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The inverse square law

  1. Nov 30, 2004 #1
    I curious about if the invese square law is valid for stars that are very very far away from the earth. like quasars. If the universe was flat and infinite. I could understand that the inverse square law would be valid when looking at objects like quasars, however, if the universe is closed and finite for example, then wouldn't their be some other law that would desrcibe how the intensity of light from a star varies as you move away from it??
     
    Last edited: Nov 30, 2004
  2. jcsd
  3. Nov 30, 2004 #2
    That's a very interesting question. If the universe is closed, then very distant objects may be just as far in one direction as they are in the opposite direction. Would the inverse square law still be in effect then?
     
  4. Nov 30, 2004 #3

    Chronos

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    If it is closed, it is mighty big. Scientists have already checked WMAP results for the kind of mirror imaging expected if it that were the case. None were found.
    http://arxiv.org/abs/astro-ph/0310233
     
  5. Nov 30, 2004 #4

    marcus

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    yes

    if the universe had positive spatial curvature
    then the brightness of a distant object would not fall off quite as fast
    as predicted by the inverse square
    and, as you say, there would be some other law

    as Chronos suggests it is almost spatially flat

    I think there may be more to discuss here but i have to go.
    back later
     
  6. Nov 30, 2004 #5

    marcus

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    oh something else damoclark,

    when you try to make sense of astronomy data there is this thing called "luminosity distance" which basically is "dimmer means farther" using approximately the inversesquare law (depending on what parameters you put in for curvature etc.)

    but you have to be careful because we normally think of brightness as watts per sq. meter which is raw power per area (regardless of wavelength) and there is the redshift to think about too

    stretching out the lightwaves makes them have less watts

    so sometimes when you read a "luminosity distance" it isnt what you expected because of redshift. at least I have had that experience, of being surprised.

    but basically the inversesquare is a very good law and used for all sorts of distance estimates, you just have to be aware that little corrections get into it
     
  7. Dec 1, 2004 #6
    Does dimmer necessarily mean farther? If we made the assumption that the universe was shaped topologically similar to say a hypersphere,for example, and jumped into our hyperthetical space ship, and started to move away from a star. we could expect that the star would appear less bright as we move away from it, however is it possible that after we have travelled a long way from the star that it actually starts to get brighter the further we move away from it?, In effect the curvature of the universe might act like a lens, and causes the light from star to convere back on itself.

    Considering this concept in two dimensions.
    If there are two ducks swimming on a flat lake,one close to the center and one near the edge, and a large rock from a nearby volcano blasted into the air and landed in the middle of the lake, causing a circular wave font to propagate outwards, from the point of impact, then the duck near the center of the lake would experience a bigger wave, than the duck at the edge of the lake. In effect the intensity of the outward propagating wave would decrease inversely with the distance from the center of the lake I=1/r.

    Now applying this concept on a curved two dimensional medium.....

    If there was a huge flood and the entire earth was covered by water, and two ships were postioned 2000 km from the south pole, and 2000 thousand km from the north pole,
    and a large meteor hit the ocean at the north pole, then what would be expected....
    A wave would propogate outward from the near pole, its intensity would fall off until it reached the equator, but as it passed the equator traveling towards the south pole it would converge back on itself. So the two ships would experience a wave of equal intensity. The people on the ship near the south pole might assume the the impact occured close to them, when infact that was not the case. They also might be suprised to find that as the wave converged to a point at the south pole, it would come back at them.

    In the hyperspherical universe might not the same principles apply?? IF we blasted off in a super fast rocket and headed towards, lets say a quasar, it is possible that to begin with as we travelled towards it, it started to get dimmer, before it got brighter??

    In a universe, that is not hyperspherical, but curved in some other way, is the inverse square valid? Is the inverse square law only truly valid in euclidean flat infinite space??
     
    Last edited: Dec 1, 2004
  8. Dec 1, 2004 #7

    marcus

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    we've got more expertise focused on this than we need. Both Chronos and you are in full command of the topic so i will bow out.

    suggest sometime you check Ned Wright on google and go to his
    UCLA website and try his "cosmological calculator" which gives
    luminosity distances for various redshifts (where you put in the cosmological parameters that determine curvature etc)

    you may have done it. if not, I believe it would intrigue you
     
  9. Dec 2, 2004 #8

    Garth

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    Exactly right damoclark, your two examples are very informative. These curvature effects have to be and are taken into account when turning apparent magnitudes into estimates of distance. Of course one has to be careful how distance is defined in a curved and expanding universe too!

    Garth
     
  10. Dec 3, 2004 #9

    hellfire

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    The observed bolometric flux, i.e. the energy of the received photons ([tex]\inline N_0[/tex]) per unit time ([tex]\inline \Delta t_0[/tex]) and unit area ([tex]\inline A[/tex]) in a detector is:

    [tex]\Phi = \frac {h \nu_0 N_0} {\Delta t_0 A}[/tex]

    The received number of photons can be written in terms of the total number of emitted photons ([tex]\inline N[/tex]):

    [tex]N_0 = \frac {A} {4 \pi r^2} N[/tex]

    Where r is a the radial coordinate which provides the surface area at the receiver, but it is not necessarily the radial distance to the observer (I think it is sometimes called "area distance").

    The relation between this area distance r and the actual distance D ([tex]\inline r = f(D)[/tex]) depends on the curvature of space. Actually, r may decrease for increasing D in a closed 3-sphere, same as in the examples you mentioned. Your examples are very intuitive, but I think I have successfully proven this for a closed Robertson-Walker metric.

    Thus, the flux can be written as:

    [tex]\Phi = \frac {h \nu_0 N} {\Delta t_0 4 \pi f^2(D)}[/tex]

    Now, expressing the time and the frequency at the receiver in terms of the same quantities at the emitter (subscript e, z: redshift):

    [tex]\Phi = \frac {h \nu_e N} {\Delta t_e 4 \pi f^2(D) (1 + z)^2}[/tex]

    [tex]\Phi = \frac {L} {4 \pi f^2(D) (1 + z)^2}[/tex]

    Where L is the luminosity.

    There is a inverse square dependence with the distance D in a spacetime which is flat (where r = D) and static (z = 0).
     
    Last edited: Dec 3, 2004
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