## Definition

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

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

## Spin

Seeds are indestructible so the foregoing definition of a graviton implies that

$N^{\sf{U}} \left( \sf{\Gamma} \vphantom{\gamma} \right) = N^{\sf{U}} \left(\gamma\right) + N^{\sf{U}} \left(\overline{\gamma}\right)$ and $N^{\sf{D}} \left( \sf{\Gamma} \vphantom{\gamma} \right) = N^{\sf{D}} \left(\gamma\right) + N^{\sf{D}} \left(\overline{\gamma}\right)$

Swapping quarks with anti-quarks doesn't change seed counts, so $N^{\sf{U}} \left(\gamma\right) = N^{\sf{U}} \left(\overline{\gamma}\right)$ and $N^{\sf{D}} \left(\gamma\right) = N^{\sf{D}} \left(\overline{\gamma}\right)$. Then the seed coefficients of the graviton are given by

$N^{\sf{U}} \left(\sf{\Gamma} \vphantom{\gamma} \right) = 2 N^{\sf{U}} \left(\gamma\right)$ and $N^{\sf{D}} \left( \sf{\Gamma} \vphantom{\gamma} \right) = 2 N^{\sf{D}} \left(\gamma\right)$

Using these values, the spin of a graviton is found as

$\begin{align} \sigma ^{\sf{\Gamma}} \equiv \frac{ \, \left| \, N^{\mathsf{U}} \left(\sf{\Gamma} \vphantom{\gamma} \right) - N^{\mathsf{D}} \left(\sf{\Gamma} \vphantom{\gamma} \right) \, \right| \, }{8} = \frac{ \, \left| \, 2 N^{\mathsf{U}} \left(\sf{\gamma} \vphantom{\gamma} \right) - 2 N^{\mathsf{D}} \left(\sf{\gamma} \vphantom{\gamma} \right) \, \right| \, }{8} = 2 \frac{ \, \left| \, N^{\mathsf{U}} \left(\sf{\gamma} \vphantom{\gamma} \right) - N^{\mathsf{D}} \left(\sf{\gamma} \vphantom{\gamma} \right) \, \right| \, }{8} =2 \, \sigma ^{\sf{\gamma}} \end{align}$

But by definition the spin of *any* photon is one. So for all gravitons

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

## Quantum Numbers

Gravitons are defined by the union of a particle $\gamma$ with its matching anti-particle $\overline{\gamma}$. So the net number of any sort of quark in a graviton is zero, including down-quarks. 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.

## Radii

Let us assess the radius vector for a graviton. 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 $\overline{\rho}^{\, \sf{\Gamma}} = (0, 0, 0)$ too. Moreover, since $\Delta n ^{\sf{\Gamma}} = 0$ for all types of quarks, including down-quarks, we know that the inner radius of any graviton will be nil as well

$\rho_{\LARGE{\bullet}} ^{\, \sf{\Gamma}} = 0$

So when a large particle absorbs a graviton, we may consider that the graviton is located at its center. This is quite different from photons where the inner radius is *always* greater than zero, and photons are vaguely located in some surrounding field.

## Mass

Since $\overline{\rho}^{\, \sf{\Gamma}} = (0, 0, 0)$, no work $W$ is required to assemble the quarks in a graviton; $W^{\sf{\Gamma}} \equiv k_{\sf{F}} \left\| \, \overline{\rho}^{\, \sf{\Gamma}} \right\| = 0$. This result is combined with the null value for the enthalpy to find the mass of a graviton as

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

## Momentum & Gravitational Force

Recall that the wavevector $\overline{\kappa}$ is defined from sums and differences of quark coefficients, and quarks are indestructible, so $\overline{\kappa}^{\, \sf{\Gamma}} = \overline{\kappa}^{\, \gamma} + \overline{\kappa}^{\, \overline{\gamma}}$. Also, the wavevector of any particle is symmetrically opposed to the wavevector of its matching anti-particle, $\overline{\kappa}^{\, \sf{\gamma}} = - \, \overline{\kappa} ^{\, \overline{\gamma}}$. So the wavevector of a graviton is

$\overline{\kappa}^{\, \sf{\Gamma}} = \overline{\kappa}^{\, \gamma} + \overline{\kappa}^{\, \overline{\gamma}} = \overline{\kappa}^{\, \gamma} - \overline{\kappa}^{\, \gamma} = (0, 0, 0)$

Then, in a frame of reference noted by F, the momentum of a graviton is given by

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

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, and this direction is determined by the character of the reference frame. Overall, gravitons are certainly "not like rain".^{1} They have no mass, there is no 'drag'. And their momenta are uniform. So **gravity**, the force they carry, is uniformly directed too.

## Energy

The foregoing results can be substituted into the definition of mechanical energy to obtain

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

So in a perfectly inertial reference frame where $\tilde{\kappa}^{ \sf{F}} = (0, 0, 0)$ gravitons carry no energy or momentum. If we assume that a frame is perfectly inertial, then we can ignore gravity. For non-inertial frames, both the momentum and energy of a graviton are directly proportional to the total number of quarks it contains.