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Positive Gravitational Force

  1. Nov 10, 2003 #1
    I know that as things "fall" towards each other, they lose potential energy, that's fine. What I want to know is when a photon falls towards a mass does the photon also lose potential energy?
    I would assume so because potons have an equivalent mass. From the equation E = hf, can I therefore assume that the frequency of a photon actually lowers when it approaches the Earth?

    If so then it seems to also follow that it's momentum also lowers from the equation E = pc, this then seems to suggest that photons lose mass when approaching other mass and that an impulse is applied to the photon to cause this loss of momentum.
    But this would mean that the impulse was acting away from the mass, ie: the photon was experiencing a repulsive force upon approaching some matter. Is this a positive gravitational force?
    Last edited: Nov 10, 2003
  2. jcsd
  3. Nov 10, 2003 #2


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    Yes. That is exactly correct.
    As the photons moves to a lower location the frequency as measured locally increases.

    But whether the frequency or energy changes depends on who is doing the measuring. Any single observer will detect neither a change in energy nor a change in frequency as a particle moves through the g-field. Such changes are meaured locally. I.e. if an observer higher in the g-field emits a photon which then travels to an observer who is lower in the g-field. The lower observer will measure a higher frequency than the frequency which left emitter. Interpret this as a decrease in potential energy which is compensated for an increase in kinetic energy. The local observer measures zero potential energy at his location and therefore all the energy he measures is kinetic energy. For a derivation see


    Note: I made an error in Eq. (7) [forgot to relable the subscripts on r)

  4. Nov 10, 2003 #3
    So you are saying that when a photon "falls" towards a mass, it's potential energy decreases while it's kinetic energy increases? I can accept that as it is true for all other particles, however this must mean that it's mass increases upon approaching a mass. Since the photon still has the same energy, it must still have the same frequency (for an observer at the point of emission of the photon), but now I am faced with another problem.
    As the equation E = mc2 shows, for any value of E there is a corresponding value for m. Since the overall energy of the photon has stayed the same, would the mass of the photon have to stay the same as well?

    Another problem, photons have mass and therefore they are attracted to things like the sun and blackholes and all other particles. If there was a photon heading straight for the Earth, wouldn't it be accelerated by the Earth's gravitational field and therefore be travelling at a velocity greaty than c?
    Since it is already travelling AT c, and it's mass is most definately not infinite, wouldn't the force due to the Earth's gravity cause it to accelerate? Any acceleration in the same direction as the photon is travelling would cause it to travel at a speed greater than c wouldn't it?
  5. Nov 11, 2003 #4


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    Yes. And that means that the total energy remains constant.
    Only when measured locally. And that, of course, is its relativistic mass that we're speaking about, not its rest mass.
    Now you're talking about a different measurement. Yes. The frequency remains constant as observed by any sinlge observer.
    That relationship does not hold in this case. mc2 is not the conserved quantity here. The contravariant 4-momentum of a particle is Pu = (E/c,p). In flat spacetime in an inertial frame of referance P0c = E = mc2 is a constant and is the energy of the photon. If the metric tensor is guv = diag(1,-1,-1,-1) and the covariant 4-momentum is Pu = (E/c,-p). So P0c = E too and also is a constant. However in a gravitational field P0 is not the same as P0. In a static gravitational field like the Earth's only the later is a constant.
    As measured locally the speed of light is always the same constant. If not measured locally, i.e. as measured by a remote observer far from the Earth the speed of light in the Earth's gravitational field is not a constant. The force of gravity of the photon will depend on the photons's location in the gravitational field. As far as the functional relation of speed vs. position.
    That depends on where in the field it is and who is doing the measuring.

    What you're failing to take into account is the fact that the acceleration of light depends on the velocity of light. E.g. at one point the speed of light starts to slow down as the light approaches a black hole.

    See also -

  6. Nov 11, 2003 #5
    The bits that looked pretty as the flew over my head:

    Is there anyway that you could dumb this down into something that I might be able to grasp a bit better?

    By the way, when I say does the frequency of light change? And you say, it depends where you are observing from, I usually mean from an infinite distance away (at 0 potential energy), where you are not affected by the gravitational field.
  7. Nov 12, 2003 #6


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    Let me get back to you on that. In the mean time you can read more on this in the article

    "On the Interpretation of the Redshift in a Static Gravitational Field," L.B. Okun, K.G. Selivanov, V.L. Telegdi, Am.J.Phys. 68 (2000) 115


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