Lecture 14 (Poynting Vector and Standing Waves) Physics 6-01 Spring 018 Douglas Fields
Reading Quiz For the wave described by E E ˆsin Max j kz t, what is the direction of the Poynting vector? A) +x direction B) -x direction C) +y direction D) +z direction E) -z direction
Review of Power in Waves Let s examine a wave moving to the right on a string. At point a, the force from the string to the left, F, can be broken down into the horizontal and vertical components using the slope of the string: And then the power at a is just given by the force times the velocity of the piece of string at that point: y x, t y x, t P x, t F x, tv x, t F Since, Then, And the power is given by: Thus, the time averaged power is given by: F x, t F y y y x, t Acos kx t, y Notice that it depends on the amplitude squared y x, t x x y x, t y x, t kasin kx t Asin kx t x t t, sin P x t Fk A kx t 1 PAvg FkA
Power in an EM Wave Let s review what we know about the energy stored in fields: So, at any point in the electromagnetic wave, But, E cb B 1 1 1 ue 0E, u B B 0 1 1 1 u ue ub 0E B E c 0 So 1 1 1 1 1 u E E E E E 0 0 0 0 0c c 0 1 0 0 0 0
Power in an EM Wave But this is the energy density at a localized point in space. We want to describe how this energy moves with the wave. Let s examine the energy (du) in a plane wave that passes through a certain area (A) within a certain time (dt). du udv ua cdt du 0E Acdt
Poynting Vector du 0E Acdt Now, let us define the energy per unit time, per unit area as: Notice that it scales with the square of the field (wave amplitude). Now, just for fun, let s rewrite this using E = cb: And put it in vector form to denote the direction of energy transmission: 1 du S c 0E A dt 1 S c 0EB EB 0 0 1 S E B Poynting Vector
The drawing shows a sinusoidal electromagnetic wave in a vacuum at one instant of time at points between x = 0 and x = l. At this instant, at which values of x does the instantaneous Poynting vector have its maximum magnitude? A. x = 0 and x = l only B. x = l/4 and x = 3l/4 only C. x = l/ only D. x = 0, x = l/, and x = l
The drawing shows a sinusoidal electromagnetic wave in a vacuum at one instant of time at points between x = 0 and x = l. At this instant, at which values of x does the instantaneous Poynting vector have its maximum magnitude? A. x = 0 and x = l only B. x = l/4 and x = 3l/4 only C. x = l/ only D. x = 0, x = l/, and x = l
Poynting Vector Remember, the Poynting vector represents the energy per time per unit area instantaneously passing through an area at a particular point in space. S energy time area power area So, if we want to know the entire instantaneous power output from a source, we have to integrate the Poynting vector over a closed surface containing the source: We ll return to this in a moment, but let s first get rid of that annoying term: instantaneous. P S da
Average Power (per unit area) The time average of the Poynting vector will depend on the its time dependence, and over what length of time you average. For instance, look at the following pulse: S r 0 c r If we integrate over the time from when the pulse just reaches r 0 until the maximum reaches r 0, we will get one answer. If we integrate over all time, we will get an answer that will approach zero So, when we speak of average power, we either have to be precise in our description of the time integration, or
Average Power (per unit area) Assume a never-ending sinusoidal plane wave. 1 S E x, t B x, t 0 1 ˆjE sin kx t kb ˆ sin kx t iˆ Max Max 0 EMaxBMax sin kx t 0 Then, because the average of sin is ½, 1 E 1 I S E B ce Max Avg Max Max 0 Max 0 0c Where, I is the intensity of the EM wave.
Example Consider a source of EM waves, say a light bulb that puts out 100W spread equally in all directions. What is the value of the intensity of light 10m from the bulb? P 100W 1W S Avg I A r 4 m 4 10m But remember that the intensity is proportional to the amplitude squared: 1 E 1 I S E B ce Max Avg Max Max 0 Max 0 0c So, as pointed out earlier, for spherical waves. E E r t kr t r 0, sin
Electromagnetic Momentum Flow Electromagnetic waves also carry momentum *, with a momentum density: And you can then consider a momentum flow rate (momentum passing through unit area in a period of time dt): Since dv Acdt dp EB S dv c c 0 1 dp EB S A dt c c 0 *see http://farside.ph.utexas.edu/teaching/em/lectures/node90.html for a discussion of how we know this.
Radiation Pressure And the average momentum flow is just: 1 dp S Avg I A dt c c Now, remember Newton s second law: F And since pressure is just force per area, If the wave is reflected, it s twice this. dp dt 1 dp F SAvg I prad A dt A c c
Example
Example p rad 3 I 1.410 W m 8 c 3.010 m s 6 4.7 10 Pa
Reflections at Conducting Surfaces What happens when a transverse matter wave strikes a fixed boundary? The wave is reflected and inverted. A similar thing happens when an EM wave strikes a conducting boundary
Reflections at Conducting Surfaces The electric field of the wave causes charges to move on the surface of the conductor. These charges move to create an electric field themselves in such a way as to keep the electric field inside the conductor zero. The moving charges also cause a magnetic field. Hence, these charges create an electromagnetic wave which is inverted from the original and travels in the opposite direction, away from the conductor. The resulting superposition of the incoming and outgoing waves is a standing wave.
Standing EM Waves Take the incoming wave (moving in the x direction) as: E x t je kx t B x t kb ˆ kx t in, ˆ cos,, cos Max Then the reflected wave must be given by: E x t je kx t B x t kb ˆ kx t ref The resulting superposition of these two is: Or, simplifying: in, ˆ cos,, cos tot Max ˆ Max,,, ˆ cos cos Etot x, t Ein x, t Eref x, t je cos kx t cos kx t, Btot x t Bin x t Bref x t kbmax kx t kx t, ˆ Max sin sin,, ˆ cos cos Etot x t j E kx t B x t k B kx t Max ref Max Max
Standing EM Waves There are then nodes in both the electric field at: And in the B-field at: l 3l x 0,, l,... Note also that the electric and magnetic field are out of phase in time tot l 3l 5l x,,... 4 4 4, ˆ Max sin sin,, ˆ cos cos Etot x t j E kx t B x t k B kx t When ωt is such that the electric field is zero everywhere, the magnetic field is maximum everywhere! Max
The drawing shows a sinusoidal electromagnetic standing wave. The average Poynting vector in this wave A.points along the x-axis. B.points along the y-axis. C. points along the z-axis. D. is zero. E. none of the above
The drawing shows a sinusoidal electromagnetic standing wave. The average Poynting vector in this wave A.points along the x-axis. B.points along the y-axis. C. points along the z-axis. D. is zero. E. none of the above
Standing EM Waves What if we put in another conducting plane, we create an electromagnetic cavity that will sustain a resonance for EM waves of wavelength: l n L, n 1,,3... n n=1 n=