From De Broglie's equation $$\lambda=\frac{h}{mv}$$ We can see that $\lambda$ depends inversely on $m$

My question is:

If the $m$ in the equation indicates the kinetic mass of the body then from this relation$$m=\frac{m_0}{√(1-v^2/c^2)}$$ $m$ would increase with $v$. Thus the $\lambda$ value should change doubly...first for the rise of $v$ and second for the rise of $m$

Or is the $m$ in De Broglie's equation representing rest mass, meaning that I am altogether wrong.

I am a layman in Physics and I stumbled on this question while understanding electron microscopy

  • $\begingroup$ The m represents the rest mass. $\endgroup$
    – Digiproc
    Oct 4, 2018 at 16:53
  • 1
    $\begingroup$ @Digiproc. No, $m$ represents the relativistic mass. The de Broglie equation actually read $\lambda = h/p$ where $p$ is the linear momentum. $\endgroup$
    – md2perpe
    Oct 4, 2018 at 17:45
  • $\begingroup$ In the rest frame of the particle $\lambda_{0}=\infty$. Related : About de Broglie relations, what exactly is E? Its energy of what? $\endgroup$
    – Frobenius
    Oct 5, 2018 at 13:14

2 Answers 2


In case you want a less mathematical answer:

what you say is not wrong but a rather upside-down way of thinking of it.

When de Broglie first came up with formula $ \lambda = \frac h p $ in 1923 he was motivated by the way quantities transform under Einstein's relativity theory.

Einstein had previously established two principles:

  • Energy is directly proportional to frequency. $ (E = hf) $
  • Energy and momentum $(E,p)$ naturally transform together when you change your reference frame

So the only way to make sense of a relativistic quantum matter wave is if you pair up $(f,\lambda)$ in a similar way, meaning that wavelength and momentum must be related by the same constant h. $ ( p = \frac h \lambda ) $

So the complexity eluded to in your question does not come from quantum theory, but rather Einstein's original relativity theory: the relation between momentum and velocity is not a simply linear relation $( p = mv) $ but a more complex one where $v$ appears twice:

$$ p = \frac {m v} { \sqrt {1 - \frac {v^2} {c^s} }} $$

[$m$ denotes rest mass here]

Reference: Weinberg, The Quantum Theory of Fields, Volume 1.


Let us restrict our studies to one spatial dimension plus time, coordinates $(t,x).$

In the frame where the particle is in rest, consider a field $\phi$ that oscillates with a frequency proportional to the rest mass/rest energy: $\phi = \sin \omega_0 t$ where $\omega_0 = 2\pi f_0 = 2\pi \, m_0 c^2/h = m_0 c^2/\hbar.$

Now we introduce a relativistic inertial frame $(t',x')$ in which the particle moves with speed $v$ in the positive $x$ direction. The Lorentz transformation between these frames is $$x' = \gamma (x + tv), \quad t' = \gamma (t + xv/c^2)$$ $$x = \gamma (x' - t'v), \quad t = \gamma (t' - x'v/c^2)$$ where $\gamma = 1/\sqrt{1-v^2/c^2}.$

Thus, $$ \phi = \sin \frac{m_0 c^2}{\hbar}t = \sin \frac{m_0 c^2}{\hbar} \gamma(t' - x'v/c^2) = \sin \frac{\gamma m_0 c^2}{\hbar} (t' - x'v/c^2) = \sin \frac{m c^2}{\hbar} (t' - x'v/c^2) \\ = \sin \frac{m c^2 t' - mv x'}{\hbar} = \sin \frac{E t' - p x'}{\hbar} = \sin (\omega t' - k x') $$

The wavelength is then given by $$ \lambda = \frac{2\pi}{k} = \frac{2\pi}{p/\hbar} = \frac{2\pi \, \hbar}{p} = \frac{h}{p}. $$


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