D'Alembert's Principle and the Derivation of the Lagrange Equations

The previous chapter ended with D'Alembert's principle,

the exact statement of dynamics in which the forces of constraint have already been eliminated. It is not yet usable: it is written in the $3N$ Cartesian displacements $\delta\vec{r}_i$, which are not independent. This chapter carries out the one substantial calculation of analytical mechanics — projecting D'Alembert's principle onto the $n$ independent generalised coordinates — and arrives at Lagrange's equations, a set of $n$ second-order ordinary differential equations that contain no reaction forces and take the same form in every coordinate system.

The calculation has three moving parts: the substitution $\delta\vec{r}_i = \sum_j (\partial\vec{r}_i/\partial q_j)\,\delta q_j$; the two kinematic identities of Chapter 1 (cancellation of dots, and the commuting of $d/dt$ with $\partial/\partial q_j$); and the observation that the resulting combinations of velocities are exactly the partial derivatives of the kinetic energy $T$. Once assembled, the applied forces appear only through the generalised force $Q_j$, and when they are conservative the whole right-hand side collapses into a single scalar function, the Lagrangian $L = T - V$.

How to read this chapter. The derivation in Sections 2–4 is worth reproducing once with pencil and paper; after that you will never do it again, because Lagrange's equations are the starting point for everything that follows. Sections 5–7 widen the class of forces the method handles — conservative, electromagnetic (velocity-dependent potential), and dissipative (Rayleigh function). Keep Chapter 1's worked systems in view: each is solved here in three or four lines.

1. The Starting Point and the Plan

Write the inertial term of D'Alembert's principle as $\sum_i \dot{\vec p}_i \cdot \delta\vec{r}_i = \sum_i m_i \ddot{\vec r}_i \cdot \delta\vec{r}_i$ and the applied term as $\sum_i \vec{F}_i^{(a)} \cdot \delta\vec{r}_i$. Insert $\delta\vec{r}_i = \sum_{j=1}^{n} \dfrac{\partial\vec{r}_i}{\partial q_j}\,\delta q_j$ into both. The applied term becomes $\sum_j Q_j\,\delta q_j$ with $Q_j$ the generalised force of Chapter 1. The inertial term becomes

The plan is to show that the bracket equals $\dfrac{d}{dt}\dfrac{\partial T}{\partial \dot q_j} - \dfrac{\partial T}{\partial q_j}$, where $T = \sum_i \tfrac12 m_i \dot{\vec r}_i^{\,2}$ is the kinetic energy expressed in terms of the $q_j$, $\dot q_j$ and $t$. Then, because the $\delta q_j$ are independent, each coefficient must vanish separately.

2. Rewriting the Inertial Term

Apply the product rule for $d/dt$ to the quantity $m_i \dot{\vec r}_i \cdot (\partial\vec{r}_i/\partial q_j)$:

Solve for the term we want:

Now use the two identities from Chapter 1. In the first term, cancellation of dots replaces $\partial\vec{r}_i/\partial q_j$ by $\partial\dot{\vec{r}}_i/\partial\dot q_j$. In the second term, the commuting identity replaces $\dfrac{d}{dt}(\partial\vec{r}_i/\partial q_j)$ by $\partial\dot{\vec{r}}_i/\partial q_j$:

3. Recognising the Kinetic Energy

The combinations on the right are derivatives of $\tfrac12 m_i \dot{\vec r}_i^{\,2}$. Since $\dfrac{\partial}{\partial\dot q_j}\left(\tfrac12 m_i \dot{\vec r}_i^{\,2}\right) = m_i \dot{\vec r}_i \cdot \dfrac{\partial\dot{\vec r}_i}{\partial\dot q_j}$ and likewise for $\partial/\partial q_j$, summing over all particles gives

This is the promised identity. It is purely kinematic — no forces have entered — and it holds for scleronomic and rheonomic holonomic systems alike, because both Chapter 1 identities do.

4. Lagrange's Equations of Motion

D'Alembert's principle is now

The $\delta q_j$ are independent, so each bracket vanishes:

These are Lagrange's equations of the second kind — valid for any generalised force, conservative or not. There are $n$ of them, one per degree of freedom, they are second order in the $q_j$, and no constraint force appears.

Worked example — plane pendulum. With $q = \theta$, the bob speed is $\ell\dot\theta$, so $T = \tfrac12 m\ell^2\dot\theta^2$. Then $\partial T/\partial\dot\theta = m\ell^2\dot\theta$, $\dfrac{d}{dt}(\cdot) = m\ell^2\ddot\theta$, and $\partial T/\partial\theta = 0$. The generalised force was $Q_\theta = -mg\ell\sin\theta$ (Chapter 1). Lagrange's equation gives $m\ell^2\ddot\theta = -mg\ell\sin\theta$, i.e. $\ddot\theta + (g/\ell)\sin\theta = 0$, in three lines and with no tension.

5. Conservative Forces and the Lagrangian

If every applied force derives from a potential, $Q_j = -\partial V/\partial q_j$ with $V = V(\vec q,t)$ independent of the $\dot q_j$. Then

where the second form is legitimate because $\partial V/\partial\dot q_j = 0$, so subtracting $V$ inside the $\dot q_j$ derivative changes nothing. Defining the Lagrangian

the equations of motion take their canonical form:

The entire dynamics of a conservative holonomic system is encoded in the single scalar function $L$. The next chapter is a catalogue of what falls out of it.

Worked example — bead on a rotating wire. From Chapter 1, $T = \tfrac12 m(\dot s^2 + s^2\omega^2)$ for a wire in the horizontal plane cranked at rate $\omega$; there is no potential ($V = 0$). So $L = \tfrac12 m\dot s^2 + \tfrac12 m s^2\omega^2$. Then $\partial L/\partial\dot s = m\dot s$, $\partial L/\partial s = m s\omega^2$, and Lagrange's equation gives $m\ddot s = m s\omega^2$, i.e. $\ddot s = \omega^2 s$: the bead is flung outward exponentially. The $s^2\omega^2$ term in $T$, from the moving constraint, has acted as an effective outward potential $-\tfrac12 m\omega^2 s^2$ — the centrifugal potential, generated automatically.

6. Velocity-Dependent Potentials

Some forces that are not conservative in the elementary sense still admit a generalised potential $U(\vec q,\dot{\vec q},t)$ such that

For such forces the Lagrangian $L = T - U$ still satisfies $\dfrac{d}{dt}\dfrac{\partial L}{\partial\dot q_j} - \dfrac{\partial L}{\partial q_j} = 0$. The decisive case is the electromagnetic force on a charge $e$: with scalar potential $\phi$ and vector potential $\vec A$,

and a short computation shows that $-\partial U/\partial x_k + \dfrac{d}{dt}(\partial U/\partial\dot x_k)$ reproduces exactly the Lorentz force $e(\vec E + \vec v\times\vec B)$ with $\vec E = -\nabla\phi - \partial\vec A/\partial t$ and $\vec B = \nabla\times\vec A$. This is why a charged particle in a magnetic field has a perfectly good Lagrangian, $L = \tfrac12 m v^2 - e\phi + e\,\vec v\cdot\vec A$, even though the magnetic force does no work.

7. Dissipative Forces and the Rayleigh Function

Forces that genuinely dissipate energy — kinetic friction, viscous drag $\vec F_i^{\text{drag}} = -k_i \vec v_i$ — have no potential. Carry them on the right-hand side as an explicit generalised force:

For drag linear in velocity, define the Rayleigh dissipation function

Then $Q_j^{\text{diss}} = -\partial\mathcal{F}/\partial\dot q_j$, and the equations of motion read $\dfrac{d}{dt}\dfrac{\partial L}{\partial\dot q_j} - \dfrac{\partial L}{\partial q_j} + \dfrac{\partial\mathcal{F}}{\partial\dot q_j} = 0$. The rate of energy loss is $dE/dt = -2\mathcal{F}$, which is the physical meaning of $\mathcal{F}$: twice the instantaneous power dissipated.

Worked example — damped oscillator. For a mass on a spring with linear drag, $L = \tfrac12 m\dot x^2 - \tfrac12 k x^2$ and $\mathcal{F} = \tfrac12 b\dot x^2$. The equation of motion is $m\ddot x + b\dot x + k x = 0$, the familiar damped harmonic oscillator, produced by the same machinery with one extra term.

8. Form Invariance and Equivalence to Newton

Lagrange's equations are covariant under point transformations: if $q_j \to Q_j(\vec q,t)$ is any invertible change of generalised coordinates, the equations in the new coordinates have the identical form $\dfrac{d}{dt}\dfrac{\partial L}{\partial\dot Q_j} - \dfrac{\partial L}{\partial Q_j} = 0$, with $L$ the same scalar re-expressed. This is proved by direct substitution using the same two dot identities, now applied to the map $q \to Q$. It is the structural reason the method is coordinate-free in a way $\vec F = m\vec a$ is not: Newton's law changes shape (picks up Christoffel-like terms) in curvilinear coordinates, whereas Lagrange's equations do not.

Equivalence to Newton runs both ways. In Cartesian coordinates with $L = \sum_i \tfrac12 m_i \dot{\vec r}_i^{\,2} - V$, Lagrange's equations reduce term by term to $m_i\ddot{\vec r}_i = -\nabla_i V$. Conversely the derivation of this chapter shows Newton's law plus ideal constraints implies Lagrange's equations. The two formulations carry the same physical content; the Lagrangian one carries it in a form built for constraints, symmetry and generalisation.

Worked example — free particle in plane polar coordinates. With $x = r\cos\phi$, $y = r\sin\phi$, $T = \tfrac12 m(\dot r^2 + r^2\dot\phi^2)$ and $V = 0$. The $r$-equation: $\dfrac{d}{dt}(m\dot r) - m r\dot\phi^2 = 0 \Rightarrow \ddot r = r\dot\phi^2$. The $\phi$-equation: $\dfrac{d}{dt}(m r^2\dot\phi) = 0 \Rightarrow m r^2\dot\phi = \text{const}$. The first is the radial equation with its centrifugal term; the second is conservation of angular momentum — both emerging without ever writing an acceleration in polar coordinates by hand.

Common Pitfalls

• Forgetting the total time derivative acts on everything. In $\dfrac{d}{dt}\dfrac{\partial L}{\partial\dot q_j}$, the inner quantity depends on $\vec q(t)$, $\dot{\vec q}(t)$ and possibly $t$; use the chain rule fully, do not treat $q_j$ as constant.
• Subtracting $V$ inside the $\dot q_j$ derivative when $V$ depends on velocity. The step $T \to T - V$ inside $\partial/\partial\dot q_j$ is valid only because $\partial V/\partial\dot q_j = 0$. For a velocity-dependent potential you must use the generalised-potential formula of Section 6.
• Using $L = T - V$ for dissipative forces. Friction and drag have no potential; put them on the right-hand side as $Q_j^{\text{diss}}$ or via the Rayleigh function.
• Expressing $T$ in mixed coordinates. $T$ must be written entirely in terms of the chosen $q_j$, $\dot q_j$ and $t$ before differentiating; a leftover Cartesian $\dot x$ will give wrong partials.
• Dropping the $\partial\vec r_i/\partial t$ piece of the velocity for a rheonomic system. It contributes the $T_1$ and $T_0$ parts of the kinetic energy and hence the centrifugal/Coriolis terms; omitting it loses real physics.
• Sign of the Lagrangian. It is $T$ minus $V$, not $T$ plus $V$. The combination $T + V$ is the energy, a different object that is (sometimes) conserved, not extremised.

Summary

• Substituting $\delta\vec r_i = \sum_j(\partial\vec r_i/\partial q_j)\,\delta q_j$ into D'Alembert's principle and using the two Chapter 1 identities turns the inertial term into $\dfrac{d}{dt}\dfrac{\partial T}{\partial\dot q_j} - \dfrac{\partial T}{\partial q_j}$.
• Independence of the $\delta q_j$ then gives Lagrange's equations $\dfrac{d}{dt}\dfrac{\partial T}{\partial\dot q_j} - \dfrac{\partial T}{\partial q_j} = Q_j$ — $n$ second-order ODEs with no constraint forces.
• For conservative forces $Q_j = -\partial V/\partial q_j$, and with the Lagrangian $L = T - V$ the equations become $\dfrac{d}{dt}\dfrac{\partial L}{\partial\dot q_j} - \dfrac{\partial L}{\partial q_j} = 0$.
• A generalised (velocity-dependent) potential $U$ with $Q_j = -\partial U/\partial q_j + \dfrac{d}{dt}\partial U/\partial\dot q_j$ extends the Lagrangian form to the electromagnetic force, $U = e\phi - e\,\vec v\cdot\vec A$.
• Dissipative linear-drag forces are handled by the Rayleigh function $\mathcal{F}$, with $Q_j^{\text{diss}} = -\partial\mathcal{F}/\partial\dot q_j$ and $dE/dt = -2\mathcal{F}$.
• Lagrange's equations are form-invariant under point transformations and fully equivalent to Newton's law plus ideal constraints; the advantage is structural, not a difference of physical content.