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Energy of light

  1. Oct 20, 2007 #1
    I have a doubt...energy of light is nothing but kinetic energy of light which is 1/2mc^2..but we know that E=mc^2..so where does the 1/2 gone?
     
  2. jcsd
  3. Oct 20, 2007 #2
    Welcome to PF, informtohagrid.
    Forget the 1/2, what is the 'm' for light?

    The energy of a photon is pc, where p is the momentum of the photon.

    More here.
     
  4. Oct 20, 2007 #3

    JesseM

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    The formula for kinetic energy is different in relativity than in Newtonian physics, instead of (1/2)mv^2 it's [tex](\gamma - 1)mc^2[/tex], where [tex]\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}[/tex]. Also, the general formula for total energy of a particle with nonzero momentum is actually [tex]E^2 = m^2 c^4 + p^2 c^2[/tex], where m is the rest mass and p is the momentum...for an object at rest this reduces to E = mc^2, for a photon m=0 so it reduces to E = pc. Also, the general formula is equivalent to the equation [tex]E = \gamma mc^2[/tex] as long as the velocity v in [tex]\gamma[/tex] is less than c (I can show the algebra for transforming the first equation into this one if you like), so if you define the "relativistic mass" M as [tex]\gamma m[/tex] then this can be written as E = Mc^2. And note that since the general formula is [tex]E = \gamma mc^2[/tex], and kinetic energy is defined as [tex]E = (\gamma - 1)mc^2[/tex], then the total energy is just the kinetic energy plus the rest energy mc^2.
     
    Last edited: Oct 20, 2007
  5. Oct 20, 2007 #4
    Thank you Neutrino...I understand that light has no rest mass but it has relativistic mass and momentum...doesn't light have kinetic energy?...do you mean since there is no rest mass it has no kinetic energy also.is this correct...
     
  6. Oct 20, 2007 #5
    It has radiation pressure which i suppose in an analog of kinetic energy. rather than exerting a force on things due to some sort of 'physical contact' of electric fields driven by intertia of mass into something else, rather photons will have their EM fields interact with another particles much like normal particles do, but there is no mass 'driving' these fields together, rather there is the self perpetuated motion of the photon.

    i think.
     
  7. Oct 20, 2007 #6

    Hi JesseM..i know the derivation of E=mc2 by Einstein..but how to get the derivation of E^2 = m^2 c^4 + p^2 c^2 ...and who derived it?
     
  8. Oct 20, 2007 #7

    JesseM

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    This page says it was Einstein who derived it:
    I don't know the exact method Einstein used to derive it or what paper it was, but you should find a derivation in any relativity textbook, and there are also two derivations on this page, look at the paragraphs above and below the equation labeled (3) about halfway down the page--that equation is equivalent to the equation for energy above (note that the author sets c=1 so it doesn't appear in the equations). Also, this paper lists 3 different approaches used in textbooks for deriving the expressions for momentum and kinetic energy in relativity, and adds a fourth one of the authors' own invention--remember that total energy is just rest energy + kinetic energy, so if you know the rest energy is mc^2 and you know the kinetic energy is [tex](\gamma - 1)mc^2[/tex], then that means the total energy must be [tex]\gamma mc^2[/tex] which a little algebra can show is equal to [tex]\sqrt{m^2 c^4 + p^2 c^2}[/tex] where [tex]p = \gamma m v[/tex].
     
  9. Oct 20, 2007 #8

    JesseM

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    Since the total energy of a particle in SR is just rest energy + kinetic energy, and a photon has zero rest energy, its kinetic energy is equal to its total energy E = pc.
     
  10. Oct 20, 2007 #9
    Thank you JesseM...
     
  11. Oct 21, 2007 #10
    The total inertial energy, E of any particle is the sum of its kinetic energy K and its rest energy E0. Therefore

    E = K + E0

    For a photon E0 = 0 so therefore the total inertial energy for a photon is

    E = K

    Best regards

    Pete
     
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