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Does light bending doubly violate the equivalence principle?

  1. Mar 8, 2005 #1
    The equivalence principle (EP) - which is the basis of general relativity – states that you cannot distinguish between an object’s behaviour in a uniform gravitational field from that in a uniformly accelerating frame.
    If light travels vertically in a gravity field it loses or gains energy, and experiments confirm that gravity pulls on light as if it possessed inertial mass m=E/c^2. So light travelling vertically obeys the EP.
    When Einstein first calculated the bending of light near the sun he got a value of 0.8 arcsec, using the above reasoning. But later corrected this to 1.75 arcsec, because light refracts in a gravitational field. So the light bends doubly.
    Consider the two cases:
    1. An astronaut is in a room inside a rocket that is accelerating. A beam of light passes horizontally across the room and strikes the opposite wall lower down by distance d.
    2. An astronaut is in a room inside a rocket that is stationary on the earth. A beam of light passes horizontally across the room and strikes the opposite wall lower down.
    Does light fall by d (gravity pulling on light’s inertial mass only)?
    Or does it fall by 2d (due to refraction and gravitational pull)?

    If it’s 2d, doesn’t that violate the EP?
  2. jcsd
  3. Mar 8, 2005 #2


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    You haven't understood the 'bending of light' in GR properly, don't worry Einstein initially made the same mistake so you are in good company!

    There are two components to the angle of deflection, the first is due to the equivalence principle, the falling of the inertial frame of reference, and the second is due to the curvature of space.

    In GR each is equal to half the total deflection.

    In your example the light 'falls' by d. As the experiment is local the curvature of space is not a factor, it only applies over an extended volume.

  4. Mar 8, 2005 #3
    When Einstein first calculated the deflection of light it was around a spherically symmetric body (not a uniform g-field). At that time GR was an incomplete theory. Later on it became apparent that the altering of space itself takes part in the delfection. So have the deflection around a star is half E/c^2 and the other half is alterered space.
    Last edited: Mar 8, 2005
  5. Mar 10, 2005 #4
    I’m not convinced that the photon doesn’t fall by 2d.
    If a horizontally travelling photon meets a vertically travelling one in the stationary room, we know from the EP that the vertical one will feel the whole pull of gravity on its inertial mass E/c^2. So why should the horizontal one feel only half the pull of gravity?
    And if there are a bunch of photons travelling horizontally shouldn’t they refract as their speed slows nearer the Earth, as well as feeling the whole pull of gravity?
  6. Mar 10, 2005 #5
    Your negeclting the space curvature. Once Einstein realized that he modified his theory so as to take that into account. So now we know that half the deflection is from spatial curvature. If there is no spacetime curvature then there is no spatial deflection since a uniform field has curvature

  7. Mar 26, 2005 #6
    Following the logic in http://www.mathpages.com/rr/s6-03/6-03.htm
    the effects of gravity on light in a uniform field must pull it down by 1d. And if an equal effect is caused by light refracting in a uniform gravitational field, then it must also refract a further 1d, making 2d in total, and not 1d.

    Gravity pull and refraction each account for 1d, which is 1/2 the total 2d.

    Now if we take an infinitely small space, the rate of change is still 2 and not 1. So the equivalence principle is false for light bending in a uniform gravity field.
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