# Extended Theory of Hawking Emmision due to Lorentz Analogue Deceleration Radiation

Discussion in 'Physics & Math' started by curvature, Aug 11, 2018.

1. ### curvatureRegistered Member

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A postulate in this post is that black holes when accelerated will obey their own radiation power law similar to an accelerated charge following the Larmor radiation law. It’s not an ad hoc hypothesis, since a micro black hole with a charge would have to obey the weak equivalence principle and give up radiation in a gravitational field in the form of the usual Hawking process. A charge in a gravitational field should give off an electromagnetic radiation. This can be formulated in the following way:

$P = \frac{2}{3} \frac{Q^2}{c^3} \frac{a^2}{(\frac{\sqrt{g_{tt}}(r)}{\sqrt{g_{tt}}(s)})}$

The metric $g_{tt}$ is made up of terms of $1 + \frac{\phi}{c^2}$. Higher terms then of the field are calculated like so:

$1 + \phi_1 + \phi_2 - \frac{1}{2}\phi^2_1 - \phi_1\phi_2... higher\ terms$

To first order it is just the difference formula

$\phi_1 - \phi_2 = \frac{m}{r_1r_2}(r_2 - r_1)$

However, employing a connection made by Motz, there is a correction to the denominator of the first equation,

https://www.worldscientific.com/doi/abs/10.1142/S0217732398000887

which he notes should be squared because there is one such term to account for. In our case it is simply the removal of the square root signs. A second difference between the approach I took and the one they have taken is that their denominator consists of only one such term of the metric (1 - \frac{R}{r}) whereas mine considered the actual ratio of the two signals in a relativistic way that satisfies the red shift.

$P = \frac{2}{3} \frac{Q^2}{c^3} \frac{a^2}{(\frac{g_tt (r)}{g_tt(s)})} = \frac{2}{3} \frac{Q^2}{c^3} (\frac{a}{(1 + z)})^2$

The hack link to Motz and Kraft's paper can be found here:

http://sci-hub.tw/https://doi.org/10.1142/S0217732398000887

His interpretation of the squared denominator ensures that the radiation increases as something falls towards a singularity. Of course, modern approaches are to involve situations which don't involve singularities at all. The Motz-Kraft model is for the calculation of the luminosity from charges accelerating in gravitational fields. Quasars are essentially black holes with a dense electrically charged material around it and they produce a large luminosity for these reasons. In our case, we are saying however, the weak equivalence applies to all kinds of systems with electric charge: Including the black hole itself. In such a case, the only way a black hole could radiate is through Hawking radiation suggesting an analogue case of the Larmor radiation ~ and did in fact satisfy such a case with an uncanny analogue to the classical electron radiation par a factor of $\frac{2}{3}$ which is only an adjustable parameter - the radiation (I derived) given up by a black hole is:

$\frac{Q^2a^2}{ c^3} + \frac{a^2}{Gc}(\frac{J}{m})^2 \leq \frac{a^2mr_g}{c}$

I obtained this nice simple solution from the known inequality satisfying a black hole with mass, charge and angular momentum (in natural units):

$Q^2 + (\frac{J}{m})^2 \leq m^2$

Notice this last term in the inequality is formally identical to the classical equation describing the radiation for an electron, with the exceptions that neither the mass or radius are associated to the electron but instead the black hole:

$P = \frac{2}{3}\frac{a^2m_er_e}{c}$

And it was at this point I realized this was a strong bit of evidence that charged black holes do give up radiation in motion. Since this too describes the power emitted it would not be wrong to set it in the same physics, in a gravitational field - doing so is childsplay, I get while adopting the coefficient for a true analogue theory:

$P = \frac{2}{3}\frac{a^2}{(1 - \frac{R}{r})^2}(\frac{Q^2}{c^3} + \frac{J^2}{Gm^2c}) \leq \frac{2}{3}\frac{mr_g}{c}\frac{a^2}{(1 - \frac{R}{r})^2}$

It is also true that the luminosity is a simple calculation which measures the power emitted by an astrophysical object, but we were looking for a more general case above for any charged object. The relationships for luminosity in literature are:

$L = \frac{f}{A(1 + z)^2} = \frac{\sigma T^4}{4 \pi R^2(1 + z)^2}$

You can calculate the two different luminosities as

$\frac{L_1}{L_2} = \frac{(\frac{\sigma T^4_1}{4 \pi R^2_1})}{(\frac{\sigma T^4_2}{4 \pi R^2_2})}= \frac{T^4_1 R^2_1}{T^4_2R^2_2}$

The total power emitted is

$P = \frac{Aj}{(1 + z)^2} = \frac{A \epsilon \sigma T^4}{(1 + z)^2} = \frac{\pi^2 A \epsilon k^4_BT^4}{60\hbar^3c^2(1 + z)^2}$

with the Stefan-Boltzmann constant taking the form:

$\sigma = \frac{2 \pi^2 k^4_B}{15 \hbar^3 c^2} = \frac{\pi^2 k^4_B}{60\hbar^3 c^2}$

Solving for temperature yields

$T =\ ^4\sqrt{\frac{j(1 + z)^2}{\epsilon \sigma}}$

Where$(\epsilon, j)$ is the emissivity and the radiant emittance. If the conditions of my formula hold true, since we have applied them only from first principles ~

$P = \frac{2}{3}\frac{a^2}{(1 - \frac{R}{r})^2}(\frac{Q^2}{c^3} + \frac{J^2}{Gm^2c}) \leq \frac{2}{3} \frac{mr_g}{c}\frac{a^2}{(1 - \frac{R}{r})^2} = \frac{2}{3} \frac{r_g}{mc}\frac{1}{(1 - \frac{R}{r})^2}(\frac{dp}{dt})^2$

The last term arises from:

$P = \frac{2}{3}\frac{Q^2}{m^2c^3} (\frac{dp}{dt})^2$

And can be found by replacing the acceleration in our equation for $(\frac{1}{m}\frac{dp}{dt})^2$ - this is Lorentz invariant but it isn't generally relativistic. The true generalized formula that is relativistic is

$P = -\frac{2}{3} \frac{Q^2}{m^2c^3}\frac{dp_{\mu}}{d\tau}\frac{dp^{\mu}}{d \tau}$

Since there are less dynamics in the luminosity relationship, the deduction from our equation (which involves the black hole mass, electric charge and angular momentum) is that a charged black hole will radiate more energy than one at rest: This partially makes sense of my stipulation that a moving black hole will appear hotter than one at rest. The only difference is the frame of reference you may be in, if you are moving at relativistic speeds, the black hole at rest will appear cooler. In units of $G = c = 1$ our equation to first order becomes:

$P = \frac{2}{3}\frac{a^2}{(1 - \Delta \phi)^2}(Q + \frac{J}{m})^2 \leq \frac{2}{3} mr_g\frac{a^2}{(1 - \frac{m}{r_1r_2}(r_2 - r_1))^2} = \frac{2}{3} \frac{r_g}{m}\frac{1}{(1 - \frac{m}{r_1r_2}(r_2 - r_1))^2}(\frac{dp}{dt})^2$

with

$\Delta \phi = \phi_1 - \phi_2$

and

$\phi_1 - \phi_2 = \frac{m}{r_1r_2}(r_2 - r_1)$

Let’s just say a bit about the standard equations you might encounter which also attempt to calculate the power emitted by black holes.

The total power emitted from a black hole is

$P = \sigma T^4 A$

with

$\sigma = \frac{\pi^2 k^4}{60\hbar^3c^2}$

And the Hawking blackbody in the Planck spectrum has a temeprature

$E = k_B T = \frac{\hbar g}{2 \pi c} = \frac{\hbar}{2 \pi c}(\frac{c^4}{4 Gm}) = \frac{\hbar c^3}{8 \pi Gm}$

So it seems one could just apply the tranformation of this temperature in the appropriate frame using $T = \gamma T_0$. Though I looked into a possible power equation (notwithstanding any adjustable constants) the Hawking black hole radiation power law is

$P = (\frac{16 \pi G^2m^2}{c^4})(\frac{\pi^2k^4}{60\hbar^3c^2})(\frac{\hbar c^3}{8 \pi k Gm})^4 = \frac{\hbar c^6}{15360 \pi G^2m^2}$

and

$A = 4 \pi r^2 = 4 \pi (\frac{2 Gm}{c^2})^2$

A last thing to note concerning my opinions on micro black holes, I liked the idea, but soon convinced myself that it is impossible for a black hole to reach a ground state. Even in our coldest chambers, much cooler than any place in space has never had he evidence to suggest it can. There are some excellent reasons why a back hole could never be stable due soley to the fact that they would violate the fourth law of thermodynamics - you'd have a sphere, but no radiation and certainly no motion, which is at odds with the Uncertainty Principle. I started to realize the laws need to be given the respect they are due; the extension to the fourth law of thermodynamics has flushed in a whole world of ideas that could all in some way refer to each other.

Last edited: Aug 11, 2018

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5. ### curvatureRegistered Member

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Well it has nothing to do with Weins law insomuch it has never been spoke about in this thread (until now), I am the first however to propose that a charged black hole will have to obey its own Larmor acceleration radiation through the usual Hawking process. If two supermassive black holes can emit radiation, one at rest and another in motion, my mode would state that the moving SMBH will decay first relative to the black hole drifting along in a more casual pace.

7. ### Q-reeusValued Senior Member

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I never claimed the two threads were on the same subject. Style is very similar. Sophisticated nonsense for the most part.

8. ### curvatureRegistered Member

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Ok you are entitled to your own opinion, and while larmor radiation and Hawking radiation isn't normally seen in any dynamical way with each other, my work has at least proven that it can - this would be an entirely a new physics for black holes. And i object for most part ''sophisticated nonsense.'' Instead of attacking me, focus on the theory at hand? That means, whatever it is you are objecting to, make it clear. If its going to be word salad, I am warning you, I'll just ignore.

9. ### Q-reeusValued Senior Member

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3,816
OK let's start at the start of your #1:
Actually, standard GR lore has it that a charge sitting in a static gravitational field is completely unaffected by the gravitational field. It has a mysterious ability to ignore curved spacetime entirely. See e.g. section 5.2 on Reissner-Nordstrom BH here: http://www.physics.uoguelph.ca/poisson/research/agr.pdf
I argued to the contrary elsewhere and won't recap here. Point is, your starting premise is wrong according to standard GR. If you are really meaning a so-called charged BH orbiting a massive body, then one should expect something like Larmor radiation. But modified by the spacetime curvature - exact details would require numerical methods. Since your above expression does not refer to the third time derivative of motion, as per Larmor formula for EM radiation - evidently you really meant sitting in a static field.

10. ### curvatureRegistered Member

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These questions can be answered through references I have given I suggest you study them, especially the Motz Kraft paper which started it all. But the fact you haven't read the material makes me wonder why you entered with such assurity is was ... what did you call it, ''sophisticated nonsense.'' I get bored in places like this with trolls who think they know a lot or have such massive ego's that they can falsify a paper which school yard talk. Your last attempt was better but I ended up disappointed in the end anyway... a lot of those questions would have been answered for you.

Last edited: Aug 11, 2018

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12. ### Q-reeusValued Senior Member

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Lots of insults there but no substance. Anyway I went to that link: https://sci-hub.tw/https://doi.org/10.1142/S0217732398000887
Some relevant parts:
So at least they admit there is polarized opinions wrt 'radiation' from a point charge sitting in a static gravitational field. If you can think logically, it should dawn on you that the very existence of a supposed 'charged BH' implies the E field of static electric charge does not and cannot 'feel' curved spacetime at all.
Later, last part in 4. Discussion and Conclusions:
So an admission their calculations yield predictions far from close to what is observed. Quasar emission iirc requires no such esoteric mechanism and is adequately explained by frictional heating of infalling ionized gas, plus synchrotron radiation via interaction with strong magnetic fields in polar jet regions. All happening well out from the supposed horizon.

13. ### RainbowSingularityValued Senior Member

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5,232
does this mean that entropy is relative to something ?

14. ### curvatureRegistered Member

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No but good question, the resolve I found was an issue that a change in volume can in theory explain an increase or decrease in the level of entropy in state.

Last edited: Aug 11, 2018
15. ### curvatureRegistered Member

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41
Temperature is also the only variable. The entroy is usually dimensionless but in this case, when set to the Boltzmann constant... it is well, constant. So I did found the right laws to desrbe those and will write them down soon.

16. ### NotEinsteinValued Senior Member

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1,986
curvature: Hello Reiku's sockpuppet. How are you today?

17. ### curvatureRegistered Member

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...summary of the power emitted through various formula including a magnetic contribution to Hawking emission.

My physicist friend is speculative with any influence of magnetic moments on black holes, but in any case, if they exist I speculate the following holds and is fully relativistic:

$P = \frac{\gamma \mathbf{m}}{J c^3}\frac{dp_{\mu}}{d\tau}\frac{dp^{\mu}}{d \tau}$

Which (should be) the radiation given up per magnetic moment. For redshift interpetations which satisfies the mass-charge-angular momentum inequality:

$P = \frac{2}{3} \frac{Q^2}{c^3} \frac{a^2}{(\frac{g_tt (r)}{g_tt(s)})} = \frac{2}{3} \frac{Q^2}{c^3} (\frac{a}{(1 + z)})^2$

The full classical equation that was possible derive turned out to be a [true] analogue of the same radiation given up by an electron:

$\frac{2}{3}(\frac{Q^2a^2}{ c^3} + \frac{a^2}{Gc}(\frac{J}{m})^2) \leq \frac{2}{3}\frac{a^2mr_g}{c}$

This should not be taken to mean that the electron is a black hole, only that the radiation law is universal for all systems that possess charge - the ease in obtaining this made a strong case in my mind for his physics. . The analogue theory was extended with adjustable parameters which is now direction-dependent (ie. does not involve two signals, one from and one to the source) ~

$P = \frac{2}{3}\frac{a^2}{(1 - \frac{R}{r})^2}(\frac{Q^2}{c^3} + \frac{J^2}{Gm^2c}) \leq \frac{2}{3}\frac{mr_g}{c}\frac{a^2}{(1 - \frac{R}{r})^2}$

This equation in terms of full general relativistic dynamics is:

$P = \frac{2}{3}\frac{1}{(1 - \frac{R}{r})^2}(\frac{Q^2}{m^2c^3} + \frac{J^2}{Gm^4c})\frac{dp_{\mu}}{d\tau}\frac{dp^{\mu}}{d \tau} \leq \frac{2}{3}\frac{r_g}{mc}\frac{1}{(1 - \frac{R}{r})^2}\frac{dp_{\mu}}{d\tau}\frac{dp^{\mu}}{d \tau}$

Which sums up, from what I know, contributes to the radiation emission of a black hole through the ordinary Hawking process. Because we found implication of the gyromagnetic radius, we implemented short hand for the charge to mass ratio.

$P = \frac{2}{3}\frac{1}{(1 - \frac{R}{r})^2}(\frac{\gamma^2}{c^3} + \frac{J^2}{Gm^4c})\frac{dp_{\mu}}{d\tau}\frac{dp^{\mu}}{d \tau} \leq \frac{2}{3}\frac{r_g}{mc}\frac{1}{(1 - \frac{R}{r})^2}\frac{dp_{\mu}}{d\tau}\frac{dp^{\mu}}{d \tau}[/math] Where \gamma is [strictly not] the function used for contractions.$

18. ### originHeading towards oblivionValued Senior Member

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No, that means that Reiku is clueless.

19. ### curvatureRegistered Member

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As I promised, I would explain how entropy changes. Earlier someone asked a brilliant question, does this make entropy relative? It seems the answer can be yes - relavity does rely on the frame of reference and the motion of a system. But if entropy is either increasing or constant, what does his say about the dynamics behind the temperature? For that matter, is temperature subject to the laws of relativity?? (short answer is yes)

If temperature is covariant, is entropy covariant? I don't think so, I think the laws have to follow the contraction of the volume of the system, just as we applied to the black hole to satisfy the Penrose theorem of rotated spheres in relativity. The temperature in such a case, has to transform in the same way as the four volume would. It seems fair to do this, since entropy can be either dimensionless or have units of the $k_B$ (the Boltzmann constant) which is well known to be a constant.

We may see how the change in entropy becomes dependent on the only covariant object which can lead to a variation in the temperature due to Lorentz contractions:

$\Delta \mathbf{S} = \log_2(\frac{V_2}{V_1}) = \log_2(2) = 1$

Or as a more compact argument,

$\Delta S = Nk_B\ \log_2(\frac{V_2}{V_1})$

Which reminds me a four-dimensional entropy equation which measured the temperature through a similar ratio. In such cases, a ratio like

$\frac{V_0}{V} = \frac{T_0}{T}$

is something I suspect can be formed. The equation I suggested not long ago for the transformation of the volume lead to a change in the entropy:

$\Delta S = Nk_B\ \log_2(\frac{V_2}{V_1})$

On my blog a while back I derived the following (with some adjustments for clarity here):

$\Delta S = Nk_B \log_2(\frac{T_2}{T_1})$

So it stands to reason the identity may hold to satisfy a covariant temperature following the same transformation laws as the four volume.

$\Delta S = Nk_B(\log_2(\frac{V_2}{V_1}) = \log_2 (\frac{T_2}{T_1}))$

The only two invariant quantities in this equation is entropy and the Boltzmann constant. The covariant temperature transforms in exactly the same way as the volume and should be approximated under the kinetic energy of the equipartition theorem (which is something I am currently working on).

Last edited: Aug 12, 2018