In 1959 Lyttleton and Bondi suggested that the expansion of the Universe could be explained il 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.
(b) Show that the velocity of expansion is proportional to the distance from the centre.

Explanation

(a) Let the Universe have a radius R. Assume that the hydrogen atoms are uniformly distributed. The expansion of the universe will start if the coulomb repulsion on a hydrogen atom at R is larger that the gravitational attraction.
The hydrogen atom contains one proton and one electron, charge on each hydrogen atom.
`e_(1p) = e_(p) + e =(-1 +y)e +e`
`=-e + ye + e`
= ye
Let E be electric field intensity at distance R, on the surface of the sphere, then according to Gauss. theorem,
`oint vecE.dvecS = q/epsilon_(0)`
`therefore E(4piR^(2)) = 4/3 (piR^(3)N|ye)/epsilon_(0)`......(1)
Let us suppose the mass of each hydrogen atom = `m_p` = mass of a proton and `G_R` = gravitational field at distance R on the sphere.
Then `=-4piR^(2)G_(R) = 4piGm_(p)(4/3piR^(3))N`
`therefore G_(B) =-4/3piGm_(p).NR`........(2)
Gravitational force on this atom is,
`F_(G) =m_(p) xx G_(R) = (-4pi)/3 Gm_(p)^(2)NR`......(3)
Coulomb force on hydrogen atom at R is,
`F_( C) = (ye)E = 1/3(y^(2)e^(2)NR)/epsilon_(0)` [From eqns (1)]
Now to expansion `F_(C) gt F_(G)` and critical value of Y to start expansion would be when `F_( C) = F_(G)`
`therefore 1/3 (Ny^(2)e^(2)R)/epsilon_(0) = (4pi)/3 Gm_(p)^(2)NR`
`therefore y^(2) = (4piepsilon_(0))G xx (m_(p)/e)^(2)`
`therefore y = sqrt(79.8 xx 10^(-38)) = 8.9 xx 10^(-19) = 10^(-18)`
Hence, `10^(-18)` is the required critical value of y corresponding to which expansion of universe would start, (b) Net force experience by the hydrogen atom is given by,
`F = F_( C) - F_(G) = 1/3 (Ny^(2)e^(2)R)/epsilon_(0) -(4pi)/3 Gm_(p)^(2)NR`
Because of this net force, the hydrogen atom experience an acceleration such that,
`m_(p) xx (d^(2)R)/(dt^(2)) = F = 1/3 (Ny^(2) e^(2)R)/epsilon_(0) - (4pi)/3 Gm_(p)^(2)NR`
`therefore (d^(2)R)/(dt^(2)) = alpha^(2)R`......(4)
where, `alpha^(2) =1/m_(p) (1/3 (Ny^(2)e^(2))/epsilon_(0) -(4pi)/3 Gm_(p)^(2)N)`
The solution of equation (4) is given by, `R = Ae^(alphat) + Be^(-alphat)`. We are looking for expansion, here so B = 0 and `R = Ae^(alphat)`
Velocity of expansion,
`v= (dR)/(dt) = d/(dT) (Ae^(alphat))`
`therefore v prop R`
Velocity of expansion is proportional to the distance from the centre.