f(x + dx, t + dt) = f(x, t) (2.1)⇔ f + fxdx + ftdt = f ⇒ fxc + ft = 0 (2.2) (2.2) ⇒ fxxc + ftx = 0 ⇒ fxxc2 + ftxc = 0 (2.3) (2.2) ⇒ fxtc + ftt = 0 (2.4) Substracting (2.4) from (2.3) and using ftx = fxt gives:
fxxc2 - ftt = 0 ⇔ fxx - (1/c2)ftt = (∂2/∂x2 - (1/c2)∂2/∂t2)f = 0 (2.5)This is the unattenuated 1D wave equation. A rather general solution (perhaps the general solution, but at least the one we are interested in) is:
f(x, t) = F(ct - x) + G(ct + x) (2.6)As can easily be verified as follows. Use U = ct - x and realize (using the chain rule):
Fxx = FUU and Ftt = c2FUU ⇒ Fxx - (1/c2)Ftt = FUU - (1/c2)c2FUU = 0 (2.7)A similar deduction can be done for G and V = ct + x. Hence F and G are both solutions of (2.5). And since (2.5) describes a lineair operator their sum is a solution too. F represents a wave traveling in the direction of increasing x at constant speed c. G represents a wave traveling in the opposite direction. Often G = 0.
fxx + fyy + fzz - (1/c2)ftt = ∇2f - (1/c2)(∂2/∂t2)f = ∂μ2f = 0. (3.1)with:
∂μ ≡ (∂/∂x1, ∂/∂x2, ∂/∂x3, ∂/∂x4) and x1 = x, x2 = y, x3 = z, x4 = jct. ⇒ ∂μ = (∇, (1/(jc))∂/∂t) (3.2)(using the chain rule: ∂/∂x4 = (∂/∂xt)(∂t/∂x4) = 1/(jc)(∂/∂t)) The operator ∂μ2 is called the d'Alembertian. Now the question is: Does (3.1) describe unattenuated 3D waves?
f(r, t) = Aejφejk(ct - x) (3.3)By definition the wave should be periodic in x with period λ, thus:
kλ = 2π ⇒ k = 2&pi/λ (3.4)Also by definition it should be periodic in t with period T, thus:
kcT = 2π (3.5)(3.4) and (3.5) ⇒ (2π/λ)cT = 2&pi ⇔ c = λ/T = λν (3.6) Which we know is true. Now let's consider a rotated frame u,v,w. Suppose a harmonic plane wave travels in the positive u direction, thus:
f(r, t) = Aejφejk(ct - u) (3.7)Let n be the unit vector in the positive u direction, e1 the unit vector in the positive x direction, e2 the unit vector in the positive y direction, e3 the unit vector in the positive z direction, θ1 the angle between n and e1, θ2 the angle between n and e2, θ3 the angle between n and e3. It is easy to see the following is true:
u = xcosθ1 + ycosθ2 + zcosθ3 = xe1 ⋅ n + ye2 ⋅ n + ze3 ⋅ n = r ⋅ n (3.8)Notice by the way that:
n ⋅ ei = cosθi = (n1e1 + n2e2 + n3e3) ⋅ ei = ni (3.9)Substituting this into (3.7) gives:
f(r, t) = Aejφejk(ct - n ⋅ r) = f(r, t) = Aejφej(kct - k ⋅ r) with k = kn (3.10)So (3.9) describes a general harmical plane wave in the direction of k. Substitution it into (3.1) learns (3.9) is a solution of (3.1) (using k2 = k12 + k22 + k32). Had we not started with (3.7) but with:
f(r, t) = g(ct - u) (3.11)We would have found:
f(r, t) = g(ct - r ⋅ n) (3.12)For (using U = ct - r ⋅ n, the chain rule, n ⋅ ei = cosθi = ni and n12 + n22 + n32 = 1):
∂t2g = c2gUU and ∂x2g = n12gUU and ∂y2g = n22gUU and ∂z2g = n32gUU ⇒ ∂μ2g = 0 (3.13)Notice (3.10) is a special case of (3.12). So far we have not considered the wave traveling in the opposite direction. It can easily be verified (using V = ct + r ⋅ n) that the following equation gives the general plane wave solution:
f(r, t) = g(ct - r ⋅ n) + h(ct + r ⋅ n) (3.14)Conclusion: Unattenuated plane waves (harmonic or not) are a solution of (3.1).
f(r, t) = g(r)ejωt (3.15)Is this a valid solution of (3.1)? Let's substitute it into (3.1) and use ∇2 in polar coordinates:
∇2 = (1/r2)∂r(r2∂r) + (1/(r2sinθ))∂θ(sinθ∂θ) + (1/(r2sin2θ))∂φ2 (3.16)(3.15) ⇒ ∂μ2f = (∇2g)ejωt + (ω2/c2)ejωt = 0 ⇔ ∇2g + k2g = 0 (using k = &omega/c) (3.17) ∇2g = (1/r2)(d/dr)(r2gr) = (1/r2)(2rgr + r2grr) = (2/r)gr + grr (3.18) Furthermore realize that:
(1/r)(d2/dr2(rg)) = (2/r)gr + grr (3.19)For:
(rg)r = g + rgr ⇒ (rg)rr = gr + gr + rgrr = 2gr + rgrr (3.20)Hence (3.17) and (3.18) and (3.19) give:
∇2g + k2g = (1/r)(rg)rr + k2g = 0 ⇔ (rg)rr + k2(rg) = 0 (3.21)The general solution of (3.21) is:
rg = ae-jkr + bejkr (3.22)With a and b being complex constants. Which can be verified by substituting it in (3.21). Hence the solution of (3.15) is:
(3.1) and (3.15) ⇒ f(r, t) = (a/r)ejk(ct - r) + (b/r)ejk(ct + r) (3.23)Which in the absence of an outwards traveling wave simplifies to (using A = |a| and φ = arga):
f(r, t) = (Aejφ/r)ejk(ct - r) (3.24)Notice these kind of waves are not really unattenuated, since they fall of (are attenuated) by 1/r. Also notice that waves of the form:
f(r, t) = (Aejφ/rn)ejk(ct - r) and n ≠ 1 (3.25)are not solutions of the wave equation (3.1). Though we consider these froms to be wave too... Example
Φ = |f|2 (3.26)with f given by (3.24). For then the amount of E passing through the sperical surface at r is then given by:
E = 4πr2|f|2 = constant (3.27)Ergo, no energy gets lost nor is created between the surfaces at r = r1 and r = r2 with r2 > r1.