While an individual photon has no rest frame, when two photons move apart, it makes sense to ask where their centre of mass is, as described here.

If I proposed that while a photon has no mass, a system of two parting photons does have mass, how would I quantify that mass, in a manner consistent with the prevailing theories of physics?

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    $\begingroup$ I do not believe that involving some symmetry changes something. Imagine that supernova has exploded and say all it's rest mass was converted into photons which fly apart evenly in a spherical fashion. Now we have exact static center reference point, which was ex-star barycenter, but this doesn't mean that somehow magically star energy converts back into rest mass just to please you that you know it's COM. $\endgroup$ Commented Jan 6 at 12:40
  • $\begingroup$ See this Q&A: physics.stackexchange.com/questions/10612/… $\endgroup$ Commented Jan 6 at 13:21
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    $\begingroup$ do you know that the pi0 decays into two photons? The pi0 has a fixed mass en.wikipedia.org/wiki/Pion#Neutral_pion_decays $\endgroup$
    – anna v
    Commented Jan 6 at 14:36
  • $\begingroup$ @AgniusVasiliauskas the part where you wrote "just to please you that you know its COM" seems unscientific. Just to be clear what you are saying: Your claim is that the system of explanding light, centered at the rest frame of the ex-star barycenter has mass 0 grammes, correct? $\endgroup$ Commented Jan 7 at 9:38
  • $\begingroup$ Yes, correct. Rest mass is zero, but energy - does not. $\endgroup$ Commented Jan 7 at 9:53

2 Answers 2


In particle physics, it is common to define the invariant mass $M$ of a system of $n$ particles with four-momenta $$p_i =(E_i, \vec{p}_i), \quad (i=1, \ldots n, \; p_i^2=E_i^2-\vec{p}_i^2=m_i^2),$$ by the Lorentz invariant quantity $$M^2:= \left(p_1+\ldots +p_n\right)^2=(E_1+\ldots +E_n)^2-(\vec{p}_1+\ldots+\vec{p}_n)^2.$$ In the case of two photons (where $E_1= |\vec{p}_1|$, $E_2=|\vec{p}_2|$), the invariant mass of this two-particle system is given by $$M^2=(p_1+p_2)^2=(|\vec{p}_1|+|\vec{p}_2|)^2-(\vec{p}_1+\vec{p}_2)^2.$$ In the reference frame where $\vec{p}_1 + \vec{p}_2=\vec{0}$ (the so-called center-of-mass frame), the invariant mass (squared) of the two-photon system becomes $$ M^2=4E_{\rm CM}^2$$ with $E_{\rm CM}:= |\vec{p}_1|=|\vec{p}_2|$.


Mass is rest energy divided by c$^2$, so you could define the mass as hf$_1$+hf$_2$. However, for it to be useful as a concept in physical analysis, mass should be constant and localized. In the case of two free photons parting in opposite directions, I don't see a use for the concept of total mass. In the case of an opaque box containing light and with infinite Q, it might be useful.

  • $\begingroup$ Thank-you. What's $Q$? Where you write "for it to be useful as a concept..." Would it be of use in order to account for the gravitational mass of this photon pair, as calculated around their centre of mass? $\endgroup$ Commented Jan 8 at 14:31
  • $\begingroup$ $Q$ is the quality factor of a cavity. ccrma.stanford.edu/~jos/fp/…. $\endgroup$
    – my2cts
    Commented Jan 8 at 14:38

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