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Gravitionel redshift – How is it working?

  1. Apr 8, 2012 #1
    Experiential data shows that light on its way out of a gravitionel field is losing energy and become redshifted.

    But what about light that moves into a gravitionel field, - Is that gaining energy (getting more blue shifted) ?

    Has we ever measured what happens with light that for example moves true a gravitionel field?

    For example we could compare light from Jupiter.

    “A” shows light that moves directly from Jupiter and to us (and hence only into the gravitational field of the Sun and the Earth)
    “B” shows that now is Jupiter on the other site of the sun.

    It is therefore possible to compare light from Jupiter, on the one hand when it moved directly to us, and therefore into the gravitational field of the Sun (exsample “A”) - and on the other hand “B” light that first moves into the Suns Gravitionel field and then out again until we revived it on Earth, on the opposite site of the Sun.

    A.) Jupiter----------- ----------> Earth Sun
    B.) Jupiter----------------------> Sun -------------> Earth

    Will light from Jupiter in these 2 cases be different redshiftet / blueshiftet ?

    Has such measurement experiments ever been executed?
  2. jcsd
  3. Apr 8, 2012 #2

    Jonathan Scott

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    Gravitational redshift (in a static gravitational field) is not something which happens to light. It is entirely due to differences of observer time rate due to gravitational time dilation. A free-falling photon has constant frequency relative to a static coordinate system. Local clocks and processes vary in frequency due to local gravitational time dilation, so when the frequency of an arriving photon is compared with one generated locally via a similar process, it will appear to be redshifted or blueshifted according to the difference in time rate.

    Photons emitted from a lower or higher gravitational potential than the observer do not lose or gain energy in flight. They start off with a lower or higher frequency and keep that frequency. Relative to a series of local observers at different potentials they will appear to have higher or lower energy than a similar photon generated locally, but if any of those observers arranges to monitor the frequency at multiple locations, the observations at each location will all show the same frequency.
  4. Apr 8, 2012 #3
    Except for Jonathon's first sentence, which might be subject to misinterpretation, I'd readily agree with his description.

    Wikipeida's opening is this:

    Notice that Jonathon says
    while Wikipedia says redshift is an observed [local measurement] phenomena...so both views are I think technically correct, but the complementary description helps to clarify what is meant. Some will argue all that matters is what is OBSERVED....that's what is 'real'.

    A quickie answer might be: different observers have different clock rates, different rates of ticks, so they will observe [measure] different frequencies....

    Simple answer:
    [a] yes,
    [c] I'm sure they have, but I don't know about your exact example. Same situation happens when CMBR passes through galaxies...but other effects like gravitational lensing may also observed....light curves....
  5. Apr 8, 2012 #4
    If we shine a light from an ordinary torch down towards a lead shield that is very close to a black hole and the light is blue shifted into the x-ray region, could that light from the torch pass through the lead shield, when ordinarily it would not in its unshifted state?
  6. Apr 8, 2012 #5

    Jonathan Scott

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    Yes if it is sufficiently shifted to reach a frequency that penetrates lead.
  7. Apr 8, 2012 #6
    Great question but I thought x-rays frequencies could NOT penetrate lead whether they were emitted au natural or blue shifted to that range?????
  8. Apr 9, 2012 #7
    I think x-rays of sufficiently high frequency will pass through a reasonably thin lead shield. If not, substitute steel sheet for lead shield.
    Last edited: Apr 9, 2012
  9. Apr 9, 2012 #8
    Or just human bodies for that matter, it's the idea thats interesting and I wasn't quibbling just curious , mea culpa ;-(
  10. Apr 9, 2012 #9
    Good one!!!!

    Jonathan: How do you explain/reconcile your earlier post:

    with your subsequent reply:
    they seem inconsistent.....

    I suspect his has to do with Schwarszchild coordinates being 'static' outside the event horizon but I have never been sure about the implications of that statement...Wikipedia is rather opaque [to me] on 'static spacetime'....

    Is your target accelerating to maintain a fixed position outside the horizon, or is the target and observer you describe free falling??
  11. Apr 9, 2012 #10

    Jonathan Scott

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    I'm assuming the target to be held in a fixed position. The target will be in a deeply time-dilated state, so the energy of the original beam relative to the target will effectively be enormously increased.
  12. Apr 9, 2012 #11
    ok...so target and observer are in a fixed position outside the BH horizon in a static gravitational field .....[All I think I know about such a static field is that it doesn't hold inside the horizon.]

    Light from the distant universe appears blue shifted [and time out there appears to be moving faster] because of the low local gravitational potential outside the BH horizon, but doesn't light also appears somewhat redshifted due to the local acceleration???

    I say this because I think a free falling observer outside the horizon measures the local speed of light as the standard 'c'....so an accelerating observer would presumably not??

    Don't these effects somewhat offset?? If so how do you know which is dominant??
  13. Apr 9, 2012 #12
    Thanks! :biggrin:
    I should of said that I intended the target to be at a fixed position relative to the gravitational field, so it would appear to be moving at high speed towards the source from the point of view of a free falling observer.

    Using Schwarzschild coordinates, clocks lower down tick at slower coordinate rate that clocks higher up and the increase in frequency can be explained in terms of different clock rates at different heights. Ordinarily we attribute the penetrating power of x-rays to the high frequency but we could just as easily attribute it to the short wavelength because high frequency and short wavelength go hand in hand when the speed of light is constant. However, in Scharwzschild coordinates, the coordinate speed of light slows down lower down, and so even though the coordinate frequency is constant, the coordinate wavelength is shorter so maybe we can attribute the increased penetration to shorter wavelength in this case. The important point I am getting at here is that while coordinate observers and local observers disagree on the frequency of the light, they both agree that the wavelength is getting shorter and should both agree on whether the light penetrates the target or not.

    From the point of view of the free falling observer, the light from the upper source is red shifted but he would see the target as moving at high velocity towards the source and should agree that the motion of the target towards the target causes a Doppler blue shift.
  14. Apr 9, 2012 #13


    Staff: Mentor

    You have to be careful here how you define "frequency". Earlier you say "frequency relative to a static coordinate system", by which I think you mean the standard Schwarzschild coordinate system. Yes, the frequency of a given freely falling photon, relative to Schwarzschild coordinate time, does remain constant.

    But "local observers at different potentials" can't directly measure the frequency of a passing light ray relative to Schwarzschild coordinate time; they can only directly measure the frequency relative to their proper time, which is "time dilated" relative to Schwarzschild coordinate time. The frequency of a given freely falling photon, measured relative to different local observers' proper time, *does* vary--it's what you refer to as the redshift/blueshift.
  15. Apr 9, 2012 #14
    Peter: Once again, very helpful........thanks!!!!!
    Getting the right perspective between physical, observable relativity effects and keeping them distinct from coordinate system effects is not so 'obvious' as I would like....
  16. Apr 9, 2012 #15
    Can anyone explain how to reconcile these two descriptions....or is one [or both] incorrect??

    yuiop posts:
    [Does PeterDonis' post suggest the above is problamatic?]

  17. Apr 9, 2012 #16

    I would suggest that you consider that space as well as time is affected by a gravitational well. Time dilation etc would affect frequency, and spacial compression etc would affect wavelength. The relation of such an effect would naturally depend on the observer, its position in spacetime relative to the gravity well, and the instrumentation used.

    As to the penetration of a target by blue shifted photons I am intrigued as to where the additional energy may come from.
  18. Apr 9, 2012 #17


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    Do you have either a) a reference or b) a thought experiment which illustrates "spatial compression" due to gravity?

    This topic comes up occasionally on the forum, but it seems hard to track the idea down to a source.

    It mostly seems to come up from laypeople who think it should be obvious, but it always seems like the references or details are lacking

    I should add that there is a reasonable amount written about "expanding space" in the context of cosmology, but the spatial expansion there isn't caused by gravity. In the end, in cosmology "expanding space" boils down to a coordinate choice.
  19. Apr 9, 2012 #18
    Perhaps I'm not understanding you correctly but I thought that gravity was equivalent to acceleration and imposed a radial spatial contraction gradient??

    Isn't contraction equivalent to compression in this context? Or is it the connotation of stress associated with compression that makes it inappropriate?
  20. Apr 9, 2012 #19
    According to Jonathon et al, the second would seem to be incorrect.
  21. Apr 9, 2012 #20
    I just had occasion to look up e-rays and it appears that energy per se is not the prime criteria. The lower part of the x-ray range, although much more energetic that light, has little penetration capability.
    Which seems to suggest that the probability of absorption is more dependent on resonant frequency correspondence than energy.
    So it is not so much an increase in energy of the source [which frequency remains the same] as it is that the dilation of the target resonant frequencies drops them out of absorption range.
    Hopefully there is some sense in here someplace.
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