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Measure of Relativistic Flattening of Electric Field

  1. Jun 1, 2012 #1
    1. The problem statement, all variables and given/known data

    As a rough measure of the relativistic "flattening" of the configuration of electric field lines from a moving charge, we might use the angle [itex]\alpha[/itex] between two conical surfaces which include between them half the total electric flux. That is, half the flux through a sphere shall be contained in the equatorial zone between [itex]\theta[/itex]' = [itex]\frac{\pi}{2}[/itex]+[itex]\frac{\alpha}{2}[/itex] and [itex]\theta[/itex]' = [itex]\frac{\pi}{2}[/itex]-[itex]\frac{\alpha}{2}[/itex]. Consider only the extreme relativistic case, with [itex]\gamma[/itex]>>1. Then only angles [itex]\theta[/itex]' such that [itex]\theta[/itex]'=[itex]\frac{\pi}{2}[/itex]-[itex]\epsilon[/itex], with [itex]\epsilon[/itex]<<1 need be considered. Show first that Eq.12 can be approximated, in this case, as

    E' = [itex]\frac{Q}{(r')^{2}}[/itex][itex]\frac{\gamma}{(1+\gamma^{2}\epsilon^{2})^{3/2}}[/itex]

    Now let [itex]\epsilon[/itex] range from [itex]\frac{-\alpha}{2}[/itex] to [itex]\frac{+\alpha}{2}[/itex] and integrate to obtain the flux through the narrow equatorial belt.

    2. Relevant equations

    Eq. 12: E' = [itex]\frac{Q}{(r')^{2}}[/itex][itex]\frac{1-\beta^{2}}{(1-\beta^{2}sin^{2}\theta')^{3/2}}[/itex]

    [itex]\gamma[/itex]=[itex]\frac{1}{\sqrt{1-\beta^{2}}}[/itex]

    Gauss' Law: Electric flux out of a sphere = 4[itex]\pi[/itex]Q

    (My book uses Gaussian units)

    3. The attempt at a solution

    I'm still stuck on approximating equation 12. I have tried the small-angle approximation for sin[itex]^{2}[/itex]([itex]\frac{\pi}{2}[/itex]-[itex]\epsilon[/itex]), and I have tried converting this to cos[itex]^{2}[/itex]([itex]\epsilon[/itex]) and approximating with a second order Taylor series, and I've tried expanding the entire equation, but I can't make Equation 12 agree with the equation they gave me.

    This is from Purcell's book, usually used for honors freshman physics, so the math probably isn't terribly advanced, but I can't see what else to try.

    I have also tried skipping that part and attacking the integral directly, but I can't seem to make it into something I can do by hand or look up in a table.

    For the integral, I know that E' is radial, so [itex]\vec{E'}[/itex][itex]\cdot[/itex][itex]\vec{dA}[/itex] will simply equal E'dA. So to find dA in terms of r' and [itex]\epsilon[/itex] I figure that the area of a thin ring of the sphere near the equator is the width r'd[itex]\epsilon[/itex] times the circumference of the ring 2[itex]\pi[/itex](r'cos[itex]\epsilon[/itex]).

    So I get the integral:

    [itex]\int[/itex][itex]^{\alpha/2}_{-\alpha/2}[/itex][itex]\frac{Q\gamma*2\pi*(r')^{2}cos\epsilon}{(r')^{2}(1+\gamma^{2}\epsilon^{2})^{3/2}}[/itex]d[itex]\epsilon[/itex]

    And I set that equal to half the total flux through the sphere, 2[itex]\pi[/itex]Q and simplify a bit to get:

    [itex]\frac{1}{\gamma}[/itex] = [itex]\int[/itex][itex]^{\alpha/2}_{-\alpha/2}[/itex][itex]\frac{cos\epsilon}{(1+\gamma^{2}\epsilon^{2})}[/itex]d[itex]\epsilon[/itex]

    But I don't know how to evaluate that integral. So someone please let me know if I have the setup wrong or if I just need an integral trick I don't know.

    The book provides the final answer by the way, and it is:

    [itex]\alpha[/itex] = [itex]\frac{2}{\sqrt{3}\gamma}[/itex]
     
    Last edited: Jun 1, 2012
  2. jcsd
  3. Jun 2, 2012 #2

    TSny

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    Gold Member

    Looks good! In your final integral, try making the approximation cos ε ≈ 1, which should be ok in the extreme relativistic limit where ε is very small.

    Also, don't forget the 3/2 power on the denominator (as you have in your next to last integral).
     
    Last edited: Jun 2, 2012
  4. Jun 9, 2012 #3
    Okay, thanks. I wasn't sure if you could approximate the numerator but leave the denominator as a variable, but I guess it's okay? And yeah, leaving out the power was just a typo. Not used to using that much latex in one post.

    So, now that the integral is taken care of, can anyone help me see how they approximated the function in the first place? I just can't figure it out.
     
    Last edited: Jun 9, 2012
  5. Jun 10, 2012 #4

    TSny

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    In the denominator of the original integral, try using a trig identity to write sin2θ in terms of cos2θ and then factor out an overall factor of 1-β2 from the expression raised to the 3/2 power.
     
  6. Jun 11, 2012 #5
    Ah! I got it! Using that trig identity was the first thing I tried before I posted, but I thought it didn't work because I ended up with an extra [itex]\beta[/itex][itex]^{2}[/itex] that wasn't in the final equation. But I just didn't notice that in the case of extreme velocities, [itex]\beta[/itex][itex]^{2}[/itex] can just be approximated as 1. I didn't think of approximating it before since it's involved in the [itex]\gamma[/itex] in the numerator, and approximating it everywhere doesn't do much good.

    So both of my difficulties involved not realizing that you can approximate an expression in one part of an equation but leave the variable alone in the rest. For some reason I kept thinking you couldn't do that.

    Thanks a lot, TSny.
     
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