# Planck Boson

## Main Question or Discussion Point

According to my research on the Standard Model, gravitation is mediated by a massless graviton with spin 2.

Upon examining the unity of comparison between cosmic time and quantum unification listed in reference 1, the unification of two forces is scaled at the rest mass (energy) of their bosons respectively, except the photon which is massless and unifies at the photon rest energy.

However, upon examining the chart provided in reference 1, the boson for TOE unification is a 'Planck Boson' at the Planck singularity rest mass (energy). According to my limited understanding of the Standard Model, a Planck mass is not a member of the particle family. For example, colliding any two particles in the SM particle family at any energy scale will not result in the production of a Planck mass Boson and there is no known Feynman diagram for such a reaction.

According to my understanding of General Relativity, the Planck mass is the gravitational Boson, however, according to the Standard Model, the graviton is the gravitational Boson.

My question is, have physicists seriously considered the possibility that even though the fine structure interaction strengths are the same for General Relativity and the Standard Model at Planck scale energies, they do not unify at Planck scale energies with a massless graviton Boson, but continue to exist as different entities within their own equation of states?

If this statement is true, then gravitons do not exist.

Reference:
http://hyperphysics.phy-astr.gsu.edu/Hbase/astro/unify.html" [Broken]

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Confirming our discussions,
Planck’s constant h is Joule-seconds, so E = h per second or hf
Since mass is E/c^2 It follows that the corresponding Planck value for mass will necessarily be 6.6 x 10^-34/c^2 or 7.37 x 10^-51 kilogramme-seconds
You noted that in Planck units, Planck mass x Planck time is 1.173 x 10^-51 kg-s.
These Planck units are based on the Schwarzschild radius of a non rotating black hole, however it is clear to see that if there is a rotation, then either the Planck length and Planck time must increase, and/or the radius reduced, otherwise a photon can escape.
The answer is very simple, if you substitute h for h-bar in the Planck unit equations, then Planck mass x Planck time = 7.37 x 10^-51 kilogramme-seconds.
Also if I put the revised Planck length in my “space curvature” algorithm I get the appropriate revised values for the Planck units.
As I said I am not prepared to discuss my theory for the quantisation of space in an open forum, so if you, or anyone else wants to discuss this you will need to contact me privately, so I know exactly who I’m talking to.

malawi_glenn
Homework Helper
hey Goccam you were the guy struggeling with that special relativity relation right? Now do you have your own theory of space-time quantization? That sounds pretty neat (irony)

Hi,

this question might get a better attention in the BtSM sub-forum.

Goccam said:
Planck’s constant h is Joule-seconds, so E = h per second or hf
Actually, the equation used for this scale is,

Planck energy:
$$E_p = \hbar \omega_p$$

Planck angular frequency:
$$\omega_p = \frac{1}{t_p} = \sqrt{\frac{c^5}{\hbar G}}$$

Integration by substitution:
$$E_p = \hbar \omega_p = \hbar \sqrt{\frac{c^5}{\hbar G}} = \sqrt{\frac{\hbar c^5}{G}}$$
$$\boxed{E_p = \sqrt{\frac{\hbar c^5}{G}}}$$

Goccam said:
It follows that the corresponding Planck value for mass will necessarily be 6.6 x 10^-34/c^2 or 7.37 x 10^-51 kilogramme-seconds
Negative, the value of Planck mass is,

Planck mass:
$$m_p = \sqrt\frac{\hbar c}{G} = \text{2.176} \cdot \text{10}^{-8} \; \text{kg}$$
$$\boxed{m_p = \text{2.176} \cdot \text{10}^{-8} \; \text{kg}}$$

Also, your SI (systeme international) units for mass is incorrect, the SI unit for mass is 'kg' (kilograms), NOT 'kg s' (kilogram seconds).

Planck mass multiplied by Planck time:
$$m_p \cdot t_p = \sqrt\frac{\hbar c}{G} \cdot \sqrt{\frac{\hbar G}{c^5}} = \frac{ \hbar}{c^2} = 1.173 \cdot 10^{-51} \; \text{kg} \cdot \text{s}}$$
$$\boxed{m_p \cdot t_p = \frac{\hbar}{c^2}}$$
$$\boxed{m_p \cdot t_p = 1.173 \cdot 10^{-51} \; \text{kg} \cdot \text{s}}$$

Goccam said:
These Planck units are based on the Schwarzschild radius of a non rotating black hole, however it is clear to see that if there is a rotation, then either the Planck length and Planck time must increase, and/or the radius reduced, otherwise a photon can escape.
If spin in included in the model, then the Kerr metric is used, which results in a model with an oblate spheroid ergosphere, however the General Relativity solutions for the Planck units remains the same for the Kerr metric.

Goccam said:
As I said I am not prepared to discuss my theory for the quantisation of space in an open forum,...
This is an unfortunate ideology, as peer review is part the the scientific process of discovery and has proven to be a very valuable research tool. Without exception, the only scientific models that can withstand the test of time are those scientific models that have been subjected to peer review, therefore I strongly encourage you to reconsider such a philosophical approach.

If the Standard Model (SM) of Quantum Chromodynamics (QCD) and General Relativity (GR) do not unify at the Planck scale (TOE) level, then the result is a Planck Boson composed of a degenerate Quark-Gluon Plasma (QGP), and gravitons do not exist.

Reference:
http://en.wikipedia.org/wiki/Planck_energy" [Broken]
http://en.wikipedia.org/wiki/Planck_angular_frequency" [Broken]
http://en.wikipedia.org/wiki/Planck_mass" [Broken]
http://en.wikipedia.org/wiki/Kerr_metric" [Broken]
http://hyperphysics.phy-astr.gsu.edu/Hbase/astro/unify.html" [Broken]

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malawi_glenn
Homework Helper
Orion, do you do research with wikipedia as source? hehe

If spin in included in the model, then the Kerr metric is used, which results in a model with an oblate spheroid ergosphere, however the General Relativity solutions for the Planck units remains the same for the Kerr metric.

Quite so, but this solution is based upon a rotation about a polar axis, thus the need for the oblate ergosphere. It occurred to me that a simpler solution was to rotate the polar axis in addition to the equatorial rotation. If these rotation rates are equal, then the locus of a point at escape velocity would be at 4pir anywhere on a spherical surface. The Planck unit of angular momentum would then be h/2pi x 2pi = h.
Substituting this in the equations,
Planck mass:
$$m_p = \sqrt\frac{\ h c}{G} = \text{4.9032}\cdot \text {10}^{9}J / {c^2} = {5.4552} \cdot \text{10}^{-8} \; \text{kg}$$
And Planck time
$$t_p = \sqrt{\frac{\ h G}{c^5}} = \text{1.351383} \cdot \text{10}^{-43} \; \text{sec}$$
So
$$m_p \cdot t_p = \sqrt\frac{\ h c}{G} \cdot \sqrt{\frac{\h G}{c^5}} = \frac{ \ h}{c^2} = 7.3724946 \cdot 10^{-51} \; \text{kg} \cdot \text{s}}$$
This is “Planck’s constant” for mass; mass is $$7.3724946 \cdot 10^{-51}$$per second.
therefore I strongly encourage you to reconsider such a philosophical approach.

It’s not that. I am still exploring the consequences of my findings, some of which are controversial. For instance I was hoping that the discussion on relativity would go into reference frames, but it didn’t. Similarly the posting on stability curves also falls out of the findings, but this has not prompted any discussion.
I was hoping to find someone to look at the implications without the knee-jerk reaction that it does not fit the conventional wisdom.