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

Total energy at zero rest mass

  1. Jun 10, 2009 #1
    According to my physics textbook, the equation [tex]E^2=\left(mc^2\right)^2+\left(pc\right)^2[/tex] suggests that a particle may have energy and momentum even when it has no rest mass, and that the total energy then is [tex]E=pc[/tex]. This strikes me as odd, since the relativistic momentum of a particle is given by [tex]p=\gamma mv[/tex], which is zero when the rest mass is zero. But that must mean that for a particle with rest mass zero, [tex]E=pc[/tex] must also be zero, and that a particle with zero rest mass can not have energy and momentum, but this is false, because photons are particles with energy and no rest mass.

    If anyone would explain this to me, I would be grateful.

    Thank you in advance.
  2. jcsd
  3. Jun 10, 2009 #2


    User Avatar
    Science Advisor
    Homework Helper

    Hi espen180! :smile:

    (have a gamma: γ :wink:)
    Why do you think that γmc = 0 when m = 0?

    At the speed of light, γ = ∞, so γmc doesn't have to be 0. :wink:
  4. Jun 10, 2009 #3


    User Avatar

    Staff: Mentor

    ...and v < c. If m = 0 and v = c, that formula gives p = 0/0 which is undefined. Therefore that formula doesn't apply to photons. The general energy-mass-momentum relationship, on the other hand, has no such trouble because it doesn't include the velocity explicitly.
  5. Jun 10, 2009 #4
    ∞ times 0 is 42, isn't it?
  6. Jun 10, 2009 #5


    User Avatar
    Science Advisor

    Wasn't that 6X9 :confused:
  7. Jun 11, 2009 #6
    I see. What does the Energy-mass-momentum relationship look like? Isn't that the formula I included above?
  8. Jun 11, 2009 #7


    User Avatar
    Homework Helper

    You mean this one?
    That's it all right...
  9. Jun 11, 2009 #8
    Not quite true.
    p = γm0v = βγm0c
    so pc = βγm0c2 = (E2 - m0c2)1/2
    so pc = E when m0c2=0
    [βγ goes to infinity as m0c2 goes to zero]
    α β γ δ ε ζ η θ ι κ λ μ ν ξ ο π ρ ς σ τ υ φ χ ψ ω
  10. Jun 11, 2009 #9


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Just to say it a bit more clearly: The formula [itex]\vec p=\gamma m\vec v[/itex] only holds for massive particles, while [itex]E^2=m^2c^4+\vec p^2c^2[/itex] holds for all particles. It even holds for tachyons ([itex]m^2<0[/itex]), but they probably don't exist anyway, so that's less relevant.
  11. Jun 12, 2009 #10
    Thanks for clearing it up, guys! :)


    But since infinity x zero is undefined, do we have to go via quantum physics and use pc=hf instead?
  12. Jun 12, 2009 #11
    You can if you wish. You could also simply write down E = pc.
  13. Jul 4, 2011 #12
    If a particle has zero rest mass, does it mean that the particle is a photon?
  14. Jul 4, 2011 #13


    User Avatar
    Science Advisor
    Gold Member

    Neutrinos "may" be massless, not sure yet. That's about all I can think of right now.
  15. Jul 4, 2011 #14


    Staff: Mentor

    Hi laeiy, welcome to PF!

    No, gluons also have 0 rest mass.
  16. Jul 4, 2011 #15


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It's not that it's undefined, it's that it's an informal verbal expression describing a limit that has to be evaluated properly. When you evaluate the limit properly, you get a well-defined real number. (Have you had calculus?) For example, [itex]\lim_{x\rightarrow 0}\left(\frac{1}{x}\right)(\sin x)[/itex] has a well-defined value of 1. You can also describe that limit informally as [itex]\infty \cdot 0[/itex], but that just means you've stripped out the information that would be needed in order to actually evaluate the limit. When you strip out the information in this way, the resulting "stripped-down" description is called an indeterminate form: http://en.wikipedia.org/wiki/Indeterminate_form

    No, you don't need quantum mechanics to resolve the issue.
  17. Jul 4, 2011 #16
    Nope. Or, strictly, no more than one of the three neutrino mass states may have a mass of zero. We know that this is necessarily the case because of neutrino mixing. There's an observable quantity known as the mixing length which depends on the difference between the squares of the masses of the mass states involved in the mixing. Different mixing processes have been show to have two significantly different mixing lengths, requiring no fewer than three different masses.

    Only in base 13. :wink:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook