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physicstrack.app/problems/7.48

Ch. 7 · Problem 7.48 — Lagrangian Gauge Invariance

Let F=F(q1,,qn)F = F(q_1, \cdots, q_n) be any function of the generalized coordinates (q1,,qn)(q_1, \cdots, q_n) of a system with Lagrangian L(q1,,qn,q˙1,,q˙n,t)\mathcal{L}(q_1, \cdots, q_n, \dot{q}_1, \cdots, \dot{q}_n, t). Prove that the two Lagrangians L\mathcal{L} and L=L+dF/dt\mathcal{L}' = \mathcal{L} + dF/dt give exactly the same equations of motion.
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Attempt 4 · Jul 15, 202699/100

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Setup. Let L(q1,,qn,q˙1,,q˙n,t)\mathcal{L}(q_1,\dots,q_n,\dot q_1,\dots,\dot q_n,t) be a Lagrangian satisfying the Euler--Lagrange equations
ddt(Lq˙i)Lqi=0,i=1,,n.(1)\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot q_i}\right) - \frac{\partial \mathcal{L}}{\partial q_i} = 0, \qquad i=1,\dots,n. \quad (1)
Let F=F(q1,,qn,t)F = F(q_1,\dots,q_n,t) be an arbitrary function of the coordinates and time only, and define
L=L+dFdt.\mathcal{L}' = \mathcal{L} + \frac{dF}{dt}.
We want to show L\mathcal{L}' satisfies the same equations of motion (1).

An excellent, rigorous, and complete proof… the student handles the chain rule, the independence of variables, and the commutativity of mixed partials with full precision.

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physicstrack.app/problems/13.5

Ch. 13 · Problem 13.5 — Bead on Helical Wire

\star\star A bead of mass mm is threaded on a frictionless wire that is bent into a helix with cylindrical polar coordinates (ρ,ϕ,z)(\rho, \phi, z) satisfying z=cϕz = c\phi and ρ=R\rho = R, with cc and RR constants. Using ϕ\phi as your generalized coordinate, write down the kinetic and potential energies, and hence the Hamiltonian H\mathcal{H}.
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Attempt 1 · Jul 19, 202695/100
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An excellent, complete solution… goes beyond the requirements with sensible limiting-case checks and dimensional analysis, showing strong physical intuition.

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physicstrack.app/problems/7.48

Ch. 7 · Problem 7.48 — Lagrangian Gauge Invariance

Attempt 3 · Jul 15, 2026

88/100
Lagrangian MechanicsEuler-Lagrange EquationsGauge InvarianceCalculus of Variations

The student correctly proves that the total time derivative of a function of coordinates only can be added to the Lagrangian without changing the equations of motion, computing both key terms and showing they cancel. The core logic is sound, though there are some notational sloppiness issues and a slightly circular framing at the start. Overall a strong and essentially correct proof.

What You Got Right

  • Correct strategy

    The student correctly recognizes that the proof amounts to showing that dF/dt contributes equally to both sides of the Euler-Lagrange equation, so its net contribution vanishes. This is exactly the right approach.

  • Both key terms computed

    The student correctly computes both the partial-derivative-with-respect-to-q term and the velocity term, showing both equal d/dt(∂F/∂q_i), which is the crux of the proof.

  • Correct use of the total time derivative operator

    Expanding dF/dt = Σ_j (∂F/∂q_j) q̇_j is correct since F depends only on the coordinates, and the use of Schwarz's theorem to swap partial derivatives is valid.

  • Clean use of the Kronecker delta

    The evaluation ∂q̇_j/∂q̇_i = δ_{ij} to collapse the sum in the velocity term is done correctly and cleanly.

Improvements

  • Circular/backwards logic in opening framing

    The student begins by asserting that L' satisfies the Euler-Lagrange equations (writing ∂L'/∂q_i = d/dt(∂L'/∂q̇_i)) as if it were already known. The logical structure of the proof should be to compute the Euler-Lagrange expression for L' and show it equals the expression for L, not to assume L' satisfies the equations and then derive that L does.

    Suggestion: Define the Euler-Lagrange operator EL_i(·) = d/dt(∂·/∂q̇_i) - ∂·/∂q_i. Show by linearity that EL_i(L') = EL_i(L) + EL_i(dF/dt), then prove EL_i(dF/dt) = 0 using exactly the two computations you already did.

  • Notational inconsistencies in the total derivative operator

    The expression d/dt = Σ_j q̇_j ∂/∂q_j + ∂/∂q̇_j + ∂/∂t is written with the sum applied inconsistently — the ∂/∂q̇_j term should carry a q̈_j coefficient (and is inside the sum), and ∂/∂t should sit outside the sum. Though harmless here since F has no velocity or explicit time dependence, the operator as written is not correct.

    Suggestion: Write d/dt = Σ_j (q̇_j ∂/∂q_j + q̈_j ∂/∂q̇_j) + ∂/∂t. Then note F depends only on coordinates so the q̈ and ∂/∂t terms drop, giving d/dt = Σ_j q̇_j ∂/∂q_j acting on F or its derivatives.

  • Missing explicit statement of the conclusion

    The student concludes the equations of motion are identical, which is correct, but could more explicitly state that because the extra term contributes zero to the Euler-Lagrange expression, setting EL_i(L') = 0 yields exactly EL_i(L) = 0.

Mathematical Accuracy

The core computations are correct: both the ∂/∂q_i(dF/dt) term and the d/dt(∂/∂q̇_i(dF/dt)) term are correctly shown to equal d/dt(∂F/∂q_i), and they properly cancel. Minor notational errors appear in the total derivative operator, and the opening lines conflate assumption with conclusion, but arithmetic and the essential manipulations are sound.

Approach Quality

The physical and mathematical reasoning is fundamentally correct and demonstrates good understanding of gauge freedom in the Lagrangian. The only weakness is the logical framing at the start, which presents the result as an assumption rather than deriving it.

Next steps: Practice writing proofs in a strictly forward logical direction — define the operator, apply linearity, then show the added piece vanishes. Also review the exact chain-rule form of the total time derivative operator to fix the notational slip.
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δS  =  δ ⁣t1t2L ⁣(qi,q˙i,t)dt  =  0\delta S \;=\; \delta\!\int_{t_1}^{t_2} \mathcal{L}\!\left(q_i,\,\dot{q}_i,\,t\right)\,dt \;=\; 0

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