Sunday, 7 June 2020

wave particle duality - How can we calculate (relativisticly correct) ratio $lambda_e/lambda_p$ if proton and electron have same speeds


This question is about group/phase velocities and also De Brogilie wavelength.


What I would like to know is how to derive ratio $\lambda_e/\lambda_p$ ($\lambda_e$ and $\lambda_p$ are De Broglie wavelength for electron and proton) if we know that electron and proton have same velocities?


I know that when we say "velocity" $v$, this velocity is the same as "group velocity" $v_g$. So for start I can calculate ratio for group velocities:


\begin{align} \boxed{v_e = v_p} \longrightarrow \frac{v_{ge}}{v_{ge}} = \frac{v_e}{v_p} = 1 \end{align}


and similarly I can do for phase velocities $v_p$:



\begin{align} \frac{v_{pe}}{v_{pp}} = \dfrac{\tfrac{c^2}{v_{ge}}}{\tfrac{c^2}{v_{gp}}} = \frac{v_{gp}}{v_{ge}} = 1 \end{align}


But when i try to calculate the ratio for wavelengths I got stuck:


\begin{align} \frac{\lambda_{e}}{\lambda_{p}} = \frac{\tfrac{h}{p_e}}{\tfrac{h}{p_p}} = \frac{p_p}{p_e} = ~\longleftarrow \substack{\text{I got stuck here where i don't know}\\\text{how to use relation $\boxed{v_p = v_e}$}} \end{align}


Can anyone give me a hint? I was trying to use relation $p = \frac{\sqrt{{E_k}^2 + 2E_0E_k}}{c}$ but I got lost...



Answer



Let's try to express the de Broglie wavelength $\lambda_j$ of a particle in terms of its mass $m_j$ and velocity $v_j$. $$ (\frac{2 \pi}{\lambda_j})^2 = (\frac{p_j}{\hbar})^2 = \frac{1}{\hbar c^2}(E_j^2 - m_j^2 c^4) $$


Where we've used the relation $E^2 = p^2 c^2 + m^2 c^4$.


From here we note that $E_j=\gamma(v_j) m_j c^2$, where $\gamma(v) = \frac{1}{\sqrt{1 - \frac{v_j^2}{c^2}}}$ to see that: $$ (\frac{2 \pi}{\lambda_j})^2 = \frac{\gamma(v_j) - 1}{\hbar c^2} m_j^2 c^4 $$


Since $v_e=v_p$, we see the only term not common between the two particles is their rest mass and so we find that: $$ \frac{\lambda_e}{\lambda_p} = \frac{m_p}{m_e} $$


Which to me is a surprising result.



No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...