Notice: this page is under construction
Notice: this page is under construction


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

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

Angular Momentum

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 angular momentum quantum number of a graviton is found as

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

But by definition the angular momentum quantum number of any photon is one. So for all gravitons

$𝘑 \, ^{\sf{\Gamma}} = 2$

Other 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.

//Bead Panel// from a baby carrier, Ngaju people. Borneo 20th century, 41 x 26 cm. From the Teo Family collection, Kuching. Photograph by D Dunlop.
Bead Panel from a baby carrier, Ngaju people. Borneo 20th century, 41 x 26 cm. From the Teo Family collection, Kuching. Photograph by D Dunlop.


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.


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

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

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

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

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, 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.


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. 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.

Examples of Gravitons


Here is a link to the most recent version of this content, including the full text.

favicon.jpeg Gravitons
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License