I have seen questions assume photons have no mass. But I have not seen any questions that directly ask whether or not photons have mass.

If photons have no mass, then how do they occupy space? How are they "there?"

Do photons have mass?

  • $\begingroup$ physics.stackexchange.com/q/34067 Probably entertained here in the sense that you're interested in. You can get a mass metric from the energy of a photon, everything else is a matter of definition of the English words. It is a universal statement to say that photons do not have rest mass. $\endgroup$ Apr 8, 2013 at 14:17
  • $\begingroup$ Coulomb's inverse square law implies that the photon does not have a mass. So if you wish find the lower limit for the photon mass, you should look at deviations from the Coulomb's law at large distances. If photon does indeed have a mass then the force between charged particles will go as en.wikipedia.org/wiki/Yukawa_potential . If you wish to discuss this formula it requires a background in Quantum field theory. As far as we know photons are point particles with no further structure and are massless $\endgroup$
    – Prathyush
    Apr 8, 2013 at 14:21
  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/4700/2451 physics.stackexchange.com/q/17939/2451 and possibly physics.stackexchange.com/q/41676/2451 $\endgroup$
    – Qmechanic
    Apr 8, 2013 at 15:30
  • 2
    $\begingroup$ Hi @Qmechanic: The first one looks very much a duplicate ;-) $\endgroup$ Apr 8, 2013 at 17:02

3 Answers 3


Photons' rest mass is almost certainly zero, using some theoretical constraints. Even purely experimentally, we know that the photon rest mass is so tiny that the corresponding wavelength is much longer than the radius of the Earth. That's because we know that the geomagnetic field obeys Maxwell's equations at these very long distance scales.

The vanishing of the rest mass, $m=0$, doesn't mean that photons don't carry any energy or "relativistic mass". Instead, it means that $$ E^2 - p^2 c^2 = m^2 c^4 =0 $$ i.e. that the energy $E$ is simply equal to $|\vec p|c$ where $\vec p$ is the momentum and $c$ is the speed of light. It's the momentum and energy that must be nonzero for us to be able to say that "something is out there" and indeed, they're nonzero for each photon around us.

You may also say that the "total relativistic mass" of the photon is nonzero and equal to $m_{\rm total} = E/c^2$. In some sense, this nonzero quantity measures the gravitational mass as well as the inertial mass (resistance to acceleration). We usually don't use the term "mass" for this quantity and prefer to talk about the energy $E=m_{\rm tot} c^2$, reserving the term "mass" for the rest mass defined above.

The fact that the photon's rest mass is zero means that there isn't any "rest frame of a photon". Such a rest frame would have to move by the speed of light relatively to any proper inertial system and with such a fast motion, all the factors such as $\gamma$ would be singular and ill-defined.

  • 2
    $\begingroup$ Excellent answer, as usual btw. $\endgroup$
    – Dani
    Apr 8, 2013 at 17:13
  • $\begingroup$ *Compton wavelength? - on line 2 $\endgroup$
    – user12345
    Apr 8, 2013 at 19:59
  • $\begingroup$ Yes excellent like related excellent answer physics.stackexchange.com/q/4700/2451 that photon mass must be zero. $\endgroup$
    – user12811
    Apr 9, 2013 at 3:19

The question is ill-posed, since it depends on your definition of "mass". In Special Relativity, the main mass-like counterpart of classical mass is "invariant mass":

$$ P^2c^2-E^2=-m^2c^4$$

The notion of "relativistic mass" is "old-fashioned" now, and it is not correct today to speak about it. If the relativistic dilation factor is defined to be:

$$ \gamma =\dfrac{1}{\sqrt{1-v^2/c^2}}$$

relativistic mass will be defined (generally speaking) by:

$$ M=m\gamma=\dfrac{m}{\sqrt{1-v^2/c^2}}$$

Answer: In Special Relativity, a photon has "invariant mass" equal to zero (like gluons and likely gravitons) due to "gauge invariance" on very general grounds. However, a photon carries energy and momentum (combined in such a way that $E=Pc$), and therefore, we can "speak" about certain kind of energy carried out/transported by photons/electromagnetic waves...And since in Special Relativity energy and mass can be swapped $$E_{tot}=Mc^2$$ we can think about a photon as having "an effective" mass

$$ M_{eff}=E/c^2$$

However, this fact does NOT mean the photon has "mass". It only means that its energetic content can be "seen" in terms of mass due to the equivalence of energy and matter. The true meaningful mass concept in Special Relativity is the invariant mass and for photons m=0. It is a pity that people does not generally understand this, and I presume those old fashioned spreading out of relativity has caused the same confusion as when unfortunately you (wrongly) hear that "(...)relativity theory says that everything is relative(...)".


First, Photons are energetic (they have momentum) mass-less particles (By mass-less, I meant their rest-mass to be zero). It's just our assumption and every single theory satisfied it, basically the wave-particle duality, relativity, etc...

Second (which is a misconception), if you're asking why photons occupy space?, then you can also ask, "why gluons are mass-less and why they occupy space?" too... It's not a rule that mass-less particles shouldn't occupy space. Due to this reason, these particles always travel at $c$. They're what "they are". But, we haven't observed one directly (while we still have some idea of their existence) and we won't..!

Please try to understand what physicists are doing. They assume something and theorize to explain (or confirm from) an observation. Once it satisfies, some try to find a way in which it's wrong, while some others try to update it. The remaining just use the theory for predicting, applications, etc...


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