laurelelizabeth said:
So if they are massless p must not =mv ... I'm assuming that p=mv is for bigger slower things.
As with kinetic energy, momentum has a more general form given by:
\vec{p}=\gamma m\vec{v}
where \gamma is again the Lorentz factor. If you use the same power series expansion as cristo described earlier, then this reduces to the classical result for momentum (just as kinetic energy did):
\vec{p}=m\vec{v}
Notice that the classical result implies a massless particle has no momentum. However, the more general and accurate equation states that the momentum is indeterminate (0\0) for massless particles iff their velocity is the speed of light. Which, as it turns out, is always the case!
Consequently, we cannot use this equation to determine the momentum of massless particles either. So where does the result p=\frac{E}{c} come from? As it turns out, there is another important equation from relativity that relates the energy of a particle to its momentum:
E^{2}=(pc)^{2}+(mc^{2})^{2}
and it is from this that we are finally able to obtain our result for massless particles. I should warn you though, that this equation actually sufferes from the same 0/0 fiasco when applied to massless particles. You just can't tell when it is written in this polished form. The actual (and rarely seen) form is:
E^{2}=(pc)^{2}+\left(\frac{mc^{2}\gamma}{\gamma}\right)^{2}
By ignoring the \frac{\gamma^{2}}{\gamma^{2}} factor, technically, we are implying that the invaraince of our 4-momentum applies to all reference frames, including even those frames moving at the speed of light. This claim is definitely overstepping our bounds. Hence this equation can only
suggest that p=\frac{E}{c} is true. It is experiments that have actually born this out (for photons and neutrinos anyways).
