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Photon Mass in E=MC^2

  1. Dec 16, 2014 #1
    A photon has “zero” rest mass, but it does have Energy,

    But with zero Mass , E=MC^2 yields E=0

    I am confused !
     
  2. jcsd
  3. Dec 16, 2014 #2

    ShayanJ

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    Last edited by a moderator: May 7, 2017
  4. Dec 16, 2014 #3

    PeroK

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    Your confusion is based on the false assumption that there is only one form of energy.

    The energy of a massive particle is actually

    ##E = \gamma mc^2##

    Whereas the energy of a massless photon is given by

    ##E= hf##
     
  5. Dec 16, 2014 #4

    PeterDonis

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    As the third article Shyan linked to says, ##E = m c^2## is not the correct general formula; the correct general formula is ##E^2 = \left( m c^2 \right)^2 + \left( p c \right)^2##. ##E = m c^2## is then the obvious special case of this for which ##p = 0##. But a photon can't have ##p = 0##, because it can never be a rest; it always moves at the speed of light, so it always has momentum. In fact, as you can see from considering the general formula when ##m = 0##, a photon's momentum is just its energy divided by ##c##.

    This is correct, but not relevant to the OP's question. The relevant formula for the OP's question, for a photon, is ##E = p c## (see above).
     
  6. Dec 16, 2014 #5

    PeroK

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    In the spirit of the OP's question:

    If ##p=\gamma mv## then the momentum of a photon would likewise be 0 (or undefined).
     
  7. Dec 16, 2014 #6

    ShayanJ

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    We have [itex] \lim_{v\rightarrow c} \gamma=\infty [/itex] and so for a photon, we'll have [itex] p=\infty \times 0 [/itex]. So this formula is compatible with [itex] p=\frac E c=\frac{hf}{c} [/itex].
     
    Last edited: Dec 16, 2014
  8. Dec 16, 2014 #7

    PeterDonis

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    But ##p = \gamma m v## is not valid for a photon. Just to be clear: the fully general formula is ##E^2 = \left( m c^2 \right)^2 + \left( p c \right)^2##. Any formula with ##\gamma## in it is not fully general; it only applies to objects with ##v < c##, for which ##\gamma## is finite.

    No, it isn't. ##\infty \times 0## is not well-defined mathematically; there's no way to make it spit out a particular finite value, much less the correct finite value for ##p## for a photon. ##p = \gamma m v## doesn't apply to a photon, period.
     
  9. Dec 16, 2014 #8

    PeroK

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    Exactly. So, how are you defining the energy or momentum of a photon? You can't use ##mc^2## or ##mv## because they both have mass in them. Yes, you can relate the two in ##E=pc## But that doesn't define either, except in terms of the other.
     
  10. Dec 16, 2014 #9

    PeterDonis

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    So what? The OP didn't ask how energy and momentum were defined for photons. He only asked how the photon's energy can be nonzero when it has zero invariant mass. That question has been answered.

    If you want to ask how energy and momentum are defined for photons, the answer is, the same way they are defined for anything else: energy is defined by the photon's ability to do work or transfer heat (for example, by being absorbed by an object and increasing its temperature), and momentum is defined by the photon's ability to exert force (for example, radiation pressure). The formula ##E = p c## then simply relates the amount of heat or work a photon can produce, to the amount of force it can produce. The more general formula I wrote down does the same thing for any particle, whether it has zero or positive invariant mass.
     
  11. Dec 16, 2014 #10

    ChrisVer

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    Well the photon has definite momentum and energy. The problem is the definition of [itex]\gamma[/itex] when you go to null-worldlines. The problem there is that [itex]\gamma[/itex] is not definite. Thus the energy can't be written as [itex]E= \gamma m [/itex] and momentum [itex]p_i= \gamma m u_i[/itex] (or in other words define them with using the rest mass).
    The photon however can have energy (and has energy) and momentum (for the whole one can check the energy-stress tensor which can be obtained from Maxwell's theory itself).
    Now let's say it has energy [itex]E[/itex] and momentum [itex]p[/itex]. It's 4 momentum is [itex]P=(E, p)[/itex] and the invariant object coming from this (as a null vector) is [itex]P^2 = E^2 - p^2 =0 \Rightarrow E= pc[/itex] (I inserted c to keep the SI units).

    If you apply the null-condition (or massless) for the four momentum, you will eventually reach the relation [itex]p_0^2=\vec{p} \cdot \vec{p}[/itex] no matter what you do. This is how a degree of freedom is lost/is not physical.
     
  12. Dec 16, 2014 #11


    Thanks, i think E=pc is more relevant answer to my question about the mass-less rest-less photon
     
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