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Photon energy in redshift

  1. Jan 20, 2010 #1
    If an atom in a distant galaxy emits a uv photon, the photon would have had enough energy to cause photoelectric emission from a negatively charged zinc plate. By the time the photon reaches me on Earth it has been sufficiently 'redshifted' so that it could not cause photoelectric emission if it hit a negatively charged zinc plate. Where, in simple terms, has the missing energy gone or, where is the flaw in the question?
  2. jcsd
  3. Jan 21, 2010 #2
    Energy in relativitiy is relative to the observer. Redshifted photon is being observed in a different frame from where it was emitted.
  4. Jan 21, 2010 #3
    Thank you for your reply. The photon reaching Earth (which how has less energy than the work function of the zinc) is not able to cause photoelectric emission. Would it have been able to cause photoelectric emission had it hit a zinc plate at the beginning of its journey?

    Can I reverse the argument and say that a 'red' photon emitted from an atom rushing towards might be able to cause photoelectric emission from the zinc plate by the time it reaches me? I can just about cope with length and time 'changes' in special relativity due to the change of reference frame but cannot yet picture what is happening to the photon.

    Any advice for a simple but clear explanation in terms of photons?
  5. Jan 21, 2010 #4
    There is nothing happening to photon.
    Note: we even dont need a photon. Exactly the same would happen to a billiard ball in Classical physics.

    Say, ball is ejected from a distant galaxy, which is receding from us at 1000km/s at 1500km/s. When it reaches us it is moving only at 1500-1000=500km/s, so it has less energy then for an observer at the fat galaxy, where it is moving at 1500km/s, so its kinetic energy is 3*3=9 times higher.

    Photon always moves at c, and the change of energy we observe as redshift.
  6. Jan 21, 2010 #5
    Also note that the color/energy of a photon is observer dependent. For us, that photon is red. For an observer from the source galaxy it is still blue!
  7. Jan 21, 2010 #6
    Thank you for your billiard ball analogy. It was very clear. However I am still confused. As seen from the distant receding galaxy, the photon is still a uv photon. What would an observer on the receding galaxy see when the photon hits the zinc plate on Earth? Would the photon, a uv photon, cause an electron to be emitted from the plate? From my perspective on Earth, the red photon did not cause an electron to be emitted from the plate. I am sure there is a simple resolution to this paradox but it escapes me at the moment. Your help would be appreciated.
  8. Jan 21, 2010 #7


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    For an observer at the distant galaxy, the zinc plate is receding. In a similar way, if we fire a UV photon at a rapidly moving zinc plate in a lab on Earth, we can set it up so that the photon will not cause electrons to be emitted. (Or if we move a zinc plate rapidly towards a source of IR photons, we can still get a photoelectric effect in a lab. It would have to move pretty fast, though.)

    Cheers -- sylas
  9. Jan 21, 2010 #8
    From a perspective of the source galaxy, uv photon is always still uv photon. However, the plate itself is receding, so you cant apply the equations for the photoeffects for the sake of simplitity are written in the frame where plate is at rest

    So you need to use more complicated equations instead. The result you will get will show that even the photon was UV, as plate was receding it failed to eject an electron.
  10. Jan 21, 2010 #9
    Thank you. From the perspective of the source galaxy, the relevant information is the hf of the (uv) photon and the work function of zinc. How would einstein's photoelectric equation be modified if the zinc plate were receding? Is the 'work function' somehow 'modified'? I appreciate the time you are spending on this and I am very grateful.
  11. Jan 21, 2010 #10
    You do not need to be confused. Energy is always E=hV, it is just what you measure for frequency. Analogy that distant galaxy see photon as uv when it reaches us is not quite good, because it really does not see it at all. It would need to be absorbed and re-emitted towards that galaxy, and by the time it reaches it again, it would be redshifted further more.
  12. Jan 21, 2010 #11


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    In a local setting, like a lab with a rapidly moving zinc plate, you simply calculate the energy of the photon in the frame of the plate, using the normal Doppler shift formula. If that energy is enough to cause electron emission, then you get emission. Otherwise... not.

    When you are comparing frames widely separated in the universe, we usually don't speak of "Doppler shift"; but there's no difference in principle. The same photon has different energy, depending on who is observing it. To see if there's a photoelectric effect, you need the energy of the photon in the frame of the zinc plate.

    Cheers -- sylas
  13. Jan 21, 2010 #12
    Thank you. Could it be argued that from the frame of reference of the distant galaxy, the uv photon arrives in a world (Earth) where all distances, times and mass are relativistically modified. Alternatively from the Earth's frame of reference, masses, distances and times are familiar but the photon has borne the change required by special relativity.

    Basically the energy 'loss' is a consequence of special relativity and the change of reference frame.

    Has anyone ever created a relativistic 'Einstein's photoelectric equation' with the zinc plate moving at high (relativistic speeds) in the lab? Does it end up with a modified work function?
  14. Jan 21, 2010 #13
    There is no energy loss. It is just how much energy photon has in relation to observing frame.
  15. Jan 21, 2010 #14
    The concept everyone is using about the relativity of the observer when it comes to photons contains a serious misconception about photons. The photon cannot be observed from the source galaxy by any means. Photons can only be seen or even detected in the single direction of their trajectory at c. The only way to detect them at all is when they interact with whatever they hit, be that our retinas, that zinc plate, or your digital camera. The fact of this can be easily understood if you think of a situation where you see me (or photograph me) while I see you. We would both be in trouble if we were seeing the same photons. I can never see the photons you call the image of me and you can't see the ones I see as you, yet amazingly these photons have actually bypassed each other without affecting each other. Once any photon is detected it is no longer a photon moving at c. Most physics books will tell you it ceases to exist when it gives up its energy to the zinc plate. I think that is also wrong but I can't establish that in a brief statement. Most light thought problems any of us have ever seen imagine the "stationary observer" seeing a light beam from a side on view in the tradition of Galileo's Ship. That is impossible! What we see when we see a beam from the side is the reflection of some of the light off of moisture and dust particles in the path of the beam. If it is in a vacuum as Einstein liked to imagine you couldn't see it at all.
  16. Jan 21, 2010 #15


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    I don't think it is a misconception, so much as a standard way we tend to talk in science. The point is that the same photon has different energies depending on the "observer", which is just a shorthand for saying energy, and wavelength, are relative to a frame of reference. You can think of "someone" in a different galaxy using a monochromatic laser and a beam splitter to test a proportion of the photons, while the rest cross space to interact with a zinc plate many billions of years later.

    But nothing about the theory really depends on having such an experimental setup. We simply mean that the same photon leaving one galaxy has a certain frequency in the frame of reference of the source, and then the same photon subsequently has a different frequency in the frame of reference of scientists here on Earth looking at the light from that galaxy. Nothing happens to the photon. It's all about giving the reference frame for which a frequency is defined. And with two different frames, you have two different frequencies.

    Cheers -- sylas
  17. Jan 21, 2010 #16
    Thank you. I now understand a bit more about the world. This was my first use of 'Physics Forums' and I am grateful that you were there with answers. Best wishes.
  18. Jan 22, 2010 #17
    I respectfully dissagree! I believe all science has a misconception about photons. If you can reguard a photon as being uv or blue or red then somthing about each photon must be oscillating. I think it is never considered how this ends up becoming a wave. I have an idea that explains that as an analogy to electric ac sine waves. Just as sine waves for voltage and amperage are the mapping of instaneous values over time of currents that are moving first one direction then the other within the conductors and not something that is actually moving through anything that looks like a wave; then photons can also be imagined to produce that sine wave that they are routinely depicted as by doing nothing other than pulsing energy in esensially straight lines 90 degrees from their propagation axes. That would mean that when the uv photon left the star it was oscillating over uv wavelength distances and whatever wavelenght it red shifted to would be its actual wavelenght upon observation. I think the "tired light" idea I have been reading about also results from the same sort of misconception. There is no energy loss if the photon behaves as I have described. Because of the inverse relationship of frequency to wavelength if you consider the consequences within my model the speed at which the energy oscillates back and forth stays constant and so does the distance traversed within any period of time. The reason we measure different energy values for different wavelength photons is because of the difference in how many can occupy any unit of area at the same time. This thinking by the way also obviates the need for expansion or dark matter or dark energy if you carry it to its logical conclusions. If photons can red shift without affecting their energy and any photon can become any frequency then it is negligent to not consider the possibility that they always redshift as they traverse vast distances.
    Last edited: Jan 22, 2010
  19. Jan 22, 2010 #18


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    You can stop right there. The role of this forum is to learn about what is the current practice in science, including the various different models considered... but not the new ideas of your own that you'd *like* to be considered.
  20. Jan 22, 2010 #19
    Sorry! I just took a break to re-read the rules that I printed out earlier. I didn't think what I wrote was that far of a deviation from mainstream thought but I appologize if I offended all science by saying they all have a misconception. But it is true that my idea is uniquely my own and hasen't been shared with the world. I don't think that makes it unworthy. Your rules would have rejected all of Einstein's work in 1905 had this kind of forum existed then but then almost all science did initially reject him. This may be the wrong forum for me if no new ideas are never welcome. I guess its a little like going into a classroom and telling the prof no no listen to me even if he is wrong and you are right. I am new to this process and will attempt to stay within the rules when and if I do fruther postings.
  21. Jan 22, 2010 #20


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    No problem; and welcome to the forums! It is a more limited focus here than a wide open discussion.

    However... we would not have missed Einstein's work, because it was all published in 1905 in the scientific literature, in a series of five shattering papers. And it wasn't rejected by scientists. In fact, it was pretty quickly recognized and accepted. Scientists quite rightly don't just roll over and accept a new idea as soon as it gets published; publication is effectively the start of the serious examination by the rest of the community, and Einstein's work was no different. Unlike most radical new ideas, Einstein's work passed muster and soon became the foundation for a lot more work and development by other scientists.

    The scientific literature is a pretty good place to be looking for the bright new ideas that will revolutionize science. The Einsteins of the future will be using the scientific literature as well, I think. But it's not easy to recognize those revolutionary ideas right away, even when published.

    Cheers -- sylas
    Last edited: Jan 22, 2010
  22. Feb 5, 2010 #21
    My understanding is that mass, energy and space are all equivalent and that the energy of the photons is lost and goes to the curvature of space.
    Probably I have this wrong, but some of the comments here don't correspond to my understanding of mainstream science - which is that to all observers, the photons lose energy. It goes to the curvature of space. The electron at the observers photoelectric effect experiment won't be emitted either for observers at earth or for those at the photon's source.
    Whilst it is true that the absolute energy of a body will be different and depend upon ones reference frame, the change in energy will be the same to all.
    If I remeber correctly, the rate of change of energy (dE/dt) is invarient and is the same for all reference frames. So if the observer on earth sees the photon losing energy, the observer on the source will see it losing energy.
    Bit like celsius and Kelvin. Whilst actual temperatures are different, any change is the same for all.
    Energy lost by the photon goes to curvature of space. Photo electron is not ejected in anyone's reference frame.
  23. Feb 5, 2010 #22


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    No, it depends on the state of motion. If you had a representative of the source at earth's location, such that she is at rest with the source, she'd measure no energy loss. Both, earth ambassador and source representative, would see the electron jump out.

    But I should mention that the idea of "at rest" is a bit fuzzy in most circumstances, and that the one I used for this explanation (parallel transport of the source velocity along the light path) is the only one with a null result.
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