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Relativistic mass of a photon

  1. Mar 3, 2005 #1
    what is the relativistic mass of a single photon
     
  2. jcsd
  3. Mar 3, 2005 #2
    Try using mass-energy equivilence.

    [tex] E = h f = m c^2[/tex]
     
  4. Mar 3, 2005 #3
    How do I plug in those variables?
     
  5. Mar 3, 2005 #4

    dextercioby

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    What do you mean...?"c" and "h" are fundamental constants.They're known.U need either the photon's energy (E),or its frequency (f),or its wavelength ([itex]\lambda[/itex]).Since the photon is a particle,the last 2 #-s refer to the classical em.wave...So the photon is described properly only by E (energy).(actually momentum & helicity in quantum treatment).

    Daniel.
     
  6. Mar 3, 2005 #5
    I said relavistic mass, not energy. Can someone give me an example? Or is that too much to ask?
     
  7. Mar 3, 2005 #6

    dextercioby

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    Relativistic mass (for a photon,at least) IS energy.In Heaviside-Lorentz units...

    Daniel.
     
  8. Mar 3, 2005 #7

    russ_watters

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    Visible light is on the order of 500nm, the speed of light is 300,000km/sec, and Planck's constant is 6.626068 × 10-34 m2 kg / s. Plug, chug.
     
  9. Mar 3, 2005 #8

    rbj

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    nothing is too much to ask. but you might want to accept it when people answer.

    the only mass property of a photon comes from

    [tex] E = m c^2 = \hbar \omega[/tex]

    resulting in

    [tex] m = \frac{E}{c^2} = \frac{\hbar \omega}{c^2}[/tex]

    that's it. the mass of a photon. because the speed of a photon in vacua is [itex] c [/itex] for all observers and because the relationship of relativistic mass to rest mass for any particle is

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

    where [itex] m_0 [/itex] is the rest mass and [itex] v [/itex] is the particle's velocity relative to some observer, if there was any non-zero rest mass, the particle's relativistic mass would be [itex] \infty [/itex] not [itex] \frac{\hbar \omega}{c^2} [/itex], so the rest mass of a photon has to be zero.

    we can't spell it out any further for you, as best as i can tell.

    r b-j
     
    Last edited: Mar 3, 2005
  10. Mar 3, 2005 #9
    Russ_Watters, you said
    As far as I know, units of speed are always in Meters/Second, so why are you using KM?
     
  11. Mar 4, 2005 #10

    dextercioby

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    There are exceptions.So-called tolerated units.In the case of velocity:km/h,km/s...For pressure mm col.Hg/Torrs,etc.


    Daniel.
     
  12. Mar 4, 2005 #11
    That's incorrect. relmass is not idetical to energy. This is equality under some specfc cases

    The relativistic mass ofa photo can be expressed in two ways one of which is m = p/c.


    pete
     
  13. Mar 4, 2005 #12

    dextercioby

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    I'm not gonna argue with you,Pete,but don't u think it's weird that u're the only person who has this opinion...?

    Daniel.
     
  14. Mar 4, 2005 #13
    How is m = p/c for a photon ??? isn't m divergent here ??? How do you renormalize this ???

    What is the second expression ???

    marlon
     
  15. Mar 4, 2005 #14
    thats a pointless equation

    <p> = d/dt(m*x) = mv

    m = mv/c, v=c, m=m
     
  16. Mar 4, 2005 #15
    http://math.ucr.edu/home/baez/physics/Relativity/SR/mass.html

    this should explain a lot
    the (invariant or rest-) mass of a photon is ZERO, c'est tout, point final
    the relativistic mass depends on the frame of reference, but since all frames of reference are equivalent, only the invariant mass matters...


    marlon
     
  17. Mar 5, 2005 #16
    No. I don't find it weird at all. If I see Rindler state this in no uncertain terms in his text then who am I to argue with Rindler?

    Besides, this is something that one can demonstrate rather easily by a simple example which I've stated. The only think weird is that nobody chose to actually look at the problem I stated and then find the energy and mass.

    You'd agree that a photon has momentum, 'p', would you not? You'd also agree that the speed of a photon is c, do you not? Then the relation m = p/c is well defined.

    The energy of a photon is related to the photons momentum, 'p', by E = pc --> p = E/c. Plug this into "m = p/c" and you'll get m = E/c^2 which is fairly obvious. The relation m = p/c is found in many texts on relativity. E.g. see French's text "Special Relativity" on page 16 where in a foot note French states

    *By inertial mass we mean the ratio of linear momentum to velocity.
     
    Last edited: Mar 5, 2005
  18. Mar 5, 2005 #17

    pervect

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    I was reviewing this whole argument, when I ran across Steve Carlip's paper

    http://lanl.arxiv.org/abs/gr-qc/9909014

    A general argument is presented that for a closed physical system to which the virial theorem applies, the total "gravitational mass" of the system is going to be the total energy of the system, E - the sum of mc^2 for the constituent particles plus the total energy E

    Carlip's "gravitational mass" appears to be identical to pmb's "relativistic mass", it's the mass that appears in the low velocity limit of p=mv or f=ma.

    Therfore I think it is quite reasonable to say that relativistic mass is just the energy of a system. The only caveat is that the system has to be closed.
     
  19. Mar 6, 2005 #18
    By the way, besides being in such texts by, say, Rindler and perhaps Mould and it can also be found in Einstein's paper on this from 1906 as well as currently in journals such as the American Journal of Physics. I recall one article called Mass renormalization in classical mechanics by Griffith. He also noted the difference between E/c2 and p/v. What I'm surprised at is that more people aren't familiar with this? Have you ever read Rindler's intro SR text?

    I created a page a while back as an example. See

    http://www.geocities.com/physics_world/sr/inertial_energy_vs_mass.htm

    Pete
     
  20. Mar 7, 2005 #19

    pervect

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    In what manner does your webpage conflict with Carlips paper, or are you not disagreeing with Carlip?
     
  21. Mar 7, 2005 #20

    russ_watters

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    I use km because its a prettier looking number. Besides, "kilo" is just a prefix, meaning "thousand" - so km are still meters (dropping the "k" and adding 3 zeroes is not a conversion) - its a very common simplification. I would hope you could reconcile the units when plugging into the equations...

    And its not polite to be argumentative when asking a question and people are trying to help you.
     
    Last edited: Mar 7, 2005
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