Lecture 16: Group and Phase Velocity Up to now we have just talked about the speed (or velocity) v of a wave. Actually there are two velocities: the Group Velocity vg and the Phase Velocity vp. They are given by the memorable formulæ: ω dω v = v = p k g dk 1. Phase velocity − 2π − A single (infinite) wave is described by the expression cos(ωt kx) or sin[ λ (x vt)] or equivalent. λ The pattern travels with a velocity (actually a speed) vp = T = fλ = ω/k vp is what matters with interference. The refractive index n is defined as c/v and this means c/vp 2. Group velocity An infinite wave is unrealistic. A real wave has to have beginning and end. The overall shape is called the envelope. Various shapes are possible - abrupt or gentle. dω vg = dk is the velocity of the envelope. 3. Illustration Consider two waves almost in step. They have ω1, k1 and ω2, k2 (and λ1, λ2...) Write the means and differences ω1+ω2 k1+k2 ω = 2 , k = 2 ω1−ω2 k1−k2 ∆ω = 2 , ∆k = 2 the original quantities can be expressed in terms of these ω1 = ω + ∆ω,ω2 = ω − ∆ω etc Adding the two waves gives a total wave ei(ω1t−k1x) + ei(ω2t−k2x) This can be written ei(ωt+∆ωt−kx−∆kx) + ei(ωt−∆ωt−kx+∆kx) Take out a common factor: ei(ωt−kx) ei(∆ωt−∆kx) + ei(−∆ωt+∆kx) iθ −iθ e +e − i(ωt−kx) Remembering cosθ = 2 this is 2cos(∆ωt ∆kx)e The first term is clearly the envelope. It has small wavenumber and frequency and so a long wavelength and period. It travels with velocity vg = ∆ω/∆k. dω This generalises: vg = dk 4. Finding vg Often vp is known from measurements or from basic principles. Take the expression for vp and write ω/k for vp in it. Turn all the λ and f etc terms into ω and k. dω Then differentiate with respect to k. This gives an expression involving dk from which vg can be extracted. As a trivial example, suppose vp is constant (i.e. independent of wavelength) with value c. Then ω = ck and dω/dk = c. Group and phase velocity are the same in this case. 5. Example: Refractive Index The velocity of light in a medium tends to depend on the wavelength. (Hence rainbows, prisms, etc.). This is called dispersion. See the previous lecture for details. People normally quote n as a function of λ rather than ω as a function of k. This contains the same information, we need to manipulate it: what follows is mere algebra: dn dn dk dλ = dk dλ Take these two differentials separately. c ck dn c ck dω First n = = so = − 2 vp ω dk ω ω dk Lecture 16 ................................................... ................Group and Phase Velocity 2π dk − 2π − k and secondly k = λ so dλ = λ2 = λ dn − k c − ck Now put these together and get dλ = λ ( ω ω2 vg) 2 dn kc ck c c = − + 2 vg = − + 2 vg dλ ωλ ω λ vpλ vpλ Rearrange to get an expression for vg 2 v λ v = p ( c + dn ) g c vpλ dλ vpλ dn λ dn vg = vp + n dλ = vp(1 + n dλ ) . − k dn − dvp c This can also be written: vg = vp(1 n dk ) = vp λ dλ = n+ω(dn/dω) . Of all these equally-valid dω alternatives, only vg = dk is memorable. Note: if n is falling with λ, vg < vp. This is called normal dispersion. If n is rising, vg > vp and this is called anomalous dispersion 6. Example Suppose some glass has a slightly different refractive index for red light and blue light: n =1.51 at 400 nm, n =1.49 at 600 nm. This is a fall of δn =0.02 in δλ = 200 nm. The mean λ is 500 nm and the mean n is 1.5. 8 8 8 8 vp =2 × 10 m/s vg =2 × 10 (1 − (500/1.5)(.02/200)) = 2 × 10 (1 − .0333) = 1.9310 m/s 7. Refractive index of X rays We saw in the last lecture that at the highest frequencies(=shortest wavelengths, hence X rays) one can write n = 1 − B/ω2 where B is a positive number containing N and all the proportionality constants. p This n is less than 1 so vp >c (!) c ck B Putting that worry on one side, let’s find the group velocity. Start from n = = = 1 − 2 vp ω q ω 2 2 c k − B Squaring: ω2 =1 ω2 Multiplying by ω2: c2k2 = ω2 − B 2 dω Then differentiate wrt k: 2c k =2ω dk 2 This gives vg = c /vp so vg <c and relativity is OK, as information (causal signals) travels with vg not vp. 2 2 In this case we have vgvp = c . The product of the group and phase velocities is equal to c . There are many cases where this turns out to be true, but it is not universal and there are some where it isn’t. dω 8. A more general picture for deriving vg = dk The earlier (standard) example just considered two waves. If you’re happy with that, fine. For a more general approach we need to bring in an extended concept of Fourier Series. A periodic function can be expressed as sum of sine and cosine terms 1 π 1 π f(θ)= aksin(kθ)+ bkcos(kθ) ak = f(θ)sin(kθ)dθ bk = f(θ)cos(kθ)dθ Pk π R−π (2)π R−π This can be extended to non-periodic functions ∞ ∞ f(x)= F (k)eikxdk where F (k)= 1 f(x)e−ikxdx R−∞ 2π R−∞ An infinite sine wave has a well-defined wavelength and thus a well-defined k. F (k) is a delta function. If F (k) is broad, the wave is made up of lots of sine waves of different wavelengths and has a short wavepacket (as the contributions all cancel away from the peak). Suppose we have a long wave packet f(x). That means that there is a small spread in k. The function F (k) will have a sharp peak about some central value k At some initial time t = 0 the wave can be written ∞ 0 f(x, 0) = F (k)eikxdk R−∞ After time t the wave has evolved to ∞ f(x, t)= F (k)ei(kx−ω(k)t)dk R−∞ where ω(k) explicitly shows that different wavelength components have different frequencies. dω − Let ω(k) = ω0 + dk (k k0). This is the Taylor expansion to 1st order, and we’re using the fact that the spread in k is small. dω i(k0x+kx−k0x−ω0t−(k−k ) t) Then f(x, t)= F (k)e o dk dk dω i(k0x−Rω0t) i(k−k0 )(x− t) f(x, t)= e F (k)e dk dk R The first part is the pure sine wave. The second part (the integral) describes the envelope. It looks − dω dω ≡ messy, but all the x and t dependence is in the x dk t. So the envelope progresses with velocity dk vg. –2–.