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I Relaxed conditions for the density: Ampère's law still valid?

  1. May 7, 2016 #1
    The most common proof that I have found of the fact that Ampère's law is entailed by the Biot-Savart law essentially uses the fact that, if ##\boldsymbol{J}:\mathbb{R}^3\to\mathbb{R}^3##, ##\boldsymbol{J}\in C_c^2(\mathbb{R}^3)##, is a compactly supported twice continuously differentiable field such that ##\nabla\cdot\boldsymbol{J}\equiv 0 ## and ##\Sigma## is a smooth surface satisfying the assumptions of Stokes' theorem, then $$\oint_{\partial^+ \Sigma}\left(\frac{\mu_0}{4\pi}\int_{\mathbb{R}^3}\frac{\boldsymbol{J}(\boldsymbol{x})\times(\boldsymbol{r}-\boldsymbol{x})}{\|\boldsymbol{r}-\boldsymbol{x}\|^3}d\mu_{\boldsymbol{x}}\right)\cdot d\boldsymbol{r}=\mu_0\int_\Sigma \boldsymbol{J}\cdot\boldsymbol{N}_e \,d\sigma\quad(1)$$where ##\mu_0## is any constant (the magnetic permeability in the physical interpretation), ##\boldsymbol{N}_e## is the surface's external normal unit vector and ##\mu_{\boldsymbol{x}}## is Lebesgue 3-dimensional measure.

    Nevertheless, common exercises and applications of Ampère's law found in books of physics use current densities ##\boldsymbol{J}\notin C_c^2(\mathbb{R}^3)##, one common example being ##\boldsymbol{J}## constant on an infinite cylinder and constantly ##\mathbf{0}## outside the infinite cylinder.

    Do mathematically rigourous formulations of Ampère's law ##(1)## exist under more relaxed assumptions on ##\boldsymbol{J}##, like the quoted case of ##\boldsymbol{J}## constant on a (bounded or unbounded) region and null outside of it, and, if they do, how can they be proved?
    I have thought about approximating such a ##\boldsymbol{J}## with ##\boldsymbol{J}_n\in C_c^2(\mathbb{R}^3)##, but it is not easy to see that the required sequence really exists.
    I ##\infty##-ly thank any answerer!
     
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  3. May 8, 2016 #2
    I always thought it was the other way around, that the Biot-Savart law was a special case pertaining to magnetostatics...whereas Maxwell's equations (i.e. Ampere's law, Faraday's law, and Gauss' laws) are still valid in those cases.
     
  4. May 8, 2016 #3
    @Megaquark Thank you for your intervention! (Sorry if I cannot speak Ferengi...:wink:) The two laws are often said to be equivalent. I know a proof of the fact that, at least if ##\boldsymbol{J}\in C_c^2(\mathbb{R}^3)##, the Biot-Savart law entails Ampère's law, but I have never seen a proof of the converse: when the etailment of the Biot-Savart law by Ampère's is desired to be proved I have really seen proofs of the fact that Ampère's law is satisfied by a magnetic field defined by ##\boldsymbol{B}(\boldsymbol{x})=\frac{\mu_0}{4\pi}\int\frac{\boldsymbol{J}\times\boldsymbol{r}}{r^3}dV####=\frac{\mu_0}{4\pi}\int_{\mathbb{R}^3}\frac{\boldsymbol{J}(\boldsymbol{x})\times(\boldsymbol{x}-\boldsymbol{y})}{\|\boldsymbol{x}-\boldsymbol{y}\|^3}d\mu_{\boldsymbol{y}}##, i.e. Biot-Savart law ##\Rightarrow## Ampère's law and not Ampère's law ##\Rightarrow## Biot-Savart law...

    Independently of whether we consider Ampère's law derived from Biot-Savart law or vice versa, I am interested in knowing whether the formula ##(1)## is valid, as many applications found in booktexts of physics seem to assume, even for ##\boldsymbol{J}\notin C_c^2(\mathbb{R}^3)##, under opportune assumptions. A common ##\boldsymbol{J}\notin C_c^2(\mathbb{R}^3)## that I find in books of physics is $$\boldsymbol{J}(\boldsymbol{x})= \begin{cases}J\mathbf{k} , &\boldsymbol{x}\in V \\ \mathbf{0}, & \boldsymbol{x}\notin V \end{cases}$$where ##\mathbf{k}## is a unit vector and ##V\subset\mathbf{V}## is a certain region, for ex. an infinite cylinder.
     
  5. May 8, 2016 #4
    I don't think I've ever seen it done in general, just for special cases...though I imagine that it probably isn't much different than deriving Coulombs law from Maxwell's equations. I know to get from Maxwell to Coulomb you have to make an assumption about spherical symmetry of the electric field around the point charge or something like that...I wouldn't be surprised if there's a similar sort of necessity when getting the Biot Savart law. If it is truly analogous, then you might have to make an assumption of circular symmetry of the magnetic field around the wire...not sure what restrictions that would place on the current density.
     
  6. May 8, 2016 #5

    lavinia

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    For a smooth vector field on a closed loop , the Biot and Savart integral gives a continuous vector field on all of ##R^3## which is ##C^∞## everywhere except at the boundary of the loop. . The vector field needs only to be defined on the loop and not on all of ##R^3##. For a steady current, i.e. a divergence free vector field ##J##, that is tangent to the surface of the loop

    ##∇×B = J##

    This means that the curl of ##B## is zero outside the current loop and equals ##J## in the current loop. Ampere's Law for a steady current follows immediately from Stokes Theorem.

    The idea of current loop can be generalized to a compact solid whose boundary is a smooth surface. So for instance, the current may flow on a knot or a two holed torus.

    It is a theorem that ##∇×B = J## if and only if ##J## is divergence free and tangent to the boundary of the loop.

    ##B## is always the curl of another vector field i.e. it always has a vector potential whether or not the current is steady.
     
    Last edited: May 9, 2016
  7. May 9, 2016 #6
    @lavinia If I correctly understand what you have written, you are saying that if ##\boldsymbol{J}:V\subset\mathbb{R}^3\to\mathbb{R}^3##, ##\boldsymbol{J}\in C^{\infty}(\mathring{V})##, where ##\mathring{V}## is the interior of the region ##V## (with what assumptions on ##V##?), is such that ##\forall\boldsymbol{x}\in \mathring{V}\quad\nabla\cdot \boldsymbol{J}(\boldsymbol{x})##, then, for any ##\boldsymbol{x}\in\mathbb{R}^3##,
    $$\nabla_x \times\left(\frac{\mu_0}{4\pi}\int_V\frac{\boldsymbol{J}(\boldsymbol{x})\times(\boldsymbol{r}-\boldsymbol{x})}{\|\boldsymbol{r}-\boldsymbol{x}\|^3}d\mu_{\boldsymbol{x}}\right)=\mu_0\boldsymbol{J}(\boldsymbol{x})$$
    Have I understood? If that is correct, how can it be proved?

    The proofs of "Biot-Savart law" ##\Rightarrow## "Ampère's law" that I have found both on line and on printed texts (Jackson's, Griffith's) usually use Dirac's ##\delta## and commute derivatives and integrals with no explanation of why that is mathematically licit and what the notations are intented to mean (ex.: ##\int d^3x## for Lebesgue integrals, symbolic notations for linear operators etc.): a very big problem for me, initially. Nevertheless, I think I have managed to understand what such proofs means (I have written my interpretation of such proofs here), with the assumption that ##\boldsymbol{J}\in C_c^2(\mathbb{R}^3)##, i.e. that ##\boldsymbol{J}:\mathbb{R}^3\to\mathbb{R}^3## is twice continuously differentiable and compactly supported. Do you know a (mathematically rigourous, by definition of proof) proof of "Biot-Savart law" ##\Rightarrow## "Ampère's law" under these more relaxed assumptions for ##\boldsymbol{J}##?

    I heartily thank you both and anybody else wanting to join the thread!
     
  8. May 9, 2016 #7

    lavinia

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    The theorem states actually states that the magnetic field ##B## is ##C^∞## on the closure of the complement of the loop. This means that you can use Stokes theorem twice to get Ampere's Law. - I think.

    I will think about the basic proof. I was only trying to show that the vector field does not need to be differentiable on all of ##R^3##.

    In general the curl of ##B## will not equal ##J##. ##J## must be steady.
     
    Last edited: May 9, 2016
  9. May 10, 2016 #8
    I've been thinking about this for a few days, and I'm pretty sure that the current only needs to be integrable. If I am given a four-current that is everywhere Lebsegue integrable, for example, I can solve for the once-differentiable electromagnetic tensor everewhere.
     
  10. May 11, 2016 #9
    @The Bill Thank you for your reply! Do you know a mathematical proof or a rigourous text giving one? I know the proof of Jackson's Classical Electrodynamics which, translated into a rigourous form, I think to mean this, but I have had to make the assumtpion that ##\boldsymbol{J}\in C_c^2(\mathbb{R}^3)##.
     
  11. May 11, 2016 #10

    lavinia

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    An arbitrary integrable current will not necessarily produce a magnetic field.

    For instance, how would you show that the curl of the vector potential, ##A(y) = 1/4π∫J(x)/|y-x|dV(x)## equals the magnetic field if ##J## is nowhere differentiable?
     
  12. May 12, 2016 #11

    lavinia

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    For Ampere's Law one needs ##∇×B## to equal the current density ##J## up to a constant ##μ_{0}##. This restricts ##J## since in general Maxwell's equation says

    ##∇×B =μ_{0}( J + ε_{0}∂E/∂t))## where ##E## is the electric field.

    Forgetting the constant ##μ_{}## and ##ε_{0}##, it turns out that ##∇×B=J## if and only if ##J## is both divergence free and tangent to the surface of the current loop.

    Here are some arguments to illustrate why these two assumptions on ##J## suffice.

    One wants ##∇×1/4π∫J(x)×y-x/|y-x|^3dVol(x)## to equal ##J(x)##.

    Setting the vector potential ##A## equal to ##1/4π∫J(x)/|y-x|dVol## this is the same as ##∇×∇×A(y) = J(y)## or using the standard formula for the iterated curl, ##∇(∇⋅A) - ∇^2A = J##.

    The first term ##∇(∇⋅A)## will be zero if A is divergence free. If ##y## lies outside of the current loop then the Divergence Theorem says that ##∇⋅A(y) = 1/4π∫J(x)/|y-x|⋅ndS(x)## where ##S## is the boundary surface of the current loop. If ##J## is tangent to ##S## then this integral is zero. So the restriction of ##J## to be tangent to the surface of the current loop guarantees that ##A## is divergence free outside of the current loop.

    If ##y## is inside the current loop then removing a small sphere centered at ##y## one approximates divergence of ##A## with the difference of two integrals one over the surface of the current loop - which we already know is zero if ##J## is assumed to be tangent to the loop - and the other over a small sphere ##S## centered at ##y##. Since |y-x| is constant on this sphere, the integral is

    ##(1/4π)1/|y-x|∫_{S}J(x).n dS(x)## and this is zero if the divergence of ##J## is zero. So the additional restriction of ##J## to be divergence free guarantees that ##A## is divergence free inside the current loop.

    * One is now left with showing that ##-∇^2A = J## and this is where the Dirac delta argument comes in.
    For the case that ##y## is outside of the current loop

    ##∇^2A(y) = 1/4π∫J(x)∇^2(1/|y-x|dVol(x)## and ##∇^2(1/|y-x|## is zero if ##y≠x## so ##∇^2A## is zero outside of the current loop as is ##J##.

    In general ##∇^2(1/|y-x|)## is 4π times the Dirac delta function ##δ(y-x)##.

    This thread discusses the Dirac delta for the Laplacian of ##1/|y-x|##

    https://www.physicsforums.com/threads/divergence-of-inverse-square-field-and-dirac-delta.678438/
     
    Last edited: May 12, 2016
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