3

I've been reading the answers to a few questions relating to force carrier particles, and it has been mentioned that massive force carriers always result in short-range forces, while massless carrier particles result in long-range forces. Why is this? Why don't the massive force carrier particles spread out in a similar manner to massless ones, just with a propagation speed slower than the speed of light?

Davis Broda
  • 130
  • 2
    If you do the calculation I outline in this post, you incur an additional factor $\mathrm{e}^{-\mu r}$ with $\mu$ as the mass of the massive vector boson. – ACuriousMind Mar 12 '15 at 14:34
  • Possible duplicates: http://physics.stackexchange.com/q/99618/2451 , http://physics.stackexchange.com/q/51827/2451 and links therein. – Qmechanic Mar 12 '15 at 16:59

1 Answers1

-1

Force carriers have a range told by the equation: $$ R≈h/4\pi mc $$ h is Planck's constant, m is mass and c is the speed of light This can be derived from the Heisenberg uncertainty principal: $$ \Delta E \Delta t ≥ h/4\pi $$ E is energy and t is time. This states that an uncertainty of energy can only be sustained over a limited amount of time depending on how big that uncertainty is, meaning you can "borrow" energy as long as you pay it back after a short period of time.

When dealing with mass carrying particles they can only exist for the time allowed by the uncertainty principal, meaning the maximum time one of these particles can exist is: $$ t=h/4 \pi E $$ $$ E=mc^2 $$ $$ t=h/4 \pi mc^2 $$ Now that we know the time, if we say that the particle moves at about the speed of light (particles with mass cannot move at exactly the speed of light, but they can get close), then using the equation for velocity we can find the approximate range of the particles: $$ v=d/t $$ $$ d=vt $$ $$ v≈c $$ $$ t=h/4 \pi mc^2 $$ $$ d≈h/4 \pi mc^2 × c $$ $$ d≈h/4 \pi mc $$ The notation uses R instead of d, but they mean the same thing.

Misc.nerdiness
  • 1,031