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Calculating how light falls under gravity

  1. Mar 2, 2012 #1
    I know that light "falls" under the presence of gravity. Imagine a pulse of light that falls a small distance h, how could i determine how far it has traveled?

    I'd really would like to know the equation that relates the kinematics of light under gravity.
     
  2. jcsd
  3. Mar 2, 2012 #2
    Don't the equations from general relativity describe this? I would think that light would simply follow the path of curved space at speed c. I don't think of this as "falling", but who am I to say? We need some help...
     
  4. Mar 3, 2012 #3

    Jonathan Scott

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    From the "Principle of equivalence" we know that locally everything appears to accelerate at the same rate in a gravitational field. This includes for example a brick and a light beam that was travelling horizontally along the brick from the point of view of an observer with the brick.

    From General Relativity, there's also an extra thing we need to take into account, which is that space is curved as well. This means that the shape of the brick and of space are both curved relative to the usual way of mapping out space (the coordinate system) as seen by a distant observer, and this imparts an extra acceleration on anything which is moving fast in the horizontal direction from the distant point of view. For the speed of light, the extra acceleration is approximately equal to the original Newtonian acceleration, so the overall acceleration is twice the Newtonian gravitational field.

    In simple cases, assuming you're not very close to a neutron star or black hole, the path of a horizontal light beam therefore either falls in exactly the same way as a brick (from a local point of view) or twice as fast (from the point of view of a distant observer who has mapped events using isotropic coordinates, as is for example the usual convention within the solar system, or a similar spherically symmetrical coordinate system). That is, the vertical distance fallen in time t starting from horizontal is simply h = (1/2) gt2 as usual from a local point of view (relative to curved space) or twice that from the distant point of view, relative to a typical practical coordinate system such as isotropic coordinates.

    The distance travelled for a horizontal light beam in time t is simply ct.

    For non-horizontal light, things get more complicated, as the coordinate speed of light decreases as one falls, so the sign of the acceleration relative to the coordinates is actually upwards. However, if one considers the coordinate relativistic momentum Ev/c2 instead of the acceleration (where c is somewhat unconventionally used to denote the speed of light within the coordinate system rather than locally) then this increases downwards with time (because of the decrease of the coordinate value of c) and a more detailed calculation shows that in isotropic coordinates the rate of change of the coordinate relativistic momentum (i.e. the effective force) is the same for horizontal or vertical light, or anything in between, and is effectively simply (1+v2/c2) times the Newtonian force for speed v.

    This factor of 2 relative to the Newtonian acceleration was famously confirmed by Eddington's expedition observing the apparent displacement of stars near to the sun during a total eclipse in 1919.
     
  5. Mar 3, 2012 #4

    pervect

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    Any test particle moves along a geodesic in general relativity. Massless test particles, like light, move along geodesics. If you are familiar with the classification of curves into spacelike, null, and timelike, the sort of geodesic that light travels along is a null geodesic.

    The geodesic equation should be in most GR textbooks, or in sources like Carroll's lecture notes. It looks something like this:

    [tex]\frac{d^2 x^\mu}{ ds^2} + \Gamma^\mu {}_{\alpha \beta} \, \frac{d x^\alpha}{ds} \, \frac{d x^\beta}{ds} = 0[/tex]

    The geodesic can be thought of as extrimizing the path-length of the space-time curve, though this approach makes more intuitive sense for cases other than light, i.e. for non-null geodesics.

    However, without a background in tensors, the notation used for the geodesic equation above (especially the Christoffel symbols [itex]\Gamma^{\mu}{}_{\alpha \beta}[/itex] may seem mysterious.

    Informally, though, you can think of geodesics as the closest thing possible to a straight line in a curved space-time.


    As far as I know, people don't usually solve optics problems in terms of "forces" anyway. For instance, try solving a simple problem in atomspheric refraction this way - for instance the way that light bends when you are surveying, due to the atomspheric temprature gradient causing the refractive index of the atmosphere to vary with height. This definitely and observably causes light to "bend", but it's not commonly though of as being due to a "force".

    Usually people use something like Fermat's theorem, which says that light travels in such a way as to minimize (actually, to extermize) optical path length to solve optics problems anyway. (See Hamiltonian optics in the wiki). So classical optics are really not that different from GR optics - both of them are based on the calculus of variations and the principle that some scalar quantity computed from the path is extremal - i.e. either a maximum, a minimum, or a saddle point.
     
    Last edited: Mar 3, 2012
  6. Mar 3, 2012 #5
    From the view of a distant observer the speed of light is reduced by the gravitational field (Shapiro delay):

    [itex]u = \frac{c}{{1 - 2 \cdot \frac{\Phi }{{c^2 }}}}[/itex]

    with

    [itex]\Phi \approx - \frac{{\gamma \cdot M}}{{\left| r \right|}}[/itex]

    The path of the light should be the same as in matter with the same optical density:

    [itex]a = 2 \cdot \frac{v}{u} \cdot \left( {v \cdot u'} \right) - u \cdot u'[/itex]
     
  7. Mar 3, 2012 #6
    Since you only want to consider the vertical speed of light we can use the abbreviated version of the Schwarzschild metric that ignores the angular velocity:

    [tex] c^2\tau^2 = (1-r_s/r) c^2dt^2 - \frac{dr^2}{(1-r_s/r)} [/tex]

    where t is the coordinate time, r is the coordinate radial (vertical) distance, [itex]r_s[/itex] is the Schwarzchild radius of the event horizon at [itex]r_s = 2GM/rc^2 [/itex] and [itex]\tau[/itex] is the proper time of a test particle. For light the proper time rate is zero so we can say for vertical light pulse:

    [tex] 0 = (1-r_s/r) c^2dt^2 - \frac{dr^2}{(1-r_s/r)} [/tex]

    and solve for the falling rate:

    [tex] \frac{dr}{dt} = (1-r_s/r) c[/tex]

    This is the rate of falling (or rising) light in coordinate terms and approaches zero as r approaches the event horizon.

    Locally the speed of light is always measured as c because lower down as light slows down, clocks also slow down and ruler lengths contract to compensate. The main problem with measuring the speed of light locally over a short distance, for example over a metre, is that the length of a metre is defined in terms of the distance light travels in a given time. If we calibrate a metre ruler by using a rod with mirror at one and time how long it takes for light to travel the length of the rod and back and divide the time by two, it is now pointless trying to time how long it takes light to fall over that distance as that is how we calibrated the rod.

    However, with sensitive enough clocks we can still detect the effects of spacetime curvature on light even over short distances. For example if we time how long it takes light to go up the rod and return back down, this time interval will be shorter than if we time how long it takes light to travel down the rod and back up again. This is because when the clock is at the top of the rod it is ticking faster than when it was at the bottom of the rod. Ultimately this means the modern metric method of defining the length of a meter in terms of light speed is flawed in highly curved spacetime because the answer depends upon whether the clock is at the top or the bottom.
     
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