# How to derive derive De-Broglie's wavelength equation?

I was only yesterday learning about the De-Broglie equation $$\lambda = h/p,$$ I thought I understood it until I came across a question similar to this

Now I always knew that photons don't have any rest/invariant mass, but this question threw me off a bit, I think it was because of how my teacher derived it:

$$E=mc^2$$ $$E=hc/\lambda$$ $$mc^2 = hc/\lambda$$ $$p = h/\lambda$$ $$\lambda = h/p$$

Now he wrote $$mc=p$$ but that doesn't really apply to photons. So is this a wrong, or maybe a child friendly derivation?

The first two equations are inconsistent with each other. The second applies to massless particles. And, indeed, $$mc=p$$ doesn't apply to anything. What your teacher has done, in essence, is an exercise in dimensional analysis. A similar way to get to de Broglie is to ask "What combinations of $$p$$ and the fundamental constants of nature leads to a quantity whose dimension is a length?"

I hope I understand your question correctly.

The basic problem is that there are two different things that are called mass: the rest mass and the kinetic mass.

The rest mass of a particle is the energy of the particle when it is not moving divided by $$c^2$$: $$m_0 = E_0/c^2$$. This is what you usually call the mass.

The term kinetic mass is a bit misleading. You define it as the whole energy of a moving particle divided by $$c^2$$: $$m = E/c^2$$, where $$E = m_0 c^2 + E_{\mathrm{kin}}$$. To avoid confusion in particle physics one does not use the term kinetic mass and only talks about the energy of a particle.

However in your question you are supposed to calculate the kinetic mass of the photon. This is simply given by the energy of the photon divided by $$c^2$$: $$m = \frac{E}{c^2} = \frac{h f}{c^2} = \frac{h \frac{c}{\lambda}}{c^2} = \frac{h}{\lambda c}$$ or $$\lambda = \frac{h}{m c}$$. So if you now insert this into the result of De-Broglie: $$p = \frac{h}{\lambda} = m \cdot c$$. So in fact since the velocity of the photon is $$v = c$$, you can write $$p = m \cdot v$$.

PS: The relation $$p = m \cdot v$$ holds for all particles in general. This is because the kinetic mass $$m$$ is defined such that this relation always holds.

• The concept of kinetic mass is problematical, and not used in modern treatments. Einstein warned against it. Among other problems, it serves as a measure of inertia only in the direction of motion. Transverse to the direction of motion, inertia is rest mass. Oct 24, 2018 at 12:55

You're over-complicating the issue. From special relativity, $$E = pc$$ for photons. From experimental observations, $$E = hf$$, from which we can obtain $$E = \frac{hc}{\lambda}$$ by substituting $$f = \frac{c}{\lambda}$$. Then, equating these 2 equations gives $$pc = \frac{hc}{\lambda}$$, therefore $$p = \frac{h}{\lambda}$$.