Index of refraction and planes of oscillating chargesby cheeseits Tags: charges, index, oscillating, planes, refraction 

#1
May1110, 11:53 AM

P: 4

Assume we have a plane of oscillating charges described by x=x_{o}e^{i(omega)t}.
The total field is given by E= [tex]\frac{\etaq}{2\epsilonc}[/tex]i[tex]\omega[/tex]x_{o}e^{i[tex]\omega[/tex](tz/c)} . (Can someone point me to a website where I can learn to type in these formulas and not not have them suck?) Ignore that weird looking x^o looking thingy. Those symbols on top of each other is supposed to be a fraction. i = sqrt(1) as usual. Can someone please explain the physical origin of the 90 degree phase shift from the oscillators. I.e. why there is an i in front of the term. If I put z=0 shouldn't I get radiation that is in phase with the oscillators? If a EM wave comes in from infinity and hits a plane of charges the charges oscillate and produce more EM waves. The new waves will not be 90 degrees out of phase with the incoming radiation, right? Is there a difference in these two situations? Why I'm thinking about this: The origin of the index of refraction seems to lay squarely on the relationship between the phases of the incoming radiation and the radiation produced by the charges that are oscillating because of the radiation. If it's not 90 degrees different then em waves won't be delayed in speed in a medium. Reference: Volume 1 lecture 31, Feynman lectures on physics. EDIT: Okay that above formula now looks even more screwed up. It's supposed to be this: E = eta*q/(2*epsilon_nought*c) * i omega x_o * exp(i omega (tz/c)) 



#2
May1110, 03:17 PM

Mentor
P: 11,230




#3
May1110, 09:21 PM

P: 166

That formula is only valid at a point far away from the plane. At that point, only the field coming from the oscillators a distance z away will be in phase with the incoming wave. However, the formula is obtained by integrating the field from all the oscillators in the plane, whose distances from the point range from z to infinity. The net result is that the field is 90 degrees out of phase.




#4
May1210, 02:53 AM

P: 4

Index of refraction and planes of oscillating charges
You say that the formula is valid only for large values of of z.
Total field at P: [tex]  \frac{q \eta}{2 \epsilon_{o} c} i \omega x_{o} e^{i \omega (t \frac{z}{c} )}[/tex] Thanks for the link to the equation tex thingy. Just so you know my source for this equation: On page 283 of Feynman lectures on physics volume 1, chapter 30 section 7 it says: "We may add, by the way, that although our derivation is valid only for distances far from the plane of oscillatory charges, it turns out that the formula (30.18) or (30.19) is correct at any distance z, even for z < [tex] \lambda [/tex]." Where 30.18 and 30.19 is the above formula. So although the derivation, or the reasoning by which he arrived at the answer by doing the integration you described would not lead us to believe that this formula is correct for all z, it turns out that it is. If you have some reason to believe it really isn't correct please say so. Otherwise I'm just trying to understand what the physical reason for the oscillating plane producing radiation that is 90 degrees out of phase of the oscillators is. So does anyone know where this phase shift is from? 



#5
May1210, 03:27 AM

P: 166

You're right. I missed that part where he says its valid for all z. Perhaps its the same reasoning as to why an infinite plane of static charge has the same field for all z. No matter how close you get, it never looks like the field from a point charge. For any finite z, you still have an interference due to charges whose distances range from z to infinity. In other words, you always have to add together fields that lag behind the applied field more and more as you include charges farther and farther away. The net result is that the field lags by 90 degrees.
Mathematically, no matter what, you still have to integrate an [itex]e^{i\omega t}[/itex], which brings down a factor of i. 


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