If light has mass, it must follow equation $m=\frac{m_0}{1-\frac{v^2}{c^2}}$. Hence must have infinite mass and must have infinite momentum. As we are not killed by the strike of light, so light must have zero mass. If so then light must not experience any force. But it doesn't escape from black hole, means experiencing force. Both the facts are contradictory so what about the mass of light.

  • $\begingroup$ Equation is Einstein's equation for variable mass $\endgroup$ Commented Dec 7, 2016 at 15:46
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    $\begingroup$ Related: physics.stackexchange.com/q/130552 , physics.stackexchange.com/q/287701 , physics.stackexchange.com/q/158481 $\endgroup$ Commented Dec 7, 2016 at 15:47
  • $\begingroup$ "As we are not killed by the strike of light, so light must have zero mass." A interesting exercise is to work out (from kinetic theory) the RMS speed of an air molecule at room temperature. It quite fast enough to be compared with that of bullets, and you are hit by them ten-to-the-many times per second without killing you... $\endgroup$ Commented Dec 7, 2016 at 16:15
  • $\begingroup$ You dared to ask for the "relativistic mass" of a photon, a scientifically correct concept which, however, is anathema to some mission-conscious wise guys on this site who accept only "rest masses". It is well known that a photon has no rest mass. Therefore my answer below is downvoted repeatedly in spite of being scientifically correct. $\endgroup$
    – freecharly
    Commented Dec 8, 2016 at 13:01

1 Answer 1


For light $m = m_0 = 0$ in general. The expression you've given, $m_0 / [1 - (v/c)^2]$ thus becomes the indeterminate form $0/0$ which could be evaluated to any number depending on how you come to it. [For example $(x-1)/(x^2 - 2x + 1)$ limits to $\infty$ at $x=1$ whereas $\sin(k (x - 1))/(x-1)$ limits to some finite $k \ne 0$ whereas $(x^2 - 2x + 1)/(x - 1)$ limits to $0$.]

Instead it is more useful to consider the relativistically correct energy-mass-momentum equation:$$E^2 = c^2 p^2 + m_0^2 c^4.$$

For massive particles moving slowly we have: $$E = m_0 c^2 \sqrt{1 + \frac{p^2}{m_0^2 c^2}} \approx m_0 c^2 + \frac{p^2}{2 m_0},$$ incorporating both the famous "${E = mc^2}$" but also the classical kinetic energy. However for massless particles or particles moving very very close to the speed of light we have:$$E \approx p c,$$which has been known since the Maxwell equations to be exactly correct for light.

In quantum mechanics this becomes even more perfect as the de Broglie wavelength for a particle is $\lambda = h/p$ and the Einstein-Planck relation for the frequency $f$ that its quantum phase oscillates is $E = h f,$ for photons the relation $f ~ \lambda = c$ we see the $h$ cancel and $E = p~c,$ confirming that all the waviness of the photon can be viewed as oscillation in its quantum phase.


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