3. Geodesic on a Sphere The shortest path between two points on a curved surface, suchas the surface of a sphere, is called a geodesic. To find a geodesic, one has first to set upan integral that gives the length of a path on the surface in question.(a)F.3 on p619 in T&M) to show that the length of a path joining two points on asphere of radius R isTo illustrate this, use spherical polar coordinates (r, 0, ø) (see Appendix1+sin2 0l(0)2 deL=R01if (01,1) and (02, P2) specify the two points and we assume that the path is ex-pressed as = (0).(b)two given points on a sphere is a great circle.Use the above result to prove that the geodesic (shortest path) betweenHint: The integrand f(o, ;0) in above result is independent of ø, so the Euler-Lagrange equation reduces to a af/od = c, a constant. This gives you as afunction of 0. You can avoid doing the final integral by the following trick: There isno loss of generality in choosing your z axis to pass through the point 1. Show thatwith this choice the constant c is necessarily zero, and describe the correspondinggeodesics.

Question
Asked Oct 29, 2019
3. Geodesic on a Sphere The shortest path between two points on a curved surface, such
as the surface of a sphere, is called a geodesic. To find a geodesic, one has first to set up
an integral that gives the length of a path on the surface in question.
(a)
F.3 on p619 in T&M) to show that the length of a path joining two points on a
sphere of radius R is
To illustrate this, use spherical polar coordinates (r, 0, ø) (see Appendix
1+sin2 0l(0)2 de
L=R
01
if (01,1) and (02, P2) specify the two points and we assume that the path is ex-
pressed as = (0).
(b)
two given points on a sphere is a great circle.
Use the above result to prove that the geodesic (shortest path) between
Hint: The integrand f(o, ;0) in above result is independent of ø, so the Euler-
Lagrange equation reduces to a af/od = c, a constant. This gives you as a
function of 0. You can avoid doing the final integral by the following trick: There is
no loss of generality in choosing your z axis to pass through the point 1. Show that
with this choice the constant c is necessarily zero, and describe the corresponding
geodesics.
help_outline

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3. Geodesic on a Sphere The shortest path between two points on a curved surface, such as the surface of a sphere, is called a geodesic. To find a geodesic, one has first to set up an integral that gives the length of a path on the surface in question. (a) F.3 on p619 in T&M) to show that the length of a path joining two points on a sphere of radius R is To illustrate this, use spherical polar coordinates (r, 0, ø) (see Appendix 1+sin2 0l(0)2 de L=R 01 if (01,1) and (02, P2) specify the two points and we assume that the path is ex- pressed as = (0). (b) two given points on a sphere is a great circle. Use the above result to prove that the geodesic (shortest path) between Hint: The integrand f(o, ;0) in above result is independent of ø, so the Euler- Lagrange equation reduces to a af/od = c, a constant. This gives you as a function of 0. You can avoid doing the final integral by the following trick: There is no loss of generality in choosing your z axis to pass through the point 1. Show that with this choice the constant c is necessarily zero, and describe the corresponding geodesics.

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check_circleExpert Solution
Step 1

(a)Write the formula for the length(L) of a curve in 3-dimension.

Vd
dydz
(1)
L =
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Vd dydz (1) L =

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Step 2

Use the spherical coordinate(r,θ,ϕ) and find the value of dx2+ dy2+ dz2 as,

=dr2 2 sin2 Odrd0* ^V
Ax
dxdydz
dx2 dydz
R2 sin2 Od
Rd0 (
dr 0 and r = R)
..
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=dr2 2 sin2 Odrd0* ^V Ax dxdydz dx2 dydz R2 sin2 Od Rd0 ( dr 0 and r = R) ..

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Step 3

Plug the value of dx2+ dy2+ dz2 in the equation (1) and solve for ...

е,
L Rsin2 Odd d0
2
ө,
d02 d0
d0
= R,/sin2
= R,/1+sin2 0
de
в
в,
=RV1+sin 04(0)° de
2
в
help_outline

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е, L Rsin2 Odd d0 2 ө, d02 d0 d0 = R,/sin2 = R,/1+sin2 0 de в в, =RV1+sin 04(0)° de 2 в

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