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Mass (both relativistic and rest) of a photon

  1. May 4, 2009 #1

    chiro

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    I have read over the definition of the derivation of the energy relationship with momentum and rest mass and I am a little confused.

    By looking at the definition of relativistic mass we have the definition

    m = m0 / SQR(1 - v^2/c^2) where m0 is the rest mass.

    Now a photon always travels at speed c (part of the relativity axioms) so the speed must be c and therefore the rest mass must be zero.

    Now enter quantum mechanics. It turns out that the energy of the photon comes in "lumps" and the energy component of a photon is given by E = hf.

    Now i'm ok with these two.

    Now I also understand that mass != matter and that you can have something with mass and no matter thats fine.

    What I don't get is how you can suddenly use the equation E^2 = p^2c^4 + m0^2c^4 and get a non-zero rest mass or even non-zero relativistic mass and as such even a non-zero energy given the mass of the photon in any state should be zero from the relativistic formula. It seems that the E=hf expression is completely incompatible with mass in the first place. So as you can see I'm a little confused as to how a photon of light can have any energy at all under standard mechanics (although i understand it when dealing with quantum mechanics).

    For example if v = c then m0 must equal 0. Now any relativistic mass should by the definition of mass equal zero since both a) the rest mass is zero and b) the photon always travels at c which results in a).

    I understand the definitions of momentum and how you use differential and integral calculus to go from momentum to force to energy etc.

    Sorry for repeating this but it seems like there is an inconsistency in the whole system to me but i'm probably missing something so I'd welcome anyone to tell me what I'm missing.

    Many thanks to any responses.
     
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  3. May 4, 2009 #2

    malawi_glenn

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    can you demonstrate how you conclude that E^2 = p^2c^4 + m0^2c^4 will give you a non-zero rest mass? If m_0 = 0, then it is = 0... since the photon is moving at v=c, one must have p = E/c, and that will give you m_0 = 0.
     
  4. May 4, 2009 #3

    Doc Al

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    If m0 = 0 and v = c, the relativistic mass formula gives an undefined value of 0/0. It does not simply equal 0.
     
  5. May 4, 2009 #4

    chiro

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    Well doesn't it have to be zero for the calculation to make sense otherwise you will get an infinite mass?
     
  6. May 4, 2009 #5

    chiro

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    Actually I thought that you must have a non-zero rest mass to begin with. Sorry for the confusion.

    What i'm confused about is where a photon can have any mass at all in any situation. I acknowledge that the photon has to absolutely have a rest mass of zero otherwise it wouldn't make sense. (Maybe I should edit my original post).

    But what I don't understand is how you can get non-zero mass through m0 = E/c^2. Thats the part that gets me when we've already said that m0 = 0.

    If rest mass is zero and nonrelativistic mass is zero then I'm happy with that but otherwise I don't understand. Sorry for the confusion earlier.
     
  7. May 4, 2009 #6

    Doc Al

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    Doesn't what have to be zero?
     
  8. May 4, 2009 #7

    Doc Al

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    That equation isn't valid; The correct version was given by malawi_glenn.
     
  9. May 4, 2009 #8

    malawi_glenn

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    where was the equation m0 = E/c^2 introduced/derived??
     
  10. May 4, 2009 #9

    JesseM

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    Yeah, you can't have m0 = E/c^2 for a moving object, the equation E = m0*c^2 only works for an object at rest. For a moving object you have E = gamma*m0*c^2, where gamma = 1/sqrt(1 - v^2/c^2). If you want to think in terms of relativistic mass mR, this equation is E =mR*c^2. Also, note that with a little algebra it's possible to show that E = gamma*m0*c^2 is equivalent to E^2 = p^2*c^2 + m0^2*c^4, given the definition of the relativistic momentum p as p=gamma*m0*v.
     
  11. May 4, 2009 #10

    jtbell

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    In this case the relativistic mass equation gives

    [tex]m_{rel} = \frac {m_0} {\sqrt{1 - v^2 / c^2}} = \frac {0}{0}[/tex]

    which is undefined, not zero. This equation simply does not apply for a particle with zero rest mass.

    I suppose another way you could look at it is that the equation above gives

    [tex]0 \cdot m_{rel} = 0[/tex]

    which is satisfied by any value for [itex]m_{rel}[/itex]!
     
  12. May 4, 2009 #11

    chiro

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    Thanks for the feedback everyone I really appreciate it. My follow up question is if the photons mass is not zero when it is moving then what is it?

    I know the photon always moves at speed c: this is a no brainer. I also realize that the photon has energy related to its frequency: again this is a no brainer.

    What I have trouble with is the problem of finding the photons mass in a given situation. At rest its zero. But what about when its moving? Is it simply just undefined? or is there a way to show that its a finite and real quantity?
     
  13. May 4, 2009 #12

    JesseM

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    The photon's rest mass is zero at all times (and it's impossible for it not to be moving). Physicists these days usually prefer to avoid the concept of "relativistic mass" altogether and just talk about energy, since relativistic mass always obeys the equation E=mc^2 (assuming no potential energy, just rest energy and kinetic energy) and therefore relativistic mass is simply E/c^2. But if you wanted to define the relativistic mass for a photon, this equation combined with E=hf from quantum physics tells you it'd be hf/c^2.
     
  14. May 4, 2009 #13

    chiro

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    Thanks for the response. So essentially the mass (relativistic) of a photon is always non-zero (since its always moving) and is proportional to the frequency of the light? This is my last question.
     
  15. May 4, 2009 #14

    JesseM

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    For relativistic mass, yes, those statements are right.
     
  16. May 4, 2009 #15

    chiro

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    Cheers thanks everyone for your help
     
  17. May 4, 2009 #16
    Is 'The Photon' the same as 'a photon'?
     
  18. May 4, 2009 #17

    JesseM

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    Who are you asking this to? Did someone use the capitalized phrase "The Photon"?
     
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