From Coordinates to Fields: the Lagrangian Density
A field has a value at every point, so the sum over coordinates becomes an integral of a Lagrangian density and the equations of motion become partial differential equations. The vibrating string is the case where the step can be checked against an answer already known.
What a learner can do afterwards
- Writes a Lagrangian density for a continuous system and derives its field equation
- Explains what replaces the index labelling a coordinate when the system becomes a field
- Recovers the wave equation from a field Lagrangian
- Finds arrival time of a spherical wave front from distance divided by wave speed
1 · Read
A field assigns a value to every point in space, so there are infinitely many parts and the label that numbered coordinates is replaced by the position itself. Where you summed over coordinates, you now integrate a Lagrangian density over space: L(t) = ∫ ℒ dx, and the action is S = ∫ dt ∫ ℒ dx. The field equation comes from demanding that S be stationary under bends of the field u(x, t) that vanish at the spatial endpoints and at the initial and final times, or have compact support. For one field the condition reads ∂ℒ/∂u − ∂/∂t(∂ℒ/∂u_t) − ∂/∂x(∂ℒ/∂u_x) = 0.
A stretched string with mass per length μ > 0, tension T > 0, and small transverse displacement u(x, t) has density ℒ = ½μ u_t² − ½T u_x². The assumptions are a uniform ideal string, small slopes so u_x stays tiny, and no external force. Then ∂ℒ/∂u = 0, ∂ℒ/∂u_t = μ u_t, and ∂ℒ/∂u_x = −T u_x, so the field equation is μ u_tt − T u_xx = 0. This is the wave equation with wave speed c = √(T/μ).
u_tt = c² u_xx. The travelling-wave solution is u(x, t) = F(x − ct) + G(x + ct), for any twice differentiable F and G. F(x − ct) travels right at speed c and G(x + ct) travels left at speed c. Released from rest with initial displacement h(x), differentiate: u_t = −cF′(x − ct) + cG′(x + ct), so zero velocity at t = 0 gives F′(x) = G′(x), hence F = G + C. F + G = h gives F = h/2 + C/2 and G = h/2 − C/2, and the constants cancel. u(x, t) = ½h(x − ct) + ½h(x + ct).
A string of length L fixed at both ends admits sine shapes that vanish at the ends. Those are sin(nπx/L) with whole numbers n, since each equals zero at x = 0 and x = L. Standing-wave solutions are A sin(nπx/L) cos(ωₙt + φ) with ωₙ = ncπ/L. A static nonzero sin shape alone is not a solution, since its second time derivative vanishes but its second space derivative does not. Always extend the initial displacement outside as an odd function about both endpoints with period 2L, and extend the initial velocity the same way, or set it to zero when the string is released from rest.
Formal check: with c = 2 and h(x) = x², u(1, 1) = ½(1 − 2)² + ½(1 + 2)² = ½(1 + 9) = 5. This polynomial example is a formal nondimensional check of the formula on the whole line, not a physical small-slope string, since x² has unbounded slope. A single time-dependent standing mode is an allowed restriction of the full field, not a general replacement for the partial differential equation. In three dimensions a point push in a uniform isotropic medium spreads as a spherical front at constant speed c with no obstacles, so arrival time is distance over speed.
Replace the coordinate list with a field, integrate the density, derive μ u_tt − T u_xx = 0 with c = √(T/μ), and check the string gives traveling waves.
Five extra questions were written for this Lagrangian density topic. No third-party text is adapted.
2 · Watch
Take it off screen
Where it sits
Where this leads
Jobs that lean on this skill. Follow one to see everything it is built on.
13 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.