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I though I would discuss the transition from radiation to matter dominated phases and from there to the dark energy phase. A fair amount of this can be discussed with just Newtonian mechanics. General relativity changes this by some subtle means, but as a coarse grained view, to borrow a stat mechanics term, Newtonian mechanics captures a lot of this.

We need to put this in the language of a scale parameter. For a radial distance $r$ we set $r = ax$, for $x$ a standard distance and a the scale parameter. We then write the first time derivative of the radius as $\mathrm dr/\mathrm dt = x ~\mathrm da/\mathrm dt$, and the second derivative $\mathrm d^2/\mathrm dt^2 = x ~\mathrm d^2a/\mathrm dt^2$. Given galaxies or matter of mass $m$ at a distance $r = xa$ the total energy in Newtonian mechanics is $$ E = \frac{1}{2}mx^2\left(\frac{\mathrm da}{\mathrm dt}\right)^2 - \frac{Gmm'}{xa}, $$ where $m'$ is all the mass-energy in the region to the radial distance $r$. We set the total energy to zero. In this way the total mass-energy of the universe is zero. This is not proven exactly, but it is a convenient assumption, and even if $E$ is a constant we can adjust the zero of the potential to make it go away, we now divide through by $m$ and we get $$ \frac{1}{2}x^2\left(\frac{da}{dt}\right)^2 - \frac{Gmm'}{xa} = 0. $$ The mass $m'$ is determined by all the mass in the volume $4\pi r^3/3$ for $r^3 = x^3a^3$ and as a result the mass is $m' = 4\pi\rho x^3a^3/3$. Inserting this in we see that $$ \frac{1}{2}x^2\left(\frac{\mathrm da}{\mathrm dt}\right)^2 - \frac{4\pi G\rho x^2a^2}{3}= 0. $$ and the ruler distance $x$ can be removed from consideration. This is a matter of the invariance of ruler measure. Our energy equation is $$ \left(\frac{\mathrm da}{\mathrm dt}\right)^2 - \frac{8\pi G\rho a^2}{3}= 0 $$ This defines the Hubble parameter $H = (\dot a/a)$ that depends on the density of mass-energy $$ \left(\frac{\dot a}{a}\right)^2 = H^2 = \frac{8\pi G\rho}{3}. $$

As a Hamiltonian ${\cal H} = 0$, which is commensurate with the ADM approach to relativity, the Hamilton equations are $$ \begin{align}\dot p = -\frac{\partial {\cal H}}{\partial a} &= \frac{16\pi G\rho a}{3} \\ \dot a &= \frac{\partial {\cal H}}{\partial p} \end{align}$$ Compare the energy equation with that for a harmonic oscillator.

Now let us consider the nature of the density. For ordinary matter we have $\rho = 3m/(4\pi a^3)$. Our $H = 0$ energy equation becomes $$ \left(\frac{da}{dt}\right)^2 - \frac{2Gm}{a} = 0 $$ We now want to find what form the scale factor $a(t)$ is and so $a(t) = bt^n$, and we get that the value of the exponent $n = 2/3$. This is the matter dominated case.

Now consider the radiation dominated situation. For radiation in a volume $V \sim a^3$, we think of it as a standing wave in a region with periodic boundary conditions. As the volume increases the energy of a photon decreases because $E = h\nu = hc/\lambda$. The wave length is then some integral fraction of the volume $V \sim a^3$. Therefore the net radiation energy $E = \rho V$ and $E \sim 1/a$ and so $\rho \sim ~ 1/a^4$. We now again find the dependency of the scale factor $a(t) = 1/2$. Finally, for the easy case we let $\rho$ = constant, and this gives the exponential solution $$ a(t) = a_0\exp\left[t \sqrt{8\pi G\rho/3}\right] $$ I now include a generic graph of these functions below. The blue curve is radiation, red is matter and the green is exponential expansion. The orange curve is the sum of the three. These are not to physical scale with the real universe.

I though I would discuss the transition from radiation to matter dominated phases and from there to the dark energy phase. A fair amount of this can be discussed with just Newtonian mechanics. General relativity changes this by some subtle means, but as a coarse grained view, to borrow a stat mechanics term, Newtonian mechanics captures a lot of this.

We need to put this in the language of a scale parameter. For a radial distance $r$ we set $r = ax$, for $x$ a standard distance and a the scale parameter. We then write the first time derivative of the radius as $\mathrm dr/\mathrm dt = x ~\mathrm da/\mathrm dt$, and the second derivative $\mathrm d^2/\mathrm dt^2 = x ~\mathrm d^2a/\mathrm dt^2$. Given galaxies or matter of mass $m$ at a distance $r = xa$ the total energy in Newtonian mechanics is $$ E = \frac{1}{2}mx^2\left(\frac{\mathrm da}{\mathrm dt}\right)^2 - \frac{Gmm'}{xa}, $$ where $m'$ is all the mass-energy in the region to the radial distance $r$. We set the total energy to zero. In this way the total mass-energy of the universe is zero. This is not proven exactly, but it is a convenient assumption, and even if $E$ is a constant we can adjust the zero of the potential to make it go away, we now divide through by $m$ and we get $$ \frac{1}{2}x^2\left(\frac{da}{dt}\right)^2 - \frac{Gm'}{xa} = 0. $$ The mass $m'$ is determined by all the mass in the volume $4\pi r^3/3$ for $r^3 = x^3a^3$ and as a result the mass is $m' = 4\pi\rho x^3a^3/3$. Inserting this in we see that $$ \frac{1}{2}x^2\left(\frac{\mathrm da}{\mathrm dt}\right)^2 - \frac{4\pi G\rho x^2a^2}{3}= 0. $$ and the ruler distance $x$ can be removed from consideration. This is a matter of the invariance of ruler measure. Our energy equation is $$ \left(\frac{\mathrm da}{\mathrm dt}\right)^2 - \frac{8\pi G\rho a^2}{3}= 0 $$ This defines the Hubble parameter $H = (\dot a/a)$ that depends on the density of mass-energy $$ \left(\frac{\dot a}{a}\right)^2 = H^2 = \frac{8\pi G\rho}{3}. $$

As a Hamiltonian ${\cal H} = 0$, which is commensurate with the ADM approach to relativity, the Hamilton equations are $$ \begin{align}\dot p = -\frac{\partial {\cal H}}{\partial a} &= \frac{16\pi G\rho a}{3} \\ \dot a &= \frac{\partial {\cal H}}{\partial p} \end{align}$$ Compare the energy equation with that for a harmonic oscillator.

Now let us consider the nature of the density. For ordinary matter we have $\rho = 3m/(4\pi a^3)$. Our $H = 0$ energy equation becomes $$ \left(\frac{da}{dt}\right)^2 - \frac{2Gm}{a} = 0 $$ We now want to find what form the scale factor $a(t)$ is and so $a(t) = bt^n$, and we get that the value of the exponent $n = 2/3$. This is the matter dominated case.

Now consider the radiation dominated situation. For radiation in a volume $V \sim a^3$, we think of it as a standing wave in a region with periodic boundary conditions. As the volume increases the energy of a photon decreases because $E = h\nu = hc/\lambda$. The wave length is then some integral fraction of the volume $V \sim a^3$. Therefore the net radiation energy $E = \rho V$ and $E \sim 1/a$ and so $\rho \sim ~ 1/a^4$. We now again find the dependency of the scale factor $a(t) = 1/2$. Finally, for the easy case we let $\rho$ = constant, and this gives the exponential solution $$ a(t) = a_0\exp\left[t \sqrt{8\pi G\rho/3}\right] $$ I now include a generic graph of these functions below. The blue curve is radiation, red is matter and the green is exponential expansion. The orange curve is the sum of the three. These are not to physical scale with the real universe.

I though I would discuss the transition from radiation to matter dominated phases and from there to the dark energy phase. A fair amount of this can be discussed with just Newtonian mechanics. General relativity changes this by some subtle means, but as a coarse grained view, to borrow a stat mechanics term, Newtonian mechanics captures a lot of this.

We need to put this in the language of a scale parameter. For a radial distance $r$ we set $r = ax$, for $x$ a standard distance and a the scale parameter. We then write the first time derivative of the radius as $dr/dt = xda/dt$, and the second derivative $d^2/dt^2 = xd^2a/dt^2$. Given galaxies or matter of mass $m$ at a distance $r = xa$ the total energy in Newtonian mechanics is $$ E = \frac{1}{2}mx^2\left(\frac{da}{dt}\right)^2 - \frac{Gmm'}{xa}, $$ where $m'$ is all the mass-energy in the region to the radial distance $r$. We set the total energy to zero. In this way the total mass-energy of the universe is zero. This is not proven exactly, but it is a convenient assumption, and even if $E$ is a constant we can adjust the zero of the potential to make it go away, we now divide through by $m$ and we get $$ \frac{1}{2}x^2\left(\frac{da}{dt}\right)^2 - \frac{Gmm'}{xa} = 0. $$ The mass $m'$ is determined by all the mass in the volume $4\pi r^3/3$ for $r^3 = x^3a^3$ and as a result the mass is $m' = 4\pi\rho x^3a^3/3$. Inserting this in we see that $$ \frac{1}{2}x^2\left(\frac{da}{dt}\right)^2 - \frac{4\pi G\rho x^2a^2}{3}= 0. $$ and the ruler distance $x$ can be removed from consideration. This is a matter of the invariance of ruler measure. Our energy equation is $$ \left(\frac{da}{dt}\right)^2 - \frac{8\pi G\rho a^2}{3}= 0 $$ This defines the Hubble parameter $H = (\dot a/a)$ that dependx on the density of mass-energy $$ \left(\frac{\dot a}{a}\right)^2 = H^2 = \frac{8\pi G\rho}{3}. $$

As a Hamiltonian ${\cal H} = 0$, which is commensurate with the ADM approach to relativity, the Hamilton equations are $$ \dot p = -\frac{\partial {\cal H}}{\partial a} = \frac{16\pi G\rho a}{3} $$ $$ \dot a = \frac{\partial {\cal H}}{\partial p} $$ Compare the energy equation with that for a harmonic oscillator.

Now let us consider the nature of the density. For ordinary matter we have $\rho = 3m/(4\pi a^3)$. Our $H = 0$ energy equation becomes $$ \left(\frac{da}{dt}\right)^2 - \frac{2Gm}{a} = 0 $$ We now want to find what form the scale factor $a(t)$ is and so $a(t) = bt^n$, and we get that the value of the exponent $n = 2/3$. This is the matter dominated case.

Now consider the radiation dominated situation. For radiation in a volume $V \sim a^3$, we think of it as a standing wave in a region with periodic boundary conditions. As the volume increases the energy of a photon decreases because $E = h\nu = hc/\lambda$. The wave length is then some integral fraction of the volume $V \sim a^3$. Therefore the net radiation energy $E = \rho V$ and $E \sim 1/a$ and so $\rho \sim ~ 1/a^4$. We now again find the dependency of the scale factor $a(t) = 1/2$. Finally, for the easy case we let $\rho$ = constant, and this gives the exponential solution $$ a(t) = a_0exp[t \sqrt{8\pi G\rho/3}] $$ I now include a generic graph of these functions below. The blue curve is radiation, red is matter and the green is exponential expansion. The orange curve is the sum of the three. These are not to physical scale with the real universe.

I though I would discuss the transition from radiation to matter dominated phases and from there to the dark energy phase. A fair amount of this can be discussed with just Newtonian mechanics. General relativity changes this by some subtle means, but as a coarse grained view, to borrow a stat mechanics term, Newtonian mechanics captures a lot of this.

We need to put this in the language of a scale parameter. For a radial distance $r$ we set $r = ax$, for $x$ a standard distance and a the scale parameter. We then write the first time derivative of the radius as $\mathrm dr/\mathrm dt = x ~\mathrm da/\mathrm dt$, and the second derivative $\mathrm d^2/\mathrm dt^2 = x ~\mathrm d^2a/\mathrm dt^2$. Given galaxies or matter of mass $m$ at a distance $r = xa$ the total energy in Newtonian mechanics is $$ E = \frac{1}{2}mx^2\left(\frac{\mathrm da}{\mathrm dt}\right)^2 - \frac{Gmm'}{xa}, $$ where $m'$ is all the mass-energy in the region to the radial distance $r$. We set the total energy to zero. In this way the total mass-energy of the universe is zero. This is not proven exactly, but it is a convenient assumption, and even if $E$ is a constant we can adjust the zero of the potential to make it go away, we now divide through by $m$ and we get $$ \frac{1}{2}x^2\left(\frac{da}{dt}\right)^2 - \frac{Gmm'}{xa} = 0. $$ The mass $m'$ is determined by all the mass in the volume $4\pi r^3/3$ for $r^3 = x^3a^3$ and as a result the mass is $m' = 4\pi\rho x^3a^3/3$. Inserting this in we see that $$ \frac{1}{2}x^2\left(\frac{\mathrm da}{\mathrm dt}\right)^2 - \frac{4\pi G\rho x^2a^2}{3}= 0. $$ and the ruler distance $x$ can be removed from consideration. This is a matter of the invariance of ruler measure. Our energy equation is $$ \left(\frac{\mathrm da}{\mathrm dt}\right)^2 - \frac{8\pi G\rho a^2}{3}= 0 $$ This defines the Hubble parameter $H = (\dot a/a)$ that depends on the density of mass-energy $$ \left(\frac{\dot a}{a}\right)^2 = H^2 = \frac{8\pi G\rho}{3}. $$

As a Hamiltonian ${\cal H} = 0$, which is commensurate with the ADM approach to relativity, the Hamilton equations are $$ \begin{align}\dot p = -\frac{\partial {\cal H}}{\partial a} &= \frac{16\pi G\rho a}{3} \\ \dot a &= \frac{\partial {\cal H}}{\partial p} \end{align}$$ Compare the energy equation with that for a harmonic oscillator.

Now let us consider the nature of the density. For ordinary matter we have $\rho = 3m/(4\pi a^3)$. Our $H = 0$ energy equation becomes $$ \left(\frac{da}{dt}\right)^2 - \frac{2Gm}{a} = 0 $$ We now want to find what form the scale factor $a(t)$ is and so $a(t) = bt^n$, and we get that the value of the exponent $n = 2/3$. This is the matter dominated case.

Now consider the radiation dominated situation. For radiation in a volume $V \sim a^3$, we think of it as a standing wave in a region with periodic boundary conditions. As the volume increases the energy of a photon decreases because $E = h\nu = hc/\lambda$. The wave length is then some integral fraction of the volume $V \sim a^3$. Therefore the net radiation energy $E = \rho V$ and $E \sim 1/a$ and so $\rho \sim ~ 1/a^4$. We now again find the dependency of the scale factor $a(t) = 1/2$. Finally, for the easy case we let $\rho$ = constant, and this gives the exponential solution $$ a(t) = a_0\exp\left[t \sqrt{8\pi G\rho/3}\right] $$ I now include a generic graph of these functions below. The blue curve is radiation, red is matter and the green is exponential expansion. The orange curve is the sum of the three. These are not to physical scale with the real universe.

I though I would discuss the transition from radiation to matter dominated phases and from there to the dark energy phase. A fair amount of this can be discussed with just Newtonian mechanics. General relativity changes this by some subtle means, but as a coarse grained view, to borrow a stat mechanics term, Newtonian mechanics captures a lot of this.

We need to put this in the language of a scale parameter. For a radial distance $r$ we set $r = ax$, for $x$ a standard distance and a the scale parameter. We then write the first time derivative of the radius as $dr/dt = xda/dt$, and the second derivative $d^2/dt^2 = xd^2a/dt^2$. Given galaxies or matter of mass $m$ at a distance $r = xa$ the total energy in Newtonian mechanics is $$ E = \frac{1}{2}mx^2\left(\frac{da}{dt}\right)^2 - \frac{Gmm'}{xa}, $$ where $m'$ is all the mass-energy in the region to the radial distance $r$. We set the total energy to zero. In this way the total mass-energy of the universe is zero. This is not proven exactly, but it is a convenient assumption, and even if $E$ is a constant we can adjust the zero of the potential to make it go away, we now divide through by $m$ and we get $$ \frac{1}{2}x^2\left(\frac{da}{dt}\right)^2 - \frac{Gmm'}{xa} = 0. $$ The mass $m'$ is determined by all the mass in the volume $4\pi r^3/3$ for $r^3 = x^3a^3$ and as a result the mass is $m' = 4\pi\rho x^3a^3/3$. Inserting this in we see that $$ \frac{1}{2}x^2\left(\frac{da}{dt}\right)^2 - \frac{4\pi G\rho x^2a^2}{3}= 0. $$ and the ruler distance $x$ can be removed from consideration. This is a matter of the invariance of ruler measure. Our energy equation is $$ \left(\frac{da}{dt}\right)^2 - \frac{8\pi Ga^2}{3}= 0 $$ This defines the Hubble parameter $H = (\dot a/a)^2$ that dependx on the density of mass-energy $$ \left(\frac{\dot a}{a}\right)^2 = H = \frac{8\pi G}{3}. $$

As a Hamiltonian $H = 0$, which is commensurate with the ADM approach to relativity, the Hamilton equations are $$ \dot p = -\frac{\partial H}{\partial a} = \frac{16\pi G\rho a}{3} $$ $$ \dot a = \frac{\partial H}{\partial p} $$ Compare the energy equation with that for a harmonic oscillator.

Now let us consider the nature of the density. For ordinary matter we have $\rho = 3m/(4\pi a^3)$. Our $H = 0$ energy equation becomes $$ \left(\frac{da}{dt}\right)^2 - \frac{2Gm}{a} = 0 $$ We now want to find what form the scale factor $a(t)$ is and so $a(t) = bt^n$, and we get that the value of the exponent $n = 2/3$. This is the matter dominated case.

Now consider the radiation dominated situation. For radiation in a volume $V \sim a^3$, we think of it as a standing wave in a region with periodic boundary conditions. As the volume increases the energy of a photon decreases because $E = h\nu = hc/\lambda$. The wave length is then some integral fraction of the volume $V \sim a^3$. Therefore the net radiation energy $E = \rho V$ and $E \sim 1/a$ and so $\rho \sim ~ 1/a^4$. We now again find the dependency of the scale factor $a(t) = 1/2$. Finally, for the easy case we let $\rho$ = constant, and this gives the exponential solution $$ a(t) = a_0exp[t \sqrt{8\pi G\rho/3}] $$ I now include a generic graph of these functions below. The blue curve is radiation, red is matter and the green is exponential expansion. The orange curve is the sum of the three. These are not to physical scale with the real universe.

I though I would discuss the transition from radiation to matter dominated phases and from there to the dark energy phase. A fair amount of this can be discussed with just Newtonian mechanics. General relativity changes this by some subtle means, but as a coarse grained view, to borrow a stat mechanics term, Newtonian mechanics captures a lot of this.

We need to put this in the language of a scale parameter. For a radial distance $r$ we set $r = ax$, for $x$ a standard distance and a the scale parameter. We then write the first time derivative of the radius as $dr/dt = xda/dt$, and the second derivative $d^2/dt^2 = xd^2a/dt^2$. Given galaxies or matter of mass $m$ at a distance $r = xa$ the total energy in Newtonian mechanics is $$ E = \frac{1}{2}mx^2\left(\frac{da}{dt}\right)^2 - \frac{Gmm'}{xa}, $$ where $m'$ is all the mass-energy in the region to the radial distance $r$. We set the total energy to zero. In this way the total mass-energy of the universe is zero. This is not proven exactly, but it is a convenient assumption, and even if $E$ is a constant we can adjust the zero of the potential to make it go away, we now divide through by $m$ and we get $$ \frac{1}{2}x^2\left(\frac{da}{dt}\right)^2 - \frac{Gmm'}{xa} = 0. $$ The mass $m'$ is determined by all the mass in the volume $4\pi r^3/3$ for $r^3 = x^3a^3$ and as a result the mass is $m' = 4\pi\rho x^3a^3/3$. Inserting this in we see that $$ \frac{1}{2}x^2\left(\frac{da}{dt}\right)^2 - \frac{4\pi G\rho x^2a^2}{3}= 0. $$ and the ruler distance $x$ can be removed from consideration. This is a matter of the invariance of ruler measure. Our energy equation is $$ \left(\frac{da}{dt}\right)^2 - \frac{8\pi G\rho a^2}{3}= 0 $$ This defines the Hubble parameter $H = (\dot a/a)$ that dependx on the density of mass-energy $$ \left(\frac{\dot a}{a}\right)^2 = H^2 = \frac{8\pi G\rho}{3}. $$

As a Hamiltonian ${\cal H} = 0$, which is commensurate with the ADM approach to relativity, the Hamilton equations are $$ \dot p = -\frac{\partial {\cal H}}{\partial a} = \frac{16\pi G\rho a}{3} $$ $$ \dot a = \frac{\partial {\cal H}}{\partial p} $$ Compare the energy equation with that for a harmonic oscillator.

Now let us consider the nature of the density. For ordinary matter we have $\rho = 3m/(4\pi a^3)$. Our $H = 0$ energy equation becomes $$ \left(\frac{da}{dt}\right)^2 - \frac{2Gm}{a} = 0 $$ We now want to find what form the scale factor $a(t)$ is and so $a(t) = bt^n$, and we get that the value of the exponent $n = 2/3$. This is the matter dominated case.

Now consider the radiation dominated situation. For radiation in a volume $V \sim a^3$, we think of it as a standing wave in a region with periodic boundary conditions. As the volume increases the energy of a photon decreases because $E = h\nu = hc/\lambda$. The wave length is then some integral fraction of the volume $V \sim a^3$. Therefore the net radiation energy $E = \rho V$ and $E \sim 1/a$ and so $\rho \sim ~ 1/a^4$. We now again find the dependency of the scale factor $a(t) = 1/2$. Finally, for the easy case we let $\rho$ = constant, and this gives the exponential solution $$ a(t) = a_0exp[t \sqrt{8\pi G\rho/3}] $$ I now include a generic graph of these functions below. The blue curve is radiation, red is matter and the green is exponential expansion. The orange curve is the sum of the three. These are not to physical scale with the real universe.