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Stargazing Effects of a massive object on light and its relation to the 1919 Eclipse

  1. May 18, 2003 #1
    [SOLVED] Effects of a massive object on light and its relation to the 1919 Eclipse

    What happens to light as it passes near a massive object and how it this principle or concept connected to the 1919 lunar eclipse where Einstein’s Photoelectric theory was proved (both the apparent and actual location of the stars were revealed around the area of the eclipse?

    Thanks for any responses in advance.
     
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  3. May 18, 2003 #2

    Integral

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    It was not Photoelectric effect that was (perhaps falsly) verified during the 1919 eclipse put General Relativity. The observations were of the position of a star that was being oculted during the eclipse. GR predicted that the postion of the star would shift as the disk of the sun passed in front of it.

    It is not clear in this day and age that the instrumentation and the methods uses were actually up to the task. It may well be that the verifing data was more wishfull thinking then sound numbers. Of course, since that time the measurements have been repeated modern equipment and methods and the theory has been verified many times over.
     
  4. May 18, 2003 #3

    marcus

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    If a ray of light passes within a distance R from an object with mass M then it is bent by an angle which can be calculuated
    using this formula from Einstein's 1916 paper on general relativity:


    angle in radians = 4GM/c2R


    It was this angle that Eddington's team was trying to check
    when they observed the Pleiades during the 1919 eclipse of the sun. They thought they verified that the formula was right---it was headline news "Einstein Proved Correct!" and so on.
     
  5. May 18, 2003 #4

    marcus

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    I hope no one is vexed by this side comment and proceeds to declare me out of the mainstream for mentioning it
    but I will take that risk.

    In natural units (c=G=hbar=1) the sun's mass is 93E36
    and a million miles is E44.

    the formula for the angle is simply 4M/R radians

    But 4M is 37E37

    So, if a ray of light passes a million miles from the sun's center,
    it is bent by the angle you get by dividing 37E37 by E44.


    37E37/E44 = 37E-7 radians-------3.7E-6-----3.7 microradians

    It seems to me that natural units, which so many physicists use nowadays in research and theoretical papers, actually help here by making the formula 4M/R so much simpler. It is a mess to do
    the formula in metric with numbers like 6.673E-11 for G and so on.

    **********

    There is a side issue of how did I know the mass of the sun is 93E36.

    that is easy to remember if you can remember that the distance to it is 93 million miles.

    Because in natural units M=RV2, where R is the radius of a small object's circular orbit and V is the orbit speed. In the case of the earth (small, in roughly circular orbit) V = E-4 and R=93E44 (93 million miles) so the mass of the sun is quick to compute just by multiplying

    M = RV2 = 93E44 x E-8 = 93E36

    Again you quite possibly cannot remember the mass of the sun in kilograms and would probably not like to calculate it from what you do know about distance and orbit speed because it would involve messy numbers like 6.673E-11 for G and so on. So natural units are somewhat more practical in this context.
    As well as increasingly mainstream (some John Baez discussion on Usenet bears on this)

    But if you can remember the mass of the sun in kilograms then of course go ahead and use the metric formula for the angle mentioned earlier

    angle = 4GM/c^2R

    instead of 4M/R
     
  6. May 18, 2003 #5
    Not Photoelectric theory (for which Einstein got Nobel Prize, by the way), but General Relativity theory.

    Basicly what you do is to take picture of starry sky at night and during eclipse and superimpose them. Distance between stars (on a photograph) increases slightly when Sun is among them - this means that starlight sligtly bends around Sun.
     
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