Rayleigh Dissipation Function

Lagrangian mechanics has a dirty secret: it describes a world that does not exist. Every pendulum it swings, swings forever; every orbit it closes, closes exactly. The formalism inherits this immortality from its foundation — Hamilton's principle only speaks about forces that come from a potential, and friction stubbornly refuses to be one. Yet look around: every clock winds down, every bridge cable hums itself quiet, every circuit warms its resistor. If analytical mechanics could not price this universal tax, engineers would have abandoned it in the nineteenth century. Lord Rayleigh's 1873 repair is a masterpiece of minimalism. For the enormous class of forces linear in velocity — Stokes drag, dashpots, electrical resistance, viscous structural damping — he packaged all of friction into one scalar function $\mathcal{F}$ , quadratic in the generalized velocities, exactly as potential energy packages all conservative forces into one scalar $U$ . Where the potential generates forces by a position-gradient, the dissipation function generates friction by a velocity-gradient, and the Euler–Lagrange equations acquire a single new term on the right-hand side. Generalized coordinates, cyclic-coordinate bookkeeping, the whole machine — everything survives. The function even audits itself: $2\mathcal{F}$ is precisely the rate at which mechanical energy is converted to heat, so the energy ledger of any damped system reads $\dot E = -2\mathcal{F}$ , no equation-solving required. That factor of two, the strict linear-in-velocity fine print, and the reason a positive $\mathcal{F}$ is the second law wearing mechanics' clothes are this article's business.

Key takeaways

Hamilton's principle handles only monogenic forces; friction linear in velocity needs one extra structure — Rayleigh's dissipation function $\mathcal{F} = \tfrac12\sum_{ij} c_{ij}\,\dot q_i\dot q_j$ , a positive-semidefinite quadratic form in the generalized velocities.
Its velocity-gradient is the frictional generalized force, $Q_j^{\text{diss}} = -\partial\mathcal{F}/\partial\dot q_j$ , exactly mirroring how $-\partial U/\partial q_j$ generates conservative forces.
The equations of motion become $\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot q_j} - \frac{\partial L}{\partial q_j} = -\frac{\partial \mathcal{F}}{\partial \dot q_j}$ — one new term, all other Lagrangian machinery intact.
The energy audit is built in: $\mathrm{d}E/\mathrm{d}t = -2\mathcal{F}$ . The dissipated power is twice the dissipation function — Euler's theorem on degree-2 homogeneous functions supplies the 2.
Scope is strict: viscous (linear) damping only. Coulomb dry friction and quadratic air drag have no quadratic $\mathcal{F}$ and must enter as bare generalized forces.
Key idea: define a positive-semidefinite quadratic form in the generalized velocities, $\mathcal{F}=\tfrac12\sum_{ij}c_{ij}\dot q_i\dot q_j$ ; its velocity-gradient is the dissipative generalized force, the equations of motion become $\tfrac{\mathrm{d}}{\mathrm{d}t}(\partial L/\partial\dot q_j)-\partial L/\partial q_j=-\partial\mathcal{F}/\partial\dot q_j$ , and $2\mathcal{F}$ is the rate of energy loss.

The core insight

Define a positive-semidefinite quadratic form in the generalized velocities, $\mathcal{F}=\tfrac12\sum_{ij}c_{ij}\dot q_i\dot q_j$ . Its velocity-gradient is the dissipative generalized force, so the equations of motion become $\tfrac{\mathrm{d}}{\mathrm{d}t}(\partial L/\partial\dot q_j)-\partial L/\partial q_j=-\partial\mathcal{F}/\partial\dot q_j$ , and $2\mathcal{F}$ is the rate of energy loss.

Where this lives today

Open any commercial finite-element package — the solvers that certify buildings against earthquakes, hulls against slamming, or launch vehicles against acoustic loads — and the damping input panel asks for two numbers: $\alpha$ and $\beta$ . Equation (7) is not a historical footnote; it is the default damping model of the global simulation industry, because proportional damping is the unique cheap ansatz that lets a million-DOF model decouple into independent damped modes. Seismic design codes effectively legislate worked example 2: engineers anchor the Rayleigh curve on the structure's dominant modes and must then check that no safety-relevant mode falls into the bathtub's under-damped sag. The same $(L, \mathcal{F})$ pair runs quieter revolutions elsewhere: MEMS gyroscope and resonator designers engineer $\mathcal{F}$ downward (quality factors above $10^6$ ) to starve the drain in equation (4), while vehicle suspension models put dashpots between coordinates exactly as in Derivation D. The deepest modern echo is conceptual. Rayleigh's function presumes that macroscopic friction is linear in velocity — and the fluctuation-dissipation theorem (1951) explains why that presumption works so widely: linear response of any system near thermal equilibrium ties the damping matrix $c_{ij}$ to the spectrum of its microscopic fluctuations. The same mathematics now appears at the frontier of machine learning for physics, where "dissipative Lagrangian" and port-Hamiltonian neural networks learn a system's $L$ and $\mathcal{F}$ separately from data — reversible physics in one scalar, irreversible physics in the other — precisely Rayleigh's 1873 division of labor, rediscovered as an inductive bias for AI.

The physical picture

Think about what the potential energy actually does for you. Before $U$ , a conservative force was three components per particle, tangled with geometry; after $U$ , it is one scalar landscape whose downhill direction is the force. Rayleigh's move is the same compression applied to friction — with one twist. Friction does not care where you are; it cares how fast you move. So the landscape must be built over velocity space, and the force must be its downhill direction there: $Q_j^{\text{diss}} = -\partial\mathcal{F}/\partial\dot q_j$ . Stand still anywhere and the landscape is flat — no drag. Move, and you climb the paraboloid; the faster you go, the steeper the wall, and the harder friction shoves you back toward the origin of velocity space, which is rest.

Figure

Where the energy goes — the mass-spring-dashpot system and its draining ledger

The paraboloid picture is worth carrying around explicitly. At any instant, evaluate $\mathcal{F}$ at the system's current velocities: that single number is half the power currently leaving the mechanical world. The gradient of the paraboloid at that point, reversed, is the entire frictional force ledger, distributed correctly over every coordinate. And because a quadratic bowl has its minimum — zero — at the origin, three facts arrive at once: friction vanishes at rest, always opposes motion, and can only ever remove energy. That last one is the second law of thermodynamics smuggled into mechanics as a convexity statement: the positive-semidefiniteness of the coefficient matrix $c_{ij}$ is what forbids a dashpot from spontaneously flinging the mass into faster motion.

Figure

The dissipation paraboloid — friction is steepest descent in velocity space

One more orientation point before the machinery. The dissipation function is not part of the Lagrangian and never will be — friction has no action principle in the ordinary sense. $\mathcal{F}$ lives beside $L$ as a second scalar, and the pair $(L, \mathcal{F})$ specifies the dynamics completely. This is the honest division of labor that d'Alembert's principle already promised: monogenic forces flow through the Lagrangian; the rest enter as generalized forces, and Rayleigh's contribution is that for linear friction those generalized forces come, once again, from a single scalar.

The dissipation function and the modified equations

Start with the definition. Take any set of generalized coordinates $q_1,\dots,q_n$ and build a quadratic form in their velocities:

\mathcal{F} = \tfrac{1}{2}\sum_{i,j} c_{ij}\,\dot q_i\,\dot q_j,\qquad c_{ij}=c_{ji},\ \ \mathbf{C}\succeq 0 \qquad (1)

where: $\mathcal{F}$ — the Rayleigh dissipation function, with units of power (watts); $c_{ij}$ — the damping coefficients, arranged as a symmetric, positive-semidefinite matrix $\mathbf{C}$ ; $\dot q_i$ — the generalized velocities. Symmetry costs nothing (only the symmetric part of $\mathbf{C}$ survives the quadratic form), and positive-semidefiniteness is physics, not mathematics: it encodes that friction never feeds energy in. For a single mass with Stokes drag $-b\dot x$ , the whole apparatus collapses to $\mathcal{F} = \tfrac12 b\dot x^2$ . The equations of motion then acquire exactly one new term. If $L = T - U$ carries all the conservative physics, the dynamics with linear friction reads

\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot q_j} - \frac{\partial L}{\partial q_j} = -\frac{\partial \mathcal{F}}{\partial \dot q_j} \qquad (2)

where: the left side is the untouched Euler–Lagrange operator and the right side is the generalized dissipative force

Q_j^{\text{diss}} = -\frac{\partial \mathcal{F}}{\partial \dot q_j} = -\sum_i c_{ji}\,\dot q_i \qquad (3)

where: $Q_j^{\text{diss}}$ — the frictional force conjugate to $q_j$ : a drag torque if $q_j$ is an angle, a drag force if a length. Read equation (2) as a division of the world into two scalars: everything reversible lives in $L$ on the left, everything irreversible lives in $\mathcal{F}$ on the right. Set $\mathcal{F}=0$ and conservative mechanics returns untouched; hold the system at rest and the right side vanishes — friction only exists in motion. Where does the recipe (3) come from? It is not a definition to memorize but a two-line consequence of d'Alembert's principle, and the derivation shows exactly which forces qualify.

Derivation AThe dissipative force as a velocity gradient

We are deriving the central recipe $Q_j^{\text{diss}} = -\partial\mathcal{F}/\partial\dot q_j$ from d'Alembert's principle, for the class of forces linear in the particle velocities. The strategy: write the linear drag law on each particle, project it onto the generalized coordinates using the standard generalized-force formula, and watch the projection reorganize itself into the velocity-gradient of one scalar. The step to watch is the "cancellation of dots" — the same lemma that powers the Euler–Lagrange derivation — which is what converts a geometric projection into a velocity derivative. The payoff of doing this honestly: you will see precisely why linear is not negotiable.

Step 1. Let each particle $a$ feel a drag force proportional to its own velocity,

\mathbf{F}_a^{\text{diss}} = -k_a\,\mathbf{v}_a

Why this is the right starting class: Stokes drag on a slow sphere, a dashpot's resistance, the effective damping of internal material friction — all are linear in velocity at small speeds, which is also the leading term of any smooth drag law expanded about rest. (Coefficients may differ per particle and even be anisotropic matrices; nothing below changes except bookkeeping.)

Step 2. Project onto the generalized coordinates with the generalized-force formula from d'Alembert's principle:

Q_j^{\text{diss}} = \sum_a \mathbf{F}_a^{\text{diss}}\cdot\frac{\partial \mathbf{r}_a}{\partial q_j} = -\sum_a k_a\,\mathbf{v}_a\cdot\frac{\partial \mathbf{r}_a}{\partial q_j}

Step 3. Apply the cancellation-of-dots identity $\partial\mathbf{r}_a/\partial q_j = \partial\mathbf{v}_a/\partial\dot q_j$ :

Q_j^{\text{diss}} = -\sum_a k_a\,\mathbf{v}_a\cdot\frac{\partial \mathbf{v}_a}{\partial \dot q_j}

Why this lemma holds: $\mathbf{v}_a = \sum_k (\partial\mathbf{r}_a/\partial q_k)\dot q_k + \partial\mathbf{r}_a/\partial t$ is linear in the $\dot q$ 's, so differentiating with respect to $\dot q_j$ simply picks out the coefficient $\partial\mathbf{r}_a/\partial q_j$ .

Step 4. Recognize a perfect velocity-derivative. Because $\mathbf{v}_a\cdot\partial\mathbf{v}_a/\partial\dot q_j = \tfrac12\,\partial(v_a^2)/\partial\dot q_j$ ,

Q_j^{\text{diss}} = -\frac{\partial}{\partial \dot q_j}\left(\tfrac12\sum_a k_a v_a^2\right) \equiv -\frac{\partial \mathcal{F}}{\partial \dot q_j}

Why linearity was essential: the drag force had to be the gradient of something quadratic in velocity, and only a linear force is. A Coulomb force of constant magnitude or a quadratic drag $\propto v^2$ leaves Step 4 with no scalar to pull out — the projection simply does not reassemble into a gradient of any smooth quadratic. *Why the coefficients inherit the right structure: substituting the scleronomic expansion $\mathbf{v}_a = \sum_k(\partial\mathbf{r}_a/\partial q_k)\dot q_k$ into $\tfrac12\sum_a k_a v_a^2$ reproduces the quadratic form (1) with $c_{ij} = \sum_a k_a\,(\partial\mathbf{r}_a/\partial q_i)\cdot(\partial\mathbf{r}_a/\partial q_j)$ — manifestly symmetric and positive-semidefinite.*

Notice what the Verified line delivered as a bonus: the coefficients $c_{ij}$ in generalized coordinates are automatically symmetric and positive-semidefinite because they descend from positive particle drags through real geometry — the same mechanism that makes the kinetic-energy mass matrix symmetric and positive. The two quadratic forms $T$ and $\mathcal{F}$ are structural twins, one built on $m_a$ , the other on $k_a$ .

The energy ledger

The dissipation function earns its name with a bookkeeping identity. Let $E$ be the Jacobi energy function $E = \sum_j \dot q_j\,\partial L/\partial\dot q_j - L$ — the quantity that conservative, time-independent Lagrangians conserve. Switch on linear friction and $E$ is no longer constant; it drains at a rate the formalism computes for you:

\frac{\mathrm{d}E}{\mathrm{d}t} = -2\mathcal{F} \qquad (4)

where: the factor 2 is not a convention but a theorem — Euler's identity for homogeneous functions applied to the degree-2 form $\mathcal{F}$ . So the number you compute when you evaluate $\mathcal{F}$ is exactly half the instantaneous power bill. Rayleigh chose the normalization (1) so that $\mathcal{F}$ mirrors the structure of kinetic energy; the price is this factor of two at the cash register, and forgetting it is the single most common error in applications.

Go deeper · Derivation B · 1 minThe energy drains at twice the dissipation function

Derivation BThe energy drains at twice the dissipation function

The goal is the exact energy law (4): for a Lagrangian without explicit time dependence, the Jacobi energy of a linearly damped system decreases at precisely $2\mathcal{F}$ , never faster, never slower. The strategy has two movements — first the general theorem, which is one application of Euler's identity for homogeneous functions; then a concrete audit on the damped oscillator, where the same result falls out of freshman calculus, confirming the general machinery against a case you can check by hand. Watch the homogeneity hypothesis: it is where the scleronomic (time-independent) assumption quietly enters.

Step 1. Differentiate the Jacobi energy along a trajectory of the modified equations (2). For $\partial L/\partial t = 0$ the standard computation leaves only the non-conservative power:

\frac{\mathrm{d}E}{\mathrm{d}t} = \sum_j \dot q_j\,Q_j^{\text{diss}} = -\sum_j \dot q_j\,\frac{\partial \mathcal{F}}{\partial \dot q_j}

Why only this survives: the conservative parts of $\dot E$ cancel by exactly the algebra that proves energy conservation in the frictionless case — the new term is the rate of working of the extra force on the right of (2).

Step 2. Apply Euler's theorem for homogeneous functions. When the coordinate transformation carries no explicit time, each $\mathbf{v}_a$ is strictly linear in the $\dot q$ 's, so $\mathcal{F}$ is homogeneous of degree 2 in the velocities, and

\sum_j \dot q_j\,\frac{\partial \mathcal{F}}{\partial \dot q_j} = 2\,\mathcal{F}

Why degree 2 means factor 2: for any function with $\mathcal{F}(\lambda\dot q) = \lambda^2\mathcal{F}(\dot q)$ , differentiate with respect to $\lambda$ and set $\lambda = 1$ . The same theorem gives $\sum \dot q_j\,\partial T/\partial\dot q_j = 2T$ for kinetic energy — the two facts are siblings.

Step 3. Combine, and read the sign:

\frac{\mathrm{d}E}{\mathrm{d}t} = -2\mathcal{F} \le 0

Why the inequality is the second law in miniature: $\mathcal{F} \ge 0$ because $\mathbf{C}$ is positive-semidefinite, so mechanical energy is monotonically non-increasing whenever any damped coordinate moves. Equality holds only at instants when the motion is entirely along undamped directions.

Step 4. Audit on the damped oscillator. Take $E = \tfrac12 m\dot x^2 + \tfrac12 kx^2$ and $\mathcal{F} = \tfrac12 b\dot x^2$ . Differentiating, $\dot E = \dot x(m\ddot x + kx)$ ; the equation of motion $m\ddot x + kx = -b\dot x$ (Derivation C below) turns this into

\frac{\mathrm{d}E}{\mathrm{d}t} = -b\,\dot x^2 = -2\mathcal{F}

Why the two approaches agree: averaged over a cycle in the lightly damped limit, $\langle b\dot x^2\rangle = (b/m)\,\langle E\rangle$ , so the energy audit $\dot E = -2\mathcal{F}$ reproduces the exponential energy decay $E \propto e^{-(b/m)t}$ — the same result obtained by solving the <a class="wiki-link" href="/wiki/aq-mech-damped-harmonic-oscillator" data-track="wiki_article.body.wiki_link">damped harmonic oscillator</a> directly.

Equation (4) is often the fastest route to physical answers. You never need to solve the motion to know the drain rate — evaluate one quadratic at the current velocities and double it. Steady states become power balances: a driven, damped system settles exactly where the drive's average power input equals $\langle 2\mathcal{F}\rangle$ , which is how the resonance curve and the quality factor $Q = m\omega_0/b$ of an oscillator fall out of the dissipation function with almost no work.

A zoo of dissipation functions

The formalism shows its reach in how little it changes from system to system. A mass on a spring with a dashpot:

L = \tfrac12 m\dot x^2 - \tfrac12 kx^2, \qquad \mathcal{F}=\tfrac12\, b\,\dot x^2 \qquad (5)

where: $b$ — the dashpot constant (N·s/m). Equation (2) delivers $m\ddot x + b\dot x + kx = 0$ , the canonical damped oscillator. An RLC circuit is the same mathematics wearing different units. With charge $Q$ as the coordinate,

L_{\text{circ}} = \tfrac12 L\dot Q^2 - \frac{Q^2}{2C}, \qquad \mathcal{F}=\tfrac12 R\,\dot Q^2 \qquad (6)

where: $L$ — inductance, playing mass; $C$ — capacitance, playing inverse stiffness; $R$ — resistance, the dissipative element; $\dot Q = I$ — the current. The modified equation reproduces $L\ddot Q + R\dot Q + Q/C = 0$ , and the energy law (4) reads $\dot E = -R I^2$ : the dissipation function is Joule heating, $2\mathcal{F} = RI^2$ . That one scalar function moves untouched between a workshop dashpot and a resistor is the formalism advertising its indifference to hardware. For structures with thousands of coordinates, nobody measures every $c_{ij}$ . Engineering practice assumes proportional — Rayleigh — damping:

\mathbf{C} = \alpha\,\mathbf{M} + \beta\,\mathbf{K} \qquad (7)

where: $\mathbf{M}, \mathbf{K}$ — the mass and stiffness matrices of the undamped structure; $\alpha$ (units s $^{-1}$ ) — mass-proportional damping, dominant at low frequency; $\beta$ (units s) — stiffness-proportional damping, dominant at high frequency. The magic of this ansatz: the modal transformation that diagonalizes $\mathbf{M}$ and $\mathbf{K}$ automatically diagonalizes $\mathbf{C}$ , so every normal mode stays an independent damped oscillator, with modal damping ratio

\zeta_n = \frac{1}{2}\left(\frac{\alpha}{\omega_n} + \beta\,\omega_n\right) \qquad (8)

where: $\omega_n$ — the $n$ -th undamped modal frequency. The two terms trace a "bathtub" in frequency: mass damping punishes slow modes, stiffness damping punishes fast ones, and the designer chooses $\alpha, \beta$ to pin acceptable damping onto the modes that matter (worked example 2 does exactly this).

Figure

Designing Rayleigh damping — the bathtub curve between two anchored modes

Which dissipation function when

situation	$\mathcal{F}$	notes
single dashpot / Stokes drag	$\tfrac12 b\dot x^2$	exact for linear drag
resistor in a circuit	$\tfrac12 R\dot Q^2$	$2\mathcal{F} = RI^2$ Joule heating
dashpot between two masses	$\tfrac12 c(\dot x_1 - \dot x_2)^2$	off-diagonal $c_{ij}$ ; Derivation D
large structure, damping unknown	$\tfrac12\dot{\mathbf q}^{\mathsf T}(\alpha\mathbf{M}+\beta\mathbf{K})\dot{\mathbf q}$	modal ratios (8); design via two anchors
dry (Coulomb) friction	none exists	use generalized force $Q_j$ directly
quadratic air drag	none (quadratic)	degree-3 form possible but $\dot E \ne -2\mathcal{F}$

Two derivations complete the toolkit: the single damped oscillator worked end to end, and the coupled pair that shows off-diagonal damping working exactly as the matrix promises.

Go deeper · Derivation C · 1 minEquation of motion for the damped oscillator

Derivation CEquation of motion for the damped oscillator

We are applying the modified Euler–Lagrange equation (2) to the simplest damped system — one mass, one spring, one dashpot — to confirm the formalism reproduces the equation every physicist already trusts. The strategy is pure computation: three derivatives from $L$ , one from $\mathcal{F}$ , assemble. The point of doing something this small carefully is calibration: after this, you can trust the same four moves on systems where you do not know the answer.

Step 1. Compute the conservative pieces from $L = \tfrac12 m\dot x^2 - \tfrac12 kx^2$ :

\frac{\partial L}{\partial \dot x} = m\dot x, \qquad \frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot x} = m\ddot x, \qquad \frac{\partial L}{\partial x} = -kx

Step 2. Compute the dissipative piece from $\mathcal{F} = \tfrac12 b\dot x^2$ :

\frac{\partial \mathcal{F}}{\partial \dot x} = b\dot x

Why the sign takes care of itself: the recipe puts $-b\dot x$ on the force side, automatically opposing whichever direction the mass currently moves — no case analysis, no arrows to flip by hand.

Step 3. Assemble equation (2):

m\ddot x + kx = -b\dot x \qquad\Longrightarrow\qquad m\ddot x + b\dot x + kx = 0

The characteristic roots $\lambda = \big(-b \pm \sqrt{b^2 - 4mk}\big)/2m$ reproduce the full solution of the <a class="wiki-link" href="/wiki/aq-mech-damped-harmonic-oscillator" data-track="wiki_article.body.wiki_link">damped harmonic oscillator</a>, covering the underdamped, critical, and overdamped regimes.

Go deeper · Derivation D · 1 minTwo coupled damped oscillators from one dissipation function

Derivation DTwo coupled damped oscillators from one dissipation function

The goal is to see the matrix structure of $\mathcal{F}$ do real work: two masses joined by a dashpot damp their relative motion, and the off-diagonal coefficients $c_{12}$ must encode that coupling with the correct signs. The strategy: write the one-line physical statement — the dashpot dissipates according to the relative velocity — as a dissipation function, expand it into the quadratic form, and differentiate. Watch Newton's third law emerge without being asked for: it is hiding in the symmetry of $\mathbf{C}$ .

Step 1. State the physics as a scalar. A dashpot of constant $c$ spanning masses 1 and 2 resists their relative sliding, so

\mathcal{F} = \tfrac12\, c\,(\dot x_1 - \dot x_2)^2

Step 2. Expand into the standard form (1):

\mathcal{F} = \tfrac12 c\,\dot x_1^2 - c\,\dot x_1\dot x_2 + \tfrac12 c\,\dot x_2^2 \qquad\Longrightarrow\qquad \mathbf{C} = \begin{pmatrix} c & -c \\ -c & c \end{pmatrix}

Why this matrix is only semi-definite: its eigenvalues are $2c$ (relative motion, damped) and $0$ (moving both masses together, undamped — the dashpot doesn't stretch). Zero modes of $\mathbf{C}$ are motions friction cannot see.

Step 3. Differentiate for each force:

Q_1^{\text{diss}} = -\frac{\partial\mathcal{F}}{\partial\dot x_1} = -c(\dot x_1 - \dot x_2), \qquad Q_2^{\text{diss}} = -\frac{\partial\mathcal{F}}{\partial\dot x_2} = +c(\dot x_1 - \dot x_2)

Why the third law came free: the two forces are equal and opposite because $\mathcal{F}$ depends on the velocities only through their difference — the symmetry of the scalar enforces the symmetry of the forces.

Step 4. Assemble the coupled equations for identical grounded springs $k$ :

m_1\ddot x_1 + kx_1 + c(\dot x_1 - \dot x_2) = 0, \qquad m_2\ddot x_2 + kx_2 - c(\dot x_1 - \dot x_2) = 0

This is the damping matrix $\mathbf{C}$ of standard structural dynamics: for $m_1 = m_2$ the in-phase normal mode is undamped and the out-of-phase mode carries damping constant $2c$ — consistent with the eigenvalues found in Step 2 and the classical treatment of damped <a class="wiki-link" href="/wiki/aq-mech-coupled-oscillators" data-track="wiki_article.body.wiki_link">coupled oscillators</a>.

Limiting cases

Case	Result	Meaning
Stokes (linear) drag	$\mathbf{F} = -b\mathbf{v} \Rightarrow \mathcal{F} = \tfrac12 bv^2$	the native regime — exact at low Reynolds number, for dashpots, and for linear resistors
zero damping $\mathcal{F}\to 0$	ordinary Euler–Lagrange equations	Jacobi energy conservation recovered; the conservative world is the $\mathbf{C} = 0$ corner
Coulomb (dry) friction	$F = -\mu N\,\mathrm{sgn}(\dot x)$	NOT representable: constant magnitude at any speed has no smooth quadratic parent
quadratic (Newtonian) drag	$F = -c\,\dot x\lvert\dot x\rvert$	NOT representable by (1); a degree-3 homogeneous form exists but then $\dot E = -3\mathcal{F}$ , not $-2\mathcal{F}$
rheonomic coordinates	$\mathcal{F}$ gains linear and constant velocity terms	homogeneity broken — the clean $\dot E = -2\mathcal{F}$ audit fails with explicitly time-dependent transformations
$\mathbf{C}$ with zero modes	some motions undamped	eigenvectors of $\mathbf{C}$ with zero eigenvalue slide friction-free (Derivation D's in-phase mode)

Common trapThe power dissipated is F — and any friction can be stuffed into it

Both halves of this sentence are wrong, and each burns a different audience. First: the dissipated power is $2\mathcal{F}$ , not $\mathcal{F}$ — Rayleigh normalized the function like a kinetic energy ( $\tfrac12$ in front), so Euler's degree-2 theorem doubles it at the energy ledger. If your simulation's energy budget is off by exactly a factor of two, this is almost certainly where. Second: the formalism is a specialist, not a generalist. It represents forces linear in velocity — dashpots, Stokes drag, resistors — and nothing else. Dry Coulomb friction ( $\lvert F\rvert$ constant, direction fighting the slide) and quadratic air drag ( $F \propto v^2$ ) have no quadratic $\mathcal{F}$ : no smooth bowl has their force as its downhill gradient. For those, go back to d'Alembert and insert the generalized force $Q_j$ by hand. The equations stay honest; you just lose the one-scalar packaging and the automatic $-2\mathcal{F}$ energy audit.

Worked examples

1 · Auditing a damped oscillator without solving it (confidence builder)

A mass $m = 0.50\ \mathrm{kg}$ hangs on a spring of stiffness $k = 200\ \mathrm{N/m}$ with a dashpot $b = 1.2\ \mathrm{kg/s}$ . (a) At an instant when the speed is $\dot x = 0.80\ \mathrm{m/s}$ , what is the dissipation function and the power being lost? (b) Released from rest at amplitude $A = 5.0\ \mathrm{cm}$ , roughly how long until half the mechanical energy is gone, and how many oscillations is that?

Step 1. The dissipation function at the given instant: $\mathcal{F} = \tfrac12 b\dot x^2 = \tfrac12 (1.2)(0.80)^2 = \boxed{0.384\ \mathrm{W}}$ . The power draining out of the mechanical energy is twice this: $2\mathcal{F} = \boxed{0.768\ \mathrm{W}}$ — the factor the warning box exists for.

Step 2. Check the regime. $\omega_0 = \sqrt{k/m} = \sqrt{400} = 20\ \mathrm{rad/s}$ and $\gamma = b/2m = 1.2\ \mathrm{rad/s}$ , so $\gamma \ll \omega_0$ : lightly damped, quality factor $Q = m\omega_0/b = (0.50)(20)/1.2 = 8.3$ .

Step 3. Energy decay without solving the ODE. Averaged over a cycle, $\langle 2\mathcal{F}\rangle = b\langle\dot x^2\rangle = (b/m)\langle E\rangle$ , so the energy obeys $\langle\dot E\rangle = -(b/m)\langle E\rangle$ : exponential decay with time constant $\tau_E = m/b = 0.50/1.2 = 0.417\ \mathrm{s}$ .

Step 4. Half-life and cycle count. $t_{1/2} = \tau_E\ln 2 = 0.417 \times 0.693 = \boxed{0.289\ \mathrm{s}}$ . The damped period is $T = 2\pi/\omega_d$ with $\omega_d = \sqrt{\omega_0^2 - \gamma^2} = 19.96\ \mathrm{rad/s}$ , so $T = 0.315\ \mathrm{s}$ : the oscillator loses half its energy in $\boxed{\text{about } 0.9\ \text{oscillations}}$ . Cross-check with $Q$ : the fractional energy loss per cycle is $2\pi/Q = 0.75$ , i.e. only a quarter of the energy survives each full swing — consistent with a half-life shorter than one period. (Initial energy, for scale: $E_0 = \tfrac12 kA^2 = \tfrac12(200)(0.05)^2 = 0.25\ \mathrm{J}$ .)

2 · Designing Rayleigh damping for a building model (exam level)

A finite-element model of a six-story frame must carry $\zeta = 2.0\%$ damping in its two seismically dominant modes, at $f_1 = 1.0\ \mathrm{Hz}$ and $f_2 = 5.0\ \mathrm{Hz}$ . (a) Find the Rayleigh coefficients $\alpha, \beta$ . (b) What damping does an intermediate 3.0 Hz mode actually receive? (c) Where is the damping minimum and how low does it go?

Step 1. Convert to angular frequencies: $\omega_1 = 2\pi(1.0) = 6.283\ \mathrm{rad/s}$ , $\omega_2 = 2\pi(5.0) = 31.42\ \mathrm{rad/s}$ .

Step 2. Impose equation (8) at both anchors. With equal target $\zeta$ at $\omega_1$ and $\omega_2$ , the pair of linear equations $\zeta = \tfrac12(\alpha/\omega + \beta\omega)$ solves to

\alpha = \frac{2\zeta\,\omega_1\omega_2}{\omega_1 + \omega_2}, \qquad \beta = \frac{2\zeta}{\omega_1 + \omega_2}

Step 3. Plug in numbers: $\omega_1\omega_2 = 197.4$ , $\omega_1 + \omega_2 = 37.70$ , so $\alpha = (0.04)(197.4)/37.70 = \boxed{0.209\ \mathrm{s^{-1}}}$ and $\beta = 0.04/37.70 = \boxed{1.061\times 10^{-3}\ \mathrm{s}}$ .

Step 4. The 3 Hz mode: $\omega = 18.85\ \mathrm{rad/s}$ , so $\zeta = \tfrac12(0.209/18.85 + 0.001061 \times 18.85) = \tfrac12(0.01111 + 0.02000) = 0.0156 = \boxed{1.56\%}$ — less than the 2% at the anchors. This sag is generic: interior modes are always under-damped relative to the anchors.

Step 5. The minimum of the bathtub. Setting $\mathrm{d}\zeta/\mathrm{d}\omega = 0$ gives $\omega^* = \sqrt{\alpha/\beta} = \sqrt{197.4} = 14.05\ \mathrm{rad/s} = 2.24\ \mathrm{Hz}$ (the geometric mean $\sqrt{\omega_1\omega_2}$ ), with $\zeta_{\min} = \sqrt{\alpha\beta} = \boxed{1.49\%}$ . Design moral: choose anchors that bracket the modes you must protect — anything between them gets slightly less damping than promised, and anything far outside gets much more (a 0.5 Hz mode here would receive $3.5\%$ ).

Origins · 1 minLord Rayleigh and how this was discovered

Lord Rayleigh

1842–1919 · Terling Place, Essex

John William Strutt inherited the barony of Rayleigh at 30 and, against every expectation attached to a Victorian estate, converted a wing of Terling Place into a private laboratory and stayed a working physicist for life. The dissipation function appeared in an 1873 paper on vibrating systems and was folded into The Theory of Sound — the two-volume masterwork he began writing on a houseboat trip up the Nile, convalescing from rheumatic fever. His career is a lesson in the power of careful second looks: re-measuring the density of nitrogen to the fourth decimal place led him (with Ramsay) to discover argon and win the 1904 Nobel Prize; thinking honestly about why the sky is blue produced Rayleigh scattering; asking when two point sources are distinguishable produced the Rayleigh criterion that still gates every telescope and microscope. He succeeded Maxwell as Cavendish Professor, standardized the ohm, ampere and volt, and served as chancellor of Cambridge — yet published his last paper in the year he died, having never really left the bench. The dissipation function is characteristic Rayleigh: no new physics claimed, just a structure so well-chosen that acousticians, circuit designers, and earthquake engineers still fill in his two coefficients a century and a half later.

History

1743

— d'Alembert's principle admits non-conservative forces into analytical mechanics as generalized forces, with no systematic packaging.

1788

— Lagrange's Mécanique analytique perfects the conservative formalism; friction remains a case-by-case intruder.

1873

— Rayleigh introduces the dissipation function in a paper on vibrations, giving linear friction its own generating scalar.

1877

— The Theory of Sound, Vol. 1 (§81) systematizes damped vibration theory around F; the book becomes the founding text of acoustics.

1951

— Callen and Welton prove the fluctuation-dissipation theorem, tying the macroscopic damping coefficients in F to microscopic thermal fluctuations.

Going further

References. Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., §1.5 and §6.5 · Landau & Lifshitz, Mechanics, 3rd ed., §25 · Morin, Introduction to Classical Mechanics, Ch. 6 · Taylor, Classical Mechanics, §7.x (generalized forces) · Rayleigh, The Theory of Sound, Vol. 1, §81 Key checkpoints in this article: the cancellation-of-dots identity $\partial\mathbf r_a/\partial q_j=\partial\dot{\mathbf r}_a/\partial\dot q_j$ ; Euler's theorem giving $\sum_j\dot q_j\partial\mathcal{F}/\partial\dot q_j=2\mathcal{F}$ for the degree-2 homogeneous form; and the Rayleigh modal damping ratio $\zeta_n=\tfrac12(\alpha/\omega_n+\beta\omega_n)$ . Worked-example values ( $2\mathcal{F} = 0.768\ \mathrm{W}$ , $t_{1/2} = 0.289\ \mathrm{s}$ , $\alpha = 0.209\ \mathrm{s^{-1}}$ , $\beta = 1.061\times10^{-3}\ \mathrm{s}$ , $\zeta_{3\,\mathrm{Hz}} = 1.56\%$ , $\zeta_{\min} = 1.49\%$ at $2.24\ \mathrm{Hz}$ ) are illustrative arithmetic computed from the boxed formulas.

Builds on: The Euler-Lagrange Equations · The Lagrangian Function · Generalized Forces and Generalized Momenta · d'Alembert's Principle · The Energy Function and Jacobi Integral Leads to: Cyclic Coordinates and Conservation of Conjugate Momenta · Normal Modes and Coupled Oscillations

Why is the dissipated power 2F and not F — and why define F with the half in the first place?

The factor traces to Euler's theorem: $\mathcal{F}$ is homogeneous of degree 2 in velocities, so $\sum_j \dot q_j\,\partial\mathcal{F}/\partial\dot q_j = 2\mathcal{F}$ , and that weighted sum is exactly the rate of frictional working. Rayleigh kept the $\tfrac12$ normalization deliberately, to make $\mathcal{F}$ a structural twin of kinetic energy $T = \tfrac12\sum m_{ij}\dot q_i\dot q_j$ : same quadratic form, mass matrix swapped for damping matrix, and force-by-differentiation with the same clean coefficients. You pay for the pretty symmetry once, at the energy ledger. Mnemonic: kinetic energy also hides its factor there — $\sum\dot q_j\,\partial T/\partial\dot q_j = 2T$ , which is why the Jacobi energy is $2T - L = T + U$ for natural systems.

If F is so useful, why can't we just fold it into the Lagrangian and use Hamilton's principle?

Because friction breaks the variational structure. Hamilton's principle demands monogenic forces — derivable from a potential $U(q)$ or a velocity-dependent potential $U(q,\dot q)$ like the electromagnetic one covered in velocity-dependent potentials. A linear drag $-b\dot x$ satisfies neither: any candidate potential you try either produces the wrong sign on the return swing or fails the Euler–Lagrange consistency conditions. The honest statement is that the dynamics needs two scalars, $L$ and $\mathcal{F}$ , feeding different sides of equation (2). There are formal workarounds — Bateman's trick doubles the coordinates with a time-reversed mirror system so the pair conserves an action — but they add ghost degrees of freedom rather than genuinely putting friction inside one Lagrangian.

What do I actually do for Coulomb friction or quadratic air drag, since F can't hold them?

Fall back one level of generality: compute the generalized force directly, $Q_j = \sum_a \mathbf{F}_a\cdot\partial\mathbf{r}_a/\partial q_j$ , and put it on the right side of the Euler–Lagrange equations in place of $-\partial\mathcal{F}/\partial\dot q_j$ . That route is exact for any force whatsoever — it's just d'Alembert's principle without the packaging. For quadratic drag there is a half-generalization: a degree-3 homogeneous function $\mathcal{F}_3 = \tfrac13\sum c_a v_a^3$ whose velocity-gradient gives $-c v^2$ forces, but Euler's theorem then yields $\dot E = -3\mathcal{F}_3$ , and the tidy "twice the function" audit is lost. For Coulomb friction even that fails — $\mathrm{sgn}(\dot x)$ is not the gradient of any smooth homogeneous form, and the force must additionally switch to static friction at sticking, which no single formula captures.

Is the dissipation function unique for a given physical system?

The generalized forces only see the velocity-gradient of $\mathcal{F}$ , so two functions differing by anything velocity-independent — a constant, or any function of coordinates and time alone — produce identical dynamics. Within the quadratic class the answer is sharper: only the symmetric part of the coefficient matrix survives the quadratic form (an antisymmetric part cancels identically in $\dot q_i\dot q_j$ ), so the symmetric $\mathbf{C}$ appearing in the equations of motion is unique. What is genuinely conventional is the coordinate representation: change generalized coordinates and the matrix transforms like a rank-2 tensor built on velocity space, exactly as the kinetic-energy mass matrix does — same geometry, different numbers.

Does F break Noether's theorem and all the conservation laws I rely on?

It modifies them in a controlled, quantitative way rather than destroying them. For a cyclic coordinate, the exact statement becomes $\dot p_j = -\partial\mathcal{F}/\partial\dot q_j$ : the conjugate momentum decays at a rate the dissipation function computes for you, and it is still conserved if $\mathcal{F}$ happens not to involve that velocity — Derivation D's in-phase mode is a live example, sliding through the dashpot untouched. The energy story is equation (4) itself: conservation degrades into the monotone audit $\dot E = -2\mathcal{F}$ . This "broken-but-budgeted" structure is often more useful than strict conservation, because it tells you not just that quantities decay but exactly how fast, mode by mode.

Where does the lost energy actually go — and is a velocity-linear force even consistent with microscopic physics being reversible?

It goes into heat: disordered kinetic energy of the fluid, the dashpot oil, the resistor's electron gas — degrees of freedom the generalized coordinates deliberately ignore. Friction is what reversible microphysics looks like after you coarse-grain away $10^{23}$ bath coordinates. The deep consistency check is the fluctuation-dissipation theorem (Callen–Welton, 1951): near thermal equilibrium, the same coupling to the bath that damps your coordinate also kicks it randomly, and the damping coefficient in $\mathcal{F}$ is locked to the spectrum of those kicks (Einstein's relation $D = k_BT/b$ for Brownian motion is the most famous instance). So Rayleigh's function is the deterministic half of a matched pair — write down $\mathcal{F}$ for a nanoscale system and thermodynamics has already fixed the noise you must add to be honest.

Can the dissipation function be negative — what would that even model?

A genuinely negative-definite $\mathcal{F}$ inverts equation (4) into $\dot E = +2\lvert\mathcal{F}\rvert$ : energy growth linear in the quadratic form, which is anti-damping. That is not exotic — it is how self-excited oscillations start. Aeroelastic flutter of a wing, the bowed violin string in its stick-slip regime, and negative-resistance electronic oscillators all behave, near their operating point, like systems with a negative effective damping coefficient feeding a mode. The formalism handles it perfectly well as long as you remember what broke: the second-law guarantee. Some external reservoir (the airstream, the bow, the power supply) is paying the bill, and the linear model is only valid until the growing amplitude leaves the linear regime — real flutter saturates or the wing fails.

In structural engineering, why assume C = αM + βK instead of using the true damping matrix?

Because the true damping matrix essentially cannot be known: damping in a real structure comes from micro-slip in ten thousand bolted joints, material hysteresis, cladding, and soil — none of it computable from blueprints the way $\mathbf{M}$ and $\mathbf{K}$ are. What is measurable is a handful of modal damping ratios. Proportional damping is the minimal model consistent with those measurements that also preserves the undamped mode shapes, so the million-DOF problem still decouples into scalar damped oscillators — computationally decisive for time-history earthquake analysis. Its known artifact is the bathtub of worked example 2 (interior modes slightly under-damped, far modes over-damped); when that matters, engineers use Caughey's series generalization or specify $\zeta_n$ mode-by-mode, both still living inside Rayleigh's quadratic- $\mathcal{F}$ framework.

Rayleigh Dissipation Function

Where this lives today

The physical picture

The dissipation function and the modified equations

The energy ledger

A zoo of dissipation functions

Which dissipation function when

Limiting cases

Worked examples

1 · Auditing a damped oscillator without solving it (confidence builder)

2 · Designing Rayleigh damping for a building model (exam level)

Going further

Before this

Where this leads