# What's wrong with de Broglie?

1. Aug 5, 2014

### americast

Hi all,

De Broglie had proposed the theory of wave particle duality. But I don't understand how the relation λ=h/(mu) holds true. I will tell you why:

De Broglie's derivation:
E=mc2 and E=hv
So mc2=hv
Thus, λ=h/(mc) and for particles its h/(mu).

I have a problem in the last line. All of us know that according to the theory of relativity, speed of light c is a constant and is independent of the frame of reference. It remains c from all reference frames i.e. if you are to travel at 2*108 m/s, you will still find light travel at c (unlike other particles which would otherwise travel at 1*108 m/s for you.) This means that the velocity of light relative to everyone is c. So h/(mc) does not need a defined reference frame.

In the above equation, when I am replacing c with u (for particles) the expression lacks a defined frame of reference. u is not an independent velocity but it is a velocity with respect to some observer. Who is this observer?

Gramercy...

2. Aug 5, 2014

### Geometry_dude

Your derivation is incorrect. The (special-relativistic) De-Broglie Einstein relations are as follows
$$p = \hbar k$$
or in coordinates
$$\begin{pmatrix} E/c \\ \vec p \end{pmatrix} = \begin{pmatrix} \hbar \omega/c \\ \hbar \vec k \end{pmatrix} \, ,$$
which makes perfect sense for photons. The problem is that you cannot use
$$E= m c^2 \dot t \, ,$$
where the dot represents differentiation with respect to the proper time of the photon. This is nonsense, since photons (considered as points) travel with $c$ and thus do not experience any time lapse. On a principle level, photons do not have any mass, so you cannot divide by it to get the Compton-Wavelength.

Last edited: Aug 5, 2014
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