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I've just tried to make sense of the De-Broglie wave relation. That is what I came up with: $E=mc^2$, but $E=hf$ as well. $\lambda=vT$ so $f=\frac{v}{\lambda}$. Equating the two equations for energy you get $mc^2=h\frac{c}{\lambda}$ because for light the velocity is $c$. $mc$ is just the non-relativistic momentum so $p=\frac {h}{\lambda}$ which is the De-Broglie wave relation. But if I didn't replace $mc=p$, $m=\frac {hc}{\lambda}$ or $m=fh$, which would imply that light does have rest mass. What would be the more rigorous derivation of the De-Broglie wave relation? Do I need to use relativistic momentum $\gamma mv$? Or is the above reasoning just simply not applicable to light?

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  • $\begingroup$ Your reasoning is applicable to photons only but the De Broglie equation is applicable to all matter. $\endgroup$ – Yashas Feb 27 '17 at 17:16
  • $\begingroup$ $E = m c^2$ is the still energy of matter with mass. As for photon, which has no mass, it only have the kinetic part of energy (and its speed cannot decrease), which is $E = pc$. For matter with mass (without potential energy), the total energy is $E = \sqrt{m^2 c^4 + p^2 c^2}$. $\endgroup$ – Exhaustive Feb 27 '17 at 18:02
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$p=mc$ is NEVER true.

Like you mentioned, for a particle with relativistic velocity (v is not negligible compared to $c$), the right equation is:

$$p = \gamma m v = v\cdot \frac{m}{\sqrt{1-(\frac{v}{c})^2}} $$ Now if you try to apply it to photons, $ m=0$ and $\sqrt{1-(\frac{v}{c})^2}=\sqrt{1-1}=0$ so you get a $\frac{0}{0}$ situation - which gives us no information at all! So you can't apply this formula for photons.

But, there is another, more accurate equation:

$$E^2 = (pc)^2 + (mc^2)^2 $$

So here, for photons, $m=0$ and we get $$E^2=(pc)^2 \Rightarrow E=pc$$ Side note: that was known before Einstein, you can get this result straight from Maxwell's equations and the Lorentz force.

Now, if $E=pc$, then $p=\frac{E}{c}$ , which for a single photon means $$ p = \frac{E}{c} = \frac{h\frac{c}{\lambda}}{c} = \frac{h}{\lambda} $$

I hope it helped :)

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  • $\begingroup$ Perfect, that is what I was looking for - thank you! I didn't know that $E^2=(pc)^2$ was known before special relativity, that is pretty interesting. On the other hand doesn't SR itself follow straight from Maxwell's equations? $\endgroup$ – Jannik Pitt Feb 27 '17 at 17:47
  • $\begingroup$ If you calculate the momentum carried by an EM wave you'll see that relation popping up in the process. You should look it up in Griffith's. $\endgroup$ – Ofek Gillon Feb 27 '17 at 17:51
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    $\begingroup$ @JannikPitt De Broglie equation was discovered after SR was formulated. $\endgroup$ – Yashas Feb 27 '17 at 17:51
  • $\begingroup$ SR is consistent with Maxwell's equation, so getting this result isn't a surprise. $\endgroup$ – Ofek Gillon Feb 27 '17 at 17:53
  • $\begingroup$ @OfekGillon I don't really have access to Griffith's because I'm still in highschool and I doubt I'll understand it haha. Thanks for your help! $\endgroup$ – Jannik Pitt Feb 27 '17 at 18:05
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de Broglie suggested that a moving body behaves in certain ways as though it has a wave nature. His conjecture of dual nature of matter was based on two points : (1) dual nature of radiation, and (2) nature loves symmetry.

de Broglie deduced the connection between particle and wave properties from the Einstein-Planck expression for the energy of the electromagnetic wave and the classical result for the momentum of such a wave. The two expressions are,

$E=hf=\frac{hc}{\lambda}$

$and\space E^{2}=(m_\circ c^{2})^{2}+(Pc)^{2}=(Pc)^{2}$, because the rest mass of photon is $0$

From these equations we get,

$\lambda=\frac{h}{P}$

de Broglie said that this equation is a completely general one that is applied to a material particle as well as photons. The waves associated with the moving particle are called matter waves or de Broglie waves.

The theory of matter waves was used to justify quantization of angular momentum as proposed in Bohr's model. If we consider standing electron wave, then to maintain a standing wave over the circumference of a circular orbit, the wavelength must be an integral fraction of that circumference.

$2\pi r = n\lambda = \frac{nh}{mv}$,

And hence we get, $mvr=\frac{nh}{2\pi}$, this equation is an integral part of Bohr's atomic model.

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  • $\begingroup$ "integral fraction"? I think you should reword it as "The circumference must be an integral multiple of the wavelength." $\endgroup$ – Yashas Feb 27 '17 at 17:46
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    $\begingroup$ $E=mc^2 = pc$? $E^2 = (m_0c^2)^2 + (pc)^2$ is the full equation. A photon does not have rest mass; therefore, $E = pc$. $\endgroup$ – Yashas Feb 27 '17 at 17:50

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