Planck Particle...
Norman said:
So something of Planck mass would probably obey macroscopic laws.
If Planck mass obeys macroscopic laws, and only microscopic particles obey De-broglie waves, then it must be stated that:
Macroscopic Law:
r_p > \overline{\lambda_p} - Planck radius greater than Planck-De Broglie wavelength
Microscopic Law:
r_p = \overline{\lambda_p} - Planck radius equals Planck-De Broglie wavelength
Planck Wavelength solution:
\overline{\lambda}_p = \frac{\hbar}{m_p c} = \frac{\hbar}{c} \sqrt{\frac{G}{\hbar c}}
\overline{\lambda}_p = \sqrt{\frac{\hbar G}{c^3}}
\boxed{r_p = \overline{\lambda}_p = \sqrt{\frac{\hbar G}{c^3}}}
Although Planck mass appears to be a macroscopic entity, it is in fact a microscopic entity which obeys quantum laws and therefore, is a quantum 'particle'.
vincentchan said:
theoretical physics set no upper limit on energy of photon...
- (see reference 2)
Maxwell's equations, which the derivatives describe all electromagnetic phenomena, do not describe any theoretical limit to a photon's energy, however given that photon energy is quantizised, a possible Maxwell energy solution is:
E_n = \frac{n \hbar}{\overline{\lambda} \sqrt{\mu_o \epsilon_o}}
Given that there is no theoretical upper limit to photon energy and given that Planck mass IS a microscopic quantum particle, then reaction 3 listed above IS energetically feasible (possible).
A photon is an electromagnetic wave, and all electromagnetic waves obey the Principle of Superposition:
For two or more photons, the resultant wave function at any point is the algebraic sum of the wave functions of the individual waves.
Is reaction 1 possible through the Principle of Superposition constructive interference?
\psi (E_a)_{\gamma} + \psi (E_b)_{\gamma} = 2 \psi (2E_t)_{\gamma} \rightarrow e^+ + e^- \; \; \; E_a = E_b = m_e c^2
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Reference:
http://galileo.phys.virginia.edu/classes/252/wave_equations.html
http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/970412e.html
