Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

How to apply e=mc^2 to a photon?

  1. Jan 11, 2007 #1
    Hey can sum1 please tell me if we can apply e=mc^2 to a photon?
    Also if the energy of a photon is relative

    and speed of light isnt relative and e=mc^2*lorentz factor

    then how is energy of a photon relative in the above Case?
     
  2. jcsd
  3. Jan 11, 2007 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    We can, if we specify that the "m" is not the rest mass which is 0 for a photon. It's the the so-called "relativistic mass" (very troublesome term). The energy of a photon relative to what ? To velocity, to momentum, spin??

    Daniel.
     
  4. Jan 11, 2007 #3
    Relative to observers travelling at different velocities
     
  5. Jan 11, 2007 #4
    and can we use the lorentz factor for a photons energy?As the apparent frequency might change!
     
  6. Jan 11, 2007 #5

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Yes, the energy is of course relative to the motion of various observers trying to measure the energy of the photon. If you know what a 4 vector and and a Lorentz boost are, then the explanation is obvious.

    Daniel.
     
  7. Jan 11, 2007 #6
    No i do not about Lorentz boost and 4 vector.Could you please explain them to me?
     
  8. Jan 11, 2007 #7
    And does magnetism take place due to photon exchange?
     
  9. Jan 11, 2007 #8

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Four vectors in relativity are just a particle's space-time coordinates or energy momentum expressed as a vector. For example, the energy and momentum of a particle could be expressed thus;

    [tex]\vec{P}=\left[\begin{array}{c}
    E\\
    p_{x}C\\
    p_{y}C\\
    p_{z}C
    \end{array}\right][/tex]

    Or more succinctly;

    [tex]\vec{P}=\left[\begin{array}{c}
    E\\
    \vec{p}
    \end{array}\right][/tex]

    More information can be found here.

    The Lorentz boost is just a Lorentz transformation, which "boosts" in a given direction, For example a "boost" in the x-direction and can also be expressed in matrix form, thus;

    [tex]\begin{bmatrix} c t' \\ x' \\ y' \\ z' \end{bmatrix} = \begin{bmatrix} \gamma&-\beta \gamma &0&0\\ -\beta \gamma&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \end{bmatrix} \begin{bmatrix} c t \\ x \\ y \\ z \end{bmatrix}[/tex]

    Yes, in QED the electromagnetic exchange force arises from the exchange of virtual photons.
     
    Last edited: Jan 11, 2007
  10. Jan 11, 2007 #9
    photon, rest mass and relativistic mass

    The high frequency at which that question appears on the Forum, shows that we should not fully give up the concept of relativistic mass!
     
  11. Jan 11, 2007 #10

    ranger

    User Avatar
    Gold Member

    The large degree of confusion caused by relativistic mass proves otherwise. The only mass we should talk about is invariant mass and only speak of how energy and momentum increases with velocity.
     
  12. Jan 11, 2007 #11
    I do not understand how the lorentz factor can be applie to e=mc^2 for a photon,as the speed of a photon aint relative.And please tell me more about exhnage of virtual photon if possible!
     
  13. Jan 11, 2007 #12
    What is difference between
    m=m(0)g(V)
    E=E(0)g(V)
    and
    p=g(V)vE(0)/cc ?
     
  14. Jan 12, 2007 #13

    Hootenanny

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Again, all these threads seem to drift back to invariant vs. relativistic mass. I agree here with ranger, relativistic mass only tends to add confusion to a subject, in particular when teaching/explaining such phenomena to a semi-mathematically literate audience. I will say however, that relativistic mass is can be used in SR if applied correctly. BUT relativistic mass certainly has no place in GR.
     
    Last edited: Jan 12, 2007
  15. Jan 12, 2007 #14
    relativistic mass

    I fully aggree with that conciliatory point of view in SR. The problem is to apply correctly!
     
  16. Jan 12, 2007 #15

    ZapperZ

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor

    This has got to be THE most frequently asked question here on PF. It is obvious that even with the various FAQ's all over the 'net, this is STILL going to be asked on here.

    Can I get a volunteer to write something to address this type of question for our own FAQ in the General Physics forum? It should be tailored to the level that people who often ask this type of question can understand (see the existing FAQ there), and it should be tailored for the specific type of issues that we continue to see on here.

    This can be done either by an individual, or via collaboration. Please PM me if you're interested. This will not stop people from asking, but at least, we can point very easily to a link on here and not have to go through ALL of this "pain and suffering" again. :) At the very least we don't have to rewrite the same thing, and go on with a discussion beyond that.

    Zz.
     
  17. Jan 12, 2007 #16

    jtbell

    User Avatar

    Staff: Mentor

    The momentum-energy four-vector transforms in the same way as the position-time four-vector. In Hootenanny's matrix version of the Lorentz transformation, simply replace the position-time four-vector with the momentum-energy four-vector:

    [tex]\begin{bmatrix} E' \\ {p'}_x c \\ {p'}_y c \\ {p'}_z c \end{bmatrix} = \begin{bmatrix} \gamma&-\beta \gamma &0&0\\ -\beta \gamma&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \end{bmatrix} \begin{bmatrix} E \\ p_x c \\ p_y c \\ p_z c \end{bmatrix}[/tex]

    This works for any particle, including photons.
     
  18. Jan 12, 2007 #17
    Photons dimensions

    Is a photon 2d or 3d or 4d.As at c ,time shudnt pass for it.But when it goes thru a refracting medium,duz it enter time.Also

    please tell me about its dimensions
     
  19. Jan 12, 2007 #18
    Last edited: Jan 12, 2007
  20. Jan 12, 2007 #19
    anant, if I understand what Einstein was thinking, then the potential of the photon would have linear velocity c and angular velocity c.
     
  21. Jan 12, 2007 #20

    jtbell

    User Avatar

    Staff: Mentor

    Eh, what? :confused:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: How to apply e=mc^2 to a photon?
  1. How to derive e=mc^2 (Replies: 2)

  2. How does E=mc^2 ? (Replies: 21)

  3. E=mc^2 paradox (Replies: 2)

  4. Photon Mass in E=MC^2 (Replies: 10)

Loading...