# If photons have no rest mass, where does Planck's constant come from?

I saw two other questions asked about why photons have momentum and energy even though they do not have a rest mass. I now wanted to ask about Planck's constant. An object's action has dimensions of $$M L^2 T^{-1}$$. I do not see this being followed by photons. A constant action of $$6.6 \cdot 10^{-31} \text{g} \, \text{m}^2 \text{s}^{-1}$$, called Planck's constant $$h$$ has some connection with photons. The energy of photons is $$E = h f$$, where $$f$$ is the frequency of the photons. Where does this constant come from?

• physics.stackexchange.com/a/730230/292464 read this to understand energy in relativistic cases. As for where planck's constant came from , read about ultraviolet catastrophe.
– user292464
Oct 7, 2022 at 15:51
• Planck’s constant is the quantum of action. Asking “where it comes from” is asking why action is quantized. We can imagine a universe in which it isn’t, but it would look nothing like our universe. Oct 7, 2022 at 15:53
• Planck's constant has some connection with photons? Have you learned that the energy $E$ and frequency $f$ of a photon are related by $E=hf$? Oct 7, 2022 at 15:58
• planck's constant was discovered in the context of photons, but is a lot more fundamental a quantity than just "governing how the energy of photons is quantized" Oct 7, 2022 at 19:56
• I think it is fair to ask "Where does Planck's constant come from?", but the answer is a trivial "We don't know." right now, at least not on the level of first principles. Oct 7, 2022 at 22:09

Its size sets the dimension (scale) of quantum phenomena, whose cornerstone is the (then, a century ago) counterintuitive noncommutativity relation $$[x,p]=i\hbar$$.
The characteristic dimensions of photons, are its wavelength (λ) and its momentum (p) understood to be inversely related to each other a century ago, $$p\lambda = h .$$
This relation leads to the small scale of the photon momentum emitted in atomic energy level transitions, since $$E= c p = h c/\lambda= h f$$ .