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E and B produced by a stationary currrent

  1. Aug 25, 2012 #1
    I have a stationary electric current like this: [itex]\vec{I}=I_0\vec{e_z}[/itex] and I have to calculate the electric and magnetic fields from an inertial frame that is moving with velocity [itex]\vec{v}=v\vec{e_z}[/itex].

    First I calculate the vector potential from a rest frame (and then I'll use Maxwell equations to get B and Lorentz transformation for E' and B') and I use cilindrical coordinates, so I get:

    [itex]\vec{A}= \frac{\mu _0}{4\pi}∫\frac{dz}{\sqrt{r^2+z^2}}[/itex]

    but if I calculate this between -infinity and infinity the integral diverges.

    Can this integral be solved? Or, is there an easier way to solve the problem?

    Thank you.
     
    Last edited: Aug 25, 2012
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  3. Aug 25, 2012 #2

    phyzguy

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    Why did you choose to use cylindrical coordinates in a problem that is so clearly rectangular?
     
  4. Aug 26, 2012 #3
    I can't see that. Why is the problem clearly rectangular?
     
  5. Aug 26, 2012 #4
    Perhaps it's too early in the morning and I haven't had enough coffee yet, but what is a stationary current?
     
  6. Aug 26, 2012 #5
    I'm not sure about that but I think it's just a constant current.
     
  7. Aug 26, 2012 #6

    phyzguy

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    Well, the only variation in the problem is with z - everything is independent of x and y. When you use cylindrical coordinates, you introduce a variation in the XY plane that only serves to make the problem more complicated. Or so it seems to me. One question is, "in what direction can the fields point?"
     
  8. Aug 26, 2012 #7

    gabbagabbahey

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    I'm guessing that the current [itex]I[/itex] is confined to the z-axis, and not spread out uniformly over then entire xy-plane.
     
  9. Aug 26, 2012 #8

    gabbagabbahey

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    (Aren't you missing an [itex]I_0[/itex]?:wink:)

    The integral indeed diverges. One solution is to extract the finite part of the integral which contains the relevant physics (see here). Another is to calculate the fields directly (not a fun calculation!).

    The best method in my opinion is to imagine that the current is turned on at some time [itex]t_0[/itex]. For [itex]t - t_0 <\frac{r}{c}[/itex], the field point will not experience the effect of the current (the fields will not propagate faster than light when the current is switched on). For [itex]t - t_0 > \frac{r}{c}[/itex], the retarded time will only be greater than or equal to [itex]t_0[/itex] where [itex]|z| \leq \sqrt{c^2(t-t_0)^2-r^2}[/itex] (if the retarded time is less than [itex]t_0[/itex], then the fields that arrive at your field point will be from before the current was turned on - that is, they'll be zero). So, it is really only this range that contributes to the integral, as outside, the current [itex]I(t_r)[/itex] is zero.

    [tex]\mathbf{A}= \frac{\mu_0 I_0}{4 \pi} \int_{-\sqrt{c^2(t-t_0)^2-r^2}}^{\sqrt{c^2(t-t_0)^2-r^2}} \frac{dz}{r^2+z^2}[/tex]
     
  10. Aug 27, 2012 #9

    vanhees71

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    It's not a good idea to turn on the current abruptly at some finite time since this violates current conservation, and the integral you get cannot solve Maxwell's equations, because these necessarily need a conserved four-current as a constraint, which is closely related to gauge invariance.

    I'd simply calculate
    [tex]\vec{A}(x)=\frac{\mu_0 I_0}{4 \pi} \int_{-\Lambda}^{\Lambda} \mathrm{d} z \frac{1}{\sqrt{r^2+z^2}}.[/tex]
    As you'll then see, for [itex]\Lambda \rightarrow \infty[/itex] you can separate off the divergent part by introducing an arbitrary finite scale [itex]r_0[/itex], upon which no physical quantity, particularly the magnetic field, [itex]\vec{B}=\vec{\nabla} \times \vec{A}[/itex] depends.

    This is one of the first encounters of a kind of "regularization and renormalization theory".
     
  11. Aug 27, 2012 #10

    gabbagabbahey

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    Does it? Can you elaborate, because I don't see it.

    The amount of charge per unit time moving into any tiny section of wire is the same as the amount of charge per unit time leaving the section of wire, both before and after the current is switched on. In other words, [itex]\frac{\partial \rho}{\partial t}[/itex] is zero both before and after. The divergence of the current density [itex]\mathbf{J}=I_0 u(t-t_0) \delta(x) \delta(y) \mathbf{e}_z[/itex] is likewise zero both before and after the switch on.
     
  12. Aug 27, 2012 #11

    Born2bwire

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    What I am not clear on is whether or not this current is a point-source or is it an infinite line of current?
     
  13. Aug 27, 2012 #12

    rude man

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    I would assume the current is along an infinitely long wire coincident with the z axis.

    Thus the problem becomes degenerate: speed of observing frame irrelevant since its motion is also along the z direction; B invariant with z, i.e. the usual ampere's law formulation about the wire; E zero everywhere except inside the wire.
     
  14. Aug 28, 2012 #13

    Born2bwire

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    Yeah, that's what I thought the OP intended but then the calculation for the magnetic field is trivial as you stated.
     
  15. Aug 28, 2012 #14
    The Faraday field can be solved for directly using the magnetic field of a straight current carrying wire, suitably made covariant. Geometric algebra makes this very easy to do.

    First, we need a way to talk about the part of a vector that is perpendicular to a plane. This is in 4D spacetime, so you can't resort to dot products and normal vectors. You need a wedge product. Let there be a plane [itex]K = a \wedge b[/itex] for two vectors [itex]a,b[/itex]. This is just the plane formed by the span of [itex]a,b[/itex]. For a third vector [itex]s[/itex], the part of [itex]s[/itex] that lies entirely outside of the plane [itex]K[/itex] is denoted by [itex](s \wedge K) \cdot K^{-1}[/itex], where [itex]K^{-1} = K/(K \cdot K)[/itex].

    The general form of the Faraday field from a current carrying wire is (in [itex]\mu_0 = \epsilon_0 = c = 1[/itex] units):

    [tex]F = \frac{1}{2\pi} I \wedge \rho^{-1}[/tex]

    where [itex]I[/itex] is the current vector and [itex]\rho = [s \wedge (I \wedge u)] \cdot (I \wedge u)^{-1}[/itex] for some four-velocity of the wire [itex]u[/itex] and a position vector [itex]s[/itex]. Here's a quick check to ensure this reproduces the stationary current carrying wire: let [itex]I = I_0 e_z[/itex] and [itex]u = e_t[/itex]. Then we have

    [tex]I \wedge u = I_0 e_{zt}[/tex]

    As well as

    [tex]\rho = [(s^\rho e_\rho + s^z e_z + s^t e_t) \wedge (I_0 e_{zt})] \cdot e_{zt}/I_0 = s^\rho I_0 e_{\rho z t} \cdot e_{zt}/I_0 = s^\rho e_\rho[/tex]

    Clearly, then, [itex]\rho^{-1} = e_\rho /s^\rho[/itex], and our Faraday field is

    [tex]F= \frac{1}{2\pi s^\rho} I_0 e_{z \rho}[/tex]

    Exactly as it is in the simple, stationary, infinite current-carrying wire, and since we've done this in terms of arbitrarily oriented wires with arbitrary velocities, we know this solution should hold for all such cases. Note that if the wire goes from carrying no current to carrying some current (that is, the current is a function of time), this method will not hold.

    It should be clear that even in the particular case where the boost velocity is parallel to the extent of the wire that an electric field arises and that the magnetic field changes by a factor of gamma. This is seen simply with the transformation [itex]e_z = \gamma (e_{z'} - \beta e_{t'}[/itex], and we get

    [tex]F = \frac{1}{2\pi s^\rho} I_0 \gamma (e_{z' \rho} - \beta e_{t' \rho})[/tex]

    Which clearly has both electric and magnetic pieces.
     
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