Physics
Physics, 05.07.2019 00:10, krojas015

Consider a 2-dimensional surface with (r1,a2) = (0,ø) (the usual spherical polar angles) and metric ds2 rdo2? sin20 dd2, where ro> 0 is a constant. you are given that the only non-zero christoffel symbols are гу — га 2 2 1 cot 0 cos 0 г. - sin 12 21 22 (a) prove that if a contravariant vector a' undergoes parallel transport around the circle -п/4 from ф %—o to ф%3d 2т аnd a' — 0, a? 3d 1 at ф 3 0, then at ф — 2т we have acos(2t) sin(27) 2 a (b) show that the components ri1, r22 of the ricci tensor are r11 1 r22 -sin20 find r, the ricci scalar

answer
Answers: 1

Similar questions

Предмет
Physics, 29.06.2019 23:30, monicorequena7
Box 11.5 constructing the schwarzschild embedding diagram our goal in this box is to construct in a three-dimensional euclidean space a two- dimensional surface having the same metric as a t = const, θ = π/2 slice through schwarzschild spacetime, which is given by equation 11.14 (repeated here) dr2 (11.14r) consider a three-dimensional euclidean space described by polar coordinates r and φ on the horizontal plane and a vertical coordinate z. the metric for such a space in this coordinate system is (11.21) note that a horizontal circle in this space has circumference s = fr 2 , so r is a circumferential coordinate for such circles, as it is in the schwarzschild metric. in this three-dimensional space, we can describe a two-dimensional surface by specifying its height z as a function of the horizontal coordinates r and φ. since the schwarzschild geometry is symmetric under rotations in φ, we would expect the surface that represents it to display the same symmetry, so in this particular case, we expect the height of our surface to be a function of r alone: z = z(r). the metric on this surface can be found by substituting z(r) into equation 11.21 to eliminate dz and thus find a metric that depends only on r and φ. doing this yields dz our goal is to find the function z(r) that makes the metric of this surface match the metric in equation 11.14, that is, z(or) such that dz ar) = 1-2gmir (11.23) this is a first-order differential equation that you can solve. exercise 11.5.1. solve equation 11.23 for dzudr and integrate the resulting equation. you should find that z(r)=±/bgm(r-2gm) plus an irrelevant constant of integration that merely moves the surface up or down vertically. we choose (arbitrarily) the positive solution to represent our surface.
Answers: 3
Предмет
Physics, 02.07.2019 20:20, andy2461
Constructing the schw arzs child embedding dia r goal in this box is to construct in a three-dimensional euclidean space a t ou dimensional surface having the same metric as a t = const, θ π/2 slice through schwarzschild spacetime, which is given by equation 11.14 (repeated here): wo dr2 1-2gmir ds (11.14r) ates r and φ on the horizontal plane and a vertical coordinate z. the metric for such a onsider a three-dimensional euclidean space described by polar coordin space in this coordinate system is (11.21) note that a horizontal circle in this space has circumference s = fr = 2m, so r is a circumferential coordinate for such circles, as it is in the schwarzschild metric. in this three-dimensional space, we can describe a two-dimensional surface by specifying its heightz as a function of the horizontal coordinates r and φ. since the schwarzschild geometry is symmetric under rotations in φ, we would expect the surface that represents it to display the same symmetry, so in this particular case, we expect the height of our surface to be a function of r alone: z = z(r). the metric on this surface can be found by substituting z(r) into equation 11.21 to eliminate dz and thus find a metric that depends only on r and φ. doing this yields dz dr r'+pdが dr (11.22) our goal is to find the function z(r) that makes the metric of this surface match the metric in equation 11.14, that is, z(r) such that (11.23) this is a firs-order differential equation that you can solve. exercise 11.5.1. solve equation 11.23 for dz/dr and integrate the resulting equation. you should find that z(r)=±v8gm(r_2gm) plus an irrelevant constant of integration that merely moves the surface up or down vertically. we choose (arbitrarily) the positive solution to represent our surface.
Answers: 2
Do you know the correct answer?
Consider a 2-dimensional surface with (r1,a2) = (0,ø) (the usual spherical polar angles) and metric...

Questions in other subjects:

Konu
Mathematics, 23.08.2021 05:30