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.

Hint: Use Gauss law to find the electric field at the surface.

(a) Step 1: Find the electric field at the surface.

Let us suppose that universe is a perfect sphere of radius R and its constituent hydrogen atoms are distributed uniformly in the sphere.

As a hydrogen atom contains one proton and one electron, charge on each hydrogen atom. 

eH=eP+e=-(1+y)e+e=-ye=(ye)

If E is electric field intensity at distance R, on the surface of the sphere, then according to Gauss' theorem,

E.ds=qε0 i.e.,

 E(4πR2)=43πR3Nyeε0
E=13NyeRε0 ..............(i)

Step 2: Find the gravitational force on an atom at the surface.

Now, suppose, the mass of hydrogen atommP=Mass of a proton, GR=gravitational field at distance R on the sphere.

Then, -4πR2GR=4πG mP (43πR3)N

GR=-43πGmPNR ..............(ii)

The gravitational force on this atom is FG=mP×GR=-4π3GmP2NR .............(iii)

Coulomb force on hydrogen atom at R is FC=(ye) E=13Ny2e2Rε0              [From Eq. (i)]

Step 3: Find the value of y.

Now, to start expansion FC>FG and critical value of Y to start expansion would be when

FC=FG
13Ny2e2Rε0=4π3GmP2NR
y2=(4πε0)G(mPe)2
=19×109×(6.67×10-11)(1.66×10-27)2(1.6×10-19)2=79.8×10-38
y=79.8×10-38=8.9×10-1910-18

Thus, 10-18 is the required critical value of Y corresponding to which expansion of the universe would start.

(b) Step 4: Find the net force on the atom at the surface.

Net force experience by the hydrogen atom is given by:

F=FC-FG=13Ny2e2Rε0-4π3Gmp2NR

Step 5: Find the net acceleration of the atom.

If the acceleration of hydrogen atom is represented by d2Rdt2, then

mPd2Rdt2=F=13Ny2e2Rε0-4π3GmP2NR
=13Ny2e2ε0-4π3GmP2NR
d2Rdt2=1mP13Ny2e2ε0-4π3GmP2NR=α2R ..........(iv)

where, α2=1mP13Ny2e2ε0-4π3GmP2N

Step 6: Find the velocity of expansion.

The general solution of Eq. (iv) is given by R=Aeαt+Be-αt. We are looking for expansion, here, so B=0 and R=Aeαt.

The velocity of expansion, v=dRdt=Aeαt(α)=αAeαt=αR

Hence, vR i.e., the velocity of expansion is proportional to the distance from the centre.