The Physics
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Opus in profectus

General Relativity

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Practice

practice problem 1

How much older is your head than your feet?

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/c2)

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/s2. 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/s2)(1 m)/(3.00 × 108 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.

Δt = 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

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…
  1. the surface area of an ordinary sphere of radius R and
  2. 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…
  1. its radius of curvature (in light years) and
  2. its volume (in cubic light years)

solution

  1. 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 2πr and width ds starting at the origin and ending at the antipodes (the point farthest away from the origin).

    A = ⌠
    ⌡    		 2πr 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 dθ 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 dθ; where θ runs from 0 at the origin to π at the antipodes.

    π
    A = ⌠
    ⌡    		 2π(R sin θ)R dθ
    0
    π
    A = 2πR2 ⌠
    ⌡    		 sin θ dθ
    0
    π
    A = 2πR2 ⎡
    ⎣    		 − cos θ ⎤
    ⎦
    0
    A = − 2πR2[−1 − 1]
    A = 4πR2

    Bump everything up in dimension and watch how this ordinary derivation becomes a hyperderivation by the mere change of a few underlined words.

  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 4πr2 and thickness ds starting at the origin and ending at the antipodes (the point farthest away from the origin).

    V = ⌠
    ⌡    		 4πr2 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 dθ 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 dθ; where θ runs from 0 at the origin to π at the antipodes.

    π
    V = ⌠
    ⌡    		 4π(R sin θ)2R dθ
    0
    π
    V = 4πR3 ⌠
    ⌡    		 (sin θ)2 dθ
    0
    π
    V = 4πR3 ⎡
    ⎣    		 ½(θ − cos θ sin θ) ⎤
    ⎦
    0
    V = 4πR3[½π − 0]
    V = 2π2R3

    And that's the way it's done.

  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
    2π
    R = 
    4.5 billion light years
    2π
    R =  716 million light years

  4. Apply the formula derived in part b.

    V = 2π2R3
    V = 2π2(716 million light years)3
    V = 7.25 × 1027 cubic light years

    The 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.