The union of a photon γ and its corresponding anti-photon γ is noted by the symbol $\sf{\Gamma}$ and called a **graviton**

$\sf{\Gamma} \equiv \left\{ \, \gamma, \, \overline{\gamma} \, \right\}$

Gravitons have the following characteristics for any pair of photons. Angular momentum is conserved^{1} and the angular momentum of a photon and its corresponding anti-photon are the same so

$\overline{J}^{\, \sf{\Gamma}} = \overline{J}^{\, \gamma} + \overline{J}^{\, \overline{\gamma}} = 2 \, \overline{J}^{\, \gamma}$

But $J$ and the spin $\sigma$ are proportional to each other, so $\sigma^{\sf{\Gamma}} =2 \sigma^{\gamma}$ and the spin of a graviton is twice the spin of a photon. Then since all photons have a spin of one by definition, all gravitons have a spin of two

$\sigma ^{\sf{\Gamma}}=2$

Also by definition, the net number of quarks in a photon is zero, and since quarks are indestructible this holds for gravitons as well. Substituting this $\Delta n = 0$ condition into expressions for charge, strangeness, lepton number, baryon number and enthalpy shows that

$q^{\, \sf{\Gamma}} =0$ $S^{\, \sf{\Gamma} } =0$ $L^{\sf{\Gamma}} =0$ $B^{\, \sf{\Gamma}} =0$ $H^{\sf{\Gamma}} =0$

Then recall that the lepton-number, baryon-number and charge are conserved, so any particle may freely absorb or emit countless gravitons without altering its own values for these quantum numbers.

A tour around the quark model of an electronic graviton. |

For a specific example of a graviton, consider the accompanying movie that shows an electronic photon combined with an electronic anti-photon. Let us assess the radius vector for a particle like this. Recall that $\overline{\rho}$ is defined from sums and differences of quark coefficients, and quarks are indestructible so $\overline{\rho}^{\, \sf{\Gamma}} = \overline{\rho}^{\, \gamma} + \overline{\rho}^{\, \overline{\gamma}}$. But as discussed earlier the radius vector of a photon is $\overline{\rho}^{\, \gamma} = (0, 0, 0)$ and the same holds for an anti-photon. So $\overline{\rho}^{\, \sf{\Gamma}} = (0, 0, 0)$ and the norm is zero too $\rho^{\sf{\Gamma}}=0$. Then $W^{\sf{\Gamma}} = 0$ and no work is required to assemble the quarks in a graviton. This result is combined with the null value for the enthalpy to find the mass of a graviton as

$\begin{align} m^{\sf{\Gamma}}\equiv \frac{1}{ c^{2} } \sqrt{ \: H^{2} - W^{2} \ \vphantom{ {\Sigma^{2}}^{2} } } = 0 \end{align}$

A similar analysis of radii finds the wavevector of a graviton to be $\overline{\kappa}^{\, \sf{\Gamma}} = (0, 0, 0)$. So a graviton is somewhat like a ground-state particle because its wavenumber is always zero. In a frame of reference noted by F, the momentum of a graviton is given by

$\begin{align} \overline{p} ^{\, \sf{\Gamma}} \equiv \frac{ h }{ 2 \pi } \left( \overline{ \kappa }^{\sf{\Gamma}} \! - N^{\sf{\Gamma}} \, \widetilde{ \kappa }^{ \sf{F}} \right) = - \frac{\; h \tilde{ \kappa }^{ \sf{F}} }{ 2 \pi } N^{\sf{\Gamma}} \end{align}$

where $\, N^{\sf{\Gamma}}$ is the total number of quarks in $\sf{\Gamma}$. This expression shows that all of the gravitons in a description have their momenta pointed in the same direction. Overall, gravitons are certainly "not like rain"^{2}. They have no mass, and their momenta are uniform, so any forces they carry are uniformly directed too. They do not cause a *drag* when absorbed by other particles. The foregoing results can be substituted into the definition of mechanical energy to obtain

$\begin{align} E^{\sf{\Gamma}} \equiv \sqrt{ \, c^{2}p^{2} + m^{2}c^{4} } = c p^{\sf{\Gamma}} = \frac{ c h }{ 2 \pi } \left\| \, \tilde{ \kappa }^{ \sf{F}} \right\| \, N^{\sf{\Gamma}} \end{align}$

So in a perfectly inertial reference frame where $\tilde{\kappa}^{ \sf{F}} = (0, 0, 0)$ gravitons carry no energy or momentum.

$\sf{P}_{\LARGE{\circ}}^{\, \Gamma}= \sf{P}_{\LARGE{\bullet}}^{\, \Gamma}$

Gravitons have phase symmetry.