Lyttelton-Bondi Model for the Expansion of the Universe

Classical Mechanics Level 5

In 1959 Lyttelton and Bondi suggested that the expansion of the Universe could be explained on the basis of Newtonian mechanics if matter carried a net electric charge.Imagine a spherical volume of astronomical size and radius $R$ and containing un-ionized atomic hydrogen gas of uniform density $n$ = $10^{-26} kg.m^{-3}$ and assume that the proton charge $e'=-(1+y)*e$ where $e$ is the electron charge.

$A$ : Obtain the least value of $y$ for which the electrostatic repulsion is larger than the gravitational attraction and the gas expands.Let $y$ = $10^z$ (approximately).

$B$ : Obtain an expression for the force of repulsion on an atom which is at a distance $R$ from the centre of the spherical volume.Then say the radial velocity is proportional to $R^p$ .Let us label the proportionality constant as $H$ .Assume that the density is maintained constant by the creation of matter in space.Assume also that the value of $y$ is larger than the equilibrium value calculated in part $A$ above and hence ignore gravity.

$C$ : Calculate the numerical value of $H$ .Take the value of $y$ to be one order of magnitude larger than the equilibrium value calculated in part $A$ above.Say the value is equal to $b*10^c$ /sec (where $0<b<2$ and evaluate $b$ upto 1 place of decimal).

$D$ : Given that at time $t=0$ ,the volume of the Universe was $V$ ,then at time $t$ the volume will be $V*e^{a*H*t}$ .

Evaluate $a+z+p+b+c$ .

Finally you can think of why Lyttelton-Bondi model has been largely discarded by the scientific community.

The answer is -29.2.

1 solution

Rajdeep Brahma
Jun 15, 2018

$A$ :Equate Energies or Pressure or Force.Note that total charge $Q$ = $Nq=N(-ey)$ ,u will easily get the value of $y=0.910^{-18}=(approx)10^{-18}$ .So $z=-18$ .

$B$ :The force on the atom at a distance r is F say, and from coulombs law F= $\frac{kQ}{R^2}$ .F=m $\frac{dv}{dt}$ =mv $\frac{dv}{dr}$ and $Q=$ $\frac{n}{m'}$ $\frac{4}{3}$* $\pi$ $r^3$ $q$ where m' is mass of proton and $q=-e*y$ so v is proportional to r.Hence p=1.

$C$ :Now H can be easily determined from the previous equation and is equal to $\frac{ey}{m'}$ $(\frac{3}{2(\epsilon)})^{0.5}$ .Plug in values where $y=10^{-17}$ and you will get $H=1.810^{-17}$ ..so b=1.8 and c=(-17)

$D$ : $\frac{dV}{dt}$ =3 V H or $\frac{dV}{V}$ =3Hdt,so a=3,Hence answer = $z+p+b+c+a$ =-29.2

Finally constant H obtained is physically similar to Hubble's constant.Observed value of Hubble's constant is 2.3X $10^{-18}$ /s where as obtained H= 1.8X $10^{-17}$ /s.Further,experiments do not indicate a difference in the magnitudes of the electron and proton charge.Some theories regarding the nature of the fundamental forces and elementary particles also do not point to a difference.

Lyttelton-Bondi Model for the Expansion of the Universe

The answer is -29.2.

1 solution

$A$ :Equate Energies or Pressure or Force.Note that total charge $Q$ = $Nq=N(-ey)$ ,u will easily get the value of $y=0.910^{-18}=(approx)10^{-18}$ .So $z=-18$ .

$B$ :The force on the atom at a distance r is F say, and from coulombs law F= $\frac{kQ}{R^2}$ .F=m $\frac{dv}{dt}$ =mv $\frac{dv}{dr}$ and $Q=$ $\frac{n}{m'}$ $\frac{4}{3}$* $\pi$ $r^3$ $q$ where m' is mass of proton and $q=-e*y$ so v is proportional to r.Hence p=1.

$C$ :Now H can be easily determined from the previous equation and is equal to $\frac{ey}{m'}$ $(\frac{3}{2(\epsilon)})^{0.5}$ .Plug in values where $y=10^{-17}$ and you will get $H=1.810^{-17}$ ..so b=1.8 and c=(-17)

$D$ : $\frac{dV}{dt}$ =3 V H or $\frac{dV}{V}$ =3Hdt,so a=3,Hence answer = $z+p+b+c+a$ =-29.2

This is the 2006 INPHO 8th Subjective Question.No claim of originality is made.

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Lyttelton-Bondi Model for the Expansion of the Universe

The answer is -29.2.

1 solution

D D D : d V d t \frac{dV}{dt} d t d V ​ =3 V H or d V V \frac{dV}{V} V d V ​ =3Hdt,so a=3,Hence answer = z + p + b + c + a z+p+b+c+a z + p + b + c + a =-29.2

This is the 2006 INPHO 8th Subjective Question.No claim of originality is made.

0 pending reports

$D$ : $\frac{dV}{dt}$ =3 V H or $\frac{dV}{V}$ =3Hdt,so a=3,Hence answer = $z+p+b+c+a$ =-29.2