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The Equivalence Principle-bending of light

  1. Oct 13, 2004 #1
    The Equivalence Principle--bending of light

    Hi everyone!
    I have learned the general relativity, but I have a few problems which I don't understand. Can someone help me?
    I am recently learning the Equivalence Principle and the bending of light. I have read the following paragraphs in a website, and I was quite confused about some questions:

    "Consider a rocket undergoing constand acceleration. Light entering one side of the rocket will appear to 'fall' toward the floor as it crosses the rocket. This is because it takes a finite amount of time for light to cross the rocket, and in that time, the rocket has moved. Since it is undergoing constant acceleration, the trajectory of the light will be a parabola. "
    "Now by the Eauivalence Principle, this result applies as well as to a stationary frame with a gravitational acceleration equal to the previous rocket acceleration. So light 'falls' in a gravitaional field."

    I understand the first part of the quotation that light appears to fall toward teh floor. But, my problem is, when this happens to a stationary frame with gravitational acceleration, why does light fall??? I know that we can apply the Equivalence Principle here, but I just don't think this (light falling toward ground) is possible. I thought that light has no mass and it would not be attracted by gravitational force. Can someone explain this to me? And please do not just say that it is because of the Equivalence Principle......Thanks!

    *Quotations from http://www.phys.ufl.edu/~acosta/phy3101/lectures/relativity5.pdf
  2. jcsd
  3. Oct 13, 2004 #2


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    Two ways to look at it:(quick Answer)

    Gravity is a curvature of spacetime, and light follows a geodesic along spacetime.

    Gravity also couples with energy, and light has energy.
  4. Oct 13, 2004 #3
    Um...I still don't get it, could you please give me some further explanations? (Sorry, I am a novice to relativity...>.<)
  5. Oct 13, 2004 #4
    What you're thinking of is the fact that light has zero rest mass. But light has energy and energy has mass and therefore light is affected by gravity. Anything which has mass will be affected by gravity and the rate at which something falls is a indpendant of the quantity of mass. Take a look in the in the Feynman Lectures Vol - I, page 7-11. Section entitled Gravitation and Relativity
    The presence of spacetime curvature is not required for something to be deflected by gravity. E.g. light can be deflected in a uniform gravitational field and such a field has zero spacetime curvature.

  6. Oct 14, 2004 #5


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    I'm not sure why or what you're confused about. You might try


    but most of the points have been mentioned in the discussion, albeit briefely.
  7. Oct 14, 2004 #6
    Thanks, I kind of get the point...I really appreciate it!
  8. Oct 14, 2004 #7


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    It might be worth pointing out that when we observe the light ray of a star 'bent' around the sun in a total eclipse there are actually two components to the angle of deflection. In his original prediction of a 1915 eclipse Einstein only took the first into account and predicted just half the correct angle. That is, he only allowed for the bending due to the "falling of light" in a gravitational field that has been clearly explained above.

    However there is another part of the deflection angle caused by the curvature of space alone, and would be that angle measured by taking a space-like slice across the solar system and measuring the deflection of a chord just tangent to the Sun's surface.

    In GR both these angles equally contribute half of the total deflection angle.

    Had the First World War not been in full swing and that eclipse actually observed he would have been proved wrong and history might have been completely different!
    Just a thought!
    - Garth
    Last edited: Oct 14, 2004
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