# What does momentum of Photons mean? [duplicate]

I have already checked out Can a force stop a Photon since Photons have momentum and What does momentum mean when talking about massless particles?, but that didn't answer my query. I already know that from the De-Broglie wavelength equation we can say, $$p = \frac{h}{\lambda}$$ where $$p$$ is the momentum of the photon. But as a high school student I know that momentum is defined as $$\vec{p} = m\vec{v}$$ but photons aren't like normal particles, right ? Their rest mass is zero and they behave as both wave and particle (aka wave-particle duality). So, how can one define the momentum for massless objects ?

PS : I had also checked : Confusion regarding photons? but that didn't fully answer my query.

But as a high school student I know that momentum is defined as $$\vec{p} = m\vec{v}$$

The solution is that in high school you aren't taught a truly complete definition of momentum.

This is a useful definition in Newtonian mechanics, but it doesn't apply to photons or gluons (i.e. to massless particles). You'll also learn that the definition of momentum must be modified in relativistic physics as well.

So, how can one define the momentum for massless objects ?

At the high school level, you must simply accept that for photons, $$p=\frac{h}{\lambda}$$, as if it were an additional definition of what momentum can be.

To be honest, even with a Ph.D. in electrical engineering, specializing in optoelectronics, this is how I treat it as well. Wikipedia has a more detailed discussion than I ever encountered in 9 years of school, but as far as I can tell it boils down to "photons can transfer momentum from one massive particle to another, therefore they must carry momentum themselves". Possibly a proper physicist will submit an answer that gives deeper reasoning.

• Einstein said photons could "transfer inertia" - is that saying the same? Guess yes. If a photon is absorbed, it is none of its business any more if the electron turns its momentum into mass. What it came on with sure wasn't "mass" - refering to the question -, it seems the electron, someone else turned photon's momentum into mass; with the photon it wasn't "inertia" but energy. Electrons - who else? - not only absorb but also emit photons. Photons "exchange" (transfer, carry) energy between masses (electrons). The energy is received "as" mass, energy becomes mass within a mass (electron). Commented Nov 12, 2022 at 18:29

Photons are elementary particles, and particles obeying quatum mechanics and Lorentz transformations..

The classical definition of momentum as $$mv$$ cannot be used rationally, one has to go to the four vector space of Lorentz transformations to understand how a zero mass particle can have (actually must have if it is to exist at all,) momentum .

The classical $$m$$ cannot be used because for high velocities the m in the $$mv$$ is not defined independent of the velocity, so is not invariant in all frames .

The Lorentz four vectors defined as $$(E, p_x.p_y,p_z)$$ have as a fixed length , the invariant mass of the particle, and it is that mass that enters the definition of the four vector:

$$\sqrt{\mathstrut P\cdot P \rule{1ex}{0pt}} = \sqrt{E^2 - (pc)^2 \rule{1ex}{0pt}} = m_0c^2$$

The length of this 4-vector is the rest energy of the particle. The invariance is associated with the fact that the rest mass is the same in any inertial frame of reference.

If you look at the four vector invariant mass, when the invariant mass is zero, which is the case for the photon, $$E=pc$$ . Since the photons have energy, they inevitably also have momentum.

This is the mathematics that fits all the data we have on photons and elementary particles, thus the conclusion that photons have zero mass, but do carry momentum

For the classical electromagnetic waves, light, this experiment that measure radiation pressure experimentally measured the momentum of the zillion of photons making up the beam. In particle physics momentum conservation in interactions and decays work correctly only if the momentum of the gamma rays follows the formula above.