# Special Relativity, Louis de Broglie Equation Dilemma

While learning atomic structure I stumbled upon a very unusual doubt.

As we know that the energy of a wave is given by the equation: $$E=\frac{hc}{λ}$$ and Louis de broglie wave equation is given by the equation $$λ_{B}=\frac{h}{p}$$. My doubt is that, that is $$λ_{B}= λ$$ $$?$$. Do the $$λ_{B}, λ$$ represent the same thing $$?$$

My teacher equated $$E=\frac{hc}{λ}$$ and $$E=mc²$$ to form $$\frac{hc}{λ}=mc²$$ and rearranged to form $$λ=\frac{h}{mc}$$ and then replaced $$λ$$ by $$λ_{Β}$$ and $$c$$ by $$v$$ for general formula and derived the Louis de broglie equation. This created my doubt in first place and I created another doubt that whether the equation for energy of wave is valid for relativistic equation of $$E=mc²$$ because the $$E=mc²$$ is for particles while the former is for waves.

Is my understanding correct$$?$$ Please help and thanks in advance$$!$$

• $E = \frac{hc}{\lambda}$ represents the energy of a photon ( a packet of energy) . Oct 6 '20 at 8:43
• @Ankit, yes but isn't according to electromagnetic theory, the equation is for all em Waves? Oct 6 '20 at 8:47
• Oct 6 '20 at 8:48
• @Kshitij Kumar okay , so from your equation $\frac{hc}{\lambda} = mc^2$ photons have mass !!! Oct 6 '20 at 8:51
• @Ankit, but in question I already mentioned that I am unsure about that this relation between energy of em wave and relativistic equation. I asked to clarify whether it is correct or not Oct 6 '20 at 8:55

You don't have an 'unusual' doubt; you are simply confusing matter waves with electromagnetic waves, which stymies many readers at first. Louis de Broglie proposed in 1924that all matter exhibits wave-like behavior, with a wavelength given by (to use your notation) $$\lambda_B = h/p$$ where $$p$$ is the matter's linear momentum.

Conversely, electromagnetic radiation behaves like it is composed of tiny corpuscles called photons, each of which carries a certain quantum of energy, given by $$E = h \nu$$, where $$\nu$$ is the frequency of the radiation. Expressing $$\nu$$ as $$\nu = c/\lambda,$$ we find that $$E = hc/\lambda,$$ which you mention in your question.

In other words, $$\lambda_B$$ and $$\lambda$$ are two completely different kinds of wavelength: the former relates to moving matter, whereas the latter, to photons.

Your teacher's approach at deriving de Broglie's relation is either incorrect or incomplete. The relation $$E = mc^2$$ uses the invariant mass of massive objects, so equating that with $$hc/\lambda,$$ which is valid only for photons, suggests that photons have mass, which is emphatically wrong.

What your teacher was perhaps trying to do was assume that all the matter in a given mass $$m$$ got converted into pure electromagnetic energy (say, through nuclear fission) and then calculate the wavelength of the ensuing photons with said energy. In that case, the derivation is justified until the generalization $$c \to v,$$ which requires much more clarification, if it is not outright wrong: so far we have only mentioned photons, which always move at the speed $$c$$.

Either way, simply equating two similar-looking expressions at face-value is a dangerously unsound way to teach Physics.

• Or the strategy may have been to use $m$ to denote the relativistic mass $p/v=E/c^2$, which is nonzero even for photons.
– J.G.
Oct 6 '20 at 18:17

$$E=mc^2$$ is true for a particle at rest. For a moving particle the formula would be $$E^2-P^2 = m^2$$ where $$P$$ is the momentum (in units where $$c=1$$).

for a massless object like the photon, we get $$E=P$$. Using $$E= \frac{2\pi}{\lambda}$$ we get $$P=\frac{2\pi}{\lambda}$$ (in units where $$\bar{h}=1$$). This is equivalent to the desired $$\lambda = \frac{h}{p}$$.

• How are we getting E=2π/λ? Oct 6 '20 at 9:00
• if you set units with $\bar{h}=1$ then $h = 2\pi$ Oct 6 '20 at 9:03

My doubt is that, that is $$λ_B=λ$$ ?. Do the $$λ_B,λ$$ represent the same thing?

They are the same thing only for photons, they are not the same as we consider an object. Indeed, $$E$$ is the total energy of the object so that we have:

$$E^2=p^2c^2+{m_0}^2c^4\space.$$

For photons, we have $$m_0=0$$, where $$m_0$$ is photon's rest mass, then:

$$E^2=p^2c^2\rightarrow E=pc\space.$$

If we use $$E=h\nu=hc/\lambda$$, we reach:

$$\lambda=\frac{h}{p}\space.$$

Indeed, De Broglie, who did not know the precise nature of his so-called matter waves, argued that they should satisfy the same two last equations ($$E=h\nu$$, $$p=h/\lambda$$) as apply to light waves, where now $$E$$ is the relativistic total energy of the object and $$p$$ is its relativistic momentum. Remember that one of the differences of these equations compared to those we use for photons is that if multiply $$\nu=E/h$$ by $$\lambda=h/p$$, we reach a (phase) velocity of the order of $$E/p=c^2/v$$ that exceeds the speed of light for a matter wave. (To prove, substitute $$E\approx mc^2$$ and $$p=mv$$.) However, it also can be shown that the group velocity equals the velocity $$v$$ of the object.