Press "Enter" to skip to content

Gravitons

a spectrum-like image used as an icon for photons

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

\mathsf{\Gamma}   \equiv  \left\{ \, \boldsymbol{\gamma}, \, \overline{\boldsymbol{\gamma}} \, \rule{0px}{11px} \right\}

Let  \mathsf{U} and  \mathsf{D} note the number of up seeds and down seeds in a particle. And recall that seeds are conserved. So the foregoing definition of a graviton implies that

N^{\mathsf{U}} \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right) = N^{\mathsf{U}} \! \left( \boldsymbol{\gamma} \rule{0px}{10px} \right)  +  N^{\mathsf{U}} \! \left( \overline{\boldsymbol{\gamma}} \rule{0px}{10px} \right)

and

N^{\mathsf{D}} \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right) = N^{\mathsf{D}} \! \left( \boldsymbol{\gamma} \rule{0px}{10px} \right)  +  N^{\mathsf{D}} \! \left( \overline{\boldsymbol{\gamma}} \rule{0px}{10px} \right)

Swapping quarks with anti-quarks does not change thermodynamic seed counts, so N^{\mathsf{U}} \!  \left( \boldsymbol{\gamma} \rule{0px}{10px}\right) = N^{\mathsf{U}} \! \left( \overline{\boldsymbol{\gamma}} \rule{0px}{10px}\right) and N^{\mathsf{D}} \! \left( \boldsymbol{\gamma} \rule{0px}{10px}\right) = N^{\mathsf{D}} \! \left( \overline{\boldsymbol{\gamma}} \rule{0px}{10px}\right). Then the seed coefficients of the graviton are given by

N^{\mathsf{U}} \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right) = 2N^{\mathsf{U}} \! \left( \boldsymbol{\gamma} \rule{0px}{10px} \right)

and

N^{\mathsf{D}} \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right) = 2N^{\mathsf{D}} \! \left( \boldsymbol{\gamma} \rule{0px}{10px} \right)

Using these values, the angular momentum quantum number of a graviton is found as

    \begin{align*} \textsl{\textsf{J}} \left(  \mathsf{\Gamma} \rule{0px}{10px}\right) &\equiv \dfrac{ \,  \left|  \,  N^{\mathsf{U}}  \! \left( \mathsf{\Gamma} \rule{0px}{10px}\right) - N^{\mathsf{D}} \! \left(\mathsf{\Gamma} \rule{0px}{10px}\right) \, \right| \,  }{8}   \\   &= \dfrac{ \, \left| \, 2N^{\mathsf{U}} \! \left( \boldsymbol{\gamma} \rule{0px}{10px}\right) - 2N^{\mathsf{D}} \! \left(\boldsymbol{\gamma} \rule{0px}{10px}\right) \, \right| \, }{8}   \\   &= 2 \, \dfrac{ \, \left| \, N^{\mathsf{U}} \! \left( \boldsymbol{\gamma} \rule{0px}{10px}\right) - N^{\mathsf{D}} \! \left(\boldsymbol{\gamma} \rule{0px}{10px}\right) \, \right| \, }{8}   \\   &= 2 \, \textsl{\textsf{J}} \left( \boldsymbol{\gamma} \right) \rule{0px}{15px} \end{align*}

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

    \begin{align*} \textsl{\textsf{J}} \left( \mathsf{\Gamma} \rule{0px}{10px} \right) = 2 \end{align*}

Gravitons are defined by the union of a photon with its matching anti-photon. So the net number of any sort of quark in a graviton is zero. This specifically includes down quarks. Photons have mismatched quantities of down-quarks and down anti-quarks. But for gravitons, these imbalances cancel each other perfectly so that  \Delta n = 0 for all types of quarks. Substituting these conditions into the definitions for charge, strangeness, lepton number, baryon number and enthalpy gives

q \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right)=0

S \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right)=0

L \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right)=0

B \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right)=0

and

H \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right)=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.

The Location of a Graviton

When some particle P absorbs a graviton, we say that the graviton is located at P’s center of gravity. To be more exact, let us assess the radius vector for a graviton. Recall that  \overline{\rho} is defined from sums of quark coefficients, and quarks are conserved. So

\overline{\rho}  \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right)  =  \overline{\rho}  \! \left( \boldsymbol{\gamma} \rule{0px}{10px} \right) +   \overline{\rho}  \! \left( \overline{\boldsymbol{\gamma}} \rule{0px}{10px} \right)

But as discussed earlier, any particle and its anti-particle have symmetrically opposed radius vectors, so

\overline{\rho}  \left( \boldsymbol{\gamma} \rule{0px}{10px}  \right) = - \overline{\rho}  \left(  \overline{\boldsymbol{\gamma}}  \right)

and

\overline{\rho}  \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right) = (0, 0, 0)

Thus the graviton is located at the origin of a quark space that is centered on P. So we say that the center of gravity is at the center of P. And since  \Delta n \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right) = 0 for all types of quarks, including down-quarks, the inner radius of any graviton is nil

    \begin{align*} \rho_{in} \! \left( \mathsf{\Gamma} \rule{0px}{10px} \rule{0px}{10px} \right) = 0 \end{align*}

An inner radius of zero means the graviton can be in P’s core. So overall, the center-of-gravity, the center-of-P and the core-of-P are all understood to be in the same location.

Graviton Dynamics

Since \overline{\rho} \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right) = (0, 0, 0) no work  W is required to assemble the quarks in a graviton

    \begin{align*} W  \hspace{-3px} \left( \mathsf{\Gamma} \rule{0px}{10px} \right) \equiv k_{\mathsf{F}} \left\| \, \overline{\rho} \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right) \right\| = 0 \end{align*}

This result is combined with the null value for the enthalpy to find the mass of a graviton as

    \begin{align*} m \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right) \equiv \frac{1}{ c^{2} } \sqrt{ H^{2}-W^{2} }  = 0 \end{align*}

So gravitons have no mass, they are ethereal. And as for their motion, recall that the wavevector  \overline{\kappa} is defined from sums of quark coefficients, and quarks are conserved. So the wavevector of a graviton is the sum of the wavevectors of its component photons. But the wavevector of any particle is symmetrically opposed to the wavevector of its matching anti-particle. Thus

    \begin{align*}  \overline{\kappa}  \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right)  &=  \overline{\kappa}  \! \left( \boldsymbol{\gamma} \rule{0px}{10px} \right) +   \overline{\kappa}  \! \left( \overline{\boldsymbol{\gamma}} \rule{0px}{10px} \right)   \\   &= \overline{\kappa} \! \left( \boldsymbol{\gamma} \rule{0px}{10px} \right) - \overline{\kappa} \! \left( \boldsymbol{\gamma} \rule{0px}{10px} \right) \rule{0px}{15px} \\  &=  (0, 0, 0)\rule{0px}{14px} \end{align*}

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

    \begin{align*} \overline{p} \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right) \equiv \frac{h}{2\pi} \! \left[ \,  \overline{\kappa} \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right)  -  N \hspace{-3px} \left( \mathsf{\Gamma} \rule{0px}{10px} \right) \widetilde{\kappa}^{ \mathsf{F}} \rule{0px}{14px} \right]  =  \frac{-h \widetilde{\kappa}^{\mathsf{F}} }{2\pi} \, N \hspace{-3px} \left( \mathsf{\Gamma} \rule{0px}{10px} \right) \end{align*}

where  N  ( \mathsf{\Gamma} ) is the total number of quarks in  \mathsf{\Gamma}, including all types. 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”1J.N. Patterson Hume, On Beyond Darwin, page166, 1983.. 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 \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right)  &\equiv \sqrt{c^{2}p^{2} + m^{2}c^{4} \, }  \\  &=  cp \! \left( \mathsf{\Gamma} \rule{0px}{10px} \right) \rule{0px}{14px}  \\   &= \frac{ch}{2\pi}      \left\| \,   \widetilde{\kappa}^{\mathsf{F}} \right\| \, N \hspace{-3px} \left( \mathsf{\Gamma} \rule{0px}{10px} \right) \rule{0px}{16px}  \end{align*}

So in a perfectly inertial reference frame where \widetilde{\kappa}^{ \mathsf{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

Gravitons and their characteristics are listed in this spreadsheet screen shot.

For additional detail about these calculations, please see the Field Quanta spreadsheet. There is a lot more to be said about gravitons and gravity, but next we shift to a discussion of nuclear particles.

Gravitons are objectified from highly symmetric combinations of black sensations, as suggested by this bead panel from Borneo.
Baby Carrier panel, Ngaju people. Borneo 20th century, 41 x 26 cm. Photograph by D Dunlop.
References
1J.N. Patterson Hume, On Beyond Darwin, page166, 1983.