1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Please explain this from a text book

  1. Jul 29, 2013 #1
    See attached scanned pages of page 118(left side) and page 125(right side) of Antenna Theory by Balanis.

    On the left side, it is just some basic EM where

    [tex]\nabla\times \vec E=-\frac{\partial \vec B}{\partial t}\;\Rightarrow\;\vec E=-\nabla V-\frac{\partial \vec A}{\partial t}[/tex]

    Then to Lorentz thingy ##\nabla\cdot \vec A=-\mu\epsilon \frac{\partial V}{\partial t}##.

    And so on......

    The only thing different is it is in time harmonic form where ##\frac{\partial}{\partial t}\rightarrow\;j\omega##.

    Now going to the right side. I underlined the sentence that I have no idea where that comes from!! What is ##\frac {1}{r^n}=0## where n=,2,3,4.....

    Please explain what is that. This supposed to be very basic thing that

    [tex]\vec B=\nabla\times\vec A[/tex]
    Then for far field where it is assumed plane wave,##\vec E=\eta\hat R \times \vec H##
    This is a very basic knowledge of finding E and H from A. Where is all the what looks like numerical stuff come from?

    Please help.


    Attached Files:

    Last edited: Jul 29, 2013
  2. jcsd
  3. Jul 30, 2013 #2

    If you look at (3-58a) [itex]\tilde {E}=-j\omega \vec{A}[/itex]

    This is nothing more than
    [tex]\nabla\times \vec{E}=-\frac{\partial{\vec{B}}}{\partial{t}}=-\nabla \times \frac{\partial{\vec{A}}}{\partial{t}}\;\Rightarrow\;\nabla\times \vec{E}+\nabla \times \frac{\partial{\vec{A}}}{\partial{t}}=0\;\Rightarrow\;\nabla\times\left[\vec{E}+\frac{\partial{\vec{A}}}{\partial{t}}\right]=0[/tex]

    Why all the fuzz?
  4. Jul 30, 2013 #3


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    This seems to be a somewhat messy book, to say it friendly ;-). Anyway, what the authors means is that in the far-field approximation (i.e., the em. field in a region very far from the charge and current distributions causing this field) you can perform an expansion in powers of [itex]1/r[/itex]. This is equivalent to the decomposition into spherical waves (multipole expansion). In the leading order you are left with the pure dipole component (except for the special case that the charge-current distributions are such that the field starts at higher order, but I guess that subtlety you can neglect for the moment).
  5. Jul 30, 2013 #4
    Thanks for the reply. Under what topic can I find more info? Is it "multi pole expansion" or some other name? Do you have link for some article? I have not study this before.

    You think I can ignore all these and use what I have in post #2?

  6. Jul 30, 2013 #5


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    For harmonic time dependence, i.e.,
    [tex]\vec{E}(t,\vec{r})=\exp(\mathrm{j} \omega t) \vec{E}'(\vec{r}),[/tex]
    and for all other fields too, you have
    [tex]\vec{E}=-\vec{\nabla} \vec{\Phi}-\partial_t \vec{A}[/tex]
    [tex]\vec{E}'=-\vec{\nabla} \Phi'-\mathrm{j} \omega \vec{A}'[/tex]

    For the Lorenz (it's the Danish physicist Ludvig Lorenz who was the first to simplify Maxwell's equations by the gauge condition) gauge you have (for [itex]\mu=\epsilon=1[/itex])
    [tex]\Box \Phi=\rho, \quad \Box \vec{A}=\vec{j}[/tex]
    which translates into
    [tex](-\Delta - k^2) \Phi'=\rho', \quad (-\Delta-k^2) \vec{A}'=\vec{j}'[/tex]
    with [itex]k=\omega/c[/itex].
    The wave equation thus translates to the Helmholtz equation for harmonic time dependence.
  7. Jul 30, 2013 #6
    Thanks, I have to do some reading. Can I ignore these and just use what I posted in post #2?

    Edit: I found a chapter in Griffiths Introduction to Electrodynamics page 146 on Multipole Expansion.
    Last edited: Jul 30, 2013
  8. Jul 30, 2013 #7
    Can anyone tell me why Balanis going through all the fuzz? What's wrong with the way I did it in this quote?
  9. Jul 30, 2013 #8


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    I don't understand your reasoning. Let's do it systematically.

    You start with the homogeneous Maxwell equations (for simplicity I use Heaviside-Lorentz units with c=1 and consider all fields in the vacuum):
    [itex]\partial_t \vec{B}+\vec{\nabla} \times \vec{E}=0, \quad \vec{\nabla} \cdot \vec{B}=0. \qquad (1)[/itex]
    The last equation implies the existence of a vector potential, such that
    [tex]\vec{B}=\vec{\nabla} \times \vec{A}. \qquad (2)[/tex]
    Using this in the first equation implies
    [tex]\vec{\nabla} \times (\partial_t \vec{A}+\vec{E})=0.[/tex]
    This implies that the field in the bracket has a scalar potential (at least locally)
    [tex]\vec{E}=-\vec{\nabla} \Phi-\partial_t \vec{A}. \qquad (3).[/tex]
    For a given em. field [itex](\vec{E},\vec{B})[/itex] the potentials are not uniquely defined, but you can always use
    [tex]\vec{A}'=\vec{A}-\vec{\nabla} \chi, \quad \Phi'=\Phi+\partial_t \Phi[/tex]
    with an arbitrary scalar field [itex]\chi[/itex] (gauge invariance of Maxwell's theory).

    As we'll see in a moment, it's very convenient to (partially) "fix the gauge" by demanding the Lorenz gauge condition,
    [tex]\partial_t \Phi+\vec{\nabla} \cdot \vec{A}=0. \quad (4)[/tex]
    Now you can consider the "inhomogeneous" Maxwell equations.
    [tex]-\partial_t \vec{E}+\vec{\nabla} \times \vec{B}=\vec{j}, \quad \vec{\nabla} \cdot \vec{E}=\rho.[/tex]
    Using (2-4) you find
    [tex]\partial_t^2 \vec{A}-\Delta \vec{A}=\vec{j}, \quad \partial_t^2 \Phi-\Delta \Phi=\rho. \qquad (5)[/tex]
    For harmonic time dependence, i.e., writing all fields in the form
    [tex]F(t,\vec{x})=\exp(\mathrm{j} \omega t) F'(\vec{x}),[/tex]
    you find from (2) and (3)
    [tex]\vec{E}'=-\mathrm{j} \omega \vec{A}'-\vec{\nabla} \Phi', \quad \vec{B}'=\vec{\nabla} \times \vec{A}',[/tex]
    and from (5)
    [tex]-(\omega^2+\Delta) \vec{A}'=\vec{j}', \quad -(\omega^2+\Delta) \Phi'=\rho'.[/tex]
    This is the starting point to solve the problem of the radiation from a given charge-current distribution for harmonic time dependence. From this you get more general cases, using the technique of Fourier transform.
  10. Jul 30, 2013 #9
    Thanks, I have to take a little time to work through this.
  11. Aug 3, 2013 #10
    I reviewed multipole expansion and come back, here. I see what I did wrong.
    [tex]\vec{E}=-\nabla \Phi-\frac{\partial{\vec{A}}}{\partial{t}}=-\nabla \Phi-j\omega\vec{A}[/tex]

    How do I get to
  12. Aug 3, 2013 #11


    User Avatar
    Science Advisor
    Gold Member
    2017 Award

    I don't know, how you got there. Perhaps, you've constraint the Lorenz gauge condition further to radiation gauge, which you can do in the source-free region. The radiation gauge conditions are
    [tex]\Phi=0, \quad \vec{\nabla} \cdot \vec{A}=0.[/tex]
  13. Aug 3, 2013 #12
    This is in the attachment in the original post. I have studied the multipole expansion, it doesn't seem to help. I am at a lost. I yet have to review Gauge transformation, just barely finish the multipole today. Please help.

  14. Aug 3, 2013 #13
    what you have written is valid only in the radiation zone i.e. far field zone.In this zone only 1/r contribution survives,all other higher powers are neglected.If you will compare 3.57 to 3.56,you will immediately see the result.
  15. Aug 3, 2013 #14
    Yes, this is exclusively in the far field zone only. Problem is I don't know how to derive 3.55-3.57
  16. Aug 3, 2013 #15
    Can anyone show me how to derive 3-56 and 3-57 from 3-55? That would really help.

  17. Aug 3, 2013 #16
    To get 3.56,you have to solve helmholtz eqn i.e. the eqn which is satisfield by A.You also have to use an expansion of 1/|x-x'| to get all those 1/rn terms and then you will have to just put it into 3.15 to get E.Let me see if I can find a proper reference(I remember Jackson has treated something similar in the chapter on waveguides).
    EDIT-Somewhat useful here

    Attached Files:

    Last edited: Aug 3, 2013
  18. Aug 3, 2013 #17
    Thanks for the info. Wow, this is a lot of study for me. This is beyond what I studied. I only studied Griffiths and two other Engineering EM books which is even simpler than Griffiths in this part. They are more concentrated on transmission lines etc. instead.

    Regarding to the article, can you:

    1) give me some info how to get
    [tex]\vec A(x,t)=\frac{\mu_0}{4\pi}\int d^3x'\int dt'\frac{\vec J(x',t')}{|\vec x-\vec x'|}\delta\left(t'-t+\frac 1 c|\vec c-\vec x'|\right)[/tex]
    2) Is ##\delta\left(t'-t+\frac 1 c|\vec c-\vec x'|\right)## a Dirac Delta function? where
    ##\delta\left(t'-t+\frac 1 c|\vec c-\vec x'|\right) = \begin{cases} +\infty, & \left(t'-t+\frac 1 c|\vec c-\vec x'|\right) = 0 \\ 0, & \left(t'-t+\frac 1 c|\vec c-\vec x'|\right) \ne 0 \end{cases} ##

    3) What is ##\int d^3 x'##?

    4) Can you give me the pages of JD Jackson that talk about this subject? I have the book at home, but I am out of town so I can't get to it. I am just trying very hard to understand how to get to ##\vec E=-j\omega\vec A##. Is there an easier to get to this?

    Last edited: Aug 3, 2013
  19. Aug 3, 2013 #18
    This may not be what you're looking for, but here's how I understand the idea of far field radiation. Maybe it'll help you find what you're looking for...

    We know that the amplitude of the Poynting vector is [itex]\left| \vec{E}\times\vec{H} \right|[/itex] which is proportional to [itex]\left| \vec{E} \right|^2 [/itex]. So the power per unit area is proportional to [itex]\left| \vec{E} \right|^2 [/itex]. If we integrate the power over the surface of a sphere of radius [itex]r[/itex], we'll get a total power which is proportional to [itex]4\pi r^2 \left| \vec{E} \right|^2[/itex]. So if the amplitude of [itex]\vec{E}[/itex] is proportional to [itex]1/r^n[/itex], then the total power over a sphere of radius [itex]r[/itex] is proportional to [itex]r^2/r^{2n}[/itex].

    If you take the limit [itex]r \to \infty[/itex] (far field), you see that the total power is zero for any terms with a [itex]1/r^n[/itex] dependence where [itex]n>1[/itex]. You also see that any terms with a [itex]1/r^n[/itex] dependence where [itex]n\leq 0[/itex] make no sense because they give you infinite power in the far field. Finally, any terms with a [itex]1/r[/itex] dependence give you a finite total power at the far field. So we call these the "far-field" terms, because they're the only terms which give us a finite, non-zero total power through any closed surface which is very far from the source. I think that's all he's really saying here.

    Now, here's something I'm guessing at which will maybe help you figure out the more "rigorous" aspect of this, but I'm definitely not confident about it:

    I *suspect* that the idea of breaking down a wave into [itex]1/r^n[/itex] components isn't actually very useful in general. The reason I think this is that the more "natural" way to break things down in spherical coordinates would be to use spherical harmonics, which are based on stuff like bessel functions and legendre polynomials. They would have nice properties like orthogonality which the [itex]1/r^n[/itex] "basis" would not. Orthogonality makes it much easier to break a function down into a sum of those components. (That's the advantage of using Legendre polynomials instead of just [itex]1, x, x^2, x^3, \ldots[/itex]). So trying to find all the [itex]1/r^n[/itex] terms might not be very fun.

    That said, in the case of finding far-fields, you don't need all the terms because you're killing off everything except for the [itex]1/r[/itex] term. So in that case, all that matters is the fact that you could break things down into a [itex]1/r^n[/itex] sum. You don't need (or want) to know how to actually do that in general. Of course, the advantage of using a [itex]1/r^n[/itex] decomposition rather than a spherical harmonic decomposition for far-fields comes from what I discussed about power above. With spherical harmonics, I *think* (not 100% sure) that the disadvantage is that all the components carry some power out to infinity and lose some power along the way, so to find the field at [itex]r \to \infty[/itex], you'd have to worry about every single term. The easier way is to say "I could break things down into a [itex]1/r^n[/itex] sum," and then argue that only the [itex]1/r[/itex] term would survive at infinity, so you only have to worry about that one.

    Like I said, I'm not completely sure about that because I'm not an expert on spherical harmonics, but what I said seems reasonable to me. Maybe it'll at least help you find an explanation you can be confident in.
  20. Aug 3, 2013 #19
    I am reviewing the gauge transformation and study back what you wrote:
    [tex]\Box \vec A=\frac {1}{c^2}\frac{\partial^2 \vec A}{\partial t^2}-\nabla^2\vec A=\mu_0\vec J\;\Rightarrow\;(-\nabla-k^2) \vec{A}=\mu_0\vec{J}[/tex]
    [tex]\Box V=\frac {1}{c^2}\frac{\partial^2 V}{\partial t^2}-\nabla^2V=\frac{\rho}{\epsilon_0}\;\Rightarrow\;(-\nabla-k^2) V=\frac{\rho}{\epsilon}[/tex]

    How come you gave
    [tex](-\Delta - k^2) \Phi'=\rho', \quad (-\Delta-k^2) \vec{A}'=\vec{j}'[/tex]
    which in my terms:
    [tex](-\Delta - k^2) \Phi=\frac{\rho}{\epsilon_0}, \quad (-\Delta-k^2) \vec{A}=\mu_0 \vec{J}[/tex]

    Do you mean ##\nabla##, not ##\Delta##?
    Last edited: Aug 3, 2013
  21. Aug 3, 2013 #20
    Is this one
    [tex]-(\omega^2+\Delta) \vec{A}'=\vec{j}', \quad -(\omega^2+\Delta) \Phi'=\rho'.[/tex]
    supposed to be:
    [tex]-(\omega^2+\nabla) \vec{A}'=\vec{j}', \quad -(\omega^2+\nabla) \Phi'=\rho'.[/tex]

    I gone through this derivation. How does this bring me any close to my original question? I need to derive (3-56) and (3-57) from (3-55) in my original attachment, and ultimately (3-58). I don't see any relation to my question.

    So far, I spent two days studying multipole expansion and today on the Gauge transformation. But I don't feel I am any where closer to derive the equations!!! Can you tell me what am I missing, how do I tie them together?

    Last edited: Aug 3, 2013
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook