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The relationship between light, energy and relativity

  1. Oct 16, 2011 #1
    In high school, doing modern physics, and was struck by a thought about light, energy of a photon and the "speed limit" of relativity.

    I'm probably very wrong, but I would like to be shown why :)

    My understanding of energy and matter relations is that when you increase the energy levels, you increase the speed of the atoms within a mass, this leads to other manifestations of the energy such as sound, heat, motion, etc.

    With light it is different, speed stays the same, but frequency/wavelength and momentum change.

    Would it be possible that light/photons of higher energy levels are actually "moving" faster than those of less energy, but, because of relativity's speed limit, the speeds observed by a static observer would appear the same?

    As space/time changes to ensure that nothing goes faster than c, it might result in the length contraction (changing wavelength) and time changing so that the frequency is observed to be faster.

    thanks in advance! :)
  2. jcsd
  3. Oct 16, 2011 #2


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    In order to observe the speed of light/photons, you have to do a measurement that involves a round trip. It is not possible to observe and measure the value of the one-way speed of light/photons but a static observer can measure that all the light/photons traveling from one particular direction toward him are traveling at the same speed, independent of the speed of the source or the energy of the source or the
    wavelength of the source or anything else about the source. So that means your idea is not based in reality. Any theory that purports to explain reality must conform to these experimental measurements.
  4. Oct 16, 2011 #3
    I'm not suggesting that some light is traveling faster than other light. Obviously light's speed is constant (in vacuum). I can see how you might need a round trip to emphasize the "relativityness" of the idea, but the idea is more about what happens after you reach the speed limit and add more energy, and what one would observe from a inertial point.

    If I'm trying to do anything I'm trying to explain the mechanic behind light of different frequencies and energy levels, instead of a model where light is just accepted to have a property where higher energy = greater frequency and shorter wavelength.

    I don't think that just because I'm not doing an experiment correctly (never really mentioned an experiment...) that an idea can be instantly discredited :P

    thanks for the response though
  5. Oct 16, 2011 #4
    or actually, is there a model that explains this? I haven't really looked in depth yet, but if there is that would be helpful if someone could link to it :)
  6. Oct 16, 2011 #5
    Ouch, photons and their intrinsic energy :)

    You got the invariant light quanta, that's the origin, and that one does not change its 'energy level', ever. So why is there a red and blue shift?

    That one is defined through the relation you have toward that 'photon' annihilating in your detector, either moving away from its propagation (makes it 'red') or moving towards, meeting it ('blue shift'). But a planet doesn't 'move', well it does, relatively so, but why is all 'photons' expected to be blue shifted, no matter from where they hit us?

    That Einstein defined through the equivalence of gravity with a acceleration. In both cases you have a sort of 'motion', and if the 'photon' moves against that 'motion', as when 'in-falling', then it will blue shift, and here you have to imagine translating gravity's 'force' into the motion you (earth) would have if accelerating uniformly towards that 'photon' at one G, and measuring it the other way (outgoing photon) will make it 'red shift' as the planet then could be thought of as 'moving away'.

    The 'expansion' is slightly confusing here in that it, as I understands it, first of all treat all radiation a waves, and then assume that when there comes to be more distance 'welling up' in 'space' it sort of 'stretch out the wave', and if you stretch a wave its heights and dips will shrink, and so it will red shift. That one is okay for a wave, but to my eyes a little tricky when it comes to 'photons', as they then are expected to become 'spaced out' relative each other. But the real mystery here is that it is acting on the 'wave' itself, not as an effect of relative motion, as I see it? Or maybe I'm thinking wrong there.

    It is also so that if we speak of photons there are only two attributes, it has a momentum and a energy (and polarization/spin but, that's not important for this). When we speak of waves you will have wavelength and frequency, which could be seen as transformations of energy and momentum. So no, light as quanta never change their 'energy' or 'momentum' as I understands it. But they will give you a red or blue shift, depending, and that 'energy' will be as real as with two cars colliding towards each other, or from 'behind', well, sort of.
    Last edited: Oct 16, 2011
  7. Oct 16, 2011 #6
    I was under the assumption that photons were able to have different energy levels where E=hf, and then for the photoelectric effect you can measure the energy through E=W+kE of the ejected electron. And yeah the Doppler effect would do the red and blue shifts :)
  8. Oct 16, 2011 #7
    Not if you go from the definition of a light quanta. If you start allowing a photon to change its energy intrinsically you also invalidate the way we look at red and blue shift. You also open for the 'tired light' proposition. And as a photon in mainstream definitions is 'time less' it shouldn't be able to change 'energy', because if it did you have introduced 'intervals/durations' in its 'propagation', as defined intrinsically.

    Thinking of it, defining them that way should also make them have a 'mass' of the more 'invariant' kind, as they then would be subjects to 'times arrow' intrinsically. And the only 'stuff' that can be so is fermions, or would that be to simplistic? Do we have any definitions of any kind of bosons that have a intrinsic 'arrow of time'? I don't think we have, but we have the Higgs boson that you might, with some difficulty, define as a 'clock' of sorts, if existing.

    As for them having different 'intrinsic energies' that is true, but that gets defined at the source, sort of, and all proofs to the other that you see, is then the direct result of 'relative motion'. It's like a whole 'frame work', where everything fits together if you look at it the mainstream way. And it makes sense as with photons 'leaking energy' or gaining you could expect them to 'die' on the way, not only in the annihilation.

    The reason a 'photon' has a 'recoil' leaving a source is not as simple as it acting with a 'force', it's instead explained through the conservation of momentum aka the conservations laws.

    They and symmetries makes some sort of border defining what 'stuff' are allowed to do, and they build on observations and experiments dating very far back in history. So physics is a 'frame work' with very long roots, and even though no law can be said to be 'absolute', as there might come a totally new way to see it, making sense, invalidating what we think us to know, the best bet so far, as I know it, is that 'photons' don't lose or gain 'energy', other than relative 'relative motion'
    Last edited: Oct 16, 2011
  9. Oct 16, 2011 #8
    I see what you mean by "changing energy" and that makes sense :) Once they start from a source they stay at a constant intrinsic energy level until annihilation, or wherever they end up. Changing the energy after the creation of photons wasn't exactly my idea, but I'm glad that you sorted that out for me :)

    But when comparing two different photons/bits of light from two different sources that are shooting out light of different frequencies, initially,( and throughout I guess,) as per E = fh, the photons have different energy levels right?

    And as a result of conservation of energy/mass, the photon/wave of light with a faster frequency has a higher momentum in a collision?

    Time for light "does not exist" from an outside observer but if we were to become the photon/light wave, time for everything else would not exist/stop/be really slow as per special relativity. This might allow the light wave/photon to go at any speed it was allowed to by the energy it has. When we return to our reference frame on earth, we see that the light wave must be at/under the speed of light. All I am suggesting is that this might manifest itself as photons of higher energies having higher frequencies, shorter wavelengths and more "momentum" in a collision.

    I'm sorry if I am blatantly wrong... :P still learning the stuff :)
  10. Oct 16, 2011 #9
    just read the sub faq for relativity, apparently if I look at a photon's rest frame everything I derive from it is meaningless :P
    ah well...
  11. Oct 16, 2011 #10
    Don't be sorry man :) Those questions are good, and tricky.

    You have 'black body radiation' and 'the photoelectric effect' defining light as quanta. And the modern way of discussing it is a lot like yours, referring to photons frequency for example. I wrote this somewhere else, but it might help see how I look at it

    I'm getting very confused reading you, it's easy to define a photon. It's intrinsically timeless, massless, expends no energy in its 'propagation', always following geodesics. Has no charge, meaning that it does not bend to any electric field. Will leave a recoil as it propagates from a source, due to the demand of symmetry of 'conservation of momentum'. Does not exist until measured, that just means that if compared to a ball coming at you in the air, a photon is no ball and does - NOT - leave itself to being observed, except in its annihilation.

    It has no acceleration, can't be defined as having a 'rest frame' as it never can be 'at rest', always moving at the invariant speed of 'c'. Unchanging in, and from, any frame measured. It is one part of the wave/particle duality that signifies light/radiation. Although some, using strategically placed out clocks in a accelerating frame, may find 'c' to be questionable comparing A-B against B-A that is a expression of 'gravity' acting on the clocks, as per Einsteins GR (equivalence principle). Even if ignoring that you will find light to always show 'c' locally, when measured. And if you want to discuss the whole spectrum use 'radiation' please. Leave hypothesizes outside the definitions, you might want to make a section called 'alternative hypothesizes' if you can't control those introducing 'pilotwaves', quantum teleportations, tunnelings, entanglements and whatever..

    As Birge said "I understand what people are getting at when they want to say a photon can be expressed by a 3D k-vector and a polarization state, but nobody has offered a good explanation of why it's helpful to readers to consider the number of free parameters in a spatially infinite plane wave to be the degrees of freedom of a single photon." It would be highly theoretical, not defined experimentally, and embarrassingly stupid to put such into a factual description of what we think us to actually 'know' about a photon. A photon does not exist, except in the recoil, aka symmetry with SpaceTime, and in its subsequent annihilation. The ball can also be defined this way actually, if you consider what communicates its motion, namely 'light'. But that would just mess with peoples heads. Concentrate on what we know, not what we guess.

    "What exactly is the nature of the photon's frequency" The photon does not have a frequency. It has a energy, but that energy can through Einstein explanation of the photoelectric effect, via Planck's constant 'h', (E=hf) be presented as that 'e'nergy of a photon is proportional to a waves 'f'requency. So there is a equivalence through that. And equivalences and symmetries are important phenomena in SpaceTime, but intensity and amplitude doesn't apply to a single photon at all. If you look at Maxwell's equations light becomes a electromagnetic radiation consisting of oscillations (waves) in the electric and magnetic fields, 'perpendicular' (at a right angle) to each other. Waves describe polarization, refraction, interference (quenching and reinforcing itself, via two waves interfering) etc, but they do not tell you about photons. And that's where 'equivalences' becomes important. And so this, to me that is, is all about trying to find a common ground for the concept of photon fitting the concept of waves. This is a good description of that.

    "The frequency of the oscillations in a beam of light is proportional to the energy in each photon, as demonstrated by the photoelectric effect, and in the case of light is related to the color of the light. The intensity of the beam is proportional to the number of photons. The polarization of light (that is explained by Maxwell) is related to the quantum-mechanical concept of spin. You can see the photon as a little top spinning around an axis that coincides with the direction of propagation. But while in classical mechanics an object can spin only in one direction at a time, in quantum mechanics you have the paradoxical and counter-intuitive fact that an object can spin lets say clockwise and counterclockwise at the same time.

    It is like having two "realities" existing at the same time. It takes a while to get used to this new idea and to accept it. A photon spinning in one direction corresponds to a rotating electric field, and to what is called circular polarization. A photon that spins in both directions at the same time gives you, under the right circumstances, plane polarization, which means the electric field is oriented always in the same direction."

    Two realties huh :) Or a particle/wave duality
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