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

Uncertainty and the Speed of Light

  1. Jun 29, 2005 #1
    Does the uncertainty principle apply to the speed of light? Do photons merely average the speed of light on large scales, but fluctuate on small scales? Or if you know the exact speed of a photon, does it mean that you can't ever know its position?

    It seems to me that if a photon did, for an instant, travel at slower than the speed of light, then it would need a mass, but then again I guess there would also be uncertainty in the mass of a photon within a certain time range? Could a photon, for an instant, travel faster than light? Would it then have negative mass?
  2. jcsd
  3. Jun 29, 2005 #2
    in QM, speed has nothing to do with momentum. [tex]\Delta{\rho}\Delta{x}\geq\frac{\hbar}{2}[/tex] says nothing about speed
  4. Jun 29, 2005 #3
    Doh! Ok gotcha, so since [tex]\rho=\frac{E}{c}[/tex] then it's merely uncertainty in the energy of the photon that causes the uncertainty in the momentum? That's pretty cool I guess, since uncertainty in energy and momentum become the same thing with light.
    Last edited: Jun 29, 2005
  5. Jun 29, 2005 #4
    i would have used [tex]\rho=\frac{\hbar}{\lambda}[/tex] to get the uncertainty to be one half the wavelength of the particle in question.
  6. Jun 29, 2005 #5
    Re: Uncertainty in c

    Good question--I'll toss out my understanding of the answer, and hopefully someone can correct me if I'm wrong!

    The momentum-position uncertainty principle doesn't affect the speed of light since the momentum of light is independent of the speed. [tex]p = \frac{h}{\lambda}[/tex].

    I guess the next natural question is then why is there no uncertainty principle for the speed of light? This, I believe, is because photons are massless (as you said). Thus, they travel on null trajectories in space-time. I suppose one could try to phrase this quantum mechanically, and ask why every observable eigenstate is also an eigenstate of the [observable] velocity of light, with eigenvalue [tex]c[/tex]. I'm not sure if this is a well posed question nor how to answer it.

    Perhaps it might also be helpful to think about this from a field theory perspective. Observable particles/fields are always on-shell, that is to say that they satisfy the energy-momentum relation [tex]p^2 = m^2[/tex]. However, "internal" processes in a Feynman diagram which are *not* observed do not have to satisfy this condition. When I was introduced to Feynman diagrams before taking QFT, this was explained roughly via some energy-time Heisenberg relation; we can violate energy conservation, but only for a short time (and a little longer if we violate it a little less). In some sense perhaps we could explain that internal photons that needn't be on-shell can have a different "speed" (perhaps more transparent if we interpret the Feynman diagram in position space and we can note that positions off-shell are giving nontrivial contributions to the integral) from [tex]c[/tex].

    I hope that was somewhat helpful--mamybe someone can clear up my own foggy ideas and misconceptions? :blushing:
  7. Jun 29, 2005 #6


    User Avatar
    Science Advisor
    Homework Helper

    In quantum field theory, the speed at which photons travel (which is what I guess you mean by "speed of light"), is subject to variation, at least for virtual photons (which is what the context of your question imlies). Despite this, there is no violation of causality, as is discussed in pretty much every QFT textbook.

    Here's a link to an explanation:
    http://encyclopedia.laborlawtalk.com/propagator [Broken]

    Last edited by a moderator: May 2, 2017
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook