# Are black holes black?

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?

• From who's point of view? – Rob Jeffries Aug 19 '18 at 8:27
• Not a duplicate of that question @JohnRennie but possibly a duplicate of a question about the formation of black holes. The question here is about the existence or not of an event horizon (which is assumed in your other answer). – Rob Jeffries Aug 19 '18 at 8:32
• @RobJeffries hmm, OK, I'm not sure I agree but I have reopened the question. – John Rennie Aug 19 '18 at 8:34
• Possible duplicate of Why is a black hole black? – John Rennie Aug 19 '18 at 8:36
• "So bearing the above in mind, are black holes hiding behind even horizons, or is their light simply red shifted too far to detect?" This question might be answered here by John Rennie. – timm Aug 19 '18 at 8:36

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 $r=r_{\text{s}}$.

• That is fudge. Lifting a 1Kg mass 1 metre mdPhi = 9.82 while Phidm = 6.833*10^-9 and is a second order term which under the rules of differential calculus should be neglected. – Bruce Harvey Aug 28 '18 at 10:33
• @BruceHarvey It's not a second order ($\ne\text{d}^2$) and you know it. The Earth gravity is just a Newtonian limit. You can't neglect in a stronger field, but even if you do, it wouldn't remove the horizon. Read the last sentence in the answer. – safesphere Aug 28 '18 at 13:38
• What you quote is the differential of E = mPhi which gives dE = mdPhi +Phi * dm and that makes your fudge seem plausible. but this is m*dPhi which is quite different. In ant course on calculus – Bruce Harvey Aug 28 '18 at 22:28
• In any course on calculus, this equation of exponential decay is the simplest example of the simplest type of differential equation, the separable variable DE and it applies to many situations in nature and physics such as the voyage of a capacitor discharging through a resistor. – Bruce Harvey Aug 28 '18 at 22:36
• It doesn't matter, the result is similar showing the same behavior - the mass-energy is reduced to zero at the horizon, so nothing crosses it. I have upvoted your question, because I agree that these is no singularity. A black hole is a mere result of energy conservation. – safesphere Aug 28 '18 at 22:43