- #1
johne1618
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A homogeneous and isotropic Universe is described by the FLRW metric:
<br /> ds^2 = c^2dt^2 + a^2(t)\ d\Sigma^2,<br />
where ##a(t)## is the scale factor and ##d\Sigma## is an interval of uniformly curved co-ordinate 3-space which is independent of cosmic time ##t##.
If we set ##dt=0## then we find that the interval of proper distance, ##ds##, between two points with co-ordinate space separation, ##d\Sigma##, is given by:
<br /> ds = a(t)\ d\Sigma.<br />
Now at the present time ##t_0## we can define the scale factor ##a(t_0)=1##. Therefore, at time ##t_0##, the co-ordinate separation, ##d\Sigma##, is equal to an interval of proper distance, ##ds_0##, given by:
<br /> ds_0 = d\Sigma.<br />
Therefore we can eliminate the co-ordinate interval ##d\Sigma## in the two equations above to give:
<br /> ds = a(t)\ ds_0,<br />
where ##ds## is a proper length at time ##t## and ##ds_0## is the corresponding proper length at the present time ##t_0##.
Imagine that we have a rigid ruler of length one meter at the present time ##t_0##.
Let us transport that ruler into some future time ##t##. Since the ruler is rigid it remains one meter in length. But the corresponding length of the future ruler, at our present time ##t_0##, ##l_0##, is given by the above formula with ##ds=1## and ##l_0=ds_0## so that we have:
<br /> l_0 = \frac{1}{a(t)}\ \hbox{meters}.<br />
Energy and length are related by the Compton wavelength:
<br /> E = \frac{\hbar c}{\lambda}.<br />
Thus the energy associated with one meter measured at time ##t## by a contemporary observer is:
<br /> E = \hbar c\ \hbox{joules}.<br />
However the energy of one meter measured at time ##t##, when described in the co-ordinate system of an observer at the present time ##t_0##, is given by:
<br /> E_0 = \frac{\hbar c}{l_0}\\<br /> E_0 = a(t)\hbar c\ \hbox{joules}.<br />
When solving the FLRW equations one refers distance scales back to the present time by using ##a(t_0)=1##. Thus everything should be in terms of an observer at the present time ##t_0##.
For example the energy density of a co-moving volume of dust is conventionally taken to be given by:
<br /> \rho \propto \frac{1}{a^3}.<br />
Given the above discussion I would say that the energy density of a co-moving volume of dust (whose atoms always have a fixed size), from the perspective of an observer at the present time ##t_0##, should be given by:
<br /> \rho \propto \frac{a}{a^3}\\<br /> \rho \propto \frac{1}{a^2}.<br />
Rather than assuming that energy density is increasing, I would explain this effect as an increase in Universal energy scale from the perspective of an observer at the present time.
The Planck mass, ##M_{Pl}##, is the fundamental energy scale.
Newton's constant ##G## is related to the Planck mass by definition:
<br /> G \propto \frac{1}{M_{Pl}^2}.<br />
Instead of the Planck mass being a constant let us assume that it is proportional to ##a(t)##.
Thus, from the perspective of an observer fixed at the present time, Newton's constant is no longer constant but given by the expression:
<br /> G = \frac{G_0}{a^2(t)}.<br />
The Friedmann equation for a spatially flat Universe would then be given by:
<br /> \left(\frac{\dot{a}}{a}\right)^2 = \frac{8 \pi G_0 \rho}{3\ a^2(t)}.<br />
Instead of the density ##\rho## varying with time let us assume that it is constant ##\rho=\rho_0## and instead ##G## varies with time.
The Hubble constant at the present time, ##H_0##, is given by:
<br /> H^2_0 = \frac{8 \pi G_0 \rho_0}{3}.<br />
Therefore we have:
<br /> \left(\frac{\dot{a}}{a}\right)^2 = \frac{H^2_0}{a^2(t)}.<br />
This equation has the simple linear solution:
<br /> a(t) = H_0\ t\\<br /> a(t) = \frac{t}{t_0},<br />
where ##t_0## is the current age of the Universe.
Therefore we obtain a remarkably elegant cosmology with the following features:
I realize that current observations favor an accelerating universal expansion. But even so this model is a lot closer to observations than a conventional matter-dominated Einstein-de Sitter Universe.
<br /> ds^2 = c^2dt^2 + a^2(t)\ d\Sigma^2,<br />
where ##a(t)## is the scale factor and ##d\Sigma## is an interval of uniformly curved co-ordinate 3-space which is independent of cosmic time ##t##.
If we set ##dt=0## then we find that the interval of proper distance, ##ds##, between two points with co-ordinate space separation, ##d\Sigma##, is given by:
<br /> ds = a(t)\ d\Sigma.<br />
Now at the present time ##t_0## we can define the scale factor ##a(t_0)=1##. Therefore, at time ##t_0##, the co-ordinate separation, ##d\Sigma##, is equal to an interval of proper distance, ##ds_0##, given by:
<br /> ds_0 = d\Sigma.<br />
Therefore we can eliminate the co-ordinate interval ##d\Sigma## in the two equations above to give:
<br /> ds = a(t)\ ds_0,<br />
where ##ds## is a proper length at time ##t## and ##ds_0## is the corresponding proper length at the present time ##t_0##.
Imagine that we have a rigid ruler of length one meter at the present time ##t_0##.
Let us transport that ruler into some future time ##t##. Since the ruler is rigid it remains one meter in length. But the corresponding length of the future ruler, at our present time ##t_0##, ##l_0##, is given by the above formula with ##ds=1## and ##l_0=ds_0## so that we have:
<br /> l_0 = \frac{1}{a(t)}\ \hbox{meters}.<br />
Energy and length are related by the Compton wavelength:
<br /> E = \frac{\hbar c}{\lambda}.<br />
Thus the energy associated with one meter measured at time ##t## by a contemporary observer is:
<br /> E = \hbar c\ \hbox{joules}.<br />
However the energy of one meter measured at time ##t##, when described in the co-ordinate system of an observer at the present time ##t_0##, is given by:
<br /> E_0 = \frac{\hbar c}{l_0}\\<br /> E_0 = a(t)\hbar c\ \hbox{joules}.<br />
When solving the FLRW equations one refers distance scales back to the present time by using ##a(t_0)=1##. Thus everything should be in terms of an observer at the present time ##t_0##.
For example the energy density of a co-moving volume of dust is conventionally taken to be given by:
<br /> \rho \propto \frac{1}{a^3}.<br />
Given the above discussion I would say that the energy density of a co-moving volume of dust (whose atoms always have a fixed size), from the perspective of an observer at the present time ##t_0##, should be given by:
<br /> \rho \propto \frac{a}{a^3}\\<br /> \rho \propto \frac{1}{a^2}.<br />
Rather than assuming that energy density is increasing, I would explain this effect as an increase in Universal energy scale from the perspective of an observer at the present time.
The Planck mass, ##M_{Pl}##, is the fundamental energy scale.
Newton's constant ##G## is related to the Planck mass by definition:
<br /> G \propto \frac{1}{M_{Pl}^2}.<br />
Instead of the Planck mass being a constant let us assume that it is proportional to ##a(t)##.
Thus, from the perspective of an observer fixed at the present time, Newton's constant is no longer constant but given by the expression:
<br /> G = \frac{G_0}{a^2(t)}.<br />
The Friedmann equation for a spatially flat Universe would then be given by:
<br /> \left(\frac{\dot{a}}{a}\right)^2 = \frac{8 \pi G_0 \rho}{3\ a^2(t)}.<br />
Instead of the density ##\rho## varying with time let us assume that it is constant ##\rho=\rho_0## and instead ##G## varies with time.
The Hubble constant at the present time, ##H_0##, is given by:
<br /> H^2_0 = \frac{8 \pi G_0 \rho_0}{3}.<br />
Therefore we have:
<br /> \left(\frac{\dot{a}}{a}\right)^2 = \frac{H^2_0}{a^2(t)}.<br />
This equation has the simple linear solution:
<br /> a(t) = H_0\ t\\<br /> a(t) = \frac{t}{t_0},<br />
where ##t_0## is the current age of the Universe.
Therefore we obtain a remarkably elegant cosmology with the following features:
- The density ##\rho## is constant (in accord with the perfect cosmological principle)
- The scale factor ##a(t) \propto t##
- The Planck mass ##M_{Pl} \propto t##
I realize that current observations favor an accelerating universal expansion. But even so this model is a lot closer to observations than a conventional matter-dominated Einstein-de Sitter Universe.
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