Or better yet:
I understand how the coatings work in terms of the destructive interference aspect (the quarter-wave thickness results in the top and bottom reflections being out of phase), I just don’t understand how destroying something that is going in one direction adds to something going in the opposite direction. If those links explained it then it went way over my head. I am not a physicist (this much must be clear by now).
Think of it as conservation of energy. The electromagnetic wave carrys energy in the direction of its propogation in an isotropic material. When reflecting from a boundary, the reflected wave carries some energy in the opposite direction. Thus the transmitted light has less energy than the incident light. If there is destructive interference for the reflected wave, then there is no energy being reflected and all the energy must be transmitted. There is no gain, only conservation.
that is counterintuitive though, is it not; that because something bounces back and is destroyed in the process, it isn’t destroyed and doesn’t bounce back and thus passes through…
Mangetout, just think of the lens coating as a resonance cavity. Light goes into the cavity and depending on the cavity size, it’ll either all go out the back end or all from the front end.
it is the diagrams that show the two rays coming back out and fizzling each other out that confuse…
At the (electric field) nodes, the electric field is zero, so the Poynting vector would have to be zero. There would be energy in the magnetic fields there. Interleaved with the electric field nodes would be the magnetic field nodes, which would also have zero Poynting vector, but would have energy due to the elctric field. In between these nodes, the Poynting vector would oscillate back and forth.
The little bit going backwards after reflecting off the second interface then (partially) reflects off the first interface. This reflected part adds in phase with the original transmitted part, increasing its power.
Part of how this works out is that the power is the square of the field. So a 4 percent loss in power would correspond to a reflected field of 20 percent of the incident field, and likewise the transmitted field is 98 percent of the incident field (and 96 percent of the power).
So take the incident field amplitude to be 100, propagating to the right. Then the forward propagating field in the AR layer is 102 (98 from the transmitted incident field, and 4 from the reflection of the backwards travelling field). That backwards propagating wave has an amplitude of 20 (but remember, only 4 percent in terms of power). At the second interface, the forward traveling wave with an amplitude of 102 is what generates the backwards propagating wave. To the right of the AR layer, the amplitude of the wave is back down to 100, after the reflection. (There’s a little round-off error in the above.)
I’m sorry, but this is incorrect. You are thinking of the time averaged Poynting Vector. Do the math… A plane wave propogating in the z direction and polarized in the x direction is given by:
E+ = 1/2 Eo (e[sup]i(kz-wt)[/sup] +e[sup]-i(kz-wt)[/sup]) x
Where x represents a unit vector in the x direction.
A plane wave propagating in the minus z direction minus z direction is:
E- = 1/2 Eo (e[sup]i(kz+wt)[/sup] +e[sup]-i(kz+wt)[/sup]) x
The corresponding maginetic fields are:
H+ = 1/([symbol]m[/symbol]c) k X E = 1/2 E0 /([symbol]m[/symbol]c) (e[sup]i(kz-wt)[/sup] +e[sup]-i(kz-wt)[/sup]) y and
and likewise:
H- = 1/2 E0 /([symbol]m[/symbol]c) (e[sup]i(kz+wt)[/sup] +e[sup]-i(kz+wt)[/sup]) y
The Poynting vector is (Gaussian Units):
S=c/(4 [symbol]p[/symbol]) E X H = c/(4 [symbol]p[/symbol]) (E+ + E-) X (H+ + H-)
If you do the simplification and collect all the cross terms, you will find that the poynting vector oscillates back and forth from the positive z direction to the negative z direction at a frequency of 2w at all points in space. When taking the time average, it is indeed zero, but there is energy in the electromagnetic field, even at the nodes. This was a problem on the midterm I took last week, trust me on it.
Actually, you are right… I woke up this morning (with a slight headache from the three glasses of wine I drank) and thought “what have I been posting?” The Poynting vector is indeed identically zero at the nodes.
However, there is still electromagnetic energy at the nodes. The time rate of change of the energy density is equal to the negative divergence of the poyning vector. The divergence of the poyning vector at the nodes is not zero, and does oscillate at a frequency of 2[symbol]w[/symbol].
I’m not time averaging, but I am admittedly doing this in my head. I feel confident in my answer because this is the situation you’d have with normal incident reflection from a PEC surface. Clearly, the Poynting vector has to be zero at the PEC surface, since E = 0 and also since there is no energy flow into the surface. Likewise, every half wavelength away from the PEC surface, E = 0.
Using phasor notation (which has the e[sup]jwt[/sup] suppressed, not averaged out),
E+ = e[sup]jkz[/sup] and E- = -e[sup]-jkz[/sup].
E = e[sup]jkz[/sup] - e[sup]-jkz[/sup] = 2sin(kz)
H+ = e[sup]jkz[/sup] and H- = e[sup]-jkz[/sup]
H = e[sup]jkz[/sup] + e[sup]-jkz[/sup] = 2cos(kz)
Puting back in the e[sup]jwt[/sup] time variation,
E = 2sin(kz) * e[sup]jwt[/sup]
H = 2cos(kz) * e[sup]jwt[/sup]
I’m not saying there isn’t energy at the nodes (see my above post), just that the time varying Poynting vector s zero there.
[sigh]And on preview I see your last post. Too damn bad, I typed it in, I’m posting it! 
I don’t know much about this but doesn’t the Poynting vector always have to point to the load?
I’m not sure I understand your question. The Poynting vector points in the direction of power flow for an electromagnetic wave. This is usually the same direction as the wave propagation direction (i.e., the direction of the wave’s momentum vector, k). The exception to this being in anisotropic materials; where the direction of energy flow and the direction of the momentum vector are not necessarily the same.
I have a feeling that you are coming from the standpoint of RF Electromagnetism and antenna theory. I know nothing about this subject, but I would think that the Poynting vector would always have to point in the direction of the load (if by load you mean receiver). If the transmitted wave is emitted as a (roughly) spherical wave (i.e. radio broadcasting), the Poynting vector points racially outwards. Thus the vector would always point toward a receiver. If instead of a spherical wave, you emit some type of directional beam, you must point the beam (and with it the Poynting vector) toward the receiver or they won’t receive it. So the answer to your question is probably yes…
I suspect Sacroiliac is thinking about circuits. If you have a simple circuit with a battery and a resistor, the Poynting vector will point into the resistor, and out from the battery. Strictly speaking, the Poynting vector does not represent power flow, although it’s often useful to think of it that way. The integral of the Poynting vector over a surface enclosing a volume gives the power flow into that volume. But, e.g., a stationary charge in a non-varying magnetic field will produce a non-zero Poynting vector, implying power flow, even though nothing is moving.
Yes, I agree…
(from above, and yes I spelled it poyning, Eye will preview my posts, Eye will preview my posts… :smack: )
However, can you help me out with this?
The magnetic field produced by a current in a wire curls around the wire IIRC. The electric field created by a battery will run parallel to the wire (E=V/d?). Is the Poynting Vector still defined by EXH? I have never run across the use of the Poynting vector in circuits, how is it used and defined?
quote:
Originally posted by Ring
The ony way to get complete cancellation across the whole beam is to send the two beams out of the same source, with same intensity and opposite phase.
Can’t you bend light with gravity? You should be able to combine two beams with opposite phase. What then?
Inside the wire the E vector will be parallel to the wire, but outside the wire it will be radial. This makes sense since the majority of energy transport is via the fields outside the conductor.
The propagation speed for a 1 mhz radio wave in copper is about 400m/s. So, obviously, the energy must be transported outside the conductor.
Got it. Thanks!
Yes. This gives you the odd picture (if you picture the Poynting vector being power flow) that the energy disspated in the resistor is somehow flowing in through its sides. But I guess I don’t really know that the Poynting vector is actually used much in circuit theory. This was all from my memory of an example from Fields and Waves in Communication Electronics by Ramo, Whinnery and Van Duzer. Example 3.12a has a little picture with P pointing radially into a lossy wire, which popped into my head when I read Sacroiliac’s post.
There will be a component of E parallel to the wire outside the wire, since tangential E is continuous. And for the DC case, there doesn’t have to be a radial component.
Um, I don’t understand this statement. If the Poynting vector represents the energy flux density then it seems that the E field would have to have a radial component outside the conductor. And given that there must be a return path this would seem to be the case. Although I’m not sure about this.
Suppose you have two oppositely charged PEC parallel plates, with a lossy wire connecting them in the middle. The electric field at the wire is parallel to the wire (any radial component to the field can be made as small as you want by making the plates larger). If you believe “the Poynting vector represents the energy flux density”, you’re stuck with having the energy dissipated in the wire coming in between the plates from their edges. I can’t prove that’s wrong necessarily, but I don’t like it.