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 11258[Quote]

suppose l is the length of the string at time 0 and the angle between the x-axis and the string is θ.

why then is the angle that sweeps out the arc length equal to θ as well. i don't get it

 11261[Quote]

ℳ:=𝒯×𝒳 with 𝒯≈ℝ (time, sheaf 𝒪_𝒯), 𝒳≈S²_ℓ⊂ℝ³ (configuration).
Total sheaf 𝒮=C^∞_ℳ. Cylindrical pole ∂B_a⊂ℝ² induces the constraint subsheaf

 𝒞={f∈𝒮 | ϱ(f)=a , ‖f‖₂=ℓ }  (ϱ horizontal radius).

rk 𝒞=1 ⇒ 1 d.o.f.; choose θ:𝒯→𝕊¹. Global section

 σ_θ(t)=(a+ℓcosθ,0,ℓsinθ)

gives |∂ₜσ_θ|²=ℓ²θ̇², so sheaf-Lagrangian

 𝒧=½mℓ²θ̇²−mgℓsinθ.

World-line W=σ_θ(𝒯) is 1-rectifiable; as an integral current

 T=⟦W,ξ,1⟧∈𝔻₁(ℝ⁴), ξ=∂ₜ/|∂ₜ|.

Action functional (Federer §4.1)

 𝒜(T)=∫_T𝒧 dℋ¹ =∫_ℝ(½mℓ²θ̇²−mgℓsinθ) dt.

First–variation δ𝒜(T)=0 ⇒ θ̈+(g/ℓ)cosθ=0.

about the arc-angle:
Let C:=∂B_a⊂ℝ², parameterised by c(φ)=a(cosφ,sinφ). The map φ↦c(φ) is a bi-Lipschitz chart for the 1-rectifiable set C; its push-forward of Lebesgue measure gives the rectifiable current

 S=⟦C,τ,1⟧, τ=c′(φ)/|c′(φ)|.

Mass of the sub-current restricted to the arc φ∈[0,θ]:

 𝑀(S|_{[0,θ]}) = ∫₀^{θ}|c′(φ)| dφ = ∫₀^{θ}a dφ = aθ.

Hence the Hausdorff-length s of that arc satisfies s=aθ; in the radian convention we encode arclength per radius, so the geometric measure theoretic “mass coordinate” s/a and the central angle φ coincide.

We therefore identify the sheaf coordinate θ with this mass-per-radius variable: θ = s/a. The equality “angle sweeps the arc length” is nothing but the statement that the multiplicity-1 integral current on C is parameterised by arclength, and ℋ¹-density |c′| is constant (=a). The confusion lifts: θ is defined precisely so that the arclength functional of the restricted current equals aθ, whence θ itself.

 11264[Quote]

>>11261
what is all this. i just need help with the geometry

 11405[Quote]

>>11264
schizophrenia also known as "Geometric Measure theory". Unless you want to do explicit estimates in, say, General Relativity, you probably won't see that formalism ever again.
What he's basically saying is that there's one degree of freedom, let’s pick the arc-length angle θ (=angle that sweeps out the arc length), so ℓ(t) = ℓ₀ − aθ and the Lagrangian is ½ mℓ²θ̇² − mgℓ sin θ.
Then he derives aθ = s, i.e. θ = s/a but that is just the central angle due to the convention of denoting arclength per radius



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