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If photons have no mass, then how can they travel the speed of light?

  1. Jul 18, 2008 #1
    im compfuzzled.
     
  2. jcsd
  3. Jul 18, 2008 #2

    CompuChip

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    Short answer: Precisely because they have no mass :smile:

    Long answer: By "no mass" you probably mean "no rest mass". Your question is almost a FAQ in these forums, I'm sure if you search you'll find loads of relevant threads (I recall in particular this recent one).
    [edit]Also this and this thread may be useful.[/edit]

    [edit (2)]I didn't know the word compfuzzled, but I like it. Does it actually exist?[/edit]
     
  4. Jul 18, 2008 #3
    thnx...so ,because its moving, it has energy(kinetic energy)?
     
  5. Jul 18, 2008 #4

    CompuChip

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    I don't think kinetic energy is the correct word, it's nothing like [itex]\frac12 m v^2[/itex] (of course, because there is a mass in there). The energy of a photon depends on its wave length. I added two more threads in an edit of my first post.
     
  6. Jul 18, 2008 #5
    okay i saw that. so it has energy because e=mc2...?
     
  7. Jul 18, 2008 #6

    Fredrik

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    Yes and no. Most of us prefer to write [itex]E^2=\vec p^2c^2+m^2c^4[/itex], where E is the energy, p is the momentum and m is the mass. Some people would write m0 instead of m on the right-hand side, and define m (yes, define is the right word) by saying that this is equal to [itex]m^2c^4[/itex]. They would call m the "mass" and m0 the "rest mass".
     
  8. Jul 18, 2008 #7

    Hootenanny

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    [itex]E=mc^2[/itex] is often quoted as Einstein's energy equation, but it isn't really the full equation. The full relationship for the energy of a particle is given by:

    [tex]E^2 = \left(m_0c^2\right)^2 + \left(pc\right)^2[/tex]

    Where [itex]m_0[/itex] is the invariant or "rest" mass of the particle and [itex]p[/itex] is it's momentum. The first term of the expression is sometimes called the "rest energy" of a particle, because that is the energy that the particle has as measured from a reference frame stationary relative to it.

    Now, photons have zero mass and therefore the first term disappears, i.e. the photon has no "rest energy". We are now simply left with:

    [tex]E = pc[/tex]

    So all the photon's energy is in fact due to the photon's motion, as you have said previously.
     
  9. Jul 18, 2008 #8
    i realised that wasntthe entire equation. i just shortened it to make it easy. is that taboo around here?lol
     
  10. Jul 18, 2008 #9

    Hootenanny

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    No not at all, I just thought that the full equation clarified the explanation somewhat. In fact your equation [itex]E=mc^2[/itex] can be considered complete if one considers [itex]m[/itex] as relativistic mass rather than invariant mass.
     
    Last edited: Jul 18, 2008
  11. Jul 18, 2008 #10
    okay...im only a high schooler who studies physics in their spare time, and ill be taking it as a classfor the first time next year. i hardly know calculus and all these equations.
     
  12. Jan 6, 2011 #11
    How to photons, that have no mass, travel in a wave?

    Sorry if the question is a bit simple, im not a physicist by profession. I seek to understand how a photon can move if it has 0 mass. Is it that photons aquire some sort of "mass" and thus can move (therefore fulfilling newtons laws)? I understand that we dont know what "mass" really is but I am perplexed as to how a photon moves in spacetime. Any answers would be appreciated.


    "Edit: How "do" photons, that have no mass, travel in a wave?
     
  13. Jan 7, 2011 #12


    So photon movement is not bound to Newtons laws?
     
  14. Jan 7, 2011 #13

    Pengwuino

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    No. Infact, because of how photons behave, we know Newton's Laws are incomplete/simply approximations.
     
  15. Jan 7, 2011 #14
    That's fine. The take home point is that relativity tells us that objects without a rest mass (which is generally the mass you think of in newtonian/classical physics) still have energy.
     
  16. Jan 7, 2011 #15

    tiny-tim

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    Welcome to PF!

    Hi Kimmurial! Welcome to PF! :wink:
    Newton's laws (and waves) require momentum, not mass (second law: force = rate of change of momentum).

    Photons have energy and momentum (but no rest-mass).

    Their momentum enables them to take part in collisions. :smile:
     
  17. Jan 7, 2011 #16
    Re: Welcome to PF!

    Thanks :)

    So you are saying that a photons ability to travel at the speed of light is attributed to their momentum but newtonian physics states p=mv.
    If photons have no mass then isnt p=0?
     
  18. Jan 7, 2011 #17

    DaveC426913

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    Re: Welcome to PF!

    Newton's laws are inadequate to apply to massless photons. That's what everyone's been saying.
     
  19. Jan 8, 2011 #18
    The root of the problem can be found in the closing line of The Nature of Mass by Max Jammer:

    "Thus in spite of all the strenuous efforts of physicist and philosophers, the notion of mass, although fundamental in physics, is as we noted in the preface, still shrouded in mystery".

    It follows that photon mass or any mathematical explanation using the term mass is an empirical explanation; it is the best known current explanation, but one that needs further work.
     
  20. Jan 8, 2011 #19

    tiny-tim

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    Newton's laws of collisions only require momentum.

    Newton's laws of collisions are completely valid. They accord with reality.

    Newtonian physics of space-time is wrong (or, if you prefer, inaccurate) … there is no point in trying to apply it to objects moving at or near the speed of light.

    Newtonian physics of space-time states p=mv. That is wrong (but accurate enough for speeds nowhere near the speed of light). There is no point in asking how a photon can fulfil a law that is wrong: it can't and it doesn't!

    The correct law for momentum is p = mv/√(1 - v2/c2) …

    if m = 0, p doesn't have to be zero if v/√(1 - v2/c2) is infinite, ie if v = c. :smile:
     
  21. Jan 8, 2011 #20

    Rap

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    Newtonian physics is not exactly correct. The momentum of a massive particle in special relativity is

    [tex]p=\frac{mv}{\sqrt{1-v^2/c^2}}[/tex]

    if you make the velocity (v) get closer to c and keep the mass constant, then the term in the denominator gets smaller and smaller and the momentum gets larger and larger. If you know that the momentum when v=c is finite, then as v gets closer to c, you must have the mass get smaller and smaller. You can see that only a massless particle can travel at the speed of light without having infinite momentum. The problem is that when v=c you have for momentum 0/0 which is undefined. Newtonian mechanics is WAY wrong at or near the speed of light, so you cannot use Newtonian definitions to figure out what is happening at or near the speed of light.
     
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