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Making basic astro/cosmo calculations easy with google calculator--examples

  1. Oct 4, 2014 #1


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    The aim is partly pedagogical. I want to assemble some examples illustrating how the google calculator can play a role in learning cosmology and related. Suppose you don't remember the Schw. radius of the earth.

    Paste in "2G*mass of earth/c^2" press return, and you get 0.887 cm.

    Gravitational time dilation is an important part of understanding cosmology. How much slower does a clock on earth surface run than one "at infinity" i.e. effectively out of earth's gravitational field? Call the rate of the distant clock "1" and subtract off the rate of the slower clock. The calculator knows the radius of earth. Paste in:

    1 - (1 - .887 cm/radius of earth)^.5

    You get about 0.7 parts per billion. Instead of typing "radius of earth" I'll assume we remember it is 6371 km.

    Compared with one at sea level, how much faster does a clock run which is at an altitude of 1 km?

    Paste in:
    ((1 - .887 cm/6372 km)/(1 - .887 cm/6371 km))^.5 - 1

    You get about 1.1 x 10-13

    A cesium clock runs about 9 billion cycles per second. After 1000 seconds the high clock is about one cycle ahead of the low clock.

    What if the high clock is at an altitude of 100 meters? Paste in:
    ((1 - .887 cm/6371.1 km)/(1 - .887 cm/6371 km))^.5 -1
    You get about 1.1 x 10-14
    So after 10,000 second the high clock is about one cycle ahead.

    That is how time varies in a gravitational field. The calculator is near the limit of its accuracy but since the field is nearly uniform in that range we can extrapolate and say that if the high clock is at an altitude of 10 meters it runs faster by about one part in 1.1 x 10-15.
    So that after 100,000 seconds the high clock is about one cycle ahead. A day is 86,400 seconds so it takes a bit over a day to gain a cycle. But 10 meters is not much difference in altitude.

    That is an example of how you use the ratio of the Schwarzschild radius of the earth, to its actual radius, to calculate something interesting namely a difference in the rates of passage of time.
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  3. Oct 4, 2014 #2


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    Curvature quantities are typically in terms of reciprocal area--e.g. might be given in terms of m-2. Think of the curvature of a circle as one over its radius, and of an ovoid surface having TWO radiuses of curvature so that its combined curvature could be the product of one over the first multiplied by one over the second. So the curvature is one over an area.The universe is thinning out and we normally assume that the cosmological constant Lambda (a small positive curvature quantity) actually is constant. That means the universe is getting more and more like a deSitter universe.
    In a deSitter setting, the Hubble growth rate is constant and its square equals Λ /3
    or would except for a factor of c2.
    More precisely if we define the Hubble radius R = c/H, then we have 1/R2 = Λ /3
    So Λ = 3/R2

    And we know the longterm value of the Hubble radius for our our universe. Jorrie's calculator use it together with the present Hubble radius, as the two main model parameters. According to recent Planck mission data it is 17.3 billion light years.

    So we can calculate the cosmological constant as a m-2 quantity. A reciprocal area presumably good throughout all space for all time.
    Paste in:
    3/(17.3 billion light years)^2 in m^-2
    You get 1.12 x 10-52 m-2

    This length 17.3 billion light years has another name. It is not only the long term value of the Hubble radius for our universe (according to standard cosmic model) it is also the so called deSitter radius of a deSitter universe which has the same cosmological constant Lambda that we do. It gives the radius of the cosmological event horizon in such a universe. And the TEMPERATURE of the cosmic event horizon, and of the so called deSitter radiation (analogous to Hawking radiation from a black hole horizon) is given by a simple formula:
    TdS = 1/(2 pi RdS)

    So as another example we can use the google calculator to find the deSitter temperature corresponding to the observed value of Lambda, which presumably is the temperature to which our universe is tending as it thins out and approximates a deSitter universe in the very long term.

    As I recall it is something like 2 femtofemtokelvin :w a bit chilly. femten is the Danish word for "five-ten" or fifteen so femto = 10-15
    femtofemtokelvin is so cold they don;t even have a metric prefix for it and you have to double up the prefix.
  4. Oct 4, 2014 #3


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    WAIT! I have to introduce some constants to make the units work out right in the last equation.

    TdS = 1/(2 pi RdS)

    superficially speaking that does not make sense because it has a temperature on one side and a reciprocal length on the other. That is how you find it presented on page 17 of Andy Strominger's introduction to deSitter space that he made with a couple of other people for the Les Houches lecture series.
    But he was setting hbar, c, and Boltzmann's constant k all equal to one and omitting them. We have to re-introduce them. The google calculator knows the values of all three so it will be a cinch.

    If T is a temperature then kT is an energy. And hbar*c is equal to energy and length multiplied together. So hbar*c/(2pi RdS) is an energy.
    And we can divide it by k to get the temperature we want. Remember that the deSitter radius RdS = 17.3 billion light years. Paste this in:
    (hbar*c/k)/(2pi*17.3 billion light years) in kelvin

    That is what should give you 2 femtofemtokelvin.

    Saying "in kelvin" is what forces the google calculator to give you the answer in kelvin rather than rounding off and talking Celsius to you.
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