InterviewSolution
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InterviewSolution
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In 1959 Lyttleton and Bondi suggested that the expansion of the universe could be explained if matter carried a ent 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. ltBrgt (b) show that the velocity of expansion is proportional tot he distance from the centre. |
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Answer» SOLUTION :(a) Let us suppose that universe is a perfect sphere of radius R 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+gamma)e+e=-Ye=(Ye)` If E is electric field intenstiy at distance R, on the surface of the sphere, then ACCORDING to gauss' theorem, `ointE.ds=(q)/(epsi_(0))` i.e., `E(4piR^(2))=(4)/(3)(piR^(3)N|Ye|)/(epsi_0)` `E=(1)/(3)(N|Ye|R)/(epsi_(0))`. . .(i) Now, suppose, mass of each hydrogen atom `congm_(P)=` mass of a proton, `G_(R)=` gravitational field at distance R on the sphere, then `-4piR^(2)G_(R)=4piGm_(P)((4)/(3)piR^(3))N` `implieG_(R)=(-4)/(3)piGm_(P)NR`. . .(II) `therefore` Gravitational force on this atom is `F_(G)=m_(P)xxG_(R)=(-4pi)/(3)Gm_(P)^(2)NR`. . .(iii) Coulomb force on hydrogen atom at R is `F_(C)=(Ye)=E=(1)/(3)(NY^(2)e^(2)R)/(epsi_(0))` . . . [From Eq. (i)] Now, to start expansion `F_(C) lt F_(G)` ad CRITICAL value of Y to start expansion WOULD be when `F_(C)=F_(G)` `implies(1)/(3)(NY^(2)e^(2)R)/(epsi_(0))=(4pi)/(3)Gm_(P)^(2)NR` `impliesY^(2)=(4piepsi_(0))G((m_(P))/(e))^(2)` `=(1)/(9xx10^(9))xx(6.67xx10^(-11))(((1.66xx10^(-27))^(2))/((1.6xx10^(-19))^(2)))=79.8xx10^(-38)` `impliesY=sqrt(79.8xx10^(-38))=8.9xx106(-19)cong10^(-18)` Thus, `10^(-18)` is the required critical value of Y corresponding to which expansio 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)/(epsi_(0))-(4pi)/(3)Gm_(P)^(2)NR` If acceleration of hydrogen atom is represent by `d^2R//dt^(2)`, then `m_(P)=(d^(2)R)/(dt^(2))=F=(1)/(3)(NY^(2)e^(2)R)/(epsi_(0))-(4pi)/(3)Gm_(P)^(2)NR` `=((1)/(3)(NY^(2)e^(2))/(epsi_(0))-(4pi)/(3)Gm_(P)^(2)N)R` `therefore(d^(2)R)/(dt^(2))=(1)/(m_(P))[(1)/(3)(NY^(2)e^(2))/(epsi_(0))-(4pi)/(3)Gm_(P)^(2)N]R=alpha^(2)R`. . .(iv) where, `alpha^(2)=(1)/(m_(P))[(1)/(3)(NY^(2)e^(2))/(epsi_(0))-(4pi)/(3)Gm_(P)^(2)N]` the general solution of Eq. (iv) is given by `R=Ae^(alphat)+Be^(-alphat)` We are looking for expansion here So `B=0 and R=Ae^(alphat)` `implies` Velocity of expansion `v=(dR)/(dt)=Ae^(alphat)(alpha)=alphaAe^(alphat)=alphaR)` Hence, `v prop R` i.e., velocity of expansion is proportional to the distance from the centre. |
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