Are Angular Quantities Vectors? — Question 4

Section 10-2 · Physics pages 1398–1399

Question 4 Infinitesimal rotation addition works only to first order For a very small rotation vector δ⃗θ, a position changes by δ⃗r = δ⃗θ ×⃗r to first order. Let δ⃗θ1 = ϵˆx and δ⃗θ2 = ϵˆy, with |ϵ| ≪1. x y z⃗ r δ⃗r A small rotation moves the tip perpendicular to its original radius; the increment is enlarged for visibility. (a) Show why their combined first-order effect is independent of order. (b) For initial⃗r = Rˆz, find that first-order change and identify the order of the omitted noncommuting correction.
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Question 4 - Solution Infinitesimal rotation addition works only to first order x y z⃗ r δ⃗r (a) Derivation and calculation Apply the two first-order operations successively:⃗ r12 =⃗r + δ⃗θ1 ×⃗r + δ⃗θ2 ×⃗r + O(ϵ2). δ⃗r = (δ⃗θ1 + δ⃗θ2) ×⃗r + O(ϵ2) . The retained linear sum is independent of order. (b) Result and interpretation For⃗r = Rˆz, δ⃗r = ϵR(ˆx × ˆz + ˆy × ˆz) = ϵR(ˆx −ˆy) to first order. The leading general order difference is⃗ r12 −⃗r21 = (δ⃗θ2 × δ⃗θ1) ×⃗r + O(ϵ3). It is generally O(ϵ2), but this quadratic term vanishes for⃗r ∥ˆz. Physical check The first-order increment has zero dot product with Rˆz, so it preserves the radius to first order.

Question 4

Infinitesimal rotation addition works only to first order

For a very small rotation vector δθ→\delta\vec\theta, a position changes by δr→=δθ→×r→\delta\vec r=\delta\vec\theta\times\vec r to first order. Let δθ→1=ϵx̂\delta\vec\theta_1=\epsilon\hat x and δθ→2=ϵŷ\delta\vec\theta_2=\epsilon\hat y, with |ϵ|≪1|\epsilon|\ll1.

Diagram shown in the original SVG worksheet.

  1. Show why their combined first-order effect is independent of order.

  2. For initial r→=Rẑ\vec r=R\hat z, find that first-order change and identify the order of the omitted noncommuting correction.

Question 4 - Solution

Infinitesimal rotation addition works only to first order

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

Apply the two first-order operations successively: r→12=r→+δθ→1×r→+δθ→2×r→+O(ϵ2).\vec r_{12}=\vec r+\delta\vec\theta_1\times\vec r +\delta\vec\theta_2\times\vec r+O(\epsilon^2). δr→=(δθ→1+δθ→2)×r→+O(ϵ2).\boxed{\delta\vec r=(\delta\vec\theta_1+\delta\vec\theta_2)\times\vec r+O(\epsilon^2)}. The retained linear sum is independent of order.

(b) Result and interpretation

For r→=Rẑ\vec r=R\hat z, δr→=ϵR(x̂×ẑ+ŷ×ẑ)=ϵR(x̂−ŷ)to first order.\delta\vec r=\epsilon R(\hat x\times\hat z+\hat y\times\hat z) =\boxed{\epsilon R(\hat x-\hat y)}\quad\text{to first order}. The leading general order difference is r→12−r→21=(δθ→2×δθ→1)×r→+O(ϵ3).\vec r_{12}-\vec r_{21}=(\delta\vec\theta_2\times\delta\vec\theta_1)\times\vec r+O(\epsilon^3). It is generally O(ϵ2)O(\epsilon^2), but this quadratic term vanishes for r→∥ẑ\vec r\parallel\hat z.

Physical check

The first-order increment has zero dot product with RẑR\hat z, so it preserves the radius to first order.