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Oscillating current along antenna, electrodynamics

  1. Feb 10, 2015 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    A thin linear antenna of length ##d## is excited in such a way that it carries the sinusoidal current ##I(t) = I_0 \sin(2\pi z/d)e^{ i\omega t}##, where ##\omega = 2\pi c/d##.

    (i) Beginning with the exact solution for the vector potential in Lorenz gauge, calculate the power radiated per unit solid angle (averaged over a periodic time of the oscillation) far from the antenna and plot the angular distribution as a function of ##\theta##.
    [Hint: The power radiated into a solid angle ##d \Omega## in the direction ##\hat r## is ##dP = \langle \mathbf S \rangle \cdot \hat r dA## where ##dA = r^2d\Omega## and ##\langle \mathbf S \rangle## is the time-averaged Poynting vector. Use that ##\langle \mathbf S \rangle = \langle c \mathbf E \times \mathbf B \rangle = c \frac{1}{2} Re \mathbf E^* \times \mathbf B.] ##

    (ii) Compare your result 2(i) with that obtained from the dipole approximation, and explain where and why this approximation fails in this case.

    2. Relevant equations

    Given in the question


    3. The attempt at a solution

    I believe the required exact expression of the vector potential is $$\mathbf A(\mathbf r, t) = \frac{1}{4\pi r c} \int d^3 r' \frac{J(\mathbf r', t - \frac{|\mathbf r - \mathbf r'|}{c})}{|\mathbf r - \mathbf r'|}$$ In the far field limit, ##r \gg d## and ##\lambda \ll r \Rightarrow kr \ll 1## I am really just looking for some hints to start, I could try expanding ##J## but not quite sure what this will look like. Thanks. I have other expressions for A but they involve dipole terms which is not what we have here.
     
  2. jcsd
  3. Feb 10, 2015 #2
    Hi. I think there's something wrong with your expression for A, it looks like a mixture between the exact integral and the radiation zone formula:
    The former would not have an r outside the integral and would have an additional integral over t';
    The latter would have an additional term exp[i(kr–wt)] outside the integral and the integrand would just be J(r',t)*exp(–ikr⋅r'/r)...
    I understand the problem would have you go from the former to the latter as a first step, since you have to evaluate A "far from the antenna", so how did you get this expression for A?
     
    Last edited: Feb 10, 2015
  4. Feb 11, 2015 #3

    CAF123

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    Hi Goddar,
    Yes you are right, sorry about that.
    My starting point is the expression $$\mathbf A(\mathbf r,t) = \frac{1}{4\pi c} \int d^3 r' \frac{\mathbf J \left(\mathbf r', t - \frac{|\mathbf r - \mathbf r'|}{c}\right)}{|\mathbf r - \mathbf r'|}.$$ Why should there be an integral over t' though? The integrand is dependent on t and this is also present on the LHS.
    So I will write the above expression like $$\mathbf A(\mathbf r,t) = \frac{1}{4\pi c} \int d^3 r' \frac{\mathbf J (\mathbf r') \exp(i \omega( t - \frac{|\mathbf r - \mathbf r'|}{c})}{|\mathbf r - \mathbf r'|}.$$ Which I can reexpress like $$\frac{e^{i \omega t}}{4 \pi c} \int d^3r' \frac{\mathbf J (\mathbf r') \exp(-ik(|\mathbf r - \mathbf r'|)}{|\mathbf r - \mathbf r'|}.$$ Now expanding, $$\frac{1}{|\mathbf r - \mathbf r'|} \approx \frac{1}{r} \left( 1 + \frac{\hat r \cdot \mathbf r'}{r} \right) \text{and} \exp(-ik|\mathbf r - \mathbf r'|) \approx e^{-ikr} (1 + ik \hat r \cdot \mathbf r')$$ I obtain $$\mathbf A(\mathbf r, t) \approx. \frac{e^{-i(kr - \omega t)}}{4 \pi c r} \int d^3 r' \mathbf J(\mathbf r') [1+ (ik + \frac{1}{r} )\hat r \cdot \mathbf r')$$ I think this matches with most of what you say.
    I'm just a bit unsure of how to use this expression for the problem at hand. I have the current distribution which lies only along ##z## so the ##d^3 r'## measure can be replaced with a ##dz## I think and ##I(t) \hat z = \int \mathbf J(z) dz \Rightarrow J(z) = \frac{d}{dz} I(t)?##
     
  5. Feb 11, 2015 #4
    Ok, the A you're starting with is fine (i usually use one where t' is not integrated out yet but this is equivalent).
    Now, in the radiation zone the denominator can just be simplified to r because the correction is very small;
    In the exponential though, look carefully: the expansion of |rr'| should give you r–rr'/r so that when you bring in the harmonic time component of your current you obtain exp[i(kr–wt)] outside the integral – with k = w/c –, and exp(–ikrr'/r) inside.
    Now you still end up with an integral over J(r') , which simplifies greatly since you have a line current along z so the volume integral becomes a line integral along z from 0 to d...
     
  6. Feb 11, 2015 #5

    CAF123

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    Ok, so my integral I am evaluating when I do the expansions is $$\frac{e^{-i(kr-\omega t)}}{4 \pi c r} \int_0^d dz J(z) e^{ik \hat r \cdot \hat z}$$ But what is ##J(z)## here? I have the line current and not the density J(z). I think ##\int_0^d dz J(z) = I## so should I differentiate I to get J(z)?
     
  7. Feb 11, 2015 #6
    Your current is I(z,t) so J(r) = I(z,t)ezδ(x)δ(y) if you wish, and your infinitesimal "volume" goes from dxdydz to just dz; J is a current per unit area...
     
    Last edited: Feb 11, 2015
  8. Feb 11, 2015 #7

    CAF123

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    Oh right I see, so the integrals of the delta functions over x and y go to one and we are left with $$\frac{e^{-i(kr-wt)}}{4\pi c r} \int dz I(z,t)\hat z e^{ik \hat r \cdot \hat z} = \mathbf A(\mathbf r,t)$$ Could you explain why we can write the current density J like that though? As far I can tell, it is written like current density = current times some area.

    Continuing, the above reduces to $$\frac{e^{iwt} \cdot e^{-i(kr-wt)} \cdot I_o}{4\pi c r} \int_0^d dz \sin kz e^{ik}$$ where ##\hat r \cdot \hat z = 1##. Is it ok?
     
  9. Feb 11, 2015 #8
    To give you a quick answer (i think you'll get the picture and we'll save on wordy explanations), these are equivalent:
    J(r) d3r <=> ∫ K(a) da <=> ∫ I(s) ds, where K is a surface current, da and ds are infinitesimal area and length respectively. In other terms:
    J(r) is the amount of charge going through a cross-section of unit-area per second;
    K(a) is the amount of charge going through a cross-section of unit-width per second;
    I(s) is the amount of charge going through a point per second.
    At the end, the three integrals above have same dimensions. For example:
    With J = I(z)δ(x)δ(y): ∫ J d3r –> ∫ I dz and everything is fine.

    What you get for A is almost right: only, r' becomes zez and not just ez. Then, the time-component of I is already outside the integral.
    Certainly related, i don't know where this extra exp(iwt) is coming from in your last expression?..
     
  10. Feb 12, 2015 #9

    CAF123

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    Ok thanks, that makes sense :)
    I see. The extra time component e^(iwt) was coming from the explicit dependence of the expression for I(z,t) given in the question. I would then compute ##\mathbf B(\mathbf r,t) = \nabla \times \mathbf A(\mathbf r,t)## and ##\mathbf E(\mathbf r,t) = \frac{1}{ik} \nabla \times \mathbf B(\mathbf r,t) ##, where ##\mathbf E(\mathbf r, t) = \mathbf E(\mathbf r) e^{iwt}##
    Would it be right to say that the curl expressed in cylindrical polars would be most helpful here? Since the cylinder axis can be taken as the z axis and the problem is symmetric(Only one term is non zero, the ##\partial A_z/\partial r## term). The B field will point in the ##\hat \phi## direction and the E field in the ##\hat z## direction.
     
    Last edited: Feb 12, 2015
  11. Feb 12, 2015 #10
    No! This is what i was telling you: we used this a long time ago when you simplified the exact A and substituted t for [t – |rr'|/c]. This is where the whole exponential outside the integral is coming from (and also where k = w/t is coming from) so don't make the mistake of reinserting exp(iwt) anywhere, just go back to your first steps to make this clear.
    B = ∇ × A indeed, always. But be careful: the r in your expression is in spherical coordinates so you can't just swap it for a cylindrical radius. It may seem confusing because we used mixed coordinates but this was fine since we could get the r's outside the integral and only the z's survived inside. I would then advise you to use the spherical curl, and be careful to convert the unit vector ez to do so.

    Then, although it may sometimes be right, your expression for E is wrong in general so i would suggest you used the (always true) expression in terms of A and Φ (Φ = 0 here of course). I assume you know it.
    Finally: B(r, t) = B(r)exp(iwt) and E(r, t) = E(r)exp(iwt), yes. But this is all coming from the fact that I(r, t) = I(z)exp(iwt) so don't reinsert exp(iwt) anywhere, it's just been along for the ride since the beginning...
     
  12. Feb 12, 2015 #11

    CAF123

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    Thanks, it is clear now.
    In the end, ##\mathbf A(\mathbf r) = f(r) \hat z## so ##\nabla \times \mathbf A(\mathbf r) \propto \hat r \times \frac{\partial f}{\partial r}\hat z## I think may be a simplification.

    Could you explain why it isn't valid? I just see that it follows from Maxwell's fourth equation $$\nabla \times \mathbf B(\mathbf r, t) = \frac{1}{c} \partial \mathbf E(\mathbf r,t)/\partial t$$ if we write ##\mathbf E(\mathbf r,t) = \mathbf E(\mathbf r) e^{iwt}##. The other equation you mention i know of, but why is phi zero?

    I spoke to a tutor today about this question and it seems whether or not you do the integral from -d/2 to d/2 or from 0 to d affects the final result, which is strange. He said he would get back to me on that.

    Thanks!
     
  13. Feb 12, 2015 #12
    The curl is not that simple on the derivatives, especially since the z unit-vector must be converted in spherical coordinates... The whole thing is not very complicated though, just don't mix up notations because that would throw all your results away.

    For E, if you derived this relation from the problem at hand it is fine, i just had the impression that you quoted it from somewhere which is why using the time-derivative of A seemed safe. And Φ is zero because the net charge in your antenna is 0 and the contribution to E you're getting is from the change in current, not from a free charge distribution.

    The fields are affected by where you decide to put the antenna (bottom at 0 or –d/2), but indeed the total power radiated shouldn't be sensitive to that; i didn't go through the calculations myself but from your last post it seemed like you haven't either (or might have made some mistakes) so are you sure that your final result is affected? (you need the fields, the Poynting vector and a final integration to get there, and all without mistakes...)

    Edit: after a check on the integral that gives A, the change in placement of the antenna only changes the fields by a phase (effectively a minus sign) and won''t affect the total power radiated.
     
    Last edited: Feb 12, 2015
  14. Feb 15, 2015 #13

    CAF123

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    Ok, I see, yeah but just to check what I write is valid if the vector ##\hat z## happened to be a constant vector? From doing the integral from -d/2 to d/2 the result is $$\mathbf A(\mathbf r,t) = \frac{iI_od e^{-ikr}}{4 \pi^2 r c} \frac{\sin (\pi \cos \theta)}{\sin^2 \theta}\hat z$$ (confirmed with a tutor here) while from 0 to d it is $$\mathbf A(\mathbf r,t) = \frac{I_o d e^{-ikr}}{8\pi^2 cr} \frac{1 - e^{i2\pi \cos \theta}}{\sin^2 \theta}\hat z$$ (seen by a tutor and could see no problems with it)

    Yep, I used the fourth maxwell eqn together with the form ##\mathbf E(\mathbf r,t) = \mathbf E(\mathbf r) e^{iwt}##

    Ok.

    In spherical polars, ##\hat z = \cos \theta \hat r - \sin \theta \hat \theta## and thus use the curl expression in SP's with only derivatives in ##A_{\theta}## and ##A_{r}## terms contributing
     
  15. Feb 15, 2015 #14
    Ok, first: 1–exp(2πi cosθ) = –2i sin(πcosθ)*exp(iπ cosθ),
    So as i said earlier the difference between the two integrals is just a phase (a factor of magnitude 1 that doesn't depend on r) and won't affect the total power but to see this in detail you have to look at the final integral over dP = r2< S >⋅erdΩ (i refer you to your original post):
    Since you don't want the total power radiated to depend on r (that wouldn't make sense physically) you see that S must be proportional to 1/r2 in order to cancel the integration term. So effectively, when computing your fields (in the approximation used) you can discard all terms that die off faster than 1/r so that E*x B is indeed proportional to 1/r2. This takes care of any phase, even when it depends on θ, since you'll see that all derivatives of A with respect to θ go away and the complex conjugation of E takes care of the phase itself.

    Also: what happened to the time-dependence exp(iwt) of A? If you already took the time-average, don't express this as A(r, t).

    Then, a detail of your calculation seems wrong:
    i think the integral to get A should be: ∫dz'[sin(2πz'/d)*exp(ikz'cosθ)], which doesn't give the denominator you obtained so you might want to check that because i think you made a wrong simplification there...
    A note about your last post: Ampere/Maxwell law (in Gaussian units) gives, as you know: ∇ x B = 4πJ/c + ∂tE/c, so it's fine to use your derivation for E here far from the antenna but you can't disregard the first term in general, that was my point...

    Yes. Just make sure you have the right expression for the curl, check this for example:
    http://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates
    Many of the terms in the curl don't survive as i said earlier so the final derivative shouldn't be too complicated.
     
  16. Feb 16, 2015 #15

    CAF123

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    Ok, thanks.

    Yes, what I gave should only have been ##\mathbf A(\mathbf r)##.

    Oh, this is the same integral I computed and got the results shown in my last post up to a minus sign, with the two results only differing in a phase. Can I just check you are getting the B field to be of the form $$\mathbf B(\mathbf r) = \frac{-k e^{-ikr}I_o d \sin (\pi \cos \theta)}{4 \pi^2 cr \sin \theta} \hat \phi?$$ My instructor has the same up to a minus and has ##\sin^2 \theta## in the denominator there. I think there should only be one power of sin theta there because one of them cancels from reexpressing the unit vector z in SP's. We are using lorentz heaviside units in this class.

    Oh right, sure yeah thanks. Since we're far field, I set J=0 and that gives the other formula for E.
     
  17. Feb 16, 2015 #16
    Seems like you're right here (disregard my last comment about denominator of A): you indeed get only one power of sin(θ) in the denominator of B so at least up to a constant depending on your units, this should be right. (by the way, B simplifies further since k = 2π/d)
    I won't have time to be on the forum this week so if you have more questions i would advise you to open a new thread... Good luck!
     
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