Special Relativity, Louis de Broglie Equation Dilemma

Question

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

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$!$

Answer 1

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.

Answer 2

$E=mc^2$ 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 $\lambda = \frac{h}{p}$.

Answer 3

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.

Answer 4

Short Answer:

Your reason for doubt is justified for both cases. It took the genius of De Broglie to come up with the reasons. See below. I think most text books don't even try to do what your teacher did. They avoid it for the most part and just say De Broglie suggests that perhaps wavelength equals h/p for matter. They will then justify later when they can show that the group velocity of the wave is shown to be equal to the velocity of the mass.

Long Answer:

De Broglie's argument might be simplified to say that for a particle at rest, there is an associated wave with frequency f=E/h=mc^2/h, where m is the rest mass, E is the rest energy. The wave is a standing wave in this frame of reference since all of space vibrates at that rate with no change in phase with distance (the wave vibrates only with time). When observing this wave in an inertial frame of reference moving in the x-direction relative to the one where the mass is at rest, the wave now has a finite wavelength as measured along a horizontal x-axis ( Minkowsky graph) equal to h/p and has a wavelength along the vertical time axis (usually c times t) equal to 1/f, f=E/h, where E is now the total energy (rest mass plus kinetic energy). The dilemma he resolved at the time was that moving clocks slow down in time and they have to be measured along the world-line of the clock. (The mass particle if considered as a clock does obey the slower frequency change of a clock when it is in motion). He needed something to speed up in frequency. He resolved that by assuming a wave which allowed him to measure the frequency at a stationary point in the inertial frame of reference where he is not moving but where the mass is moving (in other words, he measured it along his vertical axis ct). Note: the slope of lines that might represent wave crests is that of the slope of the x' axis, the axis that is the distance direction for the inertial frame of the stationary mass. The slope of these lines are always faster than the speed of light. The speed of light is at a 45 degree angle on these plots. So the slope of the wave lines slow down for a faster mass. The wave moves to the right if the mass is moving to the right and vise versa. As the mass speeds up by a certain angle on this plot, the velocity of the wave slows down by the same angle. Eventually both approach the 45 degree angle as the mass gets faster and faster.

Getting back to the OP's question, the De Broglie wavelength will only approach the EM wavelength as the mass gets faster and faster, and/or as the mass is reduced. De Broglie actually thought that the photon might have a small mass. In actuality it has no mass.