2
$\begingroup$

If we were to measure the De-Broglie wavelength of a particle of mass $m$ and velocity $v$, in a given reference frame, it would be given as $$\lambda = h/p$$ (where $h$ is Planck's constant & $p$ is momentum)

Now if we were to measure the De-Broglie wavelength of the same particle from a reference frame moving with velocity $v$ the particle would seem stationary, so $p=0$ and its De-Broglie wavelength would be $$\lambda = h/0$$

So how exactly does a particle possess two different wavelengths when observed from two different reference frames?

$\endgroup$
4
  • $\begingroup$ Energy and momentum are relative quantities and wavelengths basically can be interpreted as the energy or momentum of a particle. Just like in when a particle moves in front of you it has say energy E, but if now you co-move alongside it then it's energy in that frame is 0 and your's too wrt yourself. Energy conservation is a global phenomena not local. $\endgroup$ Commented Oct 5, 2018 at 18:34
  • $\begingroup$ Makes sense, but there is no absolute reference frame, so how would you go about defining Global Phenomena? $\endgroup$
    – Kosh Rai
    Commented Oct 5, 2018 at 18:49
  • $\begingroup$ Then we need to set a reference frame. Without a reference frame, it cannot be defined in many cases. For example, It's not that gravitational potential at infinity is 0 but it is that we set the gravitational potential at infinity to be 0 but we could theoretically set it to any constant value which we like. $\endgroup$ Commented Oct 5, 2018 at 19:12
  • $\begingroup$ The de Broglie wavelength will change in different reference frames but what matters is the interaction you're interested in describing. Those other "actors" in the interaction will also experience a change concomitant with the change you describe so that, of course, the physics remain the same. Imagine a collision between two particles. If you're in the rest frame of one (and the de Broglie wavelength becomes badly defined), the other particle will have an adjusted wavelength that will result in the same outcome. $\endgroup$
    – Paul
    Commented Oct 5, 2018 at 19:19

1 Answer 1

1
$\begingroup$

The de Broglie frequency and wavelength of the "superluminar" plane phase wave accompaning a "subluminar" particle moving with speed $\:\upsilon\:$ are(1)

\begin{equation} \nu \boldsymbol{=} \dfrac{E }{h} \boldsymbol{=} \dfrac{\sqrt{\left(m_{\mathrm{o}}c^{2}\right)^{2} \boldsymbol{+}\left(pc\right)^{2}}}{h} \boldsymbol{=}\dfrac{\gamma m_{\mathrm{o}}c^{2}}{h} \tag{01}\label{01} \end{equation}

\begin{equation} \lambda \boldsymbol{=} \dfrac{h }{p} \boldsymbol{=}\dfrac{h}{\gamma m_{\mathrm{o}}v} \tag{02}\label{02} \end{equation} where $\:\gamma=\left(1\boldsymbol{-}\dfrac{\upsilon^2}{c^2}\right)^{\boldsymbol{-}1/2}\:$ and $\:m_0\:$ the rest mass.

The plane phase wave "superluminar" speed is \begin{equation} \mathrm w=\lambda\,\nu \boldsymbol{=} \dfrac{c^{2}}{\upsilon} \tag{03}\label{03} \end{equation}

In the rest frame of the particle

\begin{equation} \nu_{0} \boldsymbol{=} \dfrac{E_{0}}{h} \boldsymbol{=} \dfrac{m_{\mathrm{o}}c^{2}}{h} \tag{04}\label{04} \end{equation} \begin{equation} \lambda_{0} \boldsymbol{=} \infty \tag{05}\label{05} \end{equation} \begin{equation} \mathrm w_{0} \boldsymbol{=} \infty \tag{06}\label{06} \end{equation} that is a uniform in space periodic in time vibration with frequency $\:\nu_{0}$.


EDIT :

responds to the OP's comment

I still don't know a lot of stuff in physics, the only thing I knew in your answer was the square root of the Lorentz gamma. Do you think you maybe dumb it down a little? – Kosh Rai

De Broglie hypothesis :

$^{\prime\prime}$The fundamental idea of [my 1924 thesis] was the following : The fact that, following Einstein's introduction of photons in light waves, one knew that light contains particles which are concentrations of energy incorporated into the wave, suggests that all particles, like the electron, must be transported by a wave into which it is incorporated... My essential idea was to extend to all particles the coexistence of waves and particles discovered by Einstein in 1905 in the case of light and photons.$^{\prime\prime}$ $^{\prime\prime}$With every particle of matter with mass $m$ and velocity $\upsilon$ a real wave must be 'associated'$^{\prime\prime}$, related to the momentum by the equation: \begin{equation} \lambda \boldsymbol{=} \dfrac{h}{p}\boldsymbol{=}\dfrac{h}{\gamma m \upsilon} \nonumber \end{equation} where $\lambda$ is the wavelength, $h$ is the Planck constant, $p$ is the momentum, $m$ is the rest mass, $\upsilon$ is the velocity and $c$ is the speed of light in a vacuum.

Louis de Broglie hypothesis is based on the Special Relativity theory and the associated Lorentz transformations. Note that this picture of a particle with velocity $\upsilon$ and its associated plane phase wave with velocity $\mathrm w=c^2/\upsilon$ is Lorentz invariant, that is the same in any inertial system.


(1) See my answer here : About de Broglie relations, what exactly is E? Its energy of what?.

$\endgroup$
1
  • $\begingroup$ I still don't know a lot of stuff in physics, the only thing I knew in your answer was the square root of the Lorentz gamma. Do you think you maybe dumb it down a little? $\endgroup$
    – Kosh Rai
    Commented Oct 6, 2018 at 10:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.