In 1959 Lyttleton and Bondi suggested that the expansion of the universe could be explained if matter carried a net charge. Suppose that the universe is made up of hydrogen atoms with a number density $\mathrm{N}$, which is maintained a constant. Let the charge on the proton be $e_p=-(1+y)$ e where $e$ is the electronic charge.
(a) Find the critical value of y such that expansion may start.
(b) Show that the velocity of expansion is proportional to the distance from the centre.

Options:

None

Solution:
(a) Consider universe is a sphere with radius R. H atoms and its constituent hydrogen atoms are distributed uniformly in the sphere.
As hydrogen atom contains one proton and one electron, charge on each hydrogen atom.
$e_H=e_P+e=-(1+Y) e+e=-Y e=(Y e)$
$\because \quad$ charge on proton $\mathrm{e}_{\mathrm{P}}=-(1+\mathrm{Y}) \mathrm{e}$
If $E$ is electric field intensity at distance $R$, on the surface of the sphere, then according to Gauss' theorem,
Now suppose, mass of each hydrogen atom $\simeq m_p=$ Mass of a proton, $G_R$ (gravitational field) at distance R on the sphere.
Then, $\quad-4 \mathrm{nR}^2 \mathrm{G}_{\mathrm{R}}=4 \pi \mathrm{G} \mathrm{m}_{\mathrm{P}}\left(\frac{4}{3} \pi \mathrm{R}^3\right) \mathrm{N}$
$$
\mathrm{G}_{\mathrm{R}}=\frac{-4}{3} \pi \mathrm{Gm}_{\mathrm{P}} \mathrm{NR}
$$
So, Gravitational force on this atom is
$$
F_G=m_p \times G R=\frac{-4 \pi}{3} G_p^2 \mathrm{NR}
$$
Coulomb force on hydrogen atom at $\mathrm{R}$ is
$$
\mathrm{F}_{\mathrm{C}}=(\text { Ye }) . \mathrm{E}
$$
From equation (i) putting the value of(E),
$$
F_C=\left[\frac{1}{3} \frac{N Y^2 \mathrm{e}^2 \mathrm{R}}{\varepsilon_0}\right]
$$
So, to start expansion $\mathrm{F}_{\mathrm{C}}>\mathrm{F}_{\mathrm{G}}$ and critical value of $\mathrm{Y}$ to start expansion would be when
$$
\begin{aligned}
\mathrm{F}_{\mathrm{C}} &=\mathrm{F}_{\mathrm{G}} \\
\frac{1}{3} \frac{N \mathrm{Y}^2 \mathrm{e}^2 \mathrm{R}}{\varepsilon_0} &=\frac{4 \pi}{3} \mathrm{Gm}_{\mathrm{P}}^2 \mathrm{NR}
\end{aligned}
$$
$$
\begin{gathered}
\mathrm{Y}^2=\left(4 \pi \varepsilon_0\right) \mathrm{G}\left(\frac{\mathrm{m}_{\mathrm{p}}}{\mathrm{e}}\right)^2 \\
\mathrm{X}^2=\frac{1}{9 \times 10^9} \times\left(6.68 \times 10^{-11}\right)\left(\frac{\left(1.66 \times 10^{10^{-77}}\right)^2}{\left(1.6 \times 10^{-19}\right)^2}\right) \\
\mathrm{Y}^2=79.8 \times 10^{-38} \\
\mathrm{Y}=\sqrt{79.8 \times 10^{-38}}=8.9 \times 10^{-19} \simeq 10^{-18}
\end{gathered}
$$
So critical value of $\mathrm{Y}$ is of the order of $\left(10^{-18}\right)$ that universe start to expand.
For expansion repulsive force $\mathrm{F}_{\mathrm{C}}$ must be greater than, attractive gravitational force.
(b) Net force experience by the hydrogen atom.
So, $F=F_C-F_G$
$$
=\frac{1}{3} \frac{\mathrm{NY}^2 \mathrm{e}^2 \mathrm{R}}{\varepsilon_0}-\frac{4 \pi}{3} \mathrm{Gm}_{\mathrm{p}}^2 \mathrm{NR}
$$
When acceleration of hydrogen atom is represent by $\left(\mathrm{d}^2 \mathrm{R}\right.$ $/ \mathrm{dt}^2$ ) then
$$
\begin{aligned}
\mathrm{m}_{\mathrm{p}} \frac{\mathrm{d}^2 \mathrm{R}}{\mathrm{dt}^2}=\mathrm{F} &=\frac{1}{3} \frac{\mathrm{NY}^2 \mathrm{e}^2 \mathrm{R}}{\varepsilon_0}-\frac{4 \pi}{3} \mathrm{Gm}_{\mathrm{p}}^2 \mathrm{NR} \\
&=\left(\frac{1}{3} \frac{N \mathrm{Y}^2 \mathrm{e}^2}{\varepsilon_0}-\frac{4 \pi}{3} \mathrm{Gm}_{\mathrm{p}}^2 \mathrm{~N}\right) \mathrm{R}
\end{aligned}
$$
So, $\frac{\mathrm{d}^2 \mathrm{R}}{\mathrm{dt}^2}=\frac{1}{\mathrm{~m}_{\mathrm{P}}}\left[\frac{1}{3} \frac{\mathrm{NY}^2 \mathrm{e}^2}{\varepsilon_0}-\frac{4 \pi}{3} \mathrm{Gm}_{\mathrm{P}}^2 \mathrm{~N}\right] \mathrm{R}$
$$
\left(\text { If } \alpha^2=\frac{1}{m_P}\left[\frac{1}{3} \frac{N Y^2 e^2}{\varepsilon_0}-\frac{4 \pi}{3} \mathrm{Gm}_{\mathrm{P}}^2 \mathrm{~N}\right]\right)
$$
then, $\quad \frac{\mathrm{d}^2 \mathrm{R}}{\mathrm{dt}^2}=\alpha^2 \mathrm{R}$
The general solution of $\mathrm{Eq}$ (iv) is $\mathrm{R}-\mathrm{Ae}^{\alpha \mathrm{t}}+\mathrm{Be}^{-\alpha t}$ We are looking for expansion of universe,
So, $\mathrm{B}=0$ and $\mathrm{R}=\mathrm{Ae}^{\alpha \mathrm{t}}$
then velocity of expansion,
$$
\begin{aligned}
&\mathrm{V}=\frac{\mathrm{dR}}{\mathrm{dt}} \\
&A e^{\alpha t}(\alpha)=\alpha A e^{\alpha t}=\alpha R \\
&
\end{aligned}
$$
Thus, $v \propto R$ i.e., velocity of expansion is proportional to the distance from the centre.

Topic: Physics, Electrostatics