Physics 2400 Spring 2017 the surface area are and the volume of n-dimensional sphere spring semester 2017 http://www.phys.uconn.edu/˜rozman/Courses/P2400_17S/ Last modified: May 5, 2017 1 Introduction The volume of n-dimensional sphere of radius r is proportional to rn, n Vn(r) = v(n)r ; (1) where the proportionality constant, v(n), is the volume of the n-dimensional unit sphere. n 1 The surface area of n-dimensional sphere of radius r is proportional to r − . n 1 Sn(r) = s(n)r − ; (2) where the proportionality constant, s(n), is the surface area of the n-dimensional unit sphere. The n-dimensional sphere is a union of concentric spherical shells: dVn(r) = Sn(r)dr (3) Therefore the surface area and the volume are related as following: ZR ZR n 1 s(n) n V (R) = S (r)dr = s(n) r − dr = R : (4) n n n 0 0 Page 1 of3 Physics 2400 n-dimensional sphere Spring 2017 The surface area and the volume of the unit sphere are related as following: s(n) v(n) = : (5) n Consider the integral Z1 Z1 2 2 2 2 x x ::: xn r In = e− 1− 2− − dVn = e− dVn(r); (6) 0 −∞ where dVn is the volume element in cartesian coordinates dVn = dx1 dx2 :::dxn (7) and n 1 dVn(r) = s(n)r − dr (8) is the volume element in spherical coordinates. Since the integrand in the first integral in Eq. (6) is a product of identical gaussians of one variable each, 0 1n Z1 p B 2 C n n B x C 2 In = B e− dxC = π = π : (9) @B AC −∞ On the other hand, the second integral in Eq. (6), evaluated in spherical coordinates Z1 Z1 Z1 n 2 2 s(n) 2 1 I = e r s(n)rn 1 dr = s(n) e r rn 1 dr = e r r2 2 − dr2 (10) n − − − − 2 − 0 0 0 Z1 s(n) n 1 s(n) n = e tt 2 dt = Γ (11) 2 − − 2 2 0 Comparing Eq. (9) and Eq. (11), we obtain: n 2π 2 s(n) = : (12) n Γ 2 n n π 2 π 2 v(n) = = (13) n n n 2 Γ 2 Γ 2 + 1 Page 2 of3 Physics 2400 n-dimensional sphere Spring 2017 2 Coulomb’s law in n-dimension In three dimensions Coulomb’s law takes the form 1 Q E3(r) = ; (14) 4π0 r where E3 is the magnitude of the electric field, 0 is the electric constant, and r is the distance from the point charge of charge Q. What is Coulomb’s law look like in high dimensions? We assume the Maxwell equations hold in any dimension. Hence Gauss law still holds: Q Φ = ; (15) 0 where Φ is the electric flux through a closed surface enclosing any volume, Q is the total charge enclosed within that volume. Due to the spherical symmetry of the field created by a point charge, Φ = En(r)Sn(r); (16) where En(r) is the radial component of the electric field, Sn(r) is the surface area of n- dimensional sphere of radius r. n n 1 2π 2 n 1 S (r) = s(n)r − = r − : (17) n n Γ 2 Comparing Eq. (15) and Eq. (16), n Q Γ 2 Q En(r) = = n n 1 : (18) 0 Sn(r) 2π 2 0 r − References [1] Wikipedia, “Volume of an n-ball — Wikipedia, the free encyclopedia,” 2016. Accessed: November 25, 2016. Page 3 of3.