Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It only takes a minute to sign up.
Sign up to join this community
I tried to combine the mass–energy equivalence for a particle with mass,
$$E=\sqrt{(mc^2)^2+(pc)^2}=\sqrt{(mc^2)^2+(\gamma mvc)^2}$$
with de Broglie wavelength,
$$\lambda=\dfrac{h}{p}=\dfrac{h}{\gamma mv}. $$
I get this equation:
$$E=\dfrac{hc^2}{\lambda v}.$$
This does not seem right, since the equations suggest the energy increases as the speed slow down which is not the case. But I can't see what I did wrong, either. Can someone help me?
Your algebra is correct: $$ \frac1{\lambda v} = \frac {\gamma m}h \\ \frac{hc^2}{\lambda v} = \gamma m c^2 = E $$ Your interpretation,
the equations suggest the energy increases as the speed slow down, which is not the case
is not correct because both $v$ and $\lambda$ change as energy and velocity increase.
The wavelength $\lambda$ cannot stay constant if the velocity $v$ changes. The question is "do they change proportionally or not?" A quick examination of your deBroglie wavelength relationship shows the following: $$\lambda v = \frac{h}{\gamma m}.$$ We know that $\gamma$ is always greater than or equal to 1 and increases non-linearly with increasing $v$. Therefore, your denominator in the final expression of $E$ is decreasing non-linearly as $v$ increases, resulting in an increasing energy.
