The energy ($E$) of a photon depends on its frequency ($f$):

$$E = hf = hc/\lambda$$


  • $E$ = energy of the photon (in joules) *$h$ = the Plank constant ($6.63 \times 10^{-34}$ joules)
  • $f$ = frequency in hertz, Hz
  • $c$ = speed of light
  • $\lambda$ = wavelength

This seems to yield an invariant value of $E$.

How can that be reconciled with the inverse-square law, whereby the intensity (field strength) of a light-source reduces with increasing distance from the light-source?

  • 1
    $\begingroup$ It's not invariant, you can do a Lorentz boost and you will get a different value for the frequency and wavelength. What's invariant is the norm of the four vector, the energy is just one of the components. $\endgroup$ Commented Feb 13, 2021 at 19:40
  • 4
    $\begingroup$ The inverse square law is to the best of my knowledge a consequence of the emitted photons having to spread out when further from the source, and not individual photons loosing energy. $\endgroup$
    – jng224
    Commented Feb 13, 2021 at 20:37

2 Answers 2


The confusion comes from mixing the classical and QM theories of light. In reality, the classical theory is beautifully built up by the herd of photons as described in QM.


In your example, as you move away from the source, the same number of photons is distributed over a larger area.

Therefore, if you move away from a light source, the same number of photons is distributed over a larger area, thus decreasing the intensity. So it is not an effect of an individual photon, but of the distribution of all the photons emitted.

How does a photon lose intensity as it passes through space?

What this tells us is that the number of photons per unit area (as you say intensity) decreases as you go away from the source. Of course, because the EM wave spreads spherically from the point source, and as it does, the same EM wave has to cover a larger surface area (the expanding sphere if you will). Since the unit area is constant, and the energy content of the EM wave is constant (in your example), the number of photons per unit area has to decrease as the surface area increases.


The electromagnetic field is to photons what the Schrödinger or Klein-Gordon wave function is to electrons. It gives the probability of finding photons or their properties, such as energy and momentum. It is this probability that falls off as $r^{-2}$ for all but exact plane waves, which do not exist.

  • $\begingroup$ All electromagnetic waves are spherical, deriving from a point-like source (a star, at interstellar distances). $\endgroup$
    – Ed999
    Commented Feb 22, 2021 at 3:21

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