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R*7.1 Question from 39-1: Given: A infinitesimal length in spherical coordinates. Find: Show that the infinitesimal length ds in (2) p. 38-6 can be written in spherical coordinates as follows. $$ds^2 = 1 \cdot dr^2 + r^2 d\theta^2 + r^2(\cos\theta)^2\, d \varphi^2$$ $$\displaystyle (Equation\;7.1.1)$$

From lecture notes [[media:pea1.f11.mtg39.djvu|Mtg 39]]

Solution
The key is to start with equation (1) on 38-5

 $$\mathbf{ds} = dx_i \mathbf{e}_i \Rightarrow \ ds^2= \mathbf{ds} \cdot \mathbf{ds} = (dx_i \mathbf{e}_i) \cdot (dx_j \mathbf{e}_j) = dx_i\, dx_j(\mathbf{e}_i \cdot \mathbf{e}_j)$$ $$\displaystyle (Equation\;7.1.2)$$

Next, note the following expressions.

$$\mathbf{e}_i \cdot \mathbf{e}_j = \delta_{ij}$$ $$\delta_{ij} := \begin{cases}1, & \text{for} i = j \\ 0, & \text{for} i \ne j \end{cases}$$

The form of the infinitesimal length ds is expressed in Cartesian coordinates. $$ds^2 = dx_i dx_i = \sum_{i=1}^3(dx_i)^2$$ This can be rewritten in the following form, which is the key for derivation steps, which are rigorous with trigonometric identities. $$ds^2 = dx_1^2 + dx_2^2 + dx_3^2$$ $$\displaystyle (Equation\;7.1.3)$$ The spherical coordinates can be utilized. Simply identify the appropriate identities. Recall that we are using the "astronomical" coordinate system. $$x = r\cos {\theta} \cos {\phi} $$ $$y = r\cos {\theta} \sin {\phi} $$ $$z = r\sin {\theta}$$ Now, make the trigonometric substitutions for each term in 7.1.3.

$$ds^2 = (dr\cos{\theta} \cos{\phi} - (r\sin{\theta} \cos{\phi} d\theta + r\cos{\theta} \sin{\phi} d\phi))^2 +(dr\cos{\theta} \sin{\phi} - (r\sin{\theta} \sin{\phi} d\theta - r\cos{\theta} \cos{\phi} d\phi))^2 + (dr\sin{\theta} + r\cos{\theta} d\theta)^2$$ In such long math equations, simple variable substitutions allow easier management of squared terms (FOIL). $$\alpha = (r\sin{\theta} \cos{\phi} d\theta + r\cos{\theta} \sin{\phi} d\phi)$$ $$\beta = (r\sin{\theta} \sin{\phi} d\theta - r\cos{\theta} \cos{\phi} d\phi)$$ The substitutions then give the shortened form, which allows derivation of the squared terms. $$ds^2 = (dr\cos{\theta} \cos{\phi} - \alpha)^2 +(dr\cos{\theta} \sin{\phi} - \beta)^2 + (dr\sin{\theta} + r\cos{\theta} d\theta)^2$$ Simply apply algebraic FOIL method to the equation above and get the terms below. <p style="font-size:125%" align="center">$$ds^2 = (dr\cos{\theta} cos{\phi})^2 - 2\alpha dr\cos{\theta} \cos{\phi} + \alpha^2 + (dr\cos{\theta} \sin{\phi})^2 - 2\beta dr\cos{\theta} \sin{\phi} + \beta^2$$ The algebraic manipulation is long, but proper addition and subtraction eliminate unnecessary terms. The key is to appply trigonometric identities and grouping with the identity of $$\sin{x}^2 + \cos{x}^2 = 1$$. <p style="font-size:125%" align="center">$$ds^2 = (dr^2\cos{\theta}^2(cos{\phi}^2 + \sin{\phi}^2) - 2\alpha dr \cos{\theta} \cos{\phi} + \alpha^2 - 2\beta dr\cos{\theta} \sin{\phi} + \beta^2 + dr^2 \sin{\theta}^2 + 2rdr \sin{\theta} \cos{\theta}d\theta + r^2 \cos{\theta}^2 d{\theta}^2$$ <p style="font-size:125%" align="center">$$ds^2 = dr^2\cos{\theta}^2 + dr^2\sin{\theta}^2 - 2\alpha dr\cos{\theta} \cos{\phi} + \alpha^2 - 2\beta dr\cos{\theta} \sin{\phi} + \beta^2 + 2rdr\sin{\theta} \cos{\theta}d\theta + r^2\cos{\theta}^2 d{\theta}^2$$ <p style="font-size:125%" align="center">$$ds^2 = dr^2 - 2\alpha dr\cos{\theta} \cos{\phi} + \alpha^2 - 2\beta dr\cos{\theta} \sin{\phi} + \beta^2 + 2rdr\sin{\theta} \cos{\theta}d\theta + r^2\cos{\theta}^2 d{\theta}^2 $$ <p style="text-align:right;">$$\displaystyle (Equation\;7.1.4)$$ The next set of steps involves solutions utilizing our substitutions of $$\alpha$$ and $$\beta$$. Recall the "inner" and "outer" terms from the FOIL multiplications. Each of those terms must now be expanded and simplified.

<p style="font-size:125%" align="center">$$A = - 2\alpha dr\cos{\theta} \cos{\phi}$$ <p style="font-size:125%" align="center">$$A = - 2rdr(\sin{\theta}\cos{\theta}\cos{\phi}^2d\theta + \cos{\theta}^2 \cos{\phi} \sin{\phi}d{\phi})$$

<p style="font-size:125%" align="center">$$B = - 2\beta dr\cos{\theta} \sin{\phi}$$ <p style="font-size:125%" align="center">$$B = - 2rdr(\sin{\theta} \cos{\theta} \sin{\phi}^2 d\phi - \cos{\theta}^2 \cos{\phi} \sin{\phi} d\phi)$$ <p style="font-size:125%" align="center">$$A + B = -2rdr \sin{\theta}\cos{\theta}(\cos{\phi}^2 + \sin{\phi}^2 d\theta = -2rdr \sin{\theta} \cos{\theta}d\theta$$

Recall equation 7.1.4. The terms A and B can be put back into the equation to get the simplification. <p style="font-size:125%" align="center">$$ds^2 = dr^2 + A + \alpha^2 + B + \beta^2 + 2rdr\sin{\theta} \cos{\theta}d\theta + r^2\cos{\theta}^2 d{\theta}^2 $$ <p style="font-size:125%" align="center">$$ds^2 = dr^2 - 2rdr \sin{\theta} \cos{\theta} d\theta + \alpha^2 + \beta^2 + 2rdr\sin{\theta} \cos{\theta}d\theta + r^2\cos{\theta}^2 d{\theta}^2 $$ Next, the terms for $${\alpha}^2 + {\beta}^2$$ must be expanded. <p style="font-size:125%" align="center">$${\alpha}^2 = r^2 \sin{\theta}^2 \cos{\phi}^2 d{\theta}^2 + 2r^2 \sin{\theta} \cos{\theta} \cos{\phi} \sin{\phi} d\theta d\phi r^2 \cos{\theta}^2 \sin{\phi}^2 d{\phi}^2$$ <p style="font-size:125%" align="center">$${\beta}^2 = r^2 \sin{\theta}^2 \sin{\phi}^2 d{\theta}^2 - 2r^2 \sin{\theta} \cos{\theta} \cos{\phi} \sin{\phi} d\theta d\phi r^2 \cos{\theta}^2 \cos{\phi}^2 d{\phi}^2$$ <p style="font-size:125%" align="center">$${\alpha}^2 + {\beta}^2 = r^2\sin{\theta}^2 d{\theta}^2 (\cos{\phi}^2 + \sin{\phi}^2) + r^2 \cos{\theta}^2(\sin{\phi}^2 + \cos{\phi}^2)d{\phi}^2$$ <p style="font-size:125%" align="center">$${\alpha}^2 + {\beta}^2 = r^2\sin{\theta}^2 d{\theta}^2 + r^2 \cos{\theta}^2d{\phi}^2$$ Moving along, all inner and outer terms from the FOIL calculation are added and simplified below.

<p style="font-size:125%" align="center">$$ds^2 = dr^2 - rdr \sin{\theta} \cos{\theta} d\theta + r^2 \cos{\theta}^2 d{\theta}^2 + r^2 \sin{\theta}^2 d{\theta}^2 + r^2 \cos{\theta}^2 d{\phi}^2 + 2rdr \sin{\theta} \cos{\phi} d\theta - 2rdr \sin{\theta} \cos{\theta} d\theta$$ Finally, after addition and subtracting terms above, the solution is derived in the desired form to solve the problem. <p style="font-size:125%" align="center">$$ds^2 = dr^2 + r^2 d{\theta}^2 + r^2 \cos{\theta}^2 d{\phi}^2$$

R*7.7 Question from 41-6: Given: Parabolic coordinates. Find: Try to obtain the separated equations for the Laplace equation in parabolic coordinates.

From lecture notes [[media:pea1.f11.mtg41.djvu|Mtg 41]]

Solution
The instructor gave a wiki page for background research on this problem. The following identities will be useful.

<p style="font-size:125%" align="center">$$x = \sigma \tau \cos{\phi}$$ <p style="font-size:125%" align="center">$$x = \sigma \tau \sin{\phi}$$ <p style="font-size:125%" align="center">$$x = \frac{1/2} ({\tau}^2 - {\sigma}^2$$

References for R*7.1 Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 38) [[media:pea1.f11.mtg38.djvu|Mtg 38]] 10 Nov 2011. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 39) [[media:pea1.f11.mtg39.djvu|Mtg 39]] 15 Nov 2011. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 40) [[media:pea1.f11.mtg40.djvu|Mtg 40]] 17 Nov 2011. Vu-Quoc, L. Class Lecture: Principles of Engineering Analysis. University of Florida, Gainesville, FL, (Meeting 41) [[media:pea1.f11.mtg41.djvu|Mtg 41]] 17 Nov 2011. http://en.wikipedia.org/wiki/Parabolic_coordinates