Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A thought experiment doubt on General Relativity.

  1. Aug 23, 2009 #1
    I was reading a chapter on "General Relativity" a few days ago. (I was reading Arthur Beiser: Modern Physics)I found it very hard to digest the relation between the "Equivalence Priciple" and "The General Theory of relativity".The equivalence relation states that "An observer in a uniformly accelarating frame cannot distiguish between the effect of the accelaration and the effect of a gravitational field". This principle follows from the fact that the gravitational and inertial masses of a body are equal( proportional). What do we mean by these two masses? Further , Beiser states that , from the above Equivalence principle, we can state that, light too, is affected by gravity. How could he come to this conclusion?

    Further , I also have developed a mind experiment to support and shape my doubts.
    " Imagine a box(made of glass, so that anyone on earth can view the contents of this box) plummeting towards the earth(free falling).
    Now , imagine a laser torch flashing a laser beam from one wall of the box to the opposite one (horizontally).
    1)Will this light hit the opposite wall at exactly the same height as when it is emitted from the torch? By this , I mean that will all the bodies in the "frame of the box" be affected by the constant accelaration(including the light) towards the earth? (But according to the special theory , this property belongs only to the frames travelling with a constant velocity. Does this theory even stretch to the frames travelling with a constant accelaration?)

    Further, a person watching this phenomenon from the surface of the earth will see the beam of light bending downwards,and thus light will be affected by gravity in his frame of reference.

    2)Now, let us assume the laser torch to be fitted on the ceiling of the box, so that the beam travels from the ceiling to the floor. In this case the beam will travel vertically downwards and will be accelarated by gravity. Does this mean that the light will be accelarated and it's velocity will increase? This sounds absured, since the special relativity itself states that nothing can travel faster than light. "

    I hope these doubts aren't really trivial ...I'm really new to this branch of physics...
  2. jcsd
  3. Aug 23, 2009 #2

    Jonathan Scott

    User Avatar
    Gold Member

    If you shine a light within a box in free-fall in a gravitational field, then apart from tidal effects (due to the non-uniformity of the field and the non-zero size of the box) it behaves as "at rest" and hits the expected spot. It is this effect which means that light must be affected by gravity as well.

    From the point of view of an external observer using an isotropic background coordinate system (or similar), the reference frame of the falling observer is curved slightly from one side to the other, with the box sides being non-parallel, the top and bottom curved and the bottom being slightly smaller than the top. Relative to the isotropic background coordinate system, the speed of light appears to decrease as twice the relative potential.

    When the light beam travels horizontally, its acceleration is twice the Newtonian acceleration, the sum of the acceleration caused by the curvature of the box plus the acceleration of the box itself. When it travels vertically, it's actually slowing down relative to the coordinate system.

    However, if instead of the velocity we consider the value of v/c2 or of the momentum Ev/c2 the picture is simpler, in that the rate of change of this is simply (1+v2/c2) times the Newtonian value of the same quantity, regardless of whether it's horizontal, vertical or any other direction. The "1" term is just like Newtonian gravity, and the "v2/c2" term is the part which is due to the distortion of rulers (being curved in the horizontal direction and shrunk at lower potentials).
  4. Aug 23, 2009 #3


    User Avatar
    Science Advisor

    The weak equivalence principle is the observation that is that if you drop two different masses from a height, they will both hit the ground at the same time (assuming the two masses are much smaller than the mass of the earth).

    In Newtonian physics, Coulomb's law is m1.a1 ~ (p1.p2)/(r.r). Across all particles, there is no relation between an inertial mass m1 and its electric charge p1.

    The law of gravity has the same form as Coulomb's law m1.a1 ~ q1.q2/(r.r) - the big difference is that across all particles inertial mass m1 is proportional to gravitational charge q1. Instead of calling q1 "gravitational charge", we call it "gravitational mass". Because of the proportionality, across all different particles, a1 ~ q2/(r.r) - the acceleration depends only on the mass of the earth, and all masses hit the ground at the same time. So in Newtonian physics, the equivalence principle is modelled as inertial mass being proportional to gravitational mass.

    Since all masses fall at the same rate, if you stand on a freely falling mass, other freely falling masses will appear stationary to you - ie. the gravitational field disappears. But if you stand on a freely falling mass, and light appears to bend, then you will be able to deduce that the gravitational field has not disappeared. So if we guess that there is a strong form of the equivalence principle that when you stand on a freely falling mass your observations of other freely falling masses *and* freely falling light will appear as if there is no gravitational field - then you will predict that light will appear to go straight for an observer on a freely falling mass, and by consistency, you will also guess that light will appear to bend to an observer who is in a gravitational field, but who is not freely falling.
    Last edited: Aug 23, 2009
  5. Aug 25, 2009 #4
    It seems you have this backwards. The freefalling box has zero proper acceleration. The earth's surface is accelerating upward (in GR). It is a box at rest with earth's surface that is equivalent to an accelerating box in deep space, assuming each is accelerating at 1 G (9.8m/s^2). A freefalling box is equivalent to an unaccelerated box in deep space. (all assuming an uniform gravitational field).

    The equivalence principle never equates a freefalling box with an accelerated box.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook