1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Uncertainty principle leads to superlumina ?

  1. Feb 28, 2014 #1
    According to the uncertainty principle Δp*Δx≥h/2pi,
    now suppose we measure a particle in a very tiny area(if x is tiny enough),
    s.t. Δp ≥ h/(2xpi) ≥ mc then v > c.
    But in fact, the velocity can not be faster than light.
    So how can we compromise these two statement?
  2. jcsd
  3. Feb 28, 2014 #2


    User Avatar

    Staff: Mentor

    Are you perhaps thinking p = mv? The relativistic momentum is
    $$p = \frac{mv}{\sqrt{1 - v^2/c^2}}$$
    in which v < c always.
  4. Feb 28, 2014 #3


    User Avatar

    Staff: Mentor

    Hi for_Higgs, welcome to PF!

    First, the equation gives you the uncertainty in the momentum, not the value of the momentum.

    Second, for velocities close to the speed of light, the change in mass due to velocity has to be accounted for. You have to use the relativistic momentum, such that ##p \rightarrow \infty## is the same as ##v \rightarrow c##, so the particle never exceeds the speed of light. [Edit: jtbell beat me to it]
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook