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  • 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

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: ep = – (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.

Answers (2)

a) Let the radius of the Universe be R and assume that the hydrogen atoms are uniformly distributed. The expansion will only occur if the Coulomb repulsion is larger than the gravitational attraction at a distance R

The Hydrogen atom contains one proton and one electron charge on each hydrogen atom
e_H=e_P+e=-(1+y)e+e=-ye=\left | ye \right |

Let E be electric field intensity at distance R, on the surface of the sphere, then according to Gauss' theorem,
\\\oint \vec{E}.d\vec{S}= \frac{q_{enclosed}}{\varepsilon _0}\\ \\ \Rightarrow E(4\pi R^2)=\frac{4}{3}\frac{\pi R^3 N\left | ye \right |}{\varepsilon _0}\\ \\ \Rightarrow E=\frac{1}{3}\frac{ N\left | ye \right |R}{\varepsilon _0}.........1
 

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a) Let the radius of Universe be R and assume that the hydrogen atoms are uniformly distributed. The expansion will only occur if the coulomb repulsion is larger than the gravitational attraction at a distance R

The Hydrogen atom contains one proton and one electron charge on each hydrogen atom
e_H=e_P+e=-(1+y)e+e=-ye=\left | ye \right |

Let E be electric field intensity at distance R, on the surface of the sphere, then according to Gauss' theorem,
\\\oint \vec{E}.d\vec{S}= \frac{q_{enclosed}}{\varepsilon _0}\\ \\ \Rightarrow E(4\pi R^2)=\frac{4}{3}\frac{\pi R^3 N\left | ye \right |}{\varepsilon _0}\\ \\ \Rightarrow E=\frac{1}{3}\frac{ N\left | ye \right |R}{\varepsilon _0}                                    .........1
Let us suppose the mass of each hydrogen atom \simeq mp = mass of proton and G_R=gravitational field at distance R on the sphere.

Then

\\-4\pi R^2G_R=4\pi G m_p\left ( \frac{4}{3} \pi R^3\right )N\\ \Rightarrow G_R=\frac{-4}{3} \pi Gm_pNR                    ...........2
\therefore  Gravitational force on this atom is

    F_G=m_P\times G_R=\frac{-4 \pi}{3}Gm_{p}^{2}NR                ...........3

Column force on hydrogen atom at R is

F_C=\left | ye \right |E=\frac{1}{3}\frac{Ny^2e^2R}{\varepsilon _0}\:from \: \: the\: \: 1\: \: equation

 

(b) Net force experience by the hydrogen atom is given by

F=F_C-F_G=\frac{1}{3}\frac{Ny^2e^2 R}{\varepsilon _0}-\frac{4 \pi}{3}Gm_{p}^{2}NR

Because the this force, the hydrogen atom experiences an acceleration such as
            m_p\frac{d^2R}{dt^2}F=\frac{1}{3}\frac{Ny^2e^2 R}{\varepsilon _0}-\frac{4 \pi}{3}Gm_{p}^{2}NR

                               =\left ( \frac{1}{3}\frac{Ny^2e^2 }{\varepsilon _0}-\frac{4 \pi}{3}Gm_{p}^{2}N \right )R
\therefore                                \frac{d^2R}{dt^2}=\frac{1}{m_p}\left ( \frac{1}{3}\frac{Ny^2e^2 }{\varepsilon _0}-\frac{4 \pi}{3}Gm_{p}^{2}N \right )
  \Rightarrow                              \frac{d^2R}{dt^2}=\alpha R                            ...........4
 

Where \alpha ^2=\frac{1}{m_p}\left ( \frac{1}{3}\frac{Ny^2e^2 }{\varepsilon _0}-\frac{4 \pi}{3}Gm_{p}^{2}N \right )

The solution of equation 4 is given by R=Ae^{\alpha t}+Be^{\alpha t} We are looking for expansion, here, So B=0 \: and \: R=Ae^{\alpha t}

\Rightarrow                    Velocity of expansion,

                        \nu =\frac{dR}{dt}= Ae^{\alpha t}\left ( \alpha \right )= \alpha Ae^{\alpha t}= \alpha R

Hence, \nu\: \: \alpha \: \: R, i.e. velociy of expansion is proportional to the distance from the centre.


 

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