# General Relativity

## Practice

### practice problem 1

Write something.

#### solution

Answer it.

### practice problem 2

Write something.

#### solution

Answer it.

### practice problem 3

Write something.

#### solution

Answer it.

### practice problem 4

Finding the volume of a hypersphere should be something like finding the surface area of an ordinary sphere. Derive the equations for…

- the
*surface area*of an*ordinary sphere*of radius*R*and - the
*volume*of a*hypersphere*of radius*R*

Say you looked out into space and saw a galaxy 4.5 billion light years away that turned out to be the Milky Way as it was 4.5 billion years in the past. (This is about the time that the Earth was forming.) Given this hypothetical, hyperspherical universe, determine…

- its radius of curvature (in light years) and
- its volume (in cubic light years)

#### solution

- Pick a point to be the center of the
*sphere*. Any point will do. The locus of points a distance*s*from the origin is a*circle*of radius*r*. The*surface area*of a*sphere*is found by integrating the*area*of an infinite number of*circular bands*of*circumference*2π*r*and*width**ds*starting at the origin and ending at the antipode (the point farthest away from the origin).

On a flat surface*A*= ∫ 2π*r**ds**r*=*s*and*dr*=*ds*, but since a*sphere*has positive curvature*r*<*s*. Think of*s*and*ds*as arc lengths for angles θ and*d*θ whose vertexes are located at the center of the*sphere*. Then the radius of a*circle on the surface*would equal*R*sin θ and an infinitesimal step away from this*circle*would equal*R**d*θ; where θ runs from 0 at the origin to π at the antipode.π π *A*=⌠

⌡2π *R*sin θ*R**d*θ = 2π*R*^{2}⌠

⌡sin θ *d*θ0 0 π *A*= 2π*R*^{2}⎡

⎣− cos θ ⎤

⎦= − 2π *R*^{2}[−1 −1]0

Bump everything up in dimension and watch how this ordinary derivation becomes a hyperderivation by the mere change of a few underlined words.*A*= 4π*R*^{2} - Pick a point to be the center of the
*hypersphere*. Any point will do. The locus of points a distance*s*from the origin is a*sphere*of radius*r*. The*volume*of a*hypersphere*is found by integrating the*volume*of an infinite number of*spherical shells*of*surface area*4π*r*^{2}and*thickness**ds*starting at the origin and ending at the antipode (the point farthest away from the origin).

On a flat surface*V*= ∫ 4π*r*^{2}*ds**r*=*s*and*dr*=*ds*, but since a*hypersphere*has positive curvature*r*<*s*. Think of*s*and*ds*as arc lengths for angles θ and*d*θ whose vertexes are located at the center of the*hypersphere*. Then the radius of a*spherical shell inside the hypersphere*would equal*R*sin θ and an infinitesimal step away from this*sphere*would equal*R d*θ; where θ runs from 0 at the origin to π at the antipode.π π *A*=⌠

⌡4π ( *R*sin θ)^{2}*R**d*θ = 4π*R*^{3}⌠

⌡(sin θ) ^{2}*d*θ0 0 π *A*= 4π*R*^{3}⎡

⎣½(θ − cos θ sin θ) ⎤

⎦= 4π *R*^{3}[½π − 0]0

And that's the way it's done.*A*= 2π^{2}*R*^{3} - If we were to look out into curved space and see ourselves, the distance from here and now to here in the past would be the circumference of our hyperspherical universe. Even on a hypersphere, circumference and radius are related in the usual manner.
*R*=*C*= 4.5 billion light years 2π 2π *R*=716 million light years - Apply the formula derived in part b.
*V*= 2π^{2}*R*^{3}= 2π^{2}(716 million light years)^{3}*V*= 7.25 × 10^{27}cubic light years