Energy Density and Power Transport

Derivation

Energy Density and Power Transport

D-068 Home PU-103 Threads energy · waves Depends on The Wave Equation as a Continuum Limit, Energy Exchange and Conservation in SHM

Statement

For a transverse travelling wave \(y(x,t)=A\sin(kx-\omega t)\) on a stretched string of linear mass density \(\mu\) under tension \(T\) (wave speed \(v=\sqrt{T/\mu}\)), the local energy density is \(u=\mu\,\omega^{2}A^{2}\cos^{2}(kx-\omega t)\), with kinetic and potential parts equal at every point and instant; the instantaneous power transported past a point is \(P=-T\,\dfrac{\partial y}{\partial x}\dfrac{\partial y}{\partial t}\); and the time averages are \(\langle u\rangle=\tfrac12\mu\,\omega^{2}A^{2}\) and \(\langle P\rangle=\tfrac12\mu v\,\omega^{2}A^{2}=\tfrac12\sqrt{\mu T}\,\omega^{2}A^{2}=v\langle u\rangle\), so the transported power scales as amplitude squared times frequency squared.

Why it matters

A wave transports no matter, yet it delivers energy from source to load: this is how a plucked string drives the surrounding air, how a transmission line carries signal power, and how any linear medium moves energy without net displacement of itself. The result isolates the two knobs the source controls, amplitude and frequency, and shows both enter squared. Doubling the frequency at fixed amplitude quadruples the delivered power, which is why high-frequency waves are such efficient energy carriers.

The identity \(\langle P\rangle=v\langle u\rangle\) is the prototype of every energy-flux law in physics: the flux equals the energy density carried along at the wave speed. The same structure reappears for sound intensity, for the electromagnetic Poynting flux, and for quantum probability currents.

Assumptions

Small transverse slope, \(|\partial y/\partial x|\ll 1\).If dropped, the potential density \(\tfrac12 T(\partial y/\partial x)^{2}\) is only the leading term of the exact stretch energy \(T(\sqrt{1+(\partial y/\partial x)^{2}}-1)\); at large slope the medium turns nonlinear, harmonics are generated, and \(\langle P\rangle\propto A^{2}\) fails.

Uniform, non-dispersive string: constant \(\mu\) and \(T\), so \(v=\sqrt{T/\mu}\) is one fixed number.If \(v\) depends on frequency (dispersion), the single speed in \(\langle P\rangle=v\langle u\rangle\) must be replaced by the group velocity, and a pure sinusoid no longer keeps its shape.

A single pure travelling harmonic wave, not a standing wave or superposition.If dropped, cross terms between counter-propagating components appear; a standing wave transports zero net power although its energy density is nonzero.

No dissipation: perfectly elastic string, no air loading or internal friction.If dropped, the amplitude decays with distance, \(A\to A(x)\), and \(\langle P\rangle\) falls along the string rather than staying constant; energy density is no longer simply advected at \(v\).

Only transverse work is counted; longitudinal motion of string elements is neglected.If dropped, one must add the longitudinal energy current and the coupling between transverse and longitudinal modes, which corrects the flux at second order in the slope.

Derivation

\[ \frac{dK}{dx}=\frac12\,\mu\left(\frac{\partial y}{\partial t}\right)^{2},\qquad \frac{dU}{dx}=\frac12\,T\left(\frac{\partial y}{\partial x}\right)^{2}. \]

Kinetic density is that of a chain of masses in the continuum limit (from the loaded-string result); potential density is the elastic energy of stretching a small element, \(\tfrac12 T\,(ds-dx)/dx\approx\tfrac12 T(\partial y/\partial x)^{2}\) to leading order in slope. A

\[ y=A\sin(kx-\omega t)\ \Rightarrow\ \frac{\partial y}{\partial t}=-A\omega\cos(kx-\omega t),\qquad \frac{\partial y}{\partial x}=Ak\cos(kx-\omega t). \]

Direct differentiation of the assumed travelling-wave solution of the wave equation. A

\[ \frac{dK}{dx}=\frac12\,\mu\,A^{2}\omega^{2}\cos^{2}(kx-\omega t),\qquad \frac{dU}{dx}=\frac12\,T\,A^{2}k^{2}\cos^{2}(kx-\omega t). \]

Substitute the partials of Step 2 into the densities of Step 1. B

\[ T k^{2}=(\mu v^{2})k^{2}=\mu(vk)^{2}=\mu\,\omega^{2}\quad\Longrightarrow\quad \frac{dU}{dx}=\frac{dK}{dx}. \]

Use \(T=\mu v^{2}\) and the dispersion relation \(\omega=vk\), both inherited from the wave equation of the loaded string. Kinetic and potential densities are identically equal at every point and instant, unlike SHM where the two forms trade off. C

\[ u=\frac{dK}{dx}+\frac{dU}{dx}=\mu\,A^{2}\omega^{2}\cos^{2}(kx-\omega t). \]

Add the two equal densities from Steps 3 and 4. A

\[ \langle u\rangle=\mu A^{2}\omega^{2}\,\big\langle\cos^{2}(kx-\omega t)\big\rangle=\frac12\,\mu\,\omega^{2}A^{2}. \]

Time-average over one period using \(\langle\cos^{2}\rangle=\tfrac12\). Symbols only, no numbers yet. A

\[ P=-\,T\,\frac{\partial y}{\partial x}\,\frac{\partial y}{\partial t}. \]

Power crossing a point equals (transverse force the left part exerts on the right) \(\times\) (transverse velocity). The left segment pulls with transverse component \(F_{y}=-T\,\partial y/\partial x\); its rate of work on the right segment is \(F_{y}\,\partial y/\partial t\). This is the energy flux. C

\[ P=-T\,(Ak\cos)(-A\omega\cos)=T\,A^{2}k\omega\,\cos^{2}(kx-\omega t). \]

Insert the partials of Step 2 into Step 7. B

\[ \langle P\rangle=\frac12\,T\,A^{2}k\omega=\frac12\,(\mu v^{2})\,\frac{\omega}{v}\,A^{2}\omega=\frac12\,\mu v\,\omega^{2}A^{2}. \]

Time-average (\(\langle\cos^{2}\rangle=\tfrac12\)), then eliminate \(T\) and \(k\) with \(T=\mu v^{2}\) and \(k=\omega/v\). B

\[ \langle P\rangle=\frac12\,\mu v\,\omega^{2}A^{2}=\frac12\sqrt{\mu T}\,\omega^{2}A^{2}=v\,\langle u\rangle. \]

Rewrite \(\mu v=\mu\sqrt{T/\mu}=\sqrt{\mu T}\), and compare with Step 6: the flux is the energy density carried forward at the wave speed. B

Result

\[ \langle u\rangle=\frac12\,\mu\,\omega^{2}A^{2},\qquad \langle P\rangle=\frac12\,\mu v\,\omega^{2}A^{2}=\frac12\sqrt{\mu T}\,\omega^{2}A^{2}=v\,\langle u\rangle. \]

Reading. A travelling wave stores, per unit length, an average energy \(\tfrac12\mu\omega^{2}A^{2}\) split equally between kinetic and potential form, and carries that energy forward at the wave speed \(v\). Both the stored energy and the transported power grow as the square of the amplitude and as the square of the frequency, so frequency is the stronger lever. The transported power is the energy density times the speed at which it moves, the universal density-times-velocity flux law.

Units check. \(\langle u\rangle\): \([\mu][\omega]^{2}[A]^{2}=(\mathrm{kg\,m^{-1}})(\mathrm{s^{-1}})^{2}(\mathrm{m})^{2}=\mathrm{kg\,m\,s^{-2}}=\mathrm{J\,m^{-1}}\), energy per length. \(\langle P\rangle\): multiply by \([v]=\mathrm{m\,s^{-1}}\) to get \(\mathrm{kg\,m^{2}\,s^{-3}}=\mathrm{J\,s^{-1}}=\mathrm{W}\). Equivalently \(\sqrt{\mu T}=\sqrt{(\mathrm{kg\,m^{-1}})(\mathrm{N})}=\sqrt{\mathrm{kg^{2}\,s^{-2}}}=\mathrm{kg\,s^{-1}}\), so \(\sqrt{\mu T}\,\omega^{2}A^{2}=(\mathrm{kg\,s^{-1}})(\mathrm{s^{-2}})(\mathrm{m^{2}})=\mathrm{W}\). Consistent.

Limiting cases

Zero amplitude \(A\to0\): \(\langle u\rangle,\langle P\rangle\to0\), no wave and no transport, as required.
Zero frequency \(\omega\to0\) (static offset): both vanish; a DC displacement carries no power. Frequency is essential to transport.
Stiff or heavy limit: at fixed \(\omega\) and \(A\), \(\langle P\rangle=\tfrac12\sqrt{\mu T}\,\omega^{2}A^{2}\) rises with both tension and mass density through \(\sqrt{\mu T}\); a taut, massive rope is a better power channel.
Standing wave (equal counter-propagating amplitudes): net \(\langle P\rangle\to0\); energy sloshes locally but no average flux crosses a node.
Non-dispersive check: since \(v\) is frequency-independent, \(\langle P\rangle=v\langle u\rangle\) holds mode by mode and adds linearly over a Fourier spectrum.

Breaks when

Large slope / nonlinearity: when \(\partial y/\partial x\) is not small, the quadratic potential density is wrong, harmonics appear, and the clean \(\langle P\rangle\propto A^{2}\omega^{2}\) scaling breaks down.
Dispersive or lossy media: if \(v=v(\omega)\), energy is advected at the group velocity \(v_{g}=d\omega/dk\), not \(v=\omega/k\); with damping, \(A=A(x)\) decays and \(\langle P\rangle\) is not constant along the string.
Superposition / standing waves: for two or more overlapping waves the flux contains interference cross terms, and the single-wave formula \(\langle P\rangle=\tfrac12\sqrt{\mu T}\,\omega^{2}A^{2}\) no longer gives the net transport.
Near a reflecting termination: within a wavelength of a boundary the incident and reflected waves coexist and the local flux oscillates; the travelling-wave result applies only far from reflections.

Failure modes

Averaging \(\cos\) instead of \(\cos^{2}\): writing \(\langle\cos\rangle=0\) and concluding \(\langle P\rangle=0\). The flux depends on \(\cos^{2}\), whose average is \(\tfrac12\), and is nonzero.
Dropping the factor \(\tfrac12\): reporting \(\langle u\rangle=\mu\omega^{2}A^{2}\) (the instantaneous peak) as the average. The time average carries the \(\tfrac12\) from \(\langle\cos^{2}\rangle\).
Using peak-to-peak displacement for \(A\): if the quoted amplitude is peak-to-peak, \(A\) is half of it; forgetting this inflates the power by a factor of four.
Confusing \(f\) with \(\omega\): \(\omega=2\pi f\), so an extra factor \((2\pi)^{2}\approx39.5\) is lost if \(f\) is substituted where \(\omega\) belongs.
Assuming KE and PE oscillate out of phase (pendulum intuition): for a travelling wave they are in phase and equal everywhere; energy does not trade locally between the two forms.
Claiming a standing wave transports power because it stores energy: its net flux averages to zero at every point.

Discussion

The deepest structural fact here is the equality of kinetic and potential energy density, point by point, for a single travelling wave. In simple harmonic motion of one oscillator the two forms trade off: all kinetic at the centre, all potential at the turning point. A travelling wave is different. At a given instant a point of maximum slope is also a point of maximum transverse speed, so kinetic and potential densities peak together and vanish together. That in-phase coincidence is exactly what makes a wave a transporter of energy rather than a mere oscillator; the energy does not sit and slosh, it streams forward.

The relation \(\langle P\rangle=v\langle u\rangle\) is the string's version of a conservation law. Written locally, \(\partial u/\partial t+\partial P/\partial x=0\) in the loss-free case, it is a continuity equation for energy: whatever energy density accumulates in a segment equals the imbalance of flux entering and leaving. This is the same bookkeeping that gives charge conservation from current continuity and mass conservation in fluids. The wave equation does not merely describe motion, it guarantees a conserved energy current.

The \(\omega^{2}A^{2}\) scaling is universal to linear waves. For sound the intensity goes as \(\rho c\,\omega^{2}s_{0}^{2}\) with displacement amplitude \(s_{0}\); for electromagnetic waves the time-averaged Poynting flux goes as \(\tfrac12 c\varepsilon_{0}E_{0}^{2}\), again quadratic in amplitude. The common origin is that energy is quadratic in the field while the wave equation is linear, so doubling the source doubles the field but quadruples the delivered power. This is why "power" and "amplitude-squared" are near synonyms across wave physics, and why the decibel scale is logarithmic in amplitude squared.

At the next level of rigour the flux \(P=-T\,(\partial y/\partial x)(\partial y/\partial t)\) is the \(T^{0x}\) component of the stress-energy tensor of the string's effective field theory. The Lagrangian density \(\mathcal{L}=\tfrac12\mu\,\dot y^{2}-\tfrac12 T\,y'^{2}\) yields, through Noether's theorem for time-translation invariance, an energy density \(u=\tfrac12\mu\,\dot y^{2}+\tfrac12 T\,y'^{2}\) and a flux \(P=-T\,y'\dot y\) obeying \(\partial_{t}u+\partial_{x}P=0\) on shell. From this vantage the equality of kinetic and potential density is the statement that the mode propagates along the characteristics \(x\mp vt=\text{const}\), and the \(v\langle u\rangle\) flux is the projection of the conserved current onto those characteristics. Applied to the electromagnetic Lagrangian, the same machinery produces the Poynting vector.

Common misconceptions. A wave does not carry the medium along with the energy: each string element merely oscillates transversely about its rest position while energy streams longitudinally. And a large static displacement stores potential energy but transports nothing; without the time variation that couples force to velocity, the flux \(P=-T\,y'\dot y\) is zero. Transport requires motion, and its rate is set by frequency, not by how far the string is pulled aside.

Worked examples

Guitar-like string: \(\mu=3.0\times10^{-3}\ \mathrm{kg\,m^{-1}}\), \(T=80\ \mathrm{N}\), driven at \(f=200\ \mathrm{Hz}\) with amplitude \(A=1.0\ \mathrm{mm}\). Find the average power transported.

Target: \(\langle P\rangle=\tfrac12\mu v\,\omega^{2}A^{2}\), with \(v=\sqrt{T/\mu}\) and \(\omega=2\pi f\). A

\[ v=\sqrt{\frac{T}{\mu}}=\sqrt{\frac{80}{3.0\times10^{-3}}}=\sqrt{2.67\times10^{4}}=163\ \mathrm{m\,s^{-1}}. \]

Wave speed from tension and mass density. A

\[ \omega=2\pi(200)=1.257\times10^{3}\ \mathrm{rad\,s^{-1}},\qquad \omega^{2}=1.579\times10^{6}\ \mathrm{s^{-2}}. \]

Convert cyclic to angular frequency before squaring. A

\[ \langle P\rangle=\tfrac12(3.0\times10^{-3})(163)(1.579\times10^{6})(1.0\times10^{-3})^{2}. \]

Insert numbers into \(\tfrac12\mu v\,\omega^{2}A^{2}\); \(A=1.0\ \mathrm{mm}=1.0\times10^{-3}\ \mathrm{m}\). B

\[ \langle P\rangle\approx0.39\ \mathrm{W}. \]

Reading. Just under half a watt streams along the string. Halving the amplitude to \(0.5\ \mathrm{mm}\) would cut this to \(0.097\ \mathrm{W}\) (factor 4); doubling the frequency to \(400\ \mathrm{Hz}\) would raise it to \(1.5\ \mathrm{W}\) (also factor 4).

Heavy rope: \(\mu=0.10\ \mathrm{kg\,m^{-1}}\) carries a wave of angular frequency \(\omega=50\ \mathrm{rad\,s^{-1}}\) and amplitude \(A=2.0\ \mathrm{cm}\). Find the average energy density and the total average energy in one \(2.4\ \mathrm{m}\) length.

Target: \(\langle u\rangle=\tfrac12\mu\omega^{2}A^{2}\), then \(E=\langle u\rangle\,L\). A

\[ A=2.0\times10^{-2}\ \mathrm{m},\qquad \omega^{2}=2.5\times10^{3}\ \mathrm{s^{-2}}. \]

Convert amplitude to metres and square the frequency. A

\[ \langle u\rangle=\tfrac12(0.10)(2.5\times10^{3})(2.0\times10^{-2})^{2}=\tfrac12(0.10)(2.5\times10^{3})(4.0\times10^{-4}). \]

Insert into \(\tfrac12\mu\omega^{2}A^{2}\). B

\[ \langle u\rangle=0.050\ \mathrm{J\,m^{-1}},\qquad E=\langle u\rangle\,L=(0.050)(2.4)=0.12\ \mathrm{J}. \]

Multiply the average density by the length. B

\[ \langle u\rangle=0.050\ \mathrm{J\,m^{-1}},\qquad E\approx0.12\ \mathrm{J}. \]

Reading. Each metre of rope holds on average \(50\ \mathrm{mJ}\); the \(2.4\ \mathrm{m}\) span holds \(0.12\ \mathrm{J}\). Half of this (\(0.060\ \mathrm{J}\)) is kinetic and half potential, at every instant.

Problems

Show, without inserting numbers, that for a single travelling wave the kinetic and potential energy densities are equal at every point and instant, and identify precisely which two relations from the wave equation make this true.
Solution
With \(y=A\sin(kx-\omega t)\): \(\dfrac{dK}{dx}=\tfrac12\mu A^{2}\omega^{2}\cos^{2}\theta\) and \(\dfrac{dU}{dx}=\tfrac12 T A^{2}k^{2}\cos^{2}\theta\), where \(\theta=kx-\omega t\). Their ratio is \(\dfrac{dU/dx}{dK/dx}=\dfrac{Tk^{2}}{\mu\omega^{2}}\). Using \(T=\mu v^{2}\) and \(\omega=vk\), the numerator \(Tk^{2}=\mu v^{2}k^{2}=\mu(vk)^{2}=\mu\omega^{2}\), so the ratio is \(1\) for all \(\theta\). The two required relations are the tension-speed relation \(T=\mu v^{2}\) and the dispersion relation \(\omega=vk\).
A string carries \(0.60\ \mathrm{W}\). Keeping tension and mass density fixed, the source amplitude is halved and the frequency is tripled. What is the new average power?
Solution
\(\langle P\rangle\propto\omega^{2}A^{2}\propto f^{2}A^{2}\). Halving \(A\) gives factor \((\tfrac12)^{2}=\tfrac14\); tripling \(f\) gives factor \(3^{2}=9\). Net factor \(=9/4=2.25\). New power \(=0.60\times2.25=1.35\ \mathrm{W}\).
A wire has \(\mu=5.0\times10^{-3}\ \mathrm{kg\,m^{-1}}\) and \(T=45\ \mathrm{N}\). A transverse wave of frequency \(f=120\ \mathrm{Hz}\) must transport \(\langle P\rangle=2.0\ \mathrm{W}\). What amplitude is required?
Solution
\(v=\sqrt{T/\mu}=\sqrt{45/(5.0\times10^{-3})}=\sqrt{9.0\times10^{3}}=94.9\ \mathrm{m\,s^{-1}}\). \(\omega=2\pi(120)=754.0\ \mathrm{rad\,s^{-1}}\), \(\omega^{2}=5.685\times10^{5}\ \mathrm{s^{-2}}\). From \(\langle P\rangle=\tfrac12\mu v\,\omega^{2}A^{2}\), \(A=\sqrt{\dfrac{2\langle P\rangle}{\mu v\,\omega^{2}}}=\sqrt{\dfrac{2(2.0)}{(5.0\times10^{-3})(94.9)(5.685\times10^{5})}}\). Denominator \(=(5.0\times10^{-3})(94.9)(5.685\times10^{5})=2.698\times10^{5}\). So \(A=\sqrt{4.0/(2.698\times10^{5})}=\sqrt{1.483\times10^{-5}}=3.85\times10^{-3}\ \mathrm{m}\approx3.9\ \mathrm{mm}\).
Two strings of equal tension \(T\) are joined; the second has four times the mass density of the first (\(\mu_{2}=4\mu_{1}\)). If a wave of the same frequency and amplitude existed on each in isolation, compare the powers each would transport, and comment on why the joined-string reflection problem is more subtle.
Solution
\(\langle P\rangle=\tfrac12\sqrt{\mu T}\,\omega^{2}A^{2}\), so at fixed \(T,\omega,A\), \(\langle P\rangle\propto\sqrt{\mu}\). Then \(\dfrac{P_{2}}{P_{1}}=\sqrt{\mu_{2}/\mu_{1}}=\sqrt{4}=2\): the denser string would carry twice the power. In reality the two are joined and \(A\) is not free on both, because continuity of displacement and slope at the junction fixes the reflected and transmitted amplitudes, and energy is conserved as \(P_{\text{inc}}=P_{\text{refl}}+P_{\text{trans}}\) rather than each side being set independently. The equal-amplitude comparison is a scaling illustration, not the boundary-value answer.
A rope with \(\mu=0.20\ \mathrm{kg\,m^{-1}}\) and \(v=12\ \mathrm{m\,s^{-1}}\) carries a wave of amplitude \(A=3.0\ \mathrm{cm}\) and frequency \(f=4.0\ \mathrm{Hz}\). Find (a) the average energy density, (b) the average power, and (c) verify \(\langle P\rangle=v\langle u\rangle\).
Solution
\(\omega=2\pi(4.0)=25.13\ \mathrm{rad\,s^{-1}}\), \(\omega^{2}=631.7\ \mathrm{s^{-2}}\); \(A=0.030\ \mathrm{m}\), \(A^{2}=9.0\times10^{-4}\ \mathrm{m^{2}}\). (a) \(\langle u\rangle=\tfrac12\mu\omega^{2}A^{2}=\tfrac12(0.20)(631.7)(9.0\times10^{-4})=0.0569\ \mathrm{J\,m^{-1}}\approx0.057\ \mathrm{J\,m^{-1}}\). (b) \(\langle P\rangle=\tfrac12\mu v\,\omega^{2}A^{2}=v\langle u\rangle=(12)(0.0569)=0.68\ \mathrm{W}\). (c) Direct check: \(v\langle u\rangle=12\times0.0569=0.68\ \mathrm{W}\), matching (b). The identity holds.