Minimum Quantum Energy of the Observable Tired-Light Universe
Pavle I. Premović
Laboratory for Geochemistry, Cosmochemistry7Astrochemistry,
University of Niš, pavleipremovic@yahoo.com, Niš, Serbia
The tired-light
theory argues that the
light of nearby or distant galaxies[1] is
redshifted because it loses energy as it passes through intergalactic space. In his recent communication, Premović {3} derived basic equations related to the tired-light hypothesis. According to it,
the initial energy of photon E0 emitted by a nearby or a distant
galaxy (hereinafter galaxy) and its energy E after traveling a distance d in
the intergalactic space are related by
E = E0e-βd
where β is the energy attenuation or the rate of energy attenuation {4}. Premović {5} has also found that the
value of β is 0.071 Gly and the above equation can be
written as
E = E0e-0.071d ... (1).
Premović
[6] estimated that
the minimum quantum energy of the observable Universe, ε, is about 1.5 × 10-51
J. Eqn. (1) allow us to calculate a distance dε at which a photon
emitted by a galaxy would have a final energy equal to the minimum quantum
energy ε (= 1.5 × 10-51 J) or
ε = E0e-0.071dε ... (2).
Example
1:
A Lyman-alpha (-α) emitter is a type of distant galaxy that emits the Lyman-α line of the hydrogen atom. The
relationship between the energy of this line Eα and its wavelength λα
(= 121.6 nm) can be described by the following equation
Eα = hc/λα
where h (= 6.63 10-34 J sec) is called Planck's constant and c (= 2.99792 × 108 m sec-1) is the speed of light. Putting the given values for h, c and λα we find that Eα = 1.6 × 10-18 J.
E = Eαe-0.071d
The Lyman-α
photon would reach its minimum quantum energy of the observable Universe
ε when E =
ε (= 1.5 × 10-51 J). So, we can write after a bit of
algebra
Eα/ε = e0.071de
where
dε is the distance from the galaxy emitter at which these energies
would be equated. Logarithmizing this expression
and after a bit of algebra, we get
dε = ln(Eα/ε)/0.071.
Introducing
into this equation the above values for Eα and
ε we have
dε = ln(1.07 × 1033)/0.071
= [ln(1.070) + ln(1033)]/0.071 = 1100 Gly.
The
time of flight of this photon up to this distance away from the galaxy emitter
is about 1100 Gy.
Example 2:
GRB 190114C was an extreme gamma-ray
burst explosion whose
radiation detected is the highest energy ever observed for a GRB: 1TeV or EGRB is about 1.6 × 10-7 J. In this case, employing eqn.
(2) and similar mathematical procedure as in Example 1, one gets
dε = ln(EGRB/ε)/0.071.
If
we plug the above values for ε and EGRB we find that
dε = ln(1.07 × 1044)/0.071
= [ln(1.07) + ln(1044)]/0.071
= 1465 Gly.
The
time of flight of this photon up to this distance away from GRB 190114C
is 1465 Gy.
As we have shown in our previous communication [3], for the Hubble law case of a linear relationship between the redshift z and the distance D between the Earth and a nearby galaxy there is the following relationship
β = H0/c
where H0 is the Hubble constant [= 72 km sec-1 (Mpc)-1] and c (= 299792 km sec-1) is the speed of light. In Hubble terminology, H0/c represents the inverse of Hubble’s distance, 1/DH. Plugging 1/DH instead of β = 0.071 (Gly)-1 and D instead of d and into eqn. (1) we have
ε = E0e-D/DH.
References
[1] We define nearby galaxies as
those whose redshift z is from 0.001 to 0.1 (or 0.001 ≤ z ≤ 0.1) and distant galaxies with z > 0.1 {1, 2}. Of course, there is no sharp line between
nearby and distant galaxies.
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