Are black holes black?

Question

GR was formulated before atomic clocks and radio signals between earth and space probes allowed actual measurements to be made.

Atomic clocks count a beat frequency, but the frequency of emitted photons is actually a measure of the energy levels within an atom. As an atomic clock is lowered doing work, its slowed rate is proof that the energy to do that work came from the energy $E = mc^2$ of its matter. This leads to a simple differential equation $c^2 dm = m d\Phi$ with solution $m = m_0 e^{\Phi/c^2}$

Thus gravitational potential has a physical effect on matter and by consideration of the effect on the time delay on radio signals deduce that the lengths of rulers and the rate of clocks are affected by factors of $e^{\Phi/c^2}.$

GR coincides with this in the weak field approximation which equates to the first two terms of the exponential series. Correcting our understanding of gravity by replacing the weak field approximation with exponential function, we avoid the concepts of event horizons and singularities.

So bearing the above in mind, are black holes hiding behind even horizons, or is their light simply red shifted too far to detect?

Answer 1

Your solution is $dE=c^2 dm$ (from $E=mc^2$) combined with $dE=m\,d\Phi$ (from $E=\Phi m$) giving

$$c^2 dm=m\,d\Phi$$

Or

$$\dfrac{dm}{m}=\dfrac{d\Phi}{c^2}$$

That solves

$$m = m_0 e^{\frac{\Phi}{c^2}}\tag{1}$$

However, you have overlooked the fact that $m=m(\Phi)$, thus $dE=m\,d\Phi{\color{red}{+\Phi\,dm}}$. This gives

$$c^2 dm=m\,d\Phi+\Phi\,dm$$

Or

$$\dfrac{dm}{m}=\dfrac{d\Phi}{c^2-\Phi}$$

Solving

$$m=\dfrac{m_o}{1-\dfrac{\Phi}{c^2}}\tag{2}$$

Note that the first two terms of the Taylor series are the same for $(1)$ and $(2)$ referring to the Newtonian gravity, but not General Relativity.

None of this has to do with the existence of the event horizon, because $\Phi(r)$ is not defined above. Its definition in relativity comes from time dilation. For example, in the Schwarzchild solution with no motion

$$\dfrac{d\tau}{dt}=\sqrt{1-\dfrac{r_{\text{s}}}{r}}$$

Where $r_{\text{s}}=\dfrac{2GM}{c^2}$ is the radius of the event horizon. Accordingly

$$\dfrac{\Phi}{c^2}=1-\dfrac{1}{\sqrt{1-\dfrac{r_{\text{s}}}{r}}}$$

Note that even if your solution $(1)$ were correct, $\Phi(r)$ would still make $m=0$ at the event horizon $r=r_{\text{s}}$.