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 conserved1 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.

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