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  Gravitational energy emitted as gravitational waves from GW170814

+ 2 like - 0 dislike

According with https://arxiv.org/pdf/1105.0265.pdf it is possible to consider a binary system of black holes as a  "Gravitationally Equivalent of an Atom (GEA)". 

The energy level $E_1$ of the GEA ground state is given by

where $G$ is the gravitational constant, $M_1$ and $M_2$ are the masses of the two black holes and $h_b $ is the Planck constant.

Assuming that for the  system GW170814 :   $M_1 = 31M_{Sol}$ and $M_2 = 25M_{Sol}$ ; then we obtain

$$E_1 = -0.5181806117 × 10 ^{206} J$$

We claim that the total gravitational energy emitted as gravitational waves due to the formation of  GW170814 is 

$$0.5181806117 × 10 ^{206}J$$.

Do you agree?

asked Sep 29, 2017 in Astronomy by juancho (1,130 points) [ revision history ]

L+Virgo annouced 2.4 to 3.1 solar mass energy which is about \(5 . 10^{47} J\)

One of the many assumptions of the document is about the BH configuration : relative point-like nucleus , a small ensemble of nuclei and a classical gravitational bound. This means that if GW170814 was an atom-like of this kind, we would get a crazy first energy level. This shows that the comparison doesn't hold.

Hi @igael, many thanks for your comment.  You are right, "L+Virgo annouced 2.4 to 3.1 solar mass energy which is about 5 x 1047J"  but the question is how such quantity was determined.  I think that such quantity was not obtained experimentally.  Such quantity was obtained as a difference between the sum of the  masses of the original black holes (31+25 = 56) and the mass of the final black hole (53); 56 - 53 = 3 solar masses.  Then, the quantity  5 x 1047was obtained assuming that GW170814 is a simple black hole but the astronomical observations do not discard the possibility that GW170814 be a gravitational atom and then the quantity 0.5181806117×10206is not actually excluded.  Do you agree?  All the best.

@Juancho : hi:) is not this calculus valid only for a size 0 black hole nucleus with some atoms around ? Such E1 is higher than the universe mass estimation by a factor of \(10^{104}\) !

@igael, thanks again for your comments.  Maybe your claim is right.  Please look other possibility in the answer to the question  All the best.

@Juancho: you pointed an interesting thing, we must check 1105.0265 :)

1 Answer

+ 1 like - 0 dislike

Other possibility is as follows. 

We assume that  the total gravitational energy emitted as gravitational waves due to the formation of  GW170814 is given by  $-E_{1}$ where

and G is the gravitational constant, M1 and M2 are the masses of the two black holes and $h_b$ is the gravitational  Planck constant.  In order to compute the value of $h_b$ we assume that

Then we obtain

it is to say

$$h_b = 0.1035899036 × 10^{46}  J-s$$

Using such value we deduce that the frequency of the gravitational wave emitted due to the formation of  GW170814 is $82.50898602 $ hertz.  The corresponding period is $0.01211989200$ seconds and the corresponding  wave length is $3635.967600$ Km.

Do you agree?

Similar computations can be performed for GW150914 and GW170104 and the obtained results are very good approximations to the observed values.

answered Sep 30, 2017 by juancho (1,130 points) [ revision history ]
edited Sep 30, 2017 by juancho

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