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.