The Doppler Effect and Shock Fronts

Derivation

The Doppler Effect and Shock Fronts

D-073 Home PU-103 Threads waves · matter Depends on d'Alembert's General Solution, Longitudinal Sound Waves in a Fluid

Statement

For a scalar wave of fixed speed \(c\) in a medium at rest, a source of proper frequency \(f_s\) and an observer in uniform motion are related by \(f_o = f_s\,\dfrac{c-\vec{v}_o\cdot\hat{n}}{c-\vec{v}_s\cdot\hat{n}}\), where \(\hat{n}\) is the unit vector pointing from source to observer along the propagation ray, so only line-of-sight velocity components shift the frequency. When the source outruns the wave, \(v_s > c\), the emitted spheres acquire a common tangent envelope — a Mach cone of half-angle \(\mu\) with \(\sin\mu = c/v_s = 1/M\).

Why it matters

The Doppler shift is the single most-used diagnostic of motion in physics: it turns a frequency measurement into a velocity measurement without ever touching the moving body. Sonar, weather radar, medical blood-flow imaging, and the recession velocities of galaxies are all one equation of this form applied in different regimes.

The same wavefront bookkeeping that predicts the shift also predicts its own breakdown. Push the source to the wave speed and the forward wavelength shrinks to zero; push past it and the compressed fronts pile into a shock — the sonic boom, the bow wave of a boat, and the Cherenkov cone of a fast charge are geometrically the same object. One picture, expanding spheres from a moving emitter, contains both regimes.

Assumptions

The medium is homogeneous, isotropic and non-dispersive, at rest in the chosen frame.Then every crest expands as a sphere of radius \(c\,t\) with one speed \(c\). If dropped, \(c\) depends on frequency or direction, rays refract, and no single shift factor exists.

Motion is resolved along the line of sight through the projections \(\vec{v}\cdot\hat{n}\).Only the radial velocity component enters; a purely transverse component gives no first-order acoustic shift. If read as full speeds, the predicted shift never crosses zero at closest approach.

The propagation speed \(c\) is fixed independent of the source's later motion (a consequence of the d'Alembert solution).An already-emitted crest keeps speed \(c\) no matter what the source does next. If dropped, the "older crest still travels at \(c\)" step is unjustified and the wavelength bookkeeping fails.

Motions are non-relativistic, so time is absolute and velocities add by Galilean composition.Frames share a universal clock, so emission and arrival intervals compare directly. If dropped, a Lorentz factor appears, the source/observer asymmetry collapses, and one must use the relativistic Doppler formula.

Velocities are uniform over each emission-to-reception interval, and the medium itself is still (no wind).Then \(T_s\) and the geometry hold constant while two successive crests are launched, and \(c\) is isotropic in the lab. If dropped, the result becomes a time-varying chirp, and a wind \(\vec{u}\) must be absorbed via \(c\hat{n}\to c\hat{n}+\vec{u}\).

Derivation

\[ \frac{\partial^2 \phi}{\partial t^2} = c^2\,\nabla^2\phi \quad\Rightarrow\quad \phi=\phi\!\left(t-\tfrac{r}{c}\right)\ \text{on outgoing rays} \]

The d'Alembert solution of the wave equation propagates every disturbance at the fixed speed \(c\) relative to the medium; a crest emitted at time \(t_e\) reaches radius \(c(t-t_e)\) regardless of the source's subsequent motion. This single speed is the only dynamical input. A

\[ T_s=\frac{1}{f_s},\qquad \text{crest }n\text{ leaves at } t_n=n\,T_s \]

Definition of the proper period: the source releases one crest every \(T_s\) on its own (here universal) clock. We track two successive crests \(n\) and \(n+1\). A

\[ \lambda = c\,T_s - (\vec{v}_s\cdot\hat{n})\,T_s = \bigl(c-\vec{v}_s\cdot\hat{n}\bigr)\,T_s \]

In one period the leading crest advances \(c\,T_s\) along \(\hat{n}\), while the source itself advances \((\vec{v}_s\cdot\hat{n})\,T_s\) after it; the crest spacing frozen into the medium is their difference. Only the component of \(\vec{v}_s\) along \(\hat{n}\) compresses the spacing. B

\[ c_{\text{rel}} = c-\vec{v}_o\cdot\hat{n} \]

Crests move through the medium at \(c\); relative to the observer they close at \(c\) minus the observer's velocity component along \(\hat{n}\) (the source-to-observer direction). Absolute time makes this a plain Galilean subtraction. B

\[ f_o = \frac{c_{\text{rel}}}{\lambda} = \frac{c-\vec{v}_o\cdot\hat{n}}{\bigl(c-\vec{v}_s\cdot\hat{n}\bigr)\,T_s} \]

The observer counts crests at (closing speed)/(spacing) — the universal relation \(f=v/\lambda\) applied in the observer's frame. Wavelength is set at emission (source term); closing speed is set at reception (observer term). A

\[ \boxed{\,f_o = f_s\,\frac{c-\vec{v}_o\cdot\hat{n}}{c-\vec{v}_s\cdot\hat{n}}\,} \]

Substitute \(T_s=1/f_s\). The vector form holds for arbitrary geometry, with \(\hat{n}\) the unit vector from the emission point to the reception point. Source and observer enter through distinct factors — a genuine physical asymmetry, because the medium defines a preferred frame, not a mere relabelling. C

\[ \text{as } v_s\to c^-:\quad \lambda=(c-v_s)T_s\to 0,\quad f_o\to\infty \]

Along the forward line of sight the wavelength collapses: the source keeps pace with its own fronts, so successive crests are launched almost on top of one another. The formula signals its own breakdown — an infinite pile-up is a wavefront envelope, i.e. a shock, not a sinusoid. B

\[ v_s>c:\quad \sin\mu = \frac{c\,t}{v_s\,t} = \frac{c}{v_s} = \frac{1}{M} \]

Geometry of the envelope. A crest emitted a time \(t\) ago has radius \(c\,t\) and sits a distance \(v_s\,t\) behind the source; the common tangent to all such spheres is a cone whose half-angle satisfies \(\sin\mu=(c t)/(v_s t)\). The \(t\) cancels, so a single cone envelopes every front. With Mach number \(M=v_s/c\), it exists only for \(M>1\) and narrows as \(M\) grows. C

Result

\[ f_o = f_s\,\frac{c-\vec{v}_o\cdot\hat{n}}{c-\vec{v}_s\cdot\hat{n}}\,,\qquad\qquad \sin\mu=\frac{c}{v_s}=\frac{1}{M}\ \ (M>1) \]

Reading. The observed frequency is the emitted frequency scaled by the closing speed of the crests over their spacing in the medium. The numerator raises the pitch when the observer moves into the waves; the denominator raises it when the source chases its own sound. Only line-of-sight components count. Source and observer enter asymmetrically — only the source term can drive the wavelength to zero — so equal speeds of the two do not cancel. Once \(v_s\) reaches \(c\) (\(M=1\)) the forward formula fails and the fronts organize into a cone whose half-angle \(\mu\) narrows as the source goes faster; measuring \(\mu\) alone returns \(M\).

Units check. Each \(\vec{v}\cdot\hat{n}\) and \(c\) carry \(\mathrm{m\,s^{-1}}\), so every ratio is dimensionless and \(f_o\) inherits the units of \(f_s\) (\(\mathrm{Hz}=\mathrm{s^{-1}}\)). In \(\sin\mu=c/v_s\) both sides are dimensionless, as a sine must be, and the constraint \(|c/v_s|\le 1\) recovers exactly the condition \(v_s\ge c\) for a real cone.

Limiting cases

Both at rest (\(\vec{v}_s=\vec{v}_o=0\)): \(f_o=f_s\), no shift.
Low speed (\(|\vec{v}|\ll c\)): \(f_o\approx f_s\bigl(1+\tfrac{(\vec{v}_s-\vec{v}_o)\cdot\hat{n}}{c}\bigr)\), so \(\Delta f/f_s\approx v_{\text{rel},\parallel}/c\) — only relative line-of-sight velocity survives to first order, and the asymmetry appears at \(O(v^2/c^2)\).
Observer only (\(\vec{v}_s=0\)): \(f_o=f_s\bigl(1-\tfrac{\vec{v}_o\cdot\hat{n}}{c}\bigr)\), a bounded linear shift for any observer speed.
Source only (\(\vec{v}_o=0\)): \(f_o=f_s\,c/(c-\vec{v}_s\cdot\hat{n})\), which diverges as \(v_s\to c\) — the qualitative asymmetry with the observer case.
Transverse motion (\(\vec{v}\perp\hat{n}\)): \(\vec{v}\cdot\hat{n}=0\), so classically \(f_o=f_s\) at closest approach (contrast the relativistic transverse shift).
Exactly sonic (\(M=1\)): \(\mu\to 90^\circ\), the cone flattens into a plane wall of piled fronts moving with the source.
Hypersonic (\(M\gg 1\)): \(\mu\approx 1/M\to 0\), a needle-thin cone hugging the flight path.

Breaks when

The source reaches or exceeds the wave speed forward (\(v_s\ge c\)): the denominator \(c-\vec{v}_s\cdot\hat{n}\) vanishes or turns negative and the linear-acoustics formula is meaningless. The physics is now a finite-amplitude shock with an entropy jump; only the geometric cone relation survives, and a real nonlinear shock runs marginally ahead of the linear Mach cone.
Finite amplitude / nonlinear steepening: loud sound travels slightly faster at its compressions than at its rarefactions, so the waveform steepens into an N-wave and the single-fixed-\(c\) assumption fails even below \(M=1\); the harmonic content, and hence the perceived shift of a rich tone, distorts.
Dispersive or absorbing media: if \(c=c(\omega)\) the crests of a pulse separate and no single \(f_o\) is received; strong absorption additionally attenuates the high-frequency approach signal more than the receding one.
Accelerating source or observer: over one period the geometry changes, producing a chirp; \(f_o\) is then only the instantaneous value, and near \(v_s=c\) the pile-up is transient.

Failure modes

Symmetry error: writing \(f_o=f_s(c+v)/(c-v)\) with the same \(v\) top and bottom for a single moving body — the source and observer terms are physically distinct and merge only when both bodies actually move.
Additive-speed error: treating source and observer speeds as adding, e.g. shifting by \(v_o+v_s\); they enter as separate multiplicative ratios.
Sign flipping: memorizing "+ for toward" without fixing \(\hat{n}\). Anchor \(\hat{n}\) source-to-observer once, take real dot products, and the signs follow; any motion shortening the path raises the pitch.
Frame confusion: measuring \(\vec{v}_o,\vec{v}_s\) relative to each other instead of relative to the still medium; in wind the two differ and the relative-velocity shortcut is wrong.
Mach-angle inversion: writing \(\sin\mu=v_s/c\) (which exceeds \(1\) and has no solution) instead of \(\sin\mu=c/v_s\).
Boom-timing error: assuming the sonic boom is heard when the jet is overhead, rather than after it has flown far enough for the trailing cone to reach the ground.

Discussion

The deep reason source and observer enter asymmetrically is that the medium picks out a preferred frame. A moving observer changes only how fast crests are counted — a kinematic effect bounded by \(f_o/f_s=1-\vec{v}_o\cdot\hat{n}/c\). A moving source changes the crests themselves, imprinting a permanently shortened wavelength into the medium that every later observer will measure. This is why the source case can diverge while the observer case cannot: one edits the wave, the other only the sampling.

The shock cone is not a separate phenomenon but the analytic continuation of the diverging shift. As \(v_s\to c\) the forward wavelength collapses; the "infinite frequency" is the mathematics warning that infinitely many fronts now coincide on a surface. Past \(M=1\) that surface is the Mach cone, and its half-angle encodes the speed directly: measuring \(\mu\) gives \(M\) with no clock and no known frequency. The identical construction governs the bow wave of a boat (built from the ratio of hull speed to surface-wave speed) and Cherenkov radiation from a charge outrunning light in a dielectric, where \(\cos\theta_C=c/(n v)\) — different \(c\), same envelope.

Connecting to the priors: the d'Alembert solution is what licenses Step 3, because it guarantees each emitted disturbance propagates at the fixed \(c\) independent of the source's subsequent motion — without it the "older crest keeps speed \(c\)" argument would be unjustified. The speed of sound in a gas, \(c=\sqrt{\gamma R T/M_{\text{molar}}}\), supplies the numerical \(c\) and shows the shift and the boom are temperature-dependent: on a cold day the same aircraft speed corresponds to a higher Mach number and a narrower cone.

A subtler point is the structural distinction between the classical acoustic Doppler formula and its relativistic optical counterpart. Acoustically the medium frame is real, so the two velocities are separately meaningful and the transverse shift is exactly zero at first order. Optically there is no medium: only the relative velocity \(\beta=v/c_{\text{light}}\) is defined, the formula must reduce to a function of \(\beta\) alone, \(f_o=f_s\sqrt{(1-\beta)/(1+\beta)}\), and a genuinely transverse shift \(f_o=f_s/\gamma\) survives purely from time dilation. That the acoustic formula fails to be a function of relative velocity alone is therefore not a defect — it is direct evidence that sound has a medium and light does not, the very asymmetry probed by the Michelson–Morley experiment.

Common misconceptions. The pitch of a passing siren does not fall because it recedes; it is constant and high while approaching, constant and low while receding, and the audible "drop" is the sudden step as the vehicle passes — a step, not a glide, smeared only by the finite time spent near closest approach. And the sonic boom is not a one-off bang at the instant the sound barrier is broken: it is a continuous cone trailing a supersonic craft for its entire flight, sweeping over each ground observer exactly once as the cone crosses them.

Worked examples

\[ f_o = f_s\,\frac{c}{c-v_s}\quad(\text{source approaching, observer at rest}) \]

Ambulance siren, \(f_s=700\ \mathrm{Hz}\), moving toward a stationary listener at \(v_s=30\ \mathrm{m\,s^{-1}}\); air at \(20^\circ\mathrm{C}\) gives \(c=343\ \mathrm{m\,s^{-1}}\). Set \(\hat{n}\) source-to-observer so approach means \(\vec{v}_s\cdot\hat{n}=+30\), and \(v_o=0\). A

\[ f_o = 700\cdot\frac{343}{343-30} = 700\cdot\frac{343}{313} \]

Insert numbers after fixing the symbolic form. A

\[ f_o = 700\cdot 1.0958 = 767\ \mathrm{Hz}\qquad(\text{receding: } 700\cdot\tfrac{343}{373}=644\ \mathrm{Hz}) \]

Evaluate; after passage \(\vec{v}_s\cdot\hat{n}=-30\), so the denominator uses \(c+30\). A

\[ f_o^{\text{app}}\approx 767\ \mathrm{Hz},\qquad f_o^{\text{rec}}\approx 644\ \mathrm{Hz} \]

Reading. The pitch jumps by about \(123\ \mathrm{Hz}\) — roughly three semitones — as the ambulance passes, the familiar drop. The shift is not symmetric about \(700\ \mathrm{Hz}\): \(+67\) up, \(-56\) down, a direct fingerprint of the moving-source denominator.

Units check. Each ratio is \((\mathrm{m\,s^{-1}})/(\mathrm{m\,s^{-1}})\), dimensionless; \(f_o\) stays in \(\mathrm{Hz}\).

\[ \sin\mu=\frac{c}{v_s}=\frac{1}{M} \]

A jet cruises at \(M=1.5\) in air with \(c=343\ \mathrm{m\,s^{-1}}\). Find the Mach-cone half-angle and the delay before a ground observer \(h=1000\ \mathrm{m}\) below the path hears the boom after the jet passes overhead. A

\[ \mu=\arcsin\!\left(\frac{1}{1.5}\right)=\arcsin(0.667)=41.8^\circ \]

Direct evaluation of the cone angle. A

\[ x=\frac{h}{\tan\mu}=h\sqrt{M^2-1} \]

The boom reaches the observer when the trailing cone surface, inclined at \(\mu\) to the path, sweeps the ground point a horizontal distance \(x\) behind the aircraft; \(\tan\mu=1/\sqrt{M^2-1}\). B

\[ t=\frac{x}{v_s}=\frac{h\sqrt{M^2-1}}{M c} \]

Divide the horizontal offset by the aircraft's ground speed \(v_s=Mc\). B

\[ t=\frac{1000\cdot\sqrt{1.25}}{1.5\cdot 343}=\frac{1118}{514.5} \]

Numbers last: \(\sqrt{1.5^2-1}=\sqrt{1.25}=1.118\), so \(x=1118\ \mathrm{m}\) and \(v_s=514.5\ \mathrm{m\,s^{-1}}\). A

\[ \mu\approx 41.8^\circ,\qquad t\approx 2.17\ \mathrm{s} \]

Reading. The jet is already \(1.1\ \mathrm{km}\) downrange when its boom arrives — the observer watches it pass in silence and hears the bang about two seconds later, the everyday proof that the boom trails behind rather than marking the overhead instant.

Units check. \(\mu\) from an inverse sine is an angle; \(x\) in \(\mathrm{m}\) over \(v_s\) in \(\mathrm{m\,s^{-1}}\) gives \(\mathrm{s}\).

Problems

A cyclist rides at \(8.0\ \mathrm{m\,s^{-1}}\) straight toward a stationary \(512\ \mathrm{Hz}\) loudspeaker (\(c=343\ \mathrm{m\,s^{-1}}\)). What frequency does the cyclist hear?
Solution
Observer moving toward a stationary source. With \(\hat{n}\) source-to-observer, moving toward the source means \(\vec{v}_o\cdot\hat{n}=-8.0\): \(f_o=f_s(c+8.0)/c = 512\cdot(343+8)/343 = 512\cdot 351/343 = 512\cdot 1.0233 = \mathbf{524\ Hz}\). A modest \(12\ \mathrm{Hz}\) rise, bounded because only the observer moves.
A train sounds a \(400\ \mathrm{Hz}\) horn while moving at \(35\ \mathrm{m\,s^{-1}}\). A passenger in a car approaching the train head-on at \(20\ \mathrm{m\,s^{-1}}\) hears it as they close (\(c=343\ \mathrm{m\,s^{-1}}\)). Find the frequency.
Solution
Both approaching. Take \(\hat{n}\) source-to-observer: source toward observer \(\vec{v}_s\cdot\hat{n}=+35\), observer toward source \(\vec{v}_o\cdot\hat{n}=-20\). \(f_o=f_s(c+20)/(c-35)=400\cdot 363/308=400\cdot 1.1786=\mathbf{471\ Hz}\). Note the speeds do not simply add: a naive relative-speed formula \(400\cdot(343+55)/343=464\ \mathrm{Hz}\) is wrong by \(7\ \mathrm{Hz}\).
A stationary sonar emits \(40.0\ \mathrm{kHz}\). A submarine moves directly away at \(6.0\ \mathrm{m\,s^{-1}}\) and reflects the sound back (seawater \(c=1500\ \mathrm{m\,s^{-1}}\)). What frequency does the sonar receive?
Solution
Two Doppler steps. Outgoing, the sub is a receding observer: \(f_1=f_s(c-v)/c\). On reflection it re-emits \(f_1\) as a receding source: \(f_2=f_1\,c/(c+v)\). Combined: \(f_2=f_s(c-v)/(c+v)=40000\cdot(1494)/(1506)=40000\cdot 0.99203=\mathbf{39.68\ kHz}\). The \(319\ \mathrm{Hz}\) downshift is what the sonar's electronics convert into the target's radial speed; note the two-way factor \((c-v)/(c+v)\), not \((c-v)/c\).
An observer measures the Mach cone of a supersonic bullet to have a half-angle of \(28^\circ\) (\(c=343\ \mathrm{m\,s^{-1}}\)). Find the bullet's speed and Mach number.
Solution
\(\sin\mu=c/v_s\Rightarrow v_s=c/\sin\mu=343/\sin 28^\circ=343/0.4695=\mathbf{731\ m\,s^{-1}}\), and \(M=v_s/c=1/\sin 28^\circ=\mathbf{2.13}\). The narrow cone directly reports a supersonic speed with no timing measurement.
A supersonic aircraft flies level at \(M=2.0\) and altitude \(8.0\ \mathrm{km}\) (\(c=320\ \mathrm{m\,s^{-1}}\) at that height). How long after it passes directly overhead does a ground observer hear the boom, and how far downrange is the aircraft then?
Solution
\(\sin\mu=1/2.0\) so \(\tan\mu=1/\sqrt{M^2-1}=1/\sqrt{3}\). Horizontal offset when the cone reaches the observer: \(x=h/\tan\mu=h\sqrt{M^2-1}=8000\cdot\sqrt{3}=8000\cdot 1.732=13856\ \mathrm{m}\approx\mathbf{13.9\ km}\). Ground speed \(v_s=Mc=640\ \mathrm{m\,s^{-1}}\), so delay \(t=x/v_s=13856/640=\mathbf{21.7\ s}\). The long silent interval is why observers rarely connect the distant boom with the aircraft that made it.