be what the OP is trying to prove.
  16. Dec 28, 2007 #15


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    Can you write down such a field to prove your point?

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