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

Where's the mass of an electric field?

  1. Jul 18, 2012 #1
    Hi all,

    This is a simple question: if I have a charged particle, and there is a mass [itex]E=mc^2[/itex] associated with the energy of the electric field, where is this mass?

    Some points I'm unsure of:

    1. Presumably one can integrate the energy density over an infinite volume to get the total stored energy, and calculate the mass. However, is this finite?

    2. Part of the problem is that the electric field is infinite at [itex]r=0[/itex] --- how is this reconciled? Quantum mechanically I presume...?

    3. If there is a mass in the electric field, how much can be attributed to the particle? If I am to measure the weight of the electron, how much would it differ compared with an "uncharged electron"?

    Cheers
     
  2. jcsd
  3. Jul 18, 2012 #2

    bcrowell

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    No, it blows up to infinity.

    Yes, this is one of the issues that QED takes care of.

    I think the answer is that QED isn't self-consistent if you try to put in uncharged fermions: http://en.wikipedia.org/wiki/Introduction_to_gauge_theory#Gauge_bosons (You certainly can't find the mass difference by integrating the energy of the electric field of a point particle, because that's infinite.)

    This question would probably attract better answers than mine if moderators moved it to High Energy, Nuclear, & Particle Physics.
     
  4. Jul 18, 2012 #3

    Mentz114

    User Avatar
    Gold Member

    From a GR point of view, the energy and momentum of the electric field is distrubuted throughout the field. The energy-momentum tensor of the field of a point particle is ( spherical polar coordinates)
    [tex]
    T_{\mu\nu}= \frac{8\pi G}{c^2} \left[ \begin{array}{cccc}
    \frac{{c}^{4}\,{Q}^{2}}{2\,{r}^{4}} & 0 & 0 & 0\\\
    0 & -\frac{{c}^{2}\,{Q}^{2}}{2\,{r}^{4}} & 0 & 0\\\
    0 & 0 & \frac{{c}^{2}\,{Q}^{2}}{2\,{r}^{2}} & 0\\\
    0 & 0 & 0 & \frac{{c}^{2}\,{sin\left( \theta\right) }^{2}\,{Q}^{2}}{2\,{r}^{2}}
    \end{array} \right]
    [/tex]
    which can be interpreted as energy density = 4πc2GQ2/r4 and momentum terms.
     
  5. Jul 18, 2012 #4
    Thanks for clarifying it's infinite.

    Do you know how to ask a moderator to move the thread?
     
  6. Jul 18, 2012 #5

    bcrowell

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I've done it. The way to do it is by using the red Report button.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Where's the mass of an electric field?
  1. Higgs Field and mass? (Replies: 38)

Loading...