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Mass of a photon

  1. Nov 3, 2008 #1
    One area of quantum theory/special relativity that I have a HUGE issue with, is the wave-particle duality of light. I firmly beleive that a photon is a particle and that some fundamental force is acting on it causing it to behave as a wave. From the little I've read into string theory it sounds like we are at least on the path to explaining this phenomenon.

    My question is about the relativistic mass of a photon. When an atom jumps down to it's ground state it releases a photon of energy [tex]E=\frac{hc}{\lambda}[/tex]. Since relativity relates energy to 'mass' wouldn't that make a photon have some kind of 'mass' dependent on its energy? I understand that theoretically a photon has a rest mass of zero and that because [tex]E^2=p^2c^2+m^2_0 c^4[/tex] that the photon has momentum but not mass, but since it has no rest mass wouldn't that also set it's momentum to zero? [tex]p=\frac{m_0 v}{\sqrt{1-\frac{v^2}{c^2}}}[/tex] or in that case would the momentum equation's rest mass be equal to it's relativistic mass?

    I have one more question that is related to my last question. Suppose that you are a rocket traveling at 3/4c. [tex]m=\sqrt{1+\frac{v^2}{c^2}}m_0 [/tex]. Where is the added mass? Is it distributed along the outside of the nose of the rocket like some sort of plasma surrounding it? Is it in the middle of the rocket? Is the added mass actually just excited atoms that cause it to appear to be heavier? Is the added mass affected by gravity? electromagnetism? aerodynamics?

    Also, I have heard about the concept of energy density, would anybody like to elaborate on what that is and if it relates?

    That should be more than enough questions for on night. Thanks in advance for any help!
     
    Last edited: Nov 3, 2008
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  3. Nov 3, 2008 #2

    Mute

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    That equation for momentum won't tell you anything about the photon's momentum, since although the numerator is zero, so is the denominator: v = c for a photon, so 1 - (v/c)^2 = 0, and hence that formula tells you p = 0/0, which is ill defined. Looking at [itex]E^2=p^2c^2+m^2_0 c^4[/itex] with m_0 = 0 tells you that E = pc.

    Many people prefer to think that the mass of an object is not what's increasing relativistically, but the momentum: The term [itex]\gamma m_0[/itex] (where [itex]\gamma = (1 - (v/c)^2)^{1/2}[/itex]) only really appears in equations like [itex]p = \gamma m_0 v[/itex], so you might instead consider that it's momentum which has a funny behaviour as v -> c instead of viewing the mass as increasing. Under that interpretation, there is no added mass - the momentum of the rocket is just doing something funny compared to our non-relativistic notions of momentum.
     
  4. Nov 3, 2008 #3

    Fredrik

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    You don't need string theory for that. The theory that defines the concept of a photon and describes its properties is a quantum field theory called QED (quantum electrodynamics).

    It certainly has energy, and if you simply write E=mc2, the quantity m that's defined by this equation has dimensions of mass. You could call that the "relativistic mass" of the photon if you'd like, but it's pretty pointless to do that.

    No and no. Your third equation only holds for massive particles. The momentum of a photon is given by the first equation. That means that the two terms on the right in your second equation are independent.

    I prefer not to use the concept relativistic mass at all, and just say that in a frame in which the rocket is moving with velocity v, the kinetic energy of the rocket is different from the energy it has in its rest frame by a factor of [itex]\gamma[/itex]. Note that this is just a difference between two coordinate systems. There's no "extra mass" that gets distributed around the rocket.

    It's just a function that describes how energy is distributed in space. If you integrate it over a volume, you get the energy contained in that volume.
     
  5. Nov 3, 2008 #4

    Fredrik

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    I think it should be added that because of E2=p2c2+m2c4 (where I'm using m to denote the rest mass), any change of the momentum must be accompanied by a change of energy. It's the energy change that can be interpreted as a change of mass.
     
  6. Nov 4, 2008 #5
    Anybody can make up a hypothesis: I may believe the sun is blue. The problem arises not with hypothesis but with their implications, predictions and experimental test results.

    If it looks like a duck, waddles like a duck and quacks like a duck, it probably IS a duck.

    If your hypothesis is correct, how would you explain double slit phenomena...say, for example, how the "fundamental force" would know when to turn on an off....I'm not interested, frankly, but you should be to test your own hypothesis.
     
  7. Nov 4, 2008 #6
    naty... http://arxiv.org/pdf/0809.0616v1 . Just because a belief in physics is generally accepted doesn't make it correct. I am not the only person to believe in a particle only description of light, some notable examples are newton and feynman. As far as i'm concerned, a particle is a particle, and a wave is a wave, and light is not a wave-particle. If you have a problem with my beliefs then by all means, express your feelings, but please use facts and not half-witted remarks about how not blindly following what other scientists might currently believe somehow makes my thoughts inferior to your time tested logic of following the crowd.

    If it looks like a fox and cry's like a baby is it a fox-baby? no, its a fishercat.
     
  8. Nov 4, 2008 #7

    Hootenanny

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    Your referenced paper discusses the double slit experiment using the particle-only model of the photon, but makes no reference to your so-called fundamental force. So let's cut the crap about half-witted remarks and stick to the physics - and by physics I mean proper physics, not simply misapplying elementary formulae.
     
  9. Nov 4, 2008 #8
    Ok, so it's just the momentum that changes and not actually the mass, that helps. What would that mean for say, a supersonic jet flying at 1/3c, from the jets frame of reference the added momentum should be coming from the air surrounding it, causing more drag as it moves faster, but from a stationary observer it is the jet that gains momentum which would overtake much of the drag. Which frame of reference would be correct?
     
  10. Nov 4, 2008 #9

    Vanadium 50

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    If you read the paper, it says:

    (emphasis mine)

    So in this model, you have photons that are learning. This makes them very different from the photons that we are familiar with.

    Newton lived 300 years ago and believed many things that we no longer do - like alchemy.

    Feynman believed no such thing.
     
  11. Nov 4, 2008 #10

    Vanadium 50

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    In the jet's frame of reference its momentum is zero.
     
  12. Nov 4, 2008 #11
    —Richard Feynman, QED: The Strange Theory of Light and Matter (1985), p. 15
     
  13. Nov 4, 2008 #12
    but the air around it is moving in the opposite direction, meaning it has increased momentum in the opposite direction...or am I missing something here?

    [edit] and by 'it' i mean the air around the jet
     
  14. Nov 4, 2008 #13

    Fredrik

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    Newton was proven wrong and you're very wrong about Feynman. He didn't believe in classical particles (which is what you have in mind). He believed in quantum particles, the kind that's defined by quantum field theories, and those have wave-like properties. Physics isn't about believing things by the way. It's about finding theories and verifying that they can correctly predict the results of experiments. The quantum field theory of light has been tested to a ridiculous degree of accuracy, and has passed every test. There are no classical particle theories of light that agree with experiments. Actually, the whole concept of a classical particle is based on ideas that have been proven wrong (very wrong), e.g. that the state of the "particle" can be represented by six numbers (corresponding to position and velocity).

    Your thoughts are inferior, but not because you're not "blindly following the crowd". They're inferior because you're thinking in terms of "beliefs" rather than in terms of theories and experiments.

    The fact that a theory agrees with all the experiments that we've been able to perform doesn't mean that it's correct (in the sense that it's an exact description of the real world). There may be a better theory that's just waiting to be found, but to suggest that it's going to be a lot like the ones that have been disproved by experiments is just silly.
     
  15. Nov 4, 2008 #14

    Hootenanny

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    What increased momentum? If the jet is travelling at constant velocity, it simply has momentum - there is no increased momentum.
     
  16. Nov 4, 2008 #15
    ----|\---- air
    === > jet
    ----|/---- air

    If the jet is moving at 1/3c lets call its momentum p, from a stationary observer it has momentum p and the air around it has momentum zero, but from its frame of reference its momentum is zero and the air's momentum is -p. Stop me if I'm wrong I'm not trying to go against everything people say I'm just trying to understand this concept. Now say it accelerates to 2/3c, it's momentum has now increased from a stationary observers point of view, but from the jet's frame of reference hasn't the momentum of the air moving in the opposite direction actually increased and not it's own?
     
  17. Nov 4, 2008 #16

    Hootenanny

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    If we're assuming that the jet doesn't disurb the air around it, and ignore the ejected gasses from the engines, then yes you are correct.
     
  18. Nov 4, 2008 #17
    Then wouldn't a stationary observer measure a faster speed of the jet than someone on the jet would?
     
  19. Nov 4, 2008 #18

    Hootenanny

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    Faster speed relative to what? Of course, from the POV of the jet pilot the jet's velocity is zero, whereas the velocity of the jet from the POV of an observer on the ground is of course non-zero.
     
  20. Nov 4, 2008 #19
    ok so from the jets point of view the air would be moving faster in the opposite direction, got it. Thanks for the help hootenanny, and i wasn't trying to piss everyone off by the way, i should have put more effort into understanding the concepts of QED before passing judgement on the topic, i'm sorry. I just don't appreciate being ridiculed for having an opinion, wether or not i was justified in having it.
     
  21. Nov 5, 2008 #20

    Vanadium 50

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    The thread went downhill when you accused us of "blindly following what other scientists currently believe" and just "following the crowd". That's pretty doggone offensive. And as it happens, you not only were incorrect, but were having problems with freshman physics.

    Nothing intrinsically wrong with that, lots of people struggle with freshman physics, but most people who struggle with it don't automatically conclude that the people who did ultimately succeed and go on to get doctorates are all wrong. And not just wrong - that they act stupidly.
     
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