It's experimental fact that the relation between energy and momentum of a photon is

*consistent* with prediction of electromagnetism E = c | p | which in turn is consistent with the theory of massless particles in the intersection of quantum mechanics and special relativity. But in all those examples, momentum is strictly conserved and the photons move in straight lines. Therefore you cannot derive p = E/c from these theories or a strong analogy with a photon.

But even the best experiment can't prove a relation like the above. The currently accepted experimental upper bound on a photon's mass is $$10^{-18} \, \textrm{eV}$$ which would say a 550 nm photon might travel slower than c by as much as 931 nm / billion years, with corresponding deviation from your summary of SR by about 1 part in $$10^{37}$$. Since none of those experiments show momentum traveling in circles instead of straight lines, they do not approximately support your assumptions.

http://pdg.lbl.gov/2016/listings/rpp2016-list-photon.pdf
Back before we had confirmed atomic theory, in Einstein's 1905 paper

*Concerning an Heuristic Point of View Toward the Emission and Transformation of Light*, he basically invents the photon to explain the thermodynamics of black body radiation with confirmation in the photoelectric effect, thus

*unifying* two distinct phenomena under a single theoretical framework.

He took Planck's experimentally successful 1901 formula for energy density at a given (cyclical) frequency:

$$\rho_{\nu} \, d \nu \; = \; \frac{\color{red}{\alpha} \nu^3}{e^{\color{red}{\beta} \nu / T} - 1} \, d\nu$$

And it's asymptotic behavior at low values of $$\nu /T $$ is:

$$ \rho_{\nu} = \frac{\color{red}{\alpha}}{\color{red}{\beta}} \nu^2 T - \frac{1}{2} \alpha \nu^3 + \frac{1}{12} \alpha \beta \nu^4 / T - \dots $$

While the a thermodynamic argument suggests that

$$ \rho_{\nu} \approx \frac{R}{N_A} \frac{8 \pi}{c^3} \nu^2 T $$

and then hypothesizes that $$ \frac{R}{N_A} \frac{8 \pi}{c^3} = \frac{\color{red}{\alpha}}{\color{red}{\beta}}$$ in the low-frequency, high-temperature limit.

Nowadays, the thermodynamic gas constant is written in terms of Boltzmann's constant and Avogadro's number : $$R= k_B N_A$$.

So $$\frac{\color{red}{\alpha}}{\color{red}{\beta}} = \frac{8 \pi k_B}{c^3}$$ (Section 2)

In sections 3, 4 and 5, he argues that the thermodynamic entropy of blackbody radiation resembles that of an ideal gas.

Finally in section 6 he shows that the energy of an "atom" of that blackbody radiation needs to have energy $$E = \frac{R \color{red}{\beta}}{N_A} \nu = k_B \color{red}{\beta} \nu$$. Inventing the symbol, h, to stand for that constant of proportionality, we have:

$$ E = h \nu \\ \color{red}{\beta} = h/ k_B \approx 4.79924466 \times 10^{-11} \, \textrm{Kelvin per Hertz} \\ \color{red}{\alpha} = \frac{8 \pi h}{c^3} \approx 6.18064462 \times 10^{-58} \, \textrm{Joules per Hertz}^4 \, \textrm{per meter}^3\\ \rho_{\nu} \, d \nu \; = \; \frac{\frac{8 \pi h }{c^3} \nu^3}{e^{\frac{h \nu }{ k_B T}} - 1} \, d\nu$$

Which is the modern form of Planck's 1901 law.

https://en.wikipedia.org/wiki/Planck's_law#Spectral_energy_density_form
http://pdg.lbl.gov/2016/reviews/rpp2016-rev-phys-constants.pdf
A neat trick, but then Einstein tries to push his analogy with particles of an ideal gas until it breaks. Because the author of an idea is the easiest one to be fooled into thinking the idea is a good one. In science, we need something more than

*soi disant* experts.

In section 7, he explains that Stokes's rule of photoluminescence seems naturally consistent with his model.

https://en.wikipedia.org/wiki/Stokes_shift
In section 8, he explains that weak source photoelectric effect seems naturally consistent with his model.

In section 9, he characterizes the ionization of dilute gasses as something with a fixed energy scale, which corresponds to the modern view of the ionization of gases at the scale of ~ 10 eV, and makes a positive prediction.

Here we see the difference between theory and experiment. Experiment can't say the energy of a photon is $$h \nu$$, but only that it is close to that value for certain light, within experimental uncertainty. Likewise, it should be clear that no series of experiments can show that $$E = p c$$ for photons, but only that the photons tested have that momentum (if tested singlely) or

*average* momentum (if tested en masse). To say that $$E = p c$$ one has to go beyond summarizing experiment and make the generalization common to all physical theories.

A scientific theory is a communicable framework for describing precisely the observable behavior of a large class of related phenomena. The equals sign connecting physical quantities is a precise description, more precise that experiment can prove. Experiment can only show if a given test of a theory is consistent or not consistent with that theory.

Here, we are told that (given the electron is at rest) $$E = m_e c^2$$ is assumed, $$p = m_e c$$ is assumed, but momentum is not conserved, resulting in a helical path through space-time. This does nothing to explain the rest mass of the electron and discards the rigid framework of SR without replacement to explain why momentum is conserved in experiments.

You have failed to elucidate any principle.

Nothing is wrong with

*defining* (in SR) Kinetic energy as the difference between the energy of a body and c² times its invariant rest mass. It's consistent with your SR-derived (γ−1)mc² rule (which only applies to massive particles) as well as the equally effective $$E - \sqrt{E^2 - c^2 p^2}$$ which works even for massless particles, and the E−E₀ conceptual model and naturally extends to massless bodies which carry energy and momentum.

For a free particle, SR dictates: $$E^2 = (mc^2)^2 + (c p)^2 $$ and $$E \vec{v} = c^2 \vec{p}$$.

For a free particle, QM dictates: $$E = h \nu = \hbar \omega \\ p = h / \lambda = \hbar k $$

For a wave, we have $$v_{\textrm{phase}} = \lambda \nu = \omega / k $$ while $$v_{\textrm{group}} = \frac{\partial \omega}{\partial k}$$

Putting those together, we have $$v_{\textrm{group}} = \frac{\partial \omega}{\partial k} = \frac{\partial h \omega}{\partial h k} = \frac{\partial E}{\partial p} = \frac{\partial mc^2 \sqrt{1 + \frac{p^2}{m^2c^2}} }{\partial p} = \frac{c^2 p}{\sqrt{(mc^2)^2 + (c p)^2}} = \frac{c^2 p}{E} = v$$, $$v_{\textrm{phase}} = \frac{\omega}{k} = \frac{E}{p} = \frac{c^2}{v}$$.

Plus, as momentum and energy are properties of the excitation of the wave function, there is no conceptual stumbling block to working with massless particles that convey energy and momentum across finite distances.

That's the power of theory.