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The CMB was emitted at an energy of $E_{em}=13.6\text{ eV}$, which is the binding energy of hydrogen. This corresponds to a wavelength of $$ \lambda_{em} = \frac{hc}{E_{em}} \approx 9.12\times 10^{-8}\text{ m}$$

Redshift can be calculated by $$ 1+z = \frac{\lambda_{obs}}{\lambda_{em}} $$

If we observe blue light at 400 nm, we get a corresponding redshift of about $z_{blue} = \lambda_{blue}/\lambda_{em} -1 =3.4$. For red light at 700 nm, we get $z_{red} = 6.7$.

The scale factor of the universe (the amount by which it has expanded) is related to redshift by $$ \frac{a_\text{obs}}{a_\text{em}} = 1+z $$

For a flat, matter-dominated universe we have $\rho a^3=\text{constant}$ so that the Friedmann equation becomes

$$\left(\frac{\dot{a}_\text{obs}}{a_\text{obs}}\right)^2=\frac{8\pi G\rho_\text{obs}}{3}=\frac{8\pi G\rho_\text{em}}{3}a_\text{em}^3a_\text{obs}^{-3}=H_\text{em}^2a^3_\text{em}a_\text{obs}^{-3} $$ where we used $H^2=8\pi G \rho/3$. Solving this differential equation for $a$ and imposing that $a_\text{obs}=0$ at $t=0$ (the Big Bang) yields

$$ \left(\frac{a_\text{obs}}{a_\text{em}}\right) = \left ( \frac{3 H_\text{em} t}{2} \right ) ^{\!2/3} $$ We solve for $t$, obtaining $$ t = \frac{2}{3 H_\text{em}}\left(\frac{a_\text{obs}}{a_\text{em}}\right)^{3/2} $$ Unfortunately, this does not seem to allow us to obtain an actual number for our estimate.

The CMB was emitted at an energy of $E_{em}=13.6\text{ eV}$, which is the binding energy of hydrogen. This corresponds to a wavelength of $$ \lambda_{em} = \frac{hc}{E_{em}} \approx 9.12\times 10^{-8}\text{ m}$$

Redshift can be calculated by $$ 1+z = \frac{\lambda_{obs}}{\lambda_{em}} $$

If we observe blue light at 400 nm, we get a corresponding redshift of about $z_{blue} = \lambda_{blue}/\lambda_{em} -1 =3.4$. For red light at 700 nm, we get $z_{red} = 6.7$.

The scale factor of the universe (the amount by which it has expanded) is related to redshift by $$ \frac{a_\text{obs}}{a_\text{em}} = 1+z $$

For a flat, matter-dominated universe we have $\rho a^3=\text{constant}$ so that the Friedmann equation becomes

$$\left(\frac{\dot{a}_\text{obs}}{a_\text{obs}}\right)^2=\frac{8\pi G\rho_\text{obs}}{3}=\frac{8\pi G\rho_\text{em}}{3}a_\text{em}^3a_\text{obs}^{-3}=H_\text{em}^2a^3_\text{em}a_\text{obs}^{-3} $$ where we used $H^2=8\pi G \rho/3$. Solving this differential equation for $a$ and imposing that $a_\text{obs}=0$ at $t=0$ (the Big Bang) yields

$$ \left(\frac{a_\text{obs}}{a_\text{em}}\right) = \left ( \frac{3 H_\text{em} t}{2} \right ) ^{\!2/3} $$ We solve for $t$, obtaining $$ t = \frac{2}{3 H_\text{em}}\left(\frac{a_\text{obs}}{a_\text{em}}\right)^{3/2} $$ Unfortunately, this does not seem to allow us to obtain an actual number for our estimate, since $H_\text{em}$ is not known...

CAUTION: ANSWER STILL IN PROGRESS

The CMB was emitted at an energy of $E_{em}=13.6\text{ eV}$, which is the binding energy of hydrogen. This corresponds to a wavelength of $$ \lambda_{em} = \frac{hc}{E_{em}} \approx 9.12\times 10^{-8}\text{ m}$$

Redshift can be calculated by $$ 1+z = \frac{\lambda_{obs}}{\lambda_{em}} $$

If we observe blue light at 400 nm, we get a corresponding redshift of about $z_{blue} = \lambda_{blue}/\lambda_{em} -1 =3.4$. For red light at 700 nm, we get $z_{red} = 6.7$.

The scale factor of the universe (the amount by which it has expanded) is related to redshift by $$ \frac{a_\text{obs}}{a_\text{em}} = 1+z $$

For a flat, matter-dominated universe we have $\rho a^3=\text{constant}$ so that the Friedmann equation becomes

$$\left(\frac{\dot{a}_\text{obs}}{a_\text{obs}}\right)^2=\frac{8\pi G\rho_\text{obs}}{3}=\frac{8\pi G\rho_\text{em}}{3}a_\text{em}^3a_\text{obs}^{-3}=H_\text{em}^2a^3_\text{em}a_\text{obs}^{-3} $$ where we used $H^2=8\pi G \rho/3$. Solving this differential equation for $a$ and imposing that $a_\text{obs}=0$ at $t=0$ (the Big Bang) yields

$$ \left(\frac{a_\text{obs}}{a_\text{em}}\right) = \left ( \frac{3 H_\text{em} t}{2} \right ) ^{\!2/3} $$ We solve for $t$, obtaining $$ t = \frac{2}{3 H_\text{em}}\left(\frac{a_\text{obs}}{a_\text{em}}\right)^{3/2} $$

The CMB was emitted at an energy of $E_{em}=13.6\text{ eV}$, which is the binding energy of hydrogen. This corresponds to a wavelength of $$ \lambda_{em} = \frac{hc}{E_{em}} \approx 9.12\times 10^{-8}\text{ m}$$

Redshift can be calculated by $$ 1+z = \frac{\lambda_{obs}}{\lambda_{em}} $$

If we observe blue light at 400 nm, we get a corresponding redshift of about $z_{blue} = \lambda_{blue}/\lambda_{em} -1 =3.4$. For red light at 700 nm, we get $z_{red} = 6.7$.

The scale factor of the universe (the amount by which it has expanded) is related to redshift by $$ \frac{a_\text{obs}}{a_\text{em}} = 1+z $$

For a flat, matter-dominated universe we have $\rho a^3=\text{constant}$ so that the Friedmann equation becomes

$$\left(\frac{\dot{a}_\text{obs}}{a_\text{obs}}\right)^2=\frac{8\pi G\rho_\text{obs}}{3}=\frac{8\pi G\rho_\text{em}}{3}a_\text{em}^3a_\text{obs}^{-3}=H_\text{em}^2a^3_\text{em}a_\text{obs}^{-3} $$ where we used $H^2=8\pi G \rho/3$. Solving this differential equation for $a$ and imposing that $a_\text{obs}=0$ at $t=0$ (the Big Bang) yields

$$ \left(\frac{a_\text{obs}}{a_\text{em}}\right) = \left ( \frac{3 H_\text{em} t}{2} \right ) ^{\!2/3} $$ We solve for $t$, obtaining $$ t = \frac{2}{3 H_\text{em}}\left(\frac{a_\text{obs}}{a_\text{em}}\right)^{3/2} $$ Unfortunately, this does not seem to allow us to obtain an actual number for our estimate.

The CMB was emitted at an energy of $E_{em}=13.6\text{ eV}$, which is the binding energy of hydrogen. This corresponds to a wavelength of $$ \lambda_{em} = \frac{hc}{E_{em}} \approx 9.12\times 10^{-8}\text{ m}$$

Redshift can be calculated by $$ 1+z = \frac{\lambda_{obs}}{\lambda_{em}} $$

If we observe blue light at 400 nm, we get a corresponding redshift of about $z_{blue} = \lambda_{blue}/\lambda_{em} -1 =3.4$. For red light at 700 nm, we get $z_{red} = 6.7$.

The scale factor of the universe (the amount by which it has expanded) is related to redshift by $$ \frac{a_\text{now}}{a} = 1+z $$

For a flat, matter-dominated universe we have $\rho a^3=\text{constant}$ so that the Friedmann equation becomes

$$\left(\frac{\dot{a}}{a}\right)^2=\frac{8\pi G\rho_\text{now}}{3}a_\text{now}^3a^{-3}=H_\text{now}^2a^{-3} $$ where we set $a_\text{now}=1$, as is conventional. Solving this differential equation for $a$ and imposing that $a=0$ at $t=0$ (the Big Bang) yields

$$ a = \left ( \frac{3 H_\text{now} t}{2} \right ) ^{\!2/3} $$ We solve for $t$, obtaining $$ t = \frac{2}{3 H_0} \frac{1}{(1+z)^{3/2}} $$

We get $t(z_{red}) \approx 425$ million years, and $t(z_{blue}) \approx 980$ million years.

So according to this model the CMB was in the visible spectrum from about 13.2 - 12.8 bilion years ago.

CAUTION: ANSWER STILL IN PROGRESS

The CMB was emitted at an energy of $E_{em}=13.6\text{ eV}$, which is the binding energy of hydrogen. This corresponds to a wavelength of $$ \lambda_{em} = \frac{hc}{E_{em}} \approx 9.12\times 10^{-8}\text{ m}$$

Redshift can be calculated by $$ 1+z = \frac{\lambda_{obs}}{\lambda_{em}} $$

If we observe blue light at 400 nm, we get a corresponding redshift of about $z_{blue} = \lambda_{blue}/\lambda_{em} -1 =3.4$. For red light at 700 nm, we get $z_{red} = 6.7$.

The scale factor of the universe (the amount by which it has expanded) is related to redshift by $$ \frac{a_\text{obs}}{a_\text{em}} = 1+z $$

For a flat, matter-dominated universe we have $\rho a^3=\text{constant}$ so that the Friedmann equation becomes

$$\left(\frac{\dot{a}_\text{obs}}{a_\text{obs}}\right)^2=\frac{8\pi G\rho_\text{obs}}{3}=\frac{8\pi G\rho_\text{em}}{3}a_\text{em}^3a_\text{obs}^{-3}=H_\text{em}^2a^3_\text{em}a_\text{obs}^{-3} $$ where we used $H^2=8\pi G \rho/3$. Solving this differential equation for $a$ and imposing that $a_\text{obs}=0$ at $t=0$ (the Big Bang) yields

$$ \left(\frac{a_\text{obs}}{a_\text{em}}\right) = \left ( \frac{3 H_\text{em} t}{2} \right ) ^{\!2/3} $$ We solve for $t$, obtaining $$ t = \frac{2}{3 H_\text{em}}\left(\frac{a_\text{obs}}{a_\text{em}}\right)^{3/2} $$