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Magnetostatics -> Electrostatics?

  1. Feb 26, 2006 #1
    Magnetostatics -> Electrostatics??

    I have a questions and I am afraid that I might look very dumb for asking such questions, so forgive me first.

    I heard many times how Maxwell's equations along with Lorents Force Law tells all the story of EM dynamics. So I wanted to see if I can show (by using four Maxwell's equations) a simple questions.

    Q: given "Magnetostatics", can I show "Electrostatics" must follow? By Magnetostatics, I mean dB/dt = 0. By Electrostatics, I mean dE/dt = 0. (Not samething as saying source charges are stationary!). So, simply, how can I show dB/dt = 0 -> dE/dt = 0 using Maxwell's equaitons.

    Thanks in advance.

    p.s. Just in case, if someone out there trying to tackle this question by employing Columb's Law, I must warn you that it is not one of Maxwell's equation. (I mean, you can't use it to show electrostatics, cuz you already have to assume electrostatic to use it!) I say this here cuz many of my friend's first response was so.
    Last edited: Feb 26, 2006
  2. jcsd
  3. Feb 26, 2006 #2
    Actually, Coloumb's law is indeed one of Maxwell's equations. The one that corresnponds to Gauss's law is just an elegant version of Coloumb's law. And you can derieve magnetism from special relativity and Gauss's law. See section 12.3 in Introduction to Electrodynamics, Griffhits".

    But I don't know if I could do the reverse.
  4. Feb 26, 2006 #3
    I don't know if that is true in non-static case. The well-known derivation of Columb's law from Gauss's law (for example, from Griffith Ch2.2) presupposes electrostatic. Also, see Feynman Lectures15-15. So I am tempted to think that the popular phrase such as "Gauss's law is equivalent to Columb's law" might be true only in static case. (I can't imagine calculating force of one electron flying by me with constant velocity using Columb's law!) But, I am just being skeptic, so I need your further justification on this part.

    Here, I don't understand when you said you can derive magnetism. What do you meant by derive magnetism? My question was given dB/dt = 0, can you derive dE/dt = 0. Do you mean the converse of it?
    dE/dt = 0 -> dB/dt =0? If so, could you show me how?

    Hmmm....now that you mentioned about special relativity, I am afraid I might be missing the big point. I simply don't know relativity yet. But if Maxwell's are true and complete (I don't know if that is true without relativity, but can I just pretend it is so to seek for the answear I want?)

  5. Feb 26, 2006 #4
    According to Maxwell equations (without dropping time dependent terms, ie laws of electrodynamics), Gauss' law holds same for both electrostatics and electrodynamics:
    [tex]\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}[/tex]
    It has no time dependence.

    As for the fact, laws of electrodynamics are relavitistically correct. Magnetism is just a relativistics effect on Coloumb's law, not a fundamental law of nature. See Griffiths 12.3, it shows how to derieve magnetism from coloumb's law and SR.
  6. Feb 26, 2006 #5
    Yes, that's why I asked you to use Maxwell's equation (that includes the gauss's law) to show dB/dt = 0 implies dE/dt=0. What I disagreed to in my previous message was to your cliam that Columb's law is identically equivalent to Gauss's law, therefore, you can argue columb's law in nonstatic charge cases. That I disagreed.

    I am still not clear, what you mean by derive "magnetism" . But if you mean that one can relate electric field to magnetic field viewd in other referene frame, I agree with you. But what does it do with my question? :confused:
  7. Feb 26, 2006 #6
    Gauss's law holds for electrodynamics. Show a case where it wouldn't hold, and let's discuss it.
    "Use the Special Relativity, young Jedi" :)
    I don't mean to relate electricity and magnetism. I explicitly mean you don't need magnetism at all. Just use SR.
  8. Feb 26, 2006 #7

    I don't know if I mislead you, but I said it already and I will say it again, Gauss's law (being one of the maxwell's equation) holds true in all electrodynamics. Read carefully, what I put down above plz. :grumpy:

    I don't wanna be rude, but I am looking out for someone who can actually show the answear to my question or show me it is simply not possible. Other talks, while interesting is not the purpose of this thread.
    Last edited: Feb 26, 2006
  9. Feb 26, 2006 #8

    Claude Bile

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    I would start by taking the first time-derivative of Ampere's law. The rest seems fairly trivial.

  10. Feb 26, 2006 #9

    Physics Monkey

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    If [tex] \frac{\partial \vec{B}}{\partial t} = 0 [/tex] then Maxwell's equations for the electric field reduce to [tex] \vec{\nabla}\cdot \vec{E} = \rho/\epsilon_0 [/tex] and [tex] \vec{\nabla}\times \vec{E} = 0 [/tex]. The second equation implies that the electric field may be written as the gradient of a scalar field as [tex] \vec{E} = - \vec{\nabla} \phi [/tex]. Substituting into the first equation, you find immediately that [tex] \phi [/tex] satisfies Poisson's equation. The unique solution to this equation for the case when the field is required to vanish at infinity is [tex] \phi(\vec{r}) = \frac{1}{4 \pi \epsilon_0} \int \frac{\rho(\vec{r}')}{|\vec{r} - \vec{r}'|} \, d^3 r' [/tex]. I have therefore shown by explicit construction that a system of charges where the field is required to vanish at infinity have an electric field of precisely the form of Coulomb's law. It should be immediately clear to you that if I have a finite system of point charges, then each charge must be at rest in order for [tex] \frac{\partial \vec{B}}{\partial t} = 0 [/tex]. Of course this then implies that the electric field is time independent.

    Furthermore, it is clear that even if the electric field has explicit time dependence, it can depend at most linearly on time. If the electric field depends at most linearly on time then the charge density must also depend at most linearly on time. Here's something for you to think about, how physical is a situation where the charge density depends linearly on time? Can you dream up a system where the electric field does depend linearly on time and there is magnetic field which remains time independent?
    Last edited: Feb 26, 2006
  11. Feb 26, 2006 #10
    Thanks! I will try again!
  12. Feb 27, 2006 #11
    you certainly did, with this:

  13. Feb 27, 2006 #12
    If you have a static uniform magnetic field in the inertial frame S and then you trransform to frame S' which is in standard configuration with S then there will be both a static uniform magnetic field and a static uniform electric field. This can be shown using special relativity.

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