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How does this formula relate to particle creation?

  1. Feb 1, 2013 #1

    ΔEΔt ≥[itex]\frac{h}{4\pi}[/itex]

    How does this formula relate to particle creation? I understand that it is relevant to the uncertainty principle, but that is essentially all that I am aware of. Does this formula indicate that when a particle with energy 'E' exists for time 't' it then decays?

    Last edited: Feb 1, 2013
  2. jcsd
  3. Feb 1, 2013 #2


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    Both are related to quantum physics, but I don't see a direct relation between particle creation and this uncertainty relation.
    For short-living particles, the uncertainty relation gives them a natural width in their mass.
  4. Feb 1, 2013 #3
    What this formula means is that if a particle only exists for a short time ##\Delta t##, then its energy is necessarily uncertain by an amount ##\Delta E = h / 4 \pi \Delta t##. For example, the mass of the rho meson is nominally 770 MeV. However, the rho meson is extremely short-lived: ##\Delta t \approx 4 \times 10^{-24}## seconds, so we actually observe rho mesons with a range of masses, with the extent of that range being about ##\Delta E = 145## MeV. So it wouldn't be uncommon to observe a rho meson with mass 700 MeV, for instance.

    See http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/parlif.html
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