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Biot-Savart + Coulomb + Charge Conservation = Maxwell?

  1. Oct 14, 2007 #1
    Biot-Savart + Coulomb + Charge Conservation = Maxwell??

    Do the Biot-Savart Law, Coulomb's Law, and the Law of Charge Conservation contain the same information as Maxwell's Equations? i.e.

    d\vec{B} = \frac{\mu_o}{4\pi} \frac{I d\vec{l} \times \hat r }{r^2} \\
    \vec{E}= \frac{1}{4\pi\varepsilon_o} \frac{Q \hat r}{r^2} \\
    \nabla \cdot \vec{J} = -\frac{\partial \rho}{\partial t} ,
    \nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0} \\
    \nabla \cdot \vec{B} = 0 \\
    \nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t} \\
    \nabla \times \vec{B} = \mu_0 \vec{J} + \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t},
    Last edited: Oct 15, 2007
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  3. Oct 14, 2007 #2
  4. Oct 14, 2007 #3

    Claude Bile

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    The original question I think is better phrased as, "can we use one set of equations to derive the other?". Coulomb's law and the Biot-Savart Law can certainly be derived from Maxwell's equations, I don't think the same is true for the Law of charge conservation however, because from memory, one needed to use this law to derive the amended version of Ampere's Law (I'm going on memory though so I could be wrong).

  5. Oct 14, 2007 #4
    Yes. It can:

    [tex]\nabla \times \vec{B} = \mu_0 \vec{J} + \mu_0 \epsilon_0 \frac{\partial \vec{E}}{\partial t}[/tex]

    Take the divergence of each side:

    [tex]\nabla \cdot \nabla \times \vec{B} = \mu_0 \nabla \cdot \vec{J} + \mu_0 \epsilon_0 \frac{\partial \nabla \cdot \vec{E}}{\partial t}[/tex]

    Now, [tex]\nabla \cdot \nabla \times \vec{B} = 0[/tex] identially, while [tex]\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}[/tex].

    Thus, [tex]0 = \mu_0 \nabla \cdot \vec{J} + \mu_0 \epsilon_0 \frac{\partial \rho /\epsilon_0}{\partial t}[/tex]

    Or, more simply [tex]\nabla \cdot \vec{J} = -\frac{\partial \rho}{\partial t}[/tex]
  6. Oct 15, 2007 #5
    I know that Maxwell's Equation can be used to derive all sorts of laws (which is why they are so famous and revered). I was actually interested in knowing if we can derive full-blown Maxwell's Equations (all 4 of 'em) using the set of three laws: {Coulomb's law, Biot-Savart Law, Law of Conservation of Charge}. Is that possible??
  7. Oct 15, 2007 #6

    I don't think so, because these three laws do not contain the possibility of radiation of electromagnetic waves. If the radiation is neglected, and there are only charges and current elements interacting with each other, then, I believe, your assumption is correct.

  8. Oct 15, 2007 #7


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    it seems to me that since magnetic fields can only be generated by the movement of charges and since when charges move from one place to another, the place they move from experiences a net loss of charge (reducing, in time, the electrostatic flux emitted from that place) and the place they move to experiences a net increase of charge (increasing, in time, the electrostatic flux emitted from there), that the conservation of charge does have a role here. but i am trying to decide if it's a native ingredient or a consequence. i think the OP might have an insight of value here.

    think of a nice cylinderical capacitor with two parallel circular plates and the two leads connected outward from the plates at right angles to the plane of the plates. the B field generated around the wires going in (where there is movement of charge) is the same as the B field generated around the empty space between the plates (where there is a changing electric field because of the accumulation of charge on the plates but no moving charge).
  9. Oct 15, 2007 #8
    I'm not 100% certain about this; but, I'm thinking that, once you impose self-consistency, you might just be able to do this. Coulomb's Law and the Biot-Savart law are solutions of electrostatics and magnetostatics respectively. So, these should, alone, give rise to Gauss' Law and Ampere's Law. Again, playing the game with the divergence of Ampere's Law, and requiring charge conservation (a.k.a. continuity) will force the introduction of the displacement current term. So, these three conditions certainly allow the construction of everything other than Faraday's Law of Induction.

    My feeling here is that requiring self-consistency among the equations probably, then, leads to the requirement that Faraday's Law hold; but, I haven't quite found the argument for it yet. I know that there's a relatively simple argument for the displacement current when the rest of Maxwell's equations are already in place; so, I would think that something might parallel that.
  10. Oct 15, 2007 #9


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    The way the original equations are written down, I think it won't be possible, because Coulomb's law, for instance, is not valid in all generality (when there is radiation for instance, you can have an E-field in the radiation which is NOT given by Coulomb's law).
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