Due to $ E =m c^2 $, one can convert mass to energy. A classic example would be matter/anti-matter annihilation to produce energy (photons, etc.).

Can one do the reverse? So could one do something to photons to create mass?


The conversion between mass and energy isn't even really a conversion. It's more that mass (or "mass energy") is a name for some amount of an object's energy. But the same energy that you call the mass can actually be a different type of energy, if you look closer. For example, we say that a proton has a specific amount of mass, about $2\times 10^{-27}\text{ kg}$. But if you look into the structure of a proton, about half that mass (or more, depending on conditions) is actually kinetic energy of the gluons.

Of course, that's probably not what you had in mind. To more directly answer your question, it is possible to produce matter from two colliding photons, although the probability is not especially high. You need energetic photons, and lots of them, to create an appreciable number of detectable matter particles. Wikipedia's article on matter creation has more information and links.

  • $\begingroup$ +1 for a sensible answer that doesn't conflate mass with matter but still covers both. $\endgroup$ – Stan Liou Jul 13 '13 at 0:52
  • $\begingroup$ Concise and precise. It's a similar equivalence to that between time and space. We actually have superfluous units; masses can be measured in elektronvolts and time intervals in meters. (or energies in grams and distances in seconds e.g.) $\endgroup$ – Wouter Jul 13 '13 at 1:12
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    $\begingroup$ Where could a math literate physics enthusiast like myself find out more about this? I don't really have much of a background with physics, but I don't mind reading fairly dense math. $\endgroup$ – Benji Altman Jun 18 at 15:13

Yes, it is possible to create matter/anti-matter particles from two photon beams. The first successful experiment of this kind was carried out at the Stanford Linear Accelerator in 1997.

Here's a summary from the NY Times: Scientists Use Light to Create Particles

And here's the scientific paper: Positron Production in Multiphoton Light-by-Light Scattering


Mass is energy. It's energy observed from a "center-of-energy" frame.

So no 'conversion' is needed. A simple thought experiment: capture a photon in a box with perfectly reflecting mirrors. The box will weigh more with the mass increment equalling $E_{photon}/c^2$. This mass increment translates into an increased inertia of the box: accelerating the box with the photon inside creates an imbalance in momentum transfer to the photon in subsequent collisions.

So how come photons are considered massless? That's simply because you can't observe a free photon from its center-of-energy frame (in loose terms: "you can't keep up with a photon"). When the photon is traveling back and forth in a box, you can observe the box + photon from its "center-of-energy" frame, and the mass=energy equivalence becomes apparent.

  • $\begingroup$ This was exactly the answer I was going to give - but I suggest mentioning the following: what no-one has mentioned here is that inertia - the constant in Newton II - is what one ies fundamentally talking about when one talks of "mass". Your thought experiment illustrates this. When you're done editing your answer, I have a few words I would like to add to it, with light blue shifted at the lagging edge of the cavity, red shifted at the other. One then sees that the difference between the blue and red momentums equals $\frac{E}{c^2}$. $\endgroup$ – WetSavannaAnimal Jul 13 '13 at 4:48
  • $\begingroup$ Good suggestions. Have added a statement on inertia. Feel free to add/modify as you see fit. $\endgroup$ – Johannes Jul 13 '13 at 6:28
  • $\begingroup$ A center-of-energy frame for a light beam makes perfect sense--it just requires some null frame vectors, and is done routinely in, e.g., a null tetrad. Of course, you meant inertial frame, which would be quite correct, but perhaps not as clear as having mass be the invariant of energy-momentum ($m^2 = E^2 - p^2$), which is also why energy-momentum of light is a null vector. $\endgroup$ – Stan Liou Jul 13 '13 at 20:15

Not many experimentalists here.

Yes energy can be converted to mass and the simplest example is a photon interacting with the electric field of the atom's electrons and generating an electron positron pair. This happens with quite high probability in all interactions when photons go through matter, actually it is one of the ways we know a photon has been generated in an interaction, as in this bubble chamber picture:


Accelerated charges radiate

We see photons coming out from the parent electron and positron pair, which lose energy, and the photons generate new electron positron pairs after a neutral ( no bubbles) path interval.

The interaction is

pair production diagram

The off mass shell interaction with a field (of the atoms in the bubble chamber liquid here) is necessary for conservation of the four momentum: the photon has mass 0 whereas the pair has at least 1 Mev ( the sum of their masses).

  • $\begingroup$ Is there any interaction with the Higgs field during this change from momentum to mass? $\endgroup$ – Jitter Jul 14 '13 at 12:06
  • $\begingroup$ @Jitter The higgs field permeates everything, it is like a coordinate system change , not an interaction entering a feynman graph . see answers here physics.stackexchange.com/questions/17944/… $\endgroup$ – anna v Jul 14 '13 at 12:30

A charged particle has an electric field (containing electric energy) which itself influences the mass of that particle in the sense that the mass of the particle is higher, than it would be for a neutral particle (a particular example is that of the pions, where $\pi^{0}$ is lighter than the charged pions $\pi^{\pm}$). In that sense one could say that energy effects the mass and vice versa. As Einstein states both are strongly correlated.

  • $\begingroup$ The eletroweak process $\gamma\gamma\rightarrow \text{W}^{+}\text{W}^{-}$ where a photon pair (massless) creates a pair of massive Gauge bosons of the Weak interaction, is another example. If the conservation laws of physics are respected, every possible process in physics is allowed! $\endgroup$ – Hansenet Jul 12 '13 at 22:41

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