# General Relativity

## Practice

### practice problem 1

#### solution

This problem is easy to set up, but hard to finish. Setting up a problem is where most of the physics lies, so I'm OK with that. Start with the approximate gravitational time dilation equation for small height differences.

t′ = |
t |

√(1 − 2gΔh/c^{2}) |

Let's set this up so the numbers can be used for a person of any age. This nominal person has probably lived their entire life on or near the surface of the Earth where *g* = 9.8 m/s^{2}. Their head has probably been about one meter above their feet on average during this time. Start by computing the time that passes at their head when their feet have experienced one second?

t′ = |
1 s |

√[1 − 2(9.8 m/s^{2})(1 m)^{8} m/s)^{2}] |

Handheld calculators *do not* have enough precision to compute this number, nor do many online calculators. You get a nonsense result. Apparently, there is no difference.

*t*′ = 1 s

If one tool doesn't work, try another. With a little bit of trial and error, I found that this online calculator gave good results (after I described the calculation to it in just the right way).

*t*′ = 1.000000000000000108889 s

I think what matters to us more than what this number is, is how much bigger it is than one second. You could count all the zeros, or you could just let the calculator do the work.

Δ*t* = *t*′ − *t* = 1.08889… × 10^{−16} s

That extra bit is about one ninth of a femtosecond per second (femto = 10^{−15}) or…

1.09 × 10^{−16} s |

× 24 × 60 × 60 |

9.41 × 10^{−12} s |

9½ picoseconds per day (pico = 10^{−12}) or…

9.41 × 10^{−12} s |

× 365.25 |

3.44 × 10^{−9} s |

3½ nanoseconds per year (nano = 10^{−9}).

This is a good place to end since we tend to think of our age in years. The answer, to within an order of magnitude, is your age in years times 3½ nanoseconds. At around 30 years, this difference would add up to 100 nanoseconds.

### practice problem 2

#### solution

Answer it.

### practice problem 3

#### solution

Answer it.

### practice problem 4

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

- 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 antipodes (the point farthest away from the origin).*A*=⌠

⌡2π *r**ds*On a flat surface

*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 antipodes.π *A*=⌠

⌡2π( *R*sin θ)*R**d*θ0 π *A*= 2π*R*^{2}⌠

⌡sin θ *d*θ0 π *A*= 2π*R*^{2}⎡

⎢

⎣− cos θ ⎤

⎥

⎦0 *A*= − 2π*R*^{2}[−1 − 1]*A*= 4π*R*^{2}Bump everything up in dimension and watch how this ordinary derivation becomes a hyperderivation by the mere change of a few underlined words.

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 antipodes (the point farthest away from the origin).*V*=⌠

⌡4π *r*^{2}*ds*On a flat surface

*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 antipodes.π *V*=⌠

⌡4π( *R*sin θ)^{2}*R**d*θ0 π *V*= 4π*R*^{3}⌠

⌡(sin θ) ^{2}*d*θ0 π *V*= 4π*R*^{3}⎡

⎢

⎣½(θ − cos θ sin θ) ⎤

⎥

⎦0 *V*= 4π*R*^{3}[½π − 0]*V*= 2π^{2}*R*^{3}And that's the way it's done.

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*2π *R*=4.5 billion light years 2π *R*=716 million light years Apply the formula derived in part b.

*V*= 2π^{2}*R*^{3}*V*= 2π^{2}(716 million light years)^{3}*V*= 7.25 × 10^{27}cubic light yearsThe universe is many times bigger than this, however. No evidence has been found for a distant copy of our region of space out to 13.8 billion light years (the edge of the observable universe). Then there's the matter of the expansion. When we look out to the end of observable space we are also looking back to the beginning of time — we are looking back to an era when the universe was a fraction of its current size. Those places are now many times farther away than they appear to be. Were the universe a hypersphere, its circumference would have to be at least 156 billion light years. For all intents and purposes the universe is essentially flat and may be infinite in size.