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Gravitational blueshift and a Moving Mass

  1. May 3, 2009 #1
    Gravitational blueshift corresponds to the difference between the underlying clock rates at the emitter and at the observer. Which means that the wavelength at the instant of emission is already longer as measured by the observer's clock than the same emission as measured by the emitter's clock.

    On the other hand, without reference to clock rate differentials, as Einstein first pointed out, energy conservation requires the light beam to gain (kinetic) energy as it "falls" deeper into a gravity well. This causes the wavelength to decrease, so an observer located deeper inside the gravity well measures blueshift.

    Obviously these two effects are complementary and not cumulative: the resulting gravitational blueshift is not twice the amount of the gravitational time dilation. They are flip sides of a single coin. But can a thought experiment be constructed to enable the two effects to be examined separately?

    Let's say there is a non-gravitating emitter initially stationary at a certain distance from a massive neutron star. It emits a light beam radially toward the star. Immediately after the light is emitted, a rocket starts towing the star rapidly radially away from the emitter, quickly attaining a constant velocity of .5c. (This is the same rocket that is sometimes employed to tow tethered galaxies.) Then the star is brought to a halt (stationary relative to the emitter) just before the light beam from the emitter arrives. The emitter and the observer on the surface of the star are stationary relative to each other at the time of both emission and reception and no SR redshift is finally observed (although SR time dilation affects the synchronization of the two clocks).

    From the perspective of clock rate differential, the blueshift finally observed at the star should be exactly the same as if the star hadn't moved. The fact that the emitter finds itself to be relatively higher in the gravity well after emission should have no effect on the light beam, whose characteristics were determined entirely by the emitter's location in the gravity well at the time of emission.

    From the perspective of the gravitational acceleration experienced by the in-flight photons, the acceleration rate increased more slowly, because the light beam's progress in penetrating deeper into the gradient of the gravity well occurred later than if it hadn't needed to chase the star. It seems to me that the light beam experienced the same average acceleration rate as if it hadn't been chasing the star, but the duration during which the acceleration was applied was significantly longer. Intuitively then it seems that the lightbeam should have accumulated more absolute acceleration effect (energy gain) in the chase scenario. Would more blueshift be finally observed than if the star hadn't moved?
    Last edited: May 4, 2009
  2. jcsd
  3. May 3, 2009 #2
    It is true that photons falling in the gravity field gain kinetic energy. However, their potential energy changes too, so that the TOTAL energy stays constant. When the photon is absorbed by a detector, it is the photon's total energy that gets transferred to the detector. Therefore, the absorbed energy is the same as was emitted by the source. The photon's "frequency" depends on its total energy, so it does not change in the course of photon's propagation.
  4. May 4, 2009 #3
    The textbook I'm referring to specifically says that a Newtonian analysis (potential energy goes down as kinetic energy increases) cannot be used to accurately assess a particle's relativistic energy in freefall in a gravitational field. It says that an observer on the surface of a massive body can perform a measurement of a falling particle's relativistic energy which is valid only locally. For example the measurement is not valid for the energy measured by a very distant observer, which is invariant regardless of the freefalling particle's velocity. Unlike a very distant observer, it is valid for a surface observer to divide the particle's total energy into kinetic energy and rest energy components. The surface observer will locally measure the particle's kinetic energy, and therefore its total energy, to increase as the particle falls farther toward the surface. The difference between the total energy measured locally by the surface observer and globally by an observer at infinity is:

    [tex]\frac{E_{surface}}{E_{infinity}} = \frac{1}{\sqrt{1 - \frac{2GM}{c^{2}r}}} [/tex]

    So I think my question is valid for a surface observer. Anyone want to venture an answer?
    Last edited: May 4, 2009
  5. May 5, 2009 #4
    Oh drat, I think I can answer my question after thinking about it some more.

    I said that for a photon chasing the moving mass, the average gravitational acceleration effect was the same as for a stationary mass, but the duration of the period during which the acceleration effect applies is significantly longer.

    What I glossed over is that, during the period when the mass is moving, its clock is running slower relative to the emitter's clock due to SR time dilation. So the acceleration period is longer but the clock is running slower in the mass' frame, and the two time factors exactly cancel out.

    I think I'm convinced that under no circumstance does the motion of a mass change the amount of gravitational blueshift experienced by an infalling photon. And, that keeps things simpler.
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