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Gravitational Red-Shift?

  1. Jun 22, 2004 #1
    Hey, this is my first post. Just talking about an idea I had.

    A month ago I was pondering the effects gravity has on light. Specifically if there could be a “gravitational red-shift” of light coming from massive stars. I first started by thinking of gravity as simply causing an object, which lets say was moving away from an object, to have an acceleration toward the object. And of course it would eventually come to a stop and reverse direction. Another way of thinking about this is that the object loses momentum to the object its attracted to (atleast this is what I thought then).

    Recently in my physics class (AP Physics B) we had used an equation that would tell you the momentum of a photon. I started thinking that if a photon had momentum then why wouldn’t it, like a particle with mass, slowly stop and reverse direction when moving away from a gravity well?

    First I remembered that light would always move at the same speed in vacuum and thought that something else might happen instead of slowing down, because it didn’t seem right. So I thought it was possible that light could lose momentum without slowing down, it would simply lose energy or decrease it’s wavelength. But this would mean that there where no “true” black holes because then it would mean that black holes could emit photons, although the photons would be red-shifted to nearly an infinite wavelength. This didn’t seem right either.

    I was about to forget the whole thing and give up when I remembered that gravity was suppose to slow down time, a concept that at the time I didn’t believe in. Then I realized that thinking of light, and any other objects, losing momentum as they fell into a gravity well was completely wrong. If time where to slow down then to an observer at a distance away from a gravity well it would seem that photons would be traveling at less than the speed of light when coming from the gravity well (assuming he knew when the photon left). This could explain how a photon could not be gravitationally red-shifted.

    If you think about it, this could be thought of as analogous to what happens to light as it goes from one optically dense material to another. The velocity and wavelength of light changes as it passes though different mediums. Its like moving into more optically dense mediums as you get closer to the center of the gravity well, just assuming that the change in optical density is infinitesimal. This would explain things like gravitational “lensing.”

    So as far as I can figure light can’t really be gravitationally red-shifted because although wavelength changes so does the velocity, and the frequency does not change. So really the light never loses any energy (or momentum) from gravity and therefore gravitational red-shift doesn’t exist. After thinking about it for a while I realized that no object loses momentum from gravity if you think about it.

    Just thought this might interesting. Tell me if I’m right or just someone trying to understand things out of his league.
  2. jcsd
  3. Jun 23, 2004 #2


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    Hello there Entropy,

    First of all, red-shift in light is known to occur. It has been very well studied, at Harvard most notably, as a test of Einstein's General relativity. Let me if i may take you through your argument and point out a couple of flaws in this otherwise excellent bit of reasoning.

    Your first observation of losing momentum to a gravitational field is a good assumption but the rate of loss of momentum is not linear as when we move further away from a source the force gets weaker. This is because the gravitational field varies with with the inverse of the distance squared, so if something is travelling fast enough it can eventually escape the gravity well. This is just a minor point though.

    If we consider the formula for the momentum of a photon, [tex] p=\frac{h}{\lambda} [/tex]. As the momentum decreases, lambda must increase (i.e. a redshift occurs if the momentum of the photon is reduced). This explains why a photon will not slow down and reverse direction all that happens is that its wavelength and frequency adjust to compensate and it continues to propagate at C, the speed of light.

    Now as for black holes emitting photons, what you propose in your argument does seem reasonable until you consider that at some point there will be a gravity well strong enough to remove all the momentum from light, and then what do you have if you have zero energy? The answer is nothing, so no photons are emitted. and we can have "true" black holes.

    Now your ideas on refraction in different mediums are the classical view of what happens. But quantum mechanics tells us the light does not change its speed or therefore its wavelength, but only its direction. This is to do with an effect of a medium called absorption which accounts for the slowing down of the light in that particular medium. Gravitational lensing is explained by the geometry of the spacetime caused by a gravitational mass. the mass curves the space and the light merely follows a staright line which, in a curved region of space makes it appear to bend and be focussed to a point. Again another point of general relativity.

    Your point about time slowing down is only a means by which we can still perceive light as being at a constant velocity when we are in motion. This wouldn't counteract the red-shift to an outside observer however. So overall the light does change its wavelength so it can maintain its constant velocity of c and energy is lost from the photon in escaping the gravity well.

    Information can be found on harvard's web site about their experiments to detect redshift I believe.
    Last edited: Jun 23, 2004
  4. Jun 23, 2004 #3


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    One problem with high school physics is they don't adequately cover this stuff...but I guess that's why you are here.
    Light always travels at C. Forget the "in a vacuum part" because its very misleading: light only travels in a vacuum. When its said that the speed decreases through a medium, thats the average speed resulting from light hitting particles, being absorbed, and re-emitted. The delay makes it appear that the light has slowed down.

    With an infinite wavelength, what would the energy be? Have you taken calculus yet? Light coming from a black hole can be viewed as infinitely red-shifted. And since infinitely red-shifted means zero energy, the photon isn't emitted.
    Careful with that. An observer can only observe things that happen to him. You can't really observe something from a distance. You can (if you want) calculate that if you could see the photon at a distance it wouldn't be traveling at C according to your meter-stick and clock, but thats not all that useful. To you, in your frame (where-ever that frame may be), photons travel at C.
    The problem is these things (time dilation, constancy of C, gravitational redshift/lensing) aren't just untested predictions of a thin and speculative theory, they are hard data. They exist and require explanation.
  5. Jun 23, 2004 #4
    Doh! You're right, don't know what I was thinking there. I know some calculus by reading some stuff on my own, but unfortunately I was stuck in Pre-Cal (huge waste of time if you ask me, just Algebra II all over again) last year instead of just going into Calculus AB. One of my quarter grades apparently wasn't good enough even though I wasn't at school a lot due to surgery that 9 weeks.

    Okay that makes sense. Thanks for clearing that up.
  6. Aug 29, 2004 #5
    Kurdt: What is the formula for the amount of red shift seen by an observer at distance from the center of gravity R-1 when the source is a distance R-2 from the center of gravity? I've been searching the web for this formula, but so far I've only found it without the distance of the observer. I think they're assuming a "distant" (i.e. infinitely distant) observer. Then they say light is infinitely red-shifted when emitted from the Shwarzchild radius. I think that's true only when the observer is infinitely distant. At any finite distance, a free-falling observer would see finite red-shifting from a free-falling source at the Schwarzchild radius. Am I right?
  7. Aug 29, 2004 #6
    See - http://www.geocities.com/physics_world/red_shift.htm

  8. Aug 29, 2004 #7


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    Ray:The two links provided by pb phy should be sufficient in answering your question. I apologise but its been a while since I posted on this forum.
  9. Aug 29, 2004 #8


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    Gravitational redshift and gravitational time dilation are closely related, but it would be misleading to say there is no gravitational redshift.

    The usual equation is

    \lambda_{local} / \sqrt{g_{00}}[/tex] = constant

    Here [tex]\lambda_{local}[/tex] is the _locally_ measured wavelength of light, and g00 depends on the metric of space-time.

    Near a single massive body, one can use the Schwarzschild metric to determine g00

    [tex]g_{00} = 1 - \frac {2GM}{c^2r}[/tex]

    here M is the mass of the massive body, and r is a coordinate which measures how far away one is from the body.

    What all this means is that if one looks at the spectrum of light emitted or absorbed from a hydrogen atom, the lines will be at different frequencies for an atom deep in a gravity well and one "right next door".

    One can call this effect by different names (gravitational time dilation, gravitational red-shift), but the result is the same.

    This can be viewed as a result of energy conservation - light cannot climb out of a gravity well, and do the same amount of work it could have done (measured by local standards, such as exciting a specific atomic transistion) before it left.

    In general, GR prefers local measurments over global ones.
    Last edited: Aug 29, 2004
  10. Aug 29, 2004 #9
    This statement is not quite clear. g00 does not depend on the local metric. It is the metric and its evaluate locally. I assume this is what you meant. It was just not clear to me and perhaps not the other readers.

    g00 is to be evaluated at the same location as the wavelength is measured. Your expression is just another way of phrasing Eq. (5) in the link above, i.e. in

    [tex]\frac{\nu_2}{\nu_1} = \sqrt{\frac{g_{00}(\bold r_1)}{g_{00}(\bold r_2)}}[/tex]

    This can be written as

    [tex]\nu_2 \sqrt{g_{00}(\bold r_2)} = \nu_1 \sqrt{g_{00}(\bold r_1)}][/tex] = constant

    Changing from frequency to wavelength and rearranging terms gives pervect's result above.

    Last edited: Aug 29, 2004
  11. Aug 29, 2004 #10


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    I agree that metric would be better than local metric....I noticed a gross spelling error as well, so I'll just edit the post.
  12. Aug 29, 2004 #11
    Still, don't you think that g00 depends on the metric of space-time. is a bit clumsy?

    Don't mind me pervect - I'm just nit picking.

  13. Aug 29, 2004 #12


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    I was re-reading my post, and I realized that I hadn't addressed parts of this post, since I talked about the wavelength and Entropy mentioned that he agreed that it changed.

    The _local_ velocity of light does not change. The _local_ energy, frequency, and wavelength all follow the usual relationships

    E = h*frequency = h*c/lambda

    Thus, the local change in wavelength is associated with a local change in frequency, and of energy, by the usual relationships (given above). This is something I didn't mention explicitly.

    If this did not happen, one could in principle build a perpetual motion machine. One could drop particles and antiparticles into a gravity well, and extract work from it as it dropped, as one can from water flowing downhill. One could then convert the particles and anti-particles into photons, and then allow them to travel upwards. If the photons did not lose energy, they could then be reconstituted to form the original matter, which would be dropped again into the gravity well, completing the cycle.

    While the laws of energy conservation in GR can be a bit subtle, they don't allow for this sort of perpetual motion machine.
  14. Aug 29, 2004 #13


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    I noticed. You looking for an argument? I'm afraid I must agree that you're nit-picking :-)
  15. Aug 30, 2004 #14
    Not really. I'm just saying that if I didn't understand all this then I'd find that statement a bit confusing. But since nobody really has complained about it and you know what you mean then its no biggy.
    What can I said. I was bored. :biggrin:

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