$1)$ 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 $N$ , 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.

$2)$ 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 $N$ , which is maintained a constant. Let the charge on the proton be:

$e_{p} = - (1 + y) e$

where e is the electronic charge.
$B)$ Show that the velocity of expansion is proportional to the distance from the centre.

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Solution

$A)$ Let the Universe has a radius $R .$ Assume that the hydrogen atoms are uniformly distributed.

The charge on each hydrogen atom is,

$e_{H} = - (1 + y) e + e = - y e = | y e |$

The mass of each hydrogen atom is $\approx m_{p}$ (mass of proton).

Expansion starts if the Coulomb repulsion on a hydrogen atom, at $R$ , is larger than the gravitational attraction.

Let the Electric Field at $R$ be $\to E .$

From Gauss law:

$\oint \to E . - -- \to (d S) = \frac{q_{i n}}{ε_{o}}$

Universe is made up of hydrogen atoms with a number density $N .$

$q_{i n} = \frac{4}{3} π R^{3} N e_{H}$

$4 π R^{2} E = \frac{4}{3 ε_{0}} π R^{3} N ∣ y e |$

$\to E (R) = \frac{1}{3} \frac{N | y e |}{ε_{0}} R^r$

Let the gravitational field at $R$ be $G_{R} .$ Then,

$- 4 π R^{2} G_{R} = 4 π G m_{P} \frac{4}{3} π R^{3} N$

$G_{R} = - \frac{4}{3} π G m_{ρ} N R$

$_{R} (R) = - \frac{4}{3} π G m_{ρ} N R^r$

Thus, the Coulombic force on a hydrogen atom at $R$ is

$y e \to E (R) = \frac{1}{3} \frac{N y^{2} e^{2}}{ε_{o}} R^r$

The gravitational force on this atom is

$m_{p}_{R} (R) = - \frac{4 π}{3} G N m_{p}^{2} R^r$

The net force on the atom is

$\to F = (\frac{1}{3} \frac{N y^{2} e^{2}}{ε_{0}} R - \frac{4 π}{3} G N m_{p}^{2} R)^r$

The critical value is when

$\frac{1}{3} \frac{N y_{c}^{2} e^{2}}{ε_{o}} R = \frac{4 π}{3} G N m_{p}^{2} R$

$\Rightarrow y_{c}^{2} = 4 π ε_{0} G \frac{m_{p}^{2}}{e^{2}}$

$≃ \frac{7 \times 10^{- 11} \times {1.6}^{2} \times 10^{- 54}}{9 \times 10^{9} \times {1.6}^{2} \times 10^{- 38}}$

$≃ - 70 \times 10^{- 38}$

$∴ y_{c} ≃ 8 \times 10^{- 19}$ , i.e. the order is $10^{- 19}$

Final Answer: $y_{c} \sim 8 \times 10^{- 19}$

$B)$ Let the Universe has a radius $R .$ Assume that the hydrogen atoms are uniformly distributed.

The charge on each hydrogen atom is,

$e_{H} = - (1 + y) e + e = - y e = | y e |$

The mass of each hydrogen atom is $\approx m_{p}$ (mass of proton).

Expansion starts if the Coulomb repulsion on a hydrogen atom, at $R$ , is larger than the gravitational attraction.

Let the Electric Field at $R$ be $\to E .$

From Gauss law:

$\oint \to E . - -- \to (d S) = \frac{q_{i n}}{ε_{o}}$

Universe is made up of hydrogen atoms with a number density $N .$

$q_{i n} = \frac{4}{3} π R^{3} N e_{H}$

$4 π R^{2} E = \frac{4}{3 ε_{0}} π R^{3} N ∣ y e |$

$\to E (R) = \frac{1}{3} \frac{N | y e |}{ε_{0}} R^r$

Let the gravitational field at $R$ be $G_{R} .$ Then,

$- 4 π R^{2} G_{R} = 4 π G m_{P} \frac{4}{3} π R^{3} N$

$G_{R} = - \frac{4}{3} π G m_{ρ} N R$

$_{R} (R) = - \frac{4}{3} π G m_{ρ} N R^r$

Thus, the Coulombic force on a hydrogen atom at $R$ is

$y e \to E (R) = \frac{1}{3} \frac{N y^{2} e^{2}}{ε_{o}} R^r$

The gravitational force on this atom is

$m_{p}_{R} (R) = - \frac{4 π}{3} G N m_{p}^{2} R^r$

The net force on the atom is

$\to F = (\frac{1}{3} \frac{N y^{2} e^{2}}{ε_{0}} R - \frac{4 π}{3} G N m_{p}^{2} R)^r$

Because of the net force, the hydrogen atom experiences an acceleration such that

$m_{p} \frac{d^{2} R}{d t^{2}} = (\frac{1}{3} \frac{N y^{2} e^{2}}{ε_{0}} R - \frac{4 π}{3} G N m_{p}^{2} R)$

$\frac{d^{2} R}{d t^{2}} = a^{2} R$

where $\propto^{2} = \frac{1}{m_{p}} (\frac{1}{3} \frac{N y^{2} e^{2}}{ε_{0}} R - \frac{4 π}{3} G N m_{p}^{2} R)$

This has a solution $R = A e^{a t} + B e^{- α t}$

As we are seeking an expansion, $B = 0.$

$∴ R = A e^{a t}$

$\Rightarrow ˙ R =\propto A e^{\propto t} =\propto R$

Thus, the velocity is proportional to the distance from the centre.

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A)

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