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Photons, mass, and energy?

  1. Apr 3, 2013 #1
    I am wondering about photons having mass. The new thing on all the space shows is the sailing through space with a sail that harnesses photons as a ship harnesses wind. They did an experiment and the photons exerted a force on the sail. Am I wrong to think that E=mc^2 (E can not be 0 so M can not be 0) and since the photons exert kinetic energy on the sail I can assume that KE=1/2MV^2 therefor KE (how its exerting force on the sail) is = 1/2Mc^2. Since the energy is not 0 then the mass can not be 0, calculate the number of photons and the amount of force and you will have the mass of a photon.

    Am I wrong to think this?
  2. jcsd
  3. Apr 3, 2013 #2


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    You are wrong. E=mc^2 only applies to objects with non-zero rest mass and photons have a zero rest mass. The equation that you're looking for is [itex]E^2=(mc^2)^2 + (pc)^2[/itex] where p is the momentum and m is the rest mass; this one works even if the rest mass is zero, as it is for a photon.

    This equation allows the photon to carry momentum, which is transferred to the sail when the photon hits it.
  4. Apr 3, 2013 #3


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    Please read our FAQ subforum.

    https://www.physicsforums.com/showthread.php?t=511175 [Broken]

    Last edited by a moderator: May 6, 2017
  5. Apr 4, 2013 #4
    ... and which are stationary in the frame of reference considered.
    Last edited: Apr 4, 2013
  6. Apr 4, 2013 #5


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    even though it is deprecated in most, nearly all, current texts, there is most certainly a usage to the famous [itex]E=mc^2[/itex] where the [itex]m[/itex] refers to the (oft deprecated) "relativistic mass":

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

    and [itex]m_0[/itex] is the "rest mass" or "invariant mass" of the body. a photon is assumed to move at a speed of [itex]c[/itex] in anyone's frame of reference, so it cannot have a non-zero rest mass because

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

    and the latter factor goes to zero when [itex] v=c [/itex].

    if [itex]m[/itex] used in [itex]E=mc^2[/itex] is the expression above, then [itex]E=mc^2[/itex] is perfectly consistent with

    [tex]E^2=(m_0 c^2)^2 + (p c)^2[/tex]


    [tex] p = m v [/tex] .

    this [itex]E[/itex] is the total energy, the "rest energy" [itex]E_0=m_0 c^2[/itex] (which is the "E=mc^2" you're referring to) plus the kinetic energy (from the POV of the observer in the frame of reference).

    [tex] E = m c^2 = E_0 + T [/tex]

    the kinetic energy

    [tex] T = E - E_0 [/tex]

    goes to the classical approximation [itex] T = \frac{1}{2}m_0 v^2 [/itex] when [itex]v<<c[/itex].

    a quantity of the dimension "mass" can be derived from dividing the momentum of the photon by its speed (which is c).

    "wrong" and "currently deprecated" are not the same thing.
    Last edited: Apr 4, 2013
  7. Apr 4, 2013 #6


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    A better characterization would be "highly misleading". One need only count up the queries here on PF that were produced as a result of this confusion. Many posters have imagined, for example, that the relativistic mass increase must eventually turn the particle into a black hole.
  8. Apr 4, 2013 #7


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    point taken. a simple black hole needs to satisfy

    [tex] \frac{M}{r} > \frac{c^2}{2 G} [/tex]

    in its own reference frame.
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