0
$\begingroup$

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

enter image description here

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?

(please make your answer simple, I don't know special relativity)

$\endgroup$

3 Answers 3

2
$\begingroup$

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?"

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ 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. $\endgroup$
    – garyp
    Oct 24, 2018 at 12:55
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.