Physics 43, Exam 2_2

Image Analysis Visual Elements: Page Layout: The image appears to be a page from a textbook or notebook. Title: "6.17-1" is at the top. Text: The page contains handwritten text and mathematical equations. Diagram: A cone is drawn to illustrate the problem. Calculations: Detailed calculations involving calculus and differential equations are shown. Textual Content: 6.17-1 Find the geodesics on the cone whose equation in cylindrical polar coordinates is z = λρ. [Let the required curve have the form φ = φ(ρ).] Check your result for the case that λ → 0.   Geodesics on a cone? Let curve have form φ = φ(ρ) [z = \lambda\rho] In cylindrical coords, a small move in z-direction adds dist. Δz a small move in ρ-direction adds dist. Δρ a small move in φ-direction adds dist. ρΔφ So: [\Delta s = \sqrt{(\Delta z)^2 + (\Delta \rho)^2 + (\rho \Delta \phi)^2}] In cylindrical coords, [z = \lambda\rho \Rightarrow \Delta z = \lambda \Delta \rho] So we have: [\Delta s = \sqrt{(\lambda \Delta \rho)^2 + (\Delta \rho)^2 + (\rho \Delta \phi)^2}] [= \sqrt{(\lambda^2 + 1)(\Delta \rho)^2 + (\rho \Delta \phi)^2}] [= \Delta \rho \sqrt{(\lambda^2 + 1) + (\rho \frac{\Delta \phi}{\Delta \rho})^2}] [\Delta s = \Delta \rho \sqrt{(\lambda^2 + 1) + (\rho \phi'(ρ))^2}] A geodesic should extremize distance, i.e., [L = \int_{\rho_1}^{\rho_2} ds = \int_{\rho_1}^{\rho_2} d\rho \sqrt{(\lambda^2 + 1) + (\rho \phi'(ρ))^2}] should be stationary for geodesic. We can apply the Euler-Lagrange equations: [\frac{\partial f}{\partial \phi} - \frac{d}{d\rho} \left(\frac{\partial f}{\partial \phi'}\right) = 0] where (f(\phi, \phi'; \rho) = \sqrt{(\lambda^2 + 1) + (\rho \phi'(ρ))^2}) Observe: [\frac{\partial f}{\partial \phi} = 0] Hence, some constant [\frac{d}{d\rho} \left[\frac{\partial f}{\partial \phi'}\right] = 0 \Rightarrow \frac{\partial f}{\partial \phi'} = C] 6.17-2 [c = \frac{\partial f}{\partial \phi'} = \frac{\partial}{\partial \phi'} \left[(\lambda^2 + 1) + \rho^2(\phi')^2\right]^{1/2}] [= \frac{\rho^2 \phi'}{[(\lambda^2 + 1) + \rho^2(\phi')^2]^{1/2}}] [c^2[(\lambda^2 + 1) + \rho^2(\phi')^2] = \rho^4(\phi')^2] [(\phi')^2[\rho^4 - c^2 \rho^2] = c^2(\lambda^2 + 1)] [(\phi')^2 \rho^2 [\rho^2 - c^2] = c^2(\lambda^2 + 1)] [(\phi')^2 = \frac{c^2(\lambda^2 + 1)}{\rho^2 [\rho^2 - c^2]}] [\phi' = \frac{c\sqrt{\lambda^2 + 1}}{\rho\sqrt{\rho^2 - c^2}}] [\phi(\rho) = \phi_0 + \sqrt{\lambda^2 + 1} \int d\rho \frac{c}{\rho\sqrt{\rho^2 - c^2}}] Interpretation This is on the inverse sine book: (cos^{-1}(x)) (\phi(\rho) = \phi_0 + \sqrt{\lambda^2 + 1} cos^{-1}(\frac{c}{\rho})) For λ → 0, i.e., the cone approaches the x-y plane (z = 0): [\phi(\rho) = \phi_0 + cos^{-1}(\frac{c}{\rho})] [cos(\phi(\rho) - \phi_0) = \frac{c}{\rho}] [\rho = \frac{c}{cos(\phi(\rho) - \phi_0)}] What does this look like? If (\phi_0 = 0), i.e., (x = c cos(\phi)), we'd recognize it as a circle.