Phase, Superposition & Coherence

For a sinusoidal wave travelling in the +x direction, y(x, t) = A sin(kx − ωt + φ₀), and the whole bracket φ(x, t) = kx − ωt + φ₀ is the phase: an angle in radians saying where that point of the medium sits in its cycle, with 2π rad for one complete cycle. The absolute value of φ depends on where you put the origin of x and t, so it is a bookkeeping choice. Differences are physical. At one instant, two points a distance Δx apart on the same wave differ in phase by Δφ = kΔx = 2πΔx/λ, so points one wavelength apart are in step by definition. At one point, two instants Δt apart differ by Δφ = ωΔt. Now keep the sign straight, because it is easy to talk yourself into the wrong one. A wave that has travelled farther arrives later, and with this convention its phase is the larger one: y = A sin(k(r + Δr) − ωt) = A sin(kr − ωt + kΔr), which is the same motion delayed by Δr/v. Later in time, larger in kx − ωt. Write the phase the other way round, as ωt − kx, and every one of those signs flips, so pick one convention and stay inside it.

Two waves that leave in step and travel different distances arrive out of step purely because one had farther to go. An extra path of one wavelength puts the wave back exactly where it started in its cycle, so an extra path Δr is worth Δφ = 2πΔr/λ. Only the part beyond whole cycles matters. Take λ = 640 nm and Δr = 1.60 μm: the ratio is 1.60/0.640 = 2.50 wavelengths, so Δφ = 5π rad, which is π rad plus two full turns, and the waves arrive exactly out of step. If the sources are not themselves in step, add their phase offset: Δφ = 2πΔr/λ + Δφ₀. Two identical antennas fed by cables differing by half a period are already π rad apart before geometry contributes anything at all.

The wave equation ∂²y/∂x² = (1/v²) ∂²y/∂t² contains y only to the first power. Differentiation is linear, so if y₁ and y₂ each satisfy it, so does y₁ + y₂: the medium's response to both waves is the sum of its responses to each. Three consequences are worth naming. The displacement at a point is the algebraic sum, signs included, so a crest meeting a trough of equal size gives zero displacement there. Two pulses emerge from a crossing with their original shapes and speeds, because neither ever changed the other. And the rule has a limit: it holds while the restoring force stays proportional to the displacement. Drive the amplitude high enough — a shock wave in air, an intense beam in glass — and the medium's response bends, harmonics appear, and waves begin to affect one another.

Take equal amplitudes first, and let wave 2 be the one that travelled farther, so its phase is θ + Δφ. A sin θ + A sin(θ + Δφ) = 2A cos(Δφ/2) sin(θ + Δφ/2): still a sinusoid of the same frequency, with amplitude 2A|cos(Δφ/2)| and a phase halfway between the two. The amplitude runs from 2A at Δφ = 0 down to 0 at Δφ = π. Unequal amplitudes are easiest as phasors — draw each wave as an arrow of length A whose angle is its phase, and add the arrows nose to tail. The triangle has sides A₁ and A₂ with an included angle of π − Δφ, so the cosine rule gives A_R = √(A₁² + A₂² + 2A₁A₂ cos Δφ). With A₁ = 3.0 mm, A₂ = 2.0 mm and Δφ = 60°: A_R = √(9.0 + 4.0 + 6.0) = 4.4 mm. Notice that only cos Δφ appears, so the sign of Δφ never matters to the amplitude — swap which wave is behind and the resultant is unchanged. Complete cancellation now needs A₁ = A₂ as well as Δφ = π; otherwise the best the waves manage is |A₁ − A₂|.

Intensity goes as amplitude squared, so the cross term in A_R² is the whole story: I = I₁ + I₂ + 2√(I₁I₂) cos Δφ. For two equal coherent waves this is I = 4I₀ cos²(Δφ/2). In step, the intensity is 4I₀, four times one wave and not twice, which is the first surprise. Out of step it is exactly zero, which is the second. Neither breaks conservation of energy, because the pattern has to be added up as a whole: the mean of cos² over a cycle is ½, so the average intensity across the fringes is 2I₀, exactly the I₁ + I₂ the two sources emit between them. Interference redistributes energy out of the dark fringes and into the bright ones. A detector at a dark fringe is not watching energy be destroyed; it is standing where none arrives.

Every result above assumed Δφ is a fixed number. Real atoms radiate in wave trains lasting a coherence time τ_c ≈ 1/Δν, where Δν is the source bandwidth, and each new train starts at a random phase. A detector averaging over many trains sees ⟨cos Δφ⟩ = 0, the cross term vanishes, and the intensities simply add — no fringes. The path difference must therefore stay inside the coherence length L_c = cτ_c = λ²/Δλ. For white light near 550 nm with Δλ ≈ 300 nm, L_c ≈ 1.0 μm, a couple of wavelengths. Filter it to Δλ = 1.0 nm and L_c grows to 0.30 mm; a laboratory HeNe laser with Δν = 1.5 GHz reaches 0.20 m. In between sits partial coherence, written I = I₁ + I₂ + 2√(I₁I₂)|γ| cos Δφ, whose fringe visibility V = (Iₘₐₓ − Iₘᵢₙ)/(Iₘₐₓ + Iₘᵢₙ) equals |γ| when the two intensities match. This is why interference experiments split one beam in two rather than using two lamps.