- #1
AStaunton
- 100
- 1
Just a very quick question:
in my notes I have the velocity of a DeBroglie wave is given by:
\nu\lambda=2\pi\nu\frac{\lambda}{2\pi}=\frac{\omega}{k}=\frac{E}{p}=\frac{c^{2}}{v}
I can't figure out how he want from E/p=c^2/v.
I think the assumption is made that E is approximately equal to mc^2+(1/2)mv^2 and p=mv:
\implies\frac{E}{p}=\frac{m_{0}c^{2}+\frac{1}{2}m_{0}v^{2}}{m_{0}v}
but I still can't see how the above simplifies to c^2/v!
in my notes I have the velocity of a DeBroglie wave is given by:
\nu\lambda=2\pi\nu\frac{\lambda}{2\pi}=\frac{\omega}{k}=\frac{E}{p}=\frac{c^{2}}{v}
I can't figure out how he want from E/p=c^2/v.
I think the assumption is made that E is approximately equal to mc^2+(1/2)mv^2 and p=mv:
\implies\frac{E}{p}=\frac{m_{0}c^{2}+\frac{1}{2}m_{0}v^{2}}{m_{0}v}
but I still can't see how the above simplifies to c^2/v!
