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Where does the energy go?

  1. Mar 6, 2008 #1
    When a photon passes from a high gravity field to a low gravity field, it is redshifted. Therefore it has less energy. Where does that energy loss go to?
  2. jcsd
  3. Mar 6, 2008 #2


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    The local energy of the photon does not stay constant, however, the energy at infinity does and is furthermore constant. (Energy-at-infinity is a term used in MTW's Gravitation).

    If a photon falls into a black hole, it is not the energy of the photon that gets added to the black hole, but the energy at infinity. So when you think of conserved energy in GR, don't think of the local energy as being conserved, but rather the "energy at infinity".

    This is a very rough overview, the concept of energy in GR is quite subtle and is different from the concept in other areas of physics. There is no *general* notion of conserved energy in GR, but for important cases such as photons falling into Schwarzschild black holes, there are useful concepts of energy.

    For more on energy in GR that is not too technical, try the sci.physics.faq:


    might also be useful
  4. Mar 6, 2008 #3
    The frequency of light does not change as if moves through a gravitational field. What changes is the frequency as measured by local observers. I.e. an observer at a particular position in at a high gravitational potential will measure a frequency which is higher that an observer at a lower position will. However any particular observer will measure a constant frequency. The energy of a photon moving through a gravitational field is conserved if the field is static. Lev B. Okun published an article on this topic. It copy is located at http://arxiv.org/PS_cache/hep-ph/pdf/0010/0010120v2.pdf

    Best wishes

  5. Mar 6, 2008 #4
    My interpretation of that paper is as follows.

    Measurements of a photon made by a distant observer (at infinity):

    Coordinate frequency [tex] w = w_o [/tex] ,

    Coordinate wavelength [tex] \lambda = {\lambda_o \over g^2} [/tex] ,

    Coordinate momentum [tex] p = p_o g^2 [/tex] ,

    Coordinate speed [tex] c = {c_o \over g^2} [/tex] and

    Coordinate energy [tex] E = hw = {hc \over \lambda}= pc = E_o [/tex]

    Where [tex] g = {1 \over \sqrt{\left( 1-{GM \over R c^2}\right)}}[/tex] and [tex] w_o , p_o , c_o , \lambda_o , E_o [/tex] are measurements of the photon at infinity made by the observer at infinity.

    Measurements of a photon made by a stationary local observer:

    [tex] w = {w_o g} [/tex] ,

    [tex] \lambda = {\lambda_o \over g} [/tex] ,

    [tex] p = {p_o g} [/tex] ,

    [tex] c = c_o [/tex] and

    [tex] E = hw = pc_o = {hc_o \over \lambda} = {E_o g} [/tex]

    On the other hand, the rest energy of a particle with non zero rest mass, increases higher up in the gravity well. Conversely, this implies (from the formula given) that the rest mass of a brick lowered towards a black hole goes to zero at the event horizon. The paper seems to suggest that the rest mass represents the gravitational potential energy of the particle.
    Last edited: Mar 6, 2008
  6. Apr 3, 2008 #5
    The observed redshift of the photon has nothing to do with the photon but is due to the relative blueshift of the absorber compared to the emitter.
    Last edited: Apr 3, 2008
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