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General dipole radiation far field equation

  1. Jun 7, 2015 #1

    Dale

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    I am looking for a reference describing the far-field electric and magnetic field of a dipole. However, I want a general formula for an arbitrary scalar function, and not specifically the usual formula which assumes a continuous sinusoid:

    $$ \mathbf{B} = -\frac{\omega^2 \mu_0 p_0 }{4\pi c} \sin\theta \frac{e^{i\omega (r/c-t)}}{r} \mathbf{\hat{\phi} } $$
    $$ \mathbf{E} = c \mathbf{B} \times \hat{\mathbf{r}}
    = -\frac{\omega^2 \mu_0 p_0 }{4\pi} \sin\theta \frac{e^{i\omega (r/c-t)}}{r} \hat{\theta} $$

    I would like something similar, but where I could use say a square pulse, or a Gaussian pulse, or some other non-oscillating waveform.
     
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  3. Jun 7, 2015 #2

    vanhees71

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    This is not so simple. The reason is that these far-field equations are indeed describing the non-transient quasi-stationary states of the field of one harmonic mode of the electromagnetic field, i.e., it describes the field far from the (compact) source a long time after it was "switched on". For a wave packet, it's not clear to me what you want to achieve since, a wave packet of finite temporal extension, at one point far from the source, this wave packet will run through (more or less deformed by dispersion if it's not a plane wave in vacuo) and the non-transient limit is then simply that there is no field there anymore, because the wave packet as moved further.

    Of course you can always use the multipole expansion to build any wavepacket out of the harmonic modes you like via a Fourier integral.
     
  4. Jun 7, 2015 #3

    Dale

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    Yes, I want the transient in time, but I am only interested in the spatial dipole. I want the field of a dipole which is pulsed on and then back off.
     
  5. Jun 7, 2015 #4

    vanhees71

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    Then you can just take the Fourier transform of the dipole fields, you've written down above, i.e.,
    $$(\vec{E}(t,\vec{x}),\vec{B}(t,\vec{x}))=\int \mathrm{d} \omega/(2 \pi) \tilde{A}(\omega) (\vec{E}_{\omega}(t,\vec{x}),\vec{B}_{\omega}(t,\vec{x})).$$
    The ##\tilde{A}(\omega)## is given by the Fourier transform of the initial conditions for the fields.
     
  6. Jun 7, 2015 #5

    blue_leaf77

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    Vanhees's equation might be applicable. But if I were in your position I would go back to the starting point where you have to model a dipole which oscillates non-sinusoidally and express this oscillating current in frequency domain, which is the spectrum. Then stay in frequency domain during the derivation since dealing with linear differentiation in frequency domain is easier. Note that those expressions of E and B fields are derived under several assumptions concerning the distance, dipole length, and wavelength, you would simply need to replace the wavelength with the bandwidth of the pulse in any appearing inequality.
     
    Last edited: Jun 7, 2015
  7. Jun 7, 2015 #6

    jasonRF

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    The general retarded potential solution for a current source ## \mathbf{J}(\mathbf{r},t) ## is,
    ## \mathbf{A}(\mathbf{r},t) = \frac{\mu_0}{4 \pi} \int \frac{d^3\mathbf{r}^\prime \, \, \mathbf{J}\left(\mathbf{r}^\prime, t-\left|\mathbf{r}-\mathbf{r}^\prime\right|/c\right)}{ \left|\mathbf{r}-\mathbf{r}^\prime\right|} ##

    If I use a current density source of ## \mathbf{J}(\mathbf{r},t) = \hat{\mathbf{z}} I d\ell dt \delta(\mathbf{r}) \delta(t) ## then for the z component of the vector potential I get
    ## A_z(\mathbf{r},t) = \frac{\mu_0 I d\ell dt }{4\pi r}\delta(r-ct) ##

    This result makes intuitive sense, so I think it is okay. This can also be written in spherical coordinates,

    ## A_\theta(\mathbf{r},t) = -\frac{\mu_0 I d\ell dt }{4\pi r}\delta(r-ct) \sin\theta##
    ## A_r(\mathbf{r},t) = \frac{\mu_0 I d\ell dt }{4\pi r}\delta(r-ct) \cos\theta##

    jason
     
    Last edited: Jun 7, 2015
  8. Jun 7, 2015 #7

    jasonRF

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    I realized I had a careless error in my post (argument of delta function wrong). It should read

    ##A_z(\mathbf{r},t) = \frac{\mu_0 I d\ell dt }{4\pi r}\delta(t-r/c)##
    and likewise for the components in spherical coords.

    Also, it is hopefully clear that a more general source,
    ## \mathbf{J}(\mathbf{r},t) = \hat{\mathbf{z}} I d\ell \delta(\mathbf{r}) f(t)##
    will yield,
    ##A_z(\mathbf{r},t) = \frac{\mu_0 I d\ell }{4\pi r}f(t-r/c)##
    As always, the spherical componeents of ##\mathbf{A}## are then,
    ## A_\theta = -A_z \sin\theta ##
    ## A_r = A_z \cos\theta ##
    And the fields would be derived from ## \mathbf{B} = \nabla \times \mathbf{A}##, etc.

    jason
     
  9. Jun 7, 2015 #8

    Dale

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    This seems like what I am looking for. Do we need to worry about charge density and continuity?

    I was pursuing a similar approach, but enforcing continuity and I started getting derivatives of delta functions.
     
  10. Jun 7, 2015 #9

    jasonRF

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    Charge continuity is included here, and you can calculate the charge density as a function of time.

    I have been assuming Lorentz gauge that connects the vector and scalar potential, ##\nabla \cdot \mathbf{A(\mathbf{r},t)} = -\mu \epsilon \frac{\partial \phi(\mathbf{r},t)}{\partial t}##. So once you know ##\mathbf{A}## you can integrate to get ##\phi##, and then calculate the charge density from ##\nabla^2\phi - \mu \epsilon \frac{\partial^2 \phi(\mathbf{r},t)}{\partial t^2} = -\frac{\rho}{\epsilon}##. Or equivelantly, you calculate ##\mathbf{E} = -\nabla \phi - \frac{\partial \mathbf{A}}{\partial t}## and then get the density from ##\rho = \epsilon \nabla \cdot E ##. It can be instructive to look at a simple dipole this way.

    EDIT: I was being slow here. Just use the continuity equation to ghet ##\rho## !!!!! ##\nabla \cdot \mathbf{J} = -\frac{\partial \rho}{\partial t}##

    jason
     
  11. Jun 8, 2015 #10

    vanhees71

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    Yep, the continuity condition is mandatory. Otherwise you violate gauge invariance, and this is a no-go. With ##c \neq 1## it reads
    $$\partial_{\mu} j^{\mu}=\frac{1}{c} \partial_t (c \rho) + \vec{\nabla} \cdot \vec{j}=\partial_t \rho+\vec{\nabla} \cdot \vec{j}.$$
    So if you give ##\vec{j}## arbitrarily, you have indeed to choose ##\rho## as given in Posting #9.
     
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