# Where does mass go when energy is converted in to photons?

If matter and anti-matter annihilate each other they emit a photon with the energy that corresponds to the mass, right? This is the best example I could think of matter/energy being converted directly in to photons. Now, if a photon doesn’t have mass (because it goes at the speed of light) where does all that mass from the matter and anti-matter go? Doesn`t it break the law of conservation of mass? If photons have energy why dont they have mass?

Thank you in advance for your professional help

• There is no law of conservation of mass: it's energy that is conserved. Photons have an energy proportional to their frequency. So as you increase the masses of your particles of matter and antimatter, the resulting photons will have higher frequencies. Commented Mar 26, 2015 at 14:54
• Also, conservation of momentum forbids the creation of a single photon: they have to come in pairs. Commented Mar 26, 2015 at 14:55

From a relativistic point of view, the energy of a particle with mass $m$ and momentum $p$ is given by: $$E^2=m^2c^4+p^2c^2$$ where $c$ is the speed of light. You can clearly see that a massive particle will have some mass energy $E_0=mc^2$, but also some kinetic energy.
The photon being massless, the above equation reduces to: $$E_\gamma=pc$$ One consequence of this equation is that if a photon exists, i.e. has a non-zero energy, it must be moving.
There is no "conservation of mass" law in physics - instead, we use the more general concept of conservation of energy (since from the first equation above, mass can be understood as a form of energy). The energy of the photon is related to its frequency by the Planck-Einstein relation: $$E_\gamma=h\nu$$ where $h$ is Planck's constant and $\nu$ is the frequency of the associated electromagnetic wave (remember that the photon is "the particle of light", i.e. an excitation of a field). Since you know the speed of light, $c$, and its frequency $\nu$, you can also derive its wavelength: $$c=\lambda\nu$$ And the energy of the photon can then be expressed as: $$E_\gamma=h\nu=\dfrac{hc}{\lambda}$$ Finally we get: $$E_\gamma=pc=\dfrac{hc}{\lambda}\Longrightarrow \lambda=\dfrac{h}{p}$$ which is nothing else than the de Broglie wavelength, and the basis for quantum mechanics. Indeed, this postulates that a particle (not necessarily massless) with momentum $p$ can be represented as a wave with wavelength $\lambda$.