Rigid bodies and classical field Lagrangians

Start with two masses on a rigid rod

Join two point masses A and B, each 1 kg, with an ideal massless rigid rod of length 1 m. Hold its center at the origin of a fixed inertial X–Y plane. An ideal bearing keeps the perpendicular rotation axis fixed and exerts no torque about that axis. Exclude gravity and friction from this supplied model. Each mass stays 0.5 m from the center.

A rigid-body model keeps the distances between its material points fixed. It permits translation or rotation without changing those distances. Here the fixed center and axis remove translation and tilting: one angle specifies the rod’s orientation. Let counterclockwise be positive and write the physical angle as θ = φ rad. The numerical radian value φ is dimensionless, so it can be used in sine and cosine.

Supply φ(t) = 2t/(1 s) on 0 ≤ t ≤ (π/2) s. The positions are r_A = (0.5 m)(cos φ, sin φ) and r_B = −r_A. Their components refer to the same fixed laboratory axes. The rate dφ/dt is 2 s⁻¹; the physical angular velocity is +2 rad/s. A and B exchange positions after the half-turn, but their labels stay with the masses.

The published lessons Units, standards, and physical constants; Algebra, functions, graphs, and logarithms; and Geometry, trigonometry, and coordinate systems provide the unit, formula and angle conventions used here.

Form the fixed-axis Lagrangian

For rotation about this fixed axis, the moment of inertia is I = Σm_i r_i². The symbol Σ means add the contributions, and r_i is each mass’s perpendicular distance from the axis. Here I = (1 kg)(0.5 m)² + (1 kg)(0.5 m)² = 0.5 kg m². Each speed is r_i(dφ/dt) = 1 m/s, giving total kinetic energy K = 1 J.

A Lagrangian L is a function used to formulate the model’s motion. In this conservative rotor model L = K − V, with potential energy V chosen to be zero, so L = I(dφ/dt)²/2. Put φ̇ = dφ/dt. The angle-coordinate equation is d/dt(∂L/∂φ̇) − ∂L/∂φ = 0: a partial derivative changes the named input while holding the others fixed. Since L has no φ dependence and I is constant, this gives Iφ̈ = 0. The supplied constant rate satisfies it.

The coordinate momentum p_φ = ∂L/∂φ̇ = Iφ̇ is the signed angular-momentum component about this axis: p_φ = +1 kg m²/s = +1 J s. Its constancy follows from the zero axial torque in this model. It does not mean that each mass is force-free. Each circular motion needs inward acceleration of 2 m/s², supplied here by a 2 N inward constraint force; that radial force has zero torque about the center.

Read rotor states in fixed axes

The upper diagram uses the same X and Y metre scales in all three snapshots. A is a circle and B is a square. A cross marks the fixed pivot, and the dashed circle is a radius guide. These symbols show positions and constraints; there are no force arrows. The table gives the exact coordinates rather than measurements.

The scalar I used here is tied to the declared fixed axis. For general three-dimensional rotation, an inertia tensor relates angular velocity to rotational angular momentum about the center of mass. This tensor is a matrix, and those two vectors need not be parallel. Our one-coordinate rod is a restricted case. Ideal rigidity also excludes deformation, so the model does not predict bending, stretching or material stresses.

Replace a coordinate list with a field

Now choose a separate stretched-string model. A fixed reference coordinate x runs from 0 to 2 m along the undeformed string. Its upward transverse displacement is y(x, t). The arguments x and t are independent: at one time, different x values label different reference positions. A field supplies a value at every position in this continuum description, rather than only one angle or a finite set of particle coordinates.

Supply constant mass per reference length μ = 1 kg/m and tension T = 4 N. Include transverse motion only, with no dissipation or additional field potential. The linear string model uses a small-slope approximation: transverse kinetic energy and the leading quadratic change in tension energy determine its formulas. These are model assumptions, not an assertion that a real string is exactly linear or infinitely finely resolved.

You can first follow the profile table, derivative values and boundary balance. The final section gives an optional variational check; return to Derivatives, Taylor series, and local approximation and Integrals and multivariable calculus for its derivative and integration steps. The supplied calculations introduce the needed mechanics locally.

Distinguish density, Lagrangian and action

Use y_t = ∂y/∂t for transverse velocity at fixed x and y_x = ∂y/∂x for the local slope at fixed t. The one-dimensional Lagrangian density is ℒ = μy_t²/2 − Ty_x²/2. The first term is kinetic energy per reference length; the second is the quadratic tension-potential contribution per reference length. The slope is dimensionless, while y_t has units m/s. Both density terms therefore have units J/m.

At each time, integrate over the reference segment: L(t) = ∫₀²ᵐ ℒ dx. Integration adds the density contributions along the string and supplies one length factor, so L has units J. The action over 0 ≤ t ≤ 1 s is S = ∫₀¹ˢ L(t) dt = ∫₀¹ˢ dt ∫₀²ᵐ dx ℒ, with units J s. The integration region and measures are part of the definition; ℒ, L and S are different quantities.

The energy density for this particular time-independent quadratic string model has a plus sign: ℰ = μy_t²/2 + Ty_x²/2. Its integral E(t) is the string segment’s kinetic plus tension energy. A Lagrangian density can be zero while energy density is positive. Energy formulas for other field models require their own derivation; a generic Lagrangian is not automatically the energy.

Check an exact field and its profiles

Define χ = x/(1 m) and τ = t/(1 s), both dimensionless. On 0 ≤ x ≤ 2 m and 0 ≤ t ≤ 1 s, supply y(x, t) = (1/32 m)(χ² + 4τ²). Its derivatives are y_t = (τ/4) m/s, y_x = χ/16, y_tt = (1/4) m/s² and y_xx = (1/16) m⁻¹. The subscripts tt and xx mean take the same partial derivative twice.

The model’s field equation is μy_tt − Ty_xx = 0 in the interior. Substitution gives (1 kg/m)(1/4 m/s²) − (4 N)(1/16 m⁻¹) = 0; both terms have units N/m. The wave-speed parameter is c = √(T/μ) = 2 m/s. Across this finite domain, |y_x| ≤ 1/8 and |y_t|/c ≤ 1/8. These stated bounds make the approximation explicit; they do not give a measured error bound for an apparatus.

The lower diagram shows three exact profiles at t = 0, 0.5 and 1 s, distinguished by line styles and marker shapes. Its horizontal axis is reference position, and its vertical axis is transverse displacement. Those two metre scales differ for legibility, so the screen slope is not the physical slope y_x. Each curve is a field profile at one time, not a particle trajectory. The two panels describe separate models with different time domains.

Include the driven boundary energy

The supplied field starts with y(x, 0) = (1/32 m)χ² and y_t(x, 0) = 0. Its endpoint data are y(0, t) = (1/8 m)τ² and y(2 m, t) = (1/8 m)(1 + τ²). They move for t > 0. The field satisfies those data, but would not satisfy a problem that instead clamps both endpoints at zero. An interior equation alone does not specify a boundary-value problem.

For this field, ℰ = (τ²/32 + χ²/128) J/m. Integrating from x = 0 to 2 m gives E(t) = (τ²/16 + 1/48) J, so E(0) = 1/48 J and E(1 s) = 1/12 J. The segment gains 1/16 J. There is no dissipation in the stipulated equation, but the driven ends can still transfer energy into the segment.

The boundary balance is dE/dt = T[y_t y_x]₀²ᵐ, where the bracket means the value at the right end minus the value at the left end. Here it is +τ/8 W, matching the derivative of E(t). In general this model conserves segment energy when the net boundary-power term is zero, as with stationary clamped ends. Holding a boundary history fixed during a variation is a different condition from holding that boundary motionless in time.

Check stationarity with fixed boundary histories

For the optional action check, compare y with y + εh. The parameter ε is dimensionless and labels trial fields; it is not time. The smooth test field h has displacement units and vanishes at x = 0 and 2 m and at t = 0 and 1 s. Thus every trial preserves the supplied, possibly moving boundary histories and the temporal endpoint profiles. Take y to have continuous second partial derivatives and h smooth enough for integration by parts.

The first derivative of S[y + εh] with respect to ε at zero is δS = ∫ dt ∫ dx (μy_t h_t − Ty_x h_x). Integration by parts moves each derivative from h onto the corresponding field term. The boundary terms vanish because h is zero there, leaving δS = −∫ dt ∫ dx (μy_tt − Ty_xx)h. Our checked field equation therefore makes δS zero for every allowed h. This proves stationarity under these variations, not a universal minimum.

For a smooth density ℒ(y, y_t, y_x, x, t), the corresponding local field equation is ∂t(∂ℒ/∂y_t) + ∂x(∂ℒ/∂y_x) − ∂ℒ/∂y = 0. The first two outer derivatives act on the composed field quantities. In our density ∂ℒ/∂y_t = μy_t, ∂ℒ/∂y_x = −Ty_x and ∂ℒ/∂y = 0. More fields, spatial dimensions, constraints or different boundary conditions need their own specification. The next lesson, Dynamical systems and phase portraits, begins a different way of describing evolving states.

Worked examples

Try it yourself

1. Keep the main rotor’s masses and radius but supply φ̇ = −2 s⁻¹ instead. What are K and p_φ?

   Show the answer and reasoning

   K = 1 J and p_φ = −1 J s.

   K = Iφ̇²/2 depends on the rate squared, whereas p_φ = Iφ̇ retains the sign. This is a separate clockwise model; it does not change the positive-rate snapshots.

2. In the original rotor, A and B exchange positions at the last snapshot. Has the rigid distance changed, and must the inward forces be zero because axial torque is zero?

   Show the answer and reasoning

   Their separation stays 1 m. The inward forces are not zero: each is 2 N in this model.

   Opposite positions at radius 0.5 m remain 1 m apart. Each mass has circular acceleration 2 m/s². An inward radial force has zero moment about the center while changing the velocity direction.

3. For this one-dimensional field, give the units of ℒ, L = ∫ℒ dx and S = ∫L dt. Is ℒ an energy per volume?

   Show the answer and reasoning

   J/m, J and J s, respectively. ℒ is per reference length, not per volume.

   The spatial integral adds a length factor and the time integral adds a time factor. A three-dimensional density would use a volume measure and different density units.

4. For the supplied μ = 1 kg/m, T = 4 N, y_tt = 1/4 m/s² and y_xx = 1/16 m⁻¹, does μy_tt + Ty_xx vanish? Which sign satisfies the stated model?

   Show the answer and reasoning

   The sum is 1/2 N/m, not zero. The difference μy_tt − Ty_xx is zero.

   Both contributions are 1/4 N/m. The minus sign comes from the negative potential term in the Lagrangian density, not from an arbitrary choice of plot direction.

5. At t = 1 s, what displacements does the supplied field assign to x = 0 and 2 m? Would it satisfy boundaries clamped at zero?

   Show the answer and reasoning

   1/8 m and 1/4 m, respectively. It would not satisfy zero clamped boundaries.

   Substitute τ = 1 into the two prescribed endpoint functions. A solution must fit both its interior equation and the selected boundary data.

6. The string’s action over the supplied rectangle is zero. What is its final segment energy and its energy gain, and why are these compatible?

   Show the answer and reasoning

   E(1 s) = 1/12 J and the gain is 1/16 J. They are compatible with S = 0 J s.

   Action integrates the difference of kinetic and potential densities over time, while energy adds them at an instant. The endpoints supply the positive energy gain; neither the zero action value nor the field equation alone removes that transfer.

What to carry forward

• A rigid-body model preserves internal distances; the declared fixed-axis constraints make this rotor a one-angle case.
• Fixed-axis inertia depends on mass distribution about that axis. Zero axial torque permits constant angular momentum while radial forces still act.
• A classical field is indexed by reference position and time; a profile at one time is not a material trajectory.
• In this one-dimensional model ℒ has units J/m, L has units J and S has units J s. The energy density uses a plus sign rather than the Lagrangian density’s minus sign.
• Boundary data and boundary power matter. A driven segment can gain energy even when its conservative interior field equation is satisfied.
• Stationarity requires an explicitly allowed variation class and regularity; it does not imply that every action is minimized.