- #1

fizzle

- 46

- 1

*effective velocity*:

$$\beta = \frac{ h \nu_0 } { h \nu_0 + m_0 c^2 }$$

What's really interesting is that if you calculate the de Broglie wavelength for an electron moving at that speed, it's the same as the Doppler shifted wavelength of the incident electromagnetic wave:

$$\lambda = \sqrt{ \frac{2 h \nu_0 + m_0 c^2 }{ m_0 c^2 } } \lambda_0$$

It's seems odd that these two values are identical. Is this just numerology or can we gather any physical information from it?

Of course, another numerology result is that the de Broglie wavelength at a given speed is identical to the

*beat frequency "wavelength"*of two Doppler-shifted incident electromagnetic waves of the Compton wavelength (from the front and behind).