Inverse and Exponential Dependences of Energy on The Earth-Galaxy Distance
Pavle I. Premović,
Laboratory for Geochemistry, Cosmochemistry&Astrochemistry,
University of Niš, pavleipremovic@yahoo.com, Niš, Serbia
In one of our previous communications {1}, we emphasized that there are two general possibilities for the loss of energy of photons traveling through intergalactic space. According to the first, the photon energy would decrease inversely proportional to the photon traveling distance in the intergalactic space distance or mathematically expressed
E ∝ 1/d.
The second is that this energy would
decay exponentially or mathematically speaking
E = E0e-αd
where E0 is the initial photon energy, E is the photon’s energy after traveling a distance d and α is the rate of energy loss. In both of these cases, it is assumed that the photons lose energy owing to some known or unknown process. Let us consider the first possibility, then the second one. We found that only one possible equation in which appears E ∝ 1/d and that is
E = κE0/(d
+ κ) … (1)
where κ is a temporary, arbitrary constant expressed in Gly. If d → 0 then E → E0 and if d → ∞ then E → E0. This equation, after a bit of algebra, can be written as
E/E0 =
1/[(d/κ) + 1] … (2).
Denote with λ0 and λ the
corresponding wavelength of the photon. Substituting into this equation E/E0
with λ0/λ[1]
and after a bit of algebra, we find
(λ/λ0)
– 1 = d/κ.
If we substitute d with a distance between the Earth and
galaxy D, that equation takes the following form
z = D/κ … (3).
This formula is equivalent to
Hubble’s linear redshift-distance relation (z = constant × D) and it is only
valid is valid for small redshifts or z << 1. Therefore, eqns. (1 - 3) are
valid for this z-limit. This Hubble’s relation is usually expressed as
cz = H0D … (4)
where H0 [say about 72
km sec-1 × (Mpc)-1] is the Hubble constant and D is the
distance between the Earth and a nearby galaxy.[1]
Combining this equation and eqn. (3), we get
κ = c/H0.
In Hubble terminology, c/H0
represents the Hubble distance DH ≈ 14 Gly. Therefore, we have
κ = DH ≈
14 Gly.
The distance DH
represents, roughly speaking, the radius of
the observable Universe. For H0 = 72 km sec-1 (Mpc)-1,
we find that DH ≈ 14 Gly. This is the limit of the observable
Universe or our observable cosmic horizon.
We now write eqn. (2) as
E/E0 = 1/[(D/DH)
+ 1] … (5).
The direct distance measurement method, the “megamaser” method, has demonstrated its capability for precise distance measurement of nearby galaxies. It appears this method is suitable for very few of these galaxies - “megamaser” galaxies. For the present case, we select five of these galaxies whose redshift z and distance D from the Earth (determined by this method) are known (Table 1) and their peculiar motion is negligible.
Name of galaxy |
Redshift z |
D (Gly)]1
|
E/E0* |
NGC 1052 |
0.004930 |
0.065 |
0.995 |
UGC 3789 |
0.010679 |
0.162 |
0.99 |
NGC 6323 |
0.02592 |
0.349 |
0.975 |
NGC 5765B |
0.02754 |
0.411 |
0.97 |
NGC 6264 |
0.03384 |
0.447 |
0.97 |
*Calculated using eqn. (5).
Plugging
into eqn. (5) given the above values of D and DH, we calculated E/E0
corresponding to these galaxies, Table 1.
Their arithmetic mean is E/E0 = 0.97 (Gly)-1. In other words, on average their final
energy is only about 3 percent of their initial photon energy.
An example of the second case is the tired-light energy loss of photons traveling through intergalactic
space. In this type of energy loss,
E = E0e-βd … (6)
where β is the
energy attenuation coefficient or the rate of energy attenuation [4, 5 and
references therein].
Within
the Hubble law
β = H0/c = 1/DH =1/κ.
As E/E0 = λ0/λ
then substituting d with D (a distance between the Earth and
galaxy), and after a bit of algebra eqn. (6) becomes
λ/λ0 = eβD.
Subtracting 1 from both sides of this
expression and since z = (λ/λ0)
– 1, we
arrive after some rearrangement at
1 + z = eβD.
Applying the natural
logarithm to both sides of this equation we get
ln(1 + z) = βD ... (7).
Having in mind that 1/β = κ, eqn. (3) can be written, after a bit of algebra, as follows
z = βD … (8).
It follows from
these two last equations that
ln(1 + z) = z.
In conclusion, we propose that Hubble’s linear
redshift-distance relationship, given by eqn. (4), is a result of the inverse
relation between a redshift and the Earth-nearby galaxy distance given by eqn.
(1) or eqn. (2).
[1] From elementary physics, we know that E/E0 = hν0/hν. Multiplying the denominator and numerator of the right side of this formula with the inverse of the speed of light c we get E/E0 = (hν0/c)/(hν/c). This physics also shows that (ν0/c)/(ν/c) = λ/λ0.
[2] We define nearby galaxies as those whose redshift z is from 0.001 to 0.1 (or 0.001 ≤ z ≤ 0.1) {2}.
No comments:
Post a Comment