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Electric Radiation Field

  1. Nov 25, 2012 #1
    I am currently reading:
    http://physics.usask.ca/~hirose/p812/notes/Ch8.pdf

    How do you go from

    [itex] \vec{E} (r,t) = \frac{e}{4 \pi \epsilon_0} \frac{1}{c \kappa^3 R} \hat{n}
    \times ( \hat{n} \times \dot{ \beta}) [/itex]

    to Eq. 8.46 on page 9

    it seems to me that at t=0 we have [itex] \dot{\beta} = ( |\dot{\beta}|, 0 ,0) [/itex] in spherical co-ordinates. and the normal vector in some arbitrary direction is [itex] (1, \theta, \phi) [/itex]

    for circular motion you have the cross product between velocity and acceleration disappearing so

    [itex] \hat{n} \times (\hat{n} \times \dot{\beta}) = |\dot{\beta}| (\theta^2 - \phi^2, \theta, \phi) [/itex]

    Not Eq. 8.46 - so I must be doing something very wrong
     
    Last edited: Nov 25, 2012
  2. jcsd
  3. Nov 25, 2012 #2

    haruspex

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    It says [itex] \vec{E} (r,t) = \frac{e}{4 \pi \epsilon_0} \frac{1}{c \kappa^3 R} \hat{n}
    \times ( (\hat{n}-\beta) \times \dot{ \beta}) [/itex]
     
  4. Nov 25, 2012 #3
    In this case the cross product with [itex] \beta \times \dot{\beta} = 0 [/itex] so I remove terms with [itex] \beta [/itex] in in both equations...

    [itex] (r, \theta, \phi) = (0, - \cos \theta \cos \phi, \sin \phi) [/itex] would be the vector part in Eq. 8.46.

    Also I think I might have the order of the angles in that vector in the wrong order, but my result is still nothing like 8.46
     
  5. Nov 25, 2012 #4

    haruspex

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    The text says they're orthogonal, not collinear.
     
  6. Nov 26, 2012 #5
    OK well either way that's not the problem I can't get to 8.46, do you know how it's done


    i.e. even with [itex] \beta \times \dot{\beta} = 0 [/itex] I don't get

    [itex] n \times (n \times \dot{\beta}) = (0, -\cos \theta \cos \phi, \sin \phi) [/itex]

    which is what would happen if you removed the [itex] -\beta [/itex] from the bracket?

    (I removed terms with [itex] \beta |\dot{\beta}| [/itex] )

    I'm beginning to think that there is a mistake in these notes since this vector looks like it has no magnitude, and there are terms which look Cartesian, and it says Cartesian above 8.46.

    Do you need values of those vectors beta at t=t'? I'm not quite sure which they refer to
     
    Last edited: Nov 26, 2012
  7. Nov 26, 2012 #6
    With that:

    [itex] n=\left(
    \begin{array}{c}
    \cos \phi \sin \theta \\
    \sin \phi \sin \theta \\
    \cos \theta
    \end{array}
    \right) [/itex]


    [itex] \beta =\left(
    \begin{array}{c}
    0 \\
    0 \\
    \beta
    \end{array}
    \right) [/itex]

    [itex]
    \dot{\beta} =\left(
    \begin{array}{c}
    |\dot{\beta} | \\
    0 \\
    0
    \end{array}
    \right) [/itex]

    so for

    [itex] n\times( (n-\beta) \times \dot{\beta}) [/itex]

    I get

    [itex]

    (\cos \theta - \beta) \cos \theta \hat{e}_x + \sin^2\theta \sin \phi \cos \phi \hat{e}_y + (\beta-\cos \theta) \sin \theta \cos \phi \hat{e}_z [/itex]
     
  8. Nov 26, 2012 #7
    So does anyone know how you get to 8.46?
     
  9. Nov 26, 2012 #8

    TSny

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    I don't think this is the correct expression. There are some sign errors. In addition, the coefficient of ##\hat{e}_x## has a term missing.

    It might be better to begin by expanding the vectors ##\beta## and ##\dot{\beta}## in terms of the unit vectors ##\hat{n}##, ##\hat{e}_\theta##, and ##\hat{e}_\phi##. Then evaluate [itex] n\times( (n-\beta) \times \dot{\beta}) [/itex]. Use of a Computer Algebra System program will make life easier.
     
  10. Nov 26, 2012 #9
    I managed to get the right answer. So this is for the time t' here? This happens to be set to zero. How would I find the [itex]E_{\text{rad}} (r,t) [/itex] field for subsequent time? The function does not seem to be of time. Say if you wanted the field as some time [itex] t=t_r+\frac{R}{c} [/itex] ?

    I guess if the field is what you want at some point a distance R away, t=t' will suffice correct?

    By the way to anyone interested: you can do the cross products by doing it first in Cartesian basis and then switching to the spherical basis.
     
  11. Nov 26, 2012 #10

    TSny

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    The time dependence of equation 8.46 is in the angles ##\theta## and ##\phi## as well as the in the unit vectors ##\hat{e}_\theta## and ##\hat{e}_\phi##. These quantities are changing with time. They are determined by the position of the particle on the circle at the retarded time and the location of the observation point relative to particle. It looks very complicated to me. If you want ##\vec{E}(r,t)## at some selected observation time ##t##, you would need to determine the location of the particle on the circle at the corresponding retarded time ##t\:'##. Then you can use equation 8.46 for the particular ##\theta##, ##\phi##, ##\hat{e}_\theta## and ##\hat{e}_\phi## that would be defined for that location of the particle on the circle and the location of the observation point.
     
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