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Derivation of ElectroMagnetics Boundary Conditions

  1. Jun 24, 2013 #1
    As learning laser fundumentals, I've just reviewed the boundary conditions for electromagnetic waves.

    However, I came back to a point that confused me in the past and wanna get it clear now :)

    One of the boundary conditions, regarding the magnetic fields parallel to the medium-interface is


    where [itex]H_{1//} \; and \; H_{2//} \; \text{are magnetic field strengths and} \; J_s \; \text{is the electric current per unit area} [/itex].

    I found the derivation in my text book elements of engineering electromagnetics (6th edition) by Nannapaneni Narayana Rao: (please view the attachment to find corresponding notations):

    [itex]\oint_{C} \vec{H} \cdot d\vec{l} = \int_{S} \vec{J} \cdot d\vec{s} + \frac{d \int_{S} \vec{D} \cdot d\vec{s}}{d \, t}[/itex]

    Note that [itex]bc \approx 0 \; and \; da \approx 0[/itex] so that flux of electric fields is ignored, hence

    [itex]ab \cdot H_{1//}+bc \cdot H_{\perp}+ cd \cdot H_{2//}+ da \cdot H_{\perp}= J_s \cdot AREA_{abcd}[/itex]

    where [itex]ab=-cd, \; bc=-da[/itex], thus

    [itex]ab \cdot H_{1//}- ab \cdot H_{2//} = J_s \cdot AREA_{abcd}[/itex]

    and now, I just can't get [itex]H_{1//}-H_{2//}=J_s[/itex] because [itex]\frac{AREA_{abcd}}{ab} \not= 1[/itex].

    Would anyone help to point out where I'm wrong or share some good references?

    Attached Files:

  2. jcsd
  3. Jun 24, 2013 #2
    I think the problem you're having is that you're treating J, the surface current density, in the wrong way. Why should the total current passing through your contour be proportional to the area of the contour when the current is confined to the interface? You can make the segments bc and da bigger without trapping any new surface current; only the lengths of ab and cd determine how big the total current flux is. You will probably not make this error if you wait until the end to set bc and da to zero; the way you've drawn it makes it look like those legs won't be any problem anyway.

    BTW, your notation is a little unusual, at least to my American eye. (For example, your use of "ab" and "cd" throws me off a little because I'd only use one symbol for a length. Also on the diagram you seem to have your contour drawn with two contradicting orientation arrows.) I hope other people reading your post don't get scared away by the notation.
  4. Jun 25, 2013 #3
    Hi Jolb, thank you for the reply.

    Regarding the notations, sorry for the inconvenience, however

    1. I quoted them from the book elements of engineering electromagnetics (6th edition) by Nannapaneni Narayana Rao so as to avoid stupid mistakes while using my own notations.

    2. It's nothing about American or European or Asian style notation :) "Contour abcd" comprises of "paths ab, bc, cd, da"; Electric/magnetic fields directions should have no relationship with the integral paths whilst the paths are also directional.
    BTW where's the contradicting arrows? If you're referring to the red arrows I should say sorry for being ambiguous: red arrows should be direction marks for the magnetic fields [itex]H_{1//} \; and \; H_{2//}[/itex], I'll correct it soon.

    Now back to the topic of [itex]J_s[/itex], I think it's an easy way to treat [itex]J_s[/itex] as a "confined to the interface" current density. However if the current is only proportional to [itex]ab \; or \; da[/itex], the metric of [itex]J_s[/itex] will be changed from [itex]"A/m^2"[/itex] to [itex]"A/m"[/itex].

    I'm fine with it but the book does not mention any change of metrics in this chapter, and I'm also concerned about that whether the current density "near" the interface should be treated in this way. I'd say that if not confined to text books, I doubt why current density is continuous around the medium interface, because it makes more sense to put [itex]J_{s1}[/itex] and [itex]J_{s2}[/itex] separately in 2 mediums (with the same direction as [itex]J_s[/itex] in the figure) even if [itex]bc \approx 0 \; and \; da \approx 0[/itex] are assumed.
  5. Jun 25, 2013 #4
    Modified Figure

    Here is the modified figure for avoiding ambiguous description.

    Attached Files:

  6. Jun 25, 2013 #5
    Maybe it's an engineering notation, then. With my physics background, my instinct would be to give a separate variable to the lengths of the segments (perhaps w and l) so that I can more easily keep track of units--for example, an expression like lw2 can be easily seen to have dimensions of length3, whereas I might confuse myself with something like ab2. But aside from convenience, it really doesn't matter, and I was able to figure out what you're saying.

    You are right that the magnetic fields do not depend on the orientation of the contour used to calculate the fields, but the contour must have some orientation for the calculation, and it helps to show on the diagram which orientation you used when comparing with the algebra.

    I got confused by your diagram because I thought the red arrows were orientation arrows (I assumed they were orientation arrows since drawing the magnetic field would imply a particular sign for J, which the problem hasn't specified.) The corrected diagram is more clear.

    Well, on a basic level (magnetostatics), you don't need any fancy math like "change of metrics" (do you mean "measure" instead of "metric"?). I think you might be confusing yourself with thinking about fancy math instead of the basic vector calculus that's all over EM. (Maybe I'm totally misinterpreting you--do you mean "unit" instead of "metric"? In physics and math, the word "metric" typically has something to do with geometry.)

    Let me sketch the main ideas here, and please forgive me for using my own notation. (Also, let me work this out "in vacuum"--so I'll work with B rather than H, since we're not too concerned with the properties of the media. It is trivial to do the generalization to linear media.) First of all, let me write down the theorem that I call "Green's Theorem."
    [tex]\oint _C \vec{B}\cdot\vec{dl}=\iint_{C'}\left (\nabla\times\vec{B} \right )\cdot\vec{da}[/tex]

    Hopefully that makes sense to you. The symbol C denotes the closed contour (like the one in your diagram), while C' denotes a surface bounded by the contour C.

    From Maxwell's equations we have:
    Where we have set the permeability of free space μ to be equal to 1. Plugging this into the above equation is what you were using here:
    \oint _C \vec{B}\cdot\vec{dl}=\iint_{C'}\vec{J}\cdot\vec{da}[/tex]
    Now the problem you're having is evaluating the integral on the right hand side. Instead of worrying about measures or "metrics", the point is that the quantity on the right is just the total flux of current through the surface C'. You often see this in physics books written as:
    Since J is a surface current density, you'll want to think of it in the following way: First, we can treat the interface of the two media as a 2-dimensional sheet of conductor, infinite in extent (along both axes).***[See bottom] If we wanted to measure the magnitude of current flowing in this infinite 2D conductor, we would pick some line of length L in the conductor and measure how much current crosses that line. This would be equal to J*L. From this you can see that J would have units of current/length.

    If you think of it this way you'll find Ienclosed=(J amperes/meter)(ab meters enclosed).

    I think what you're attempting to do is to evaluate the integral [itex]\iint_{C'}\vec{J}\cdot\vec{da}[/itex] to find Ienclosed, instead of just realizing it is very trivial. That sort of integral would require some sort of measure theory/distribution theory (you might stick a dirac delta in there to show all the current is isolated on the interface, and then do the integral rigorously), but why bother? Keep in mind the example of using basically the same reasoning to calculate the magnetic field at a distance r from a current-carrying wire (1D conductor).

    *** -- Note: you are right that in real-world scenarios, the current is not confined exactly to the 2-dimensional interface of media. There are fantastically intricate things going on at the molecular scale, and there is indeed a thickness to the "surface current." But your problem deals with the basic theoretical aspects, so we are not concerned with those practicalities; plus the diagram does indicate that all the J is on the surface. In that case, the easiest way to get an answer is to do what I recommended and treat the interface as a 2D conductor.
    Last edited: Jun 25, 2013
  7. Jun 25, 2013 #6
    I just found an excellent reference for you. David J. Griffiths "Introduction to Electrodynamics", Third Edition, page 226, Example 5.8. This is exactly the problem of a 2-D conductor, and he solves it using my method, and earlier in the chapter he explains all the various arguments I invoked. To see how the 2-D conductor is related to the boundary condition problem, see p.241. That's a great textbook in general.
    Last edited: Jun 25, 2013
  8. Jun 25, 2013 #7
    Thank you Jolb ! Your explanation is very clear :) I'll take a look at the book "Introduction to Electrodynamics".

    So when I choose a contour to apply Ampere's Law, I should not care about how the flux (unit: Ampere) is calculated (A/m * m, A/m^2 * m^2), is it right ?

    P.S. I should use "unit" instead of "metric", sorry for my poor English
  9. Jun 26, 2013 #8
    You should always check your units in calculations like this, so I wouldn't necessarily say that. You should be able to figure out what the right units are based on what kind of charge/current distribution you're dealing with. The way I remember the right units for any kind of charge distribution is like this:

    Point Charge (0-D charge)-- has units of Coulombs
    Line charge (1-D charge distribution) -- has units of Coulombs/meter
    Surface charge (2-D charge distribution) -- has units of Coulombs/meter2
    Charged body (3-D charge distribution) -- has units of Coulombs/meter3
    (1 ampere) = (1 coulomb/second) -- this is the definition of ampere.
    This last one is key. Just keep in mind: "a current distribution is a charge distribution in motion."

    To show how you can use these basic facts to get the units for anything, let me do the example of a 2D current distribution. First think of what units are for a 2D charge distribution--Coulombs/meter2. Now we just imagine we put it in motion at 1 meter/second to yield the 2D current distribution. (1 coulomb/meter2)*(1 meter/second) = (1 coulomb/second)(1/meter) = (1 ampere/meter). You can repeat this line of reasoning to get the units for any current distribution.
  10. Jun 26, 2013 #9
    Thank you so much Jolb! Now the answer is good enough for me :)
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