Q.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 $e_p = -(1 + y)e$ where $e$ is the electronic charge.

The Lyttleton-Bondi model proposes that a slight charge imbalance in matter could drive the expansion of the Universe. The key idea is to compare the electrostatic repulsion between charged matter with its gravitational attraction.

Let $e_0$ be the magnitude of the elementary charge. The charge of an electron is $-e_0$.

The problem states the charge on the proton is $e_p = -(1+y)e$, where $e$ is the electronic charge. Interpreting "electronic charge" as the charge of an electron, we have $e = -e_0$.

Thus, the proton charge is $e_p = -(1+y)(-e_0) = (1+y)e_0$.

The net charge of a hydrogen atom (one proton, one electron) is $q_{net} = e_p + (-e_0) = (1+y)e_0 - e_0 = y e_0$.

The mass of a hydrogen atom is approximately the mass of a proton, $m_p$.

Consider a spherical region of radius $r$.

The total number of hydrogen atoms in this region is $N_{atoms} = \frac{4}{3}\pi r^3 N$.

The total mass within this sphere is $M = N_{atoms} m_p = \frac{4}{3}\pi r^3 N m_p$.

The total charge within this sphere is $Q = N_{atoms} q_{net} = \frac{4}{3}\pi r^3 N y e_0$.

(a) Critical value of $y$ for expansion to start:

Expansion begins when the outward electrostatic force ($F_E$) on a hydrogen atom at the edge of the sphere overcomes the inward gravitational force ($F_G$).

1. The gravitational force on a hydrogen atom (mass $m_p$) at radius $r$ due to the mass $M$ inside is: $F_G = \frac{G M m_p}{r^2} = \frac{G (\frac{4}{3}\pi r^3 N m_p) m_p}{r^2} = \frac{4}{3}\pi G N m_p^2 r$.
2. The electrostatic force on a hydrogen atom (charge $y e_0$) at radius $r$ due to the charge $Q$ inside is: $F_E = \frac{1}{4\pi\epsilon_0} \frac{Q (y e_0)}{r^2} = \frac{1}{4\pi\epsilon_0} \frac{(\frac{4}{3}\pi r^3 N y e_0) (y e_0)}{r^2} = \frac{1}{3\epsilon_0} N (y e_0)^2 r$.
3. For expansion to start, $F_E \ge F_G$: $\frac{1}{3\epsilon_0} N (y e_0)^2 r \ge \frac{4}{3}\pi G N m_p^2 r$.
4. Cancelling $N$, $r$, and $\frac{1}{3}$ (assuming $N, r > 0$): $\frac{(y e_0)^2}{\epsilon_0} \ge 4\pi G m_p^2 \implies y^2 \ge \frac{4\pi G \epsilon_0 m_p^2}{e_0^2}$. …

The problem explores a fascinating hypothesis by Lyttleton and Bondi, suggesting that a slight charge imbalance in matter could drive the expansion of the Universe. The core idea is to compare the electrostatic force between charged hydrogen atoms with their gravitational attraction. If the electrostatic repulsion is strong enough to overcome gravity, expansion can occur.

We are given that the Universe is made of hydrogen atoms with a constant number density $N$. The charge on a proton is $e_p = -(1+y)e$, where $e$ is the electronic charge (the charge of an electron). This means $e$ is a negative value (approximately $-1.6 \times 10^{-19} \text{ C}$).

First, let's determine the net charge of a single hydrogen atom:

The charge of an electron is $e$.

The charge of a proton is $e_p = -(1+y)e$.

The net charge of a hydrogen atom, $q_{atom}$, is the sum of the proton and electron charges:

$q_{atom} = e_p + e = -(1+y)e + e = (-1 - y + 1)e = -ye$.

Now we can proceed to analyze the forces.

(a) Find the critical value of $y$ such that expansion may start.

For expansion to start, the outward electrostatic repulsive force must at least balance the inward gravitational attractive force. For the critical value of $y$, these forces are exactly equal.

1. Consider a spherical region of matter:

   To calculate the forces, we imagine a spherical region of radius $r$ centered at an arbitrary point in the Universe. Due to the uniform number density $N$, this sphere contains a total number of hydrogen atoms given by:

   $N_{atoms} = N \times (\text{Volume of sphere}) = N \left(\frac{4}{3}\pi r^3\right)$.

2. Calculate the total mass within radius $r$:

   Each hydrogen atom has a mass $m_H$. So, the total mass $M(r)$ within the sphere is:

   $M(r) = N_{atoms} \times m_H = N \left(\frac{4}{3}\pi r^3\right) m_H$.

3. Calculate the total charge within radius $r$:

   Each hydrogen atom has a net charge $q_{atom} = -ye$. So, the total charge $Q(r)$ within the sphere is:

   $Q(r) = N_{atoms} \times q_{atom} = N \left(\frac{4}{3}\pi r^3\right) (-ye)$.

4. Determine the gravitational force on a test hydrogen atom at radius $r$:

   We consider a test hydrogen atom (mass $m_H$) located at the surface of this sphere (distance $r$ from the center). According to Newton's Shell Theorem, the gravitational force exerted by a spherically symmetric mass distribution on an external particle is the same as if all the mass were concentrated at the center.

   The gravitational force $F_g$ is attractive and directed inwards:

   $F_g = G \frac{M(r) m_H}{r^2}$

   Substitute $M(r)$:

   $F_g = G \frac{\left(N \frac{4}{3}\pi r^3 m_H\right) m_H}{r^2} = G N \frac{4}{3}\pi r m_H^2$.

5. Determine the electric field at radius $r$:

   We use Gauss's Law for a spherically symmetric charge distribution. The electric field $E$ at a distance $r$ from the center is given by:

   $E \cdot (4\pi r^2) = \frac{Q(r)}{\epsilon_0}$

   $E = \frac{Q(r)}{4\pi\epsilon_0 r^2}$

   Substitute $Q(r)$:

   $E = \frac{N \left(\frac{4}{3}\pi r^3\right) (-ye)}{4\pi\epsilon_0 r^2} = \frac{N (-ye) r}{3\epsilon_0}$.

   The direction of $E$ depends on the sign of $-ye$. If $-ye > 0$ (i.e., $y < 0$ since $e < 0$), the field is outward. If $-ye < 0$ (i.e., $y > 0$), the field is inward.

6. Determine the electrostatic force on a test hydrogen atom at radius $r$:

   The test hydrogen atom has charge $q_{atom} = -ye$. The electrostatic force $F_e$ on this atom is $F_e = q_{atom} E$.

   $F_e = (-ye) \left( \frac{N (-ye) r}{3\epsilon_0} \right) = \frac{N y^2 e^2 r}{3\epsilon_0}$.

   Since $y^2 e^2$ is always positive (for $y \ne 0$), this force is always repulsive and directed outwards.

7. Set up the condition for expansion to start:

   For expansion to start, the outward electrostatic force must be equal to or greater than the inward gravitational force. For the critical value of $y$, they are equal:

   $F_e = F_g$

   $\frac{N y^2 e^2 r}{3\epsilon_0} = G N \frac{4}{3}\pi r m_H^2$.

8. Solve for $y$:

   We can cancel $N$, $r$, and $\frac{1}{3}$ from both sides:

   $y^2 e^2 = G (4\pi \epsilon_0) m_H^2$.

   Recall that Coulomb's constant $k_e = \frac{1}{4\pi\epsilon_0}$, so $4\pi\epsilon_0 = \frac{1}{k_e}$.

   $y^2 e^2 = G \frac{1}{k_e} m_H^2$.

   $y^2 = \frac{G m_H^2}{k_e e^2}$.

   Taking the square root, we get the critical value for $y$:

   $y = \sqrt{\frac{G m_H^2}{k_e e^2}}$.

   (We take the positive root as $y$ represents a magnitude of deviation from neutrality; a negative $y$ would also result in a repulsive force due to $y^2$).

(b) Show that the velocity of expansion is proportional to the distance from the centre.

If $y$ is slightly greater than the critical value, the electrostatic repulsion will exceed the gravitational attraction, leading to a net outward force and thus expansion.

1. Calculate the net force on a test hydrogen atom: …

Concept: Modified Newtonian Gravity with Net Charge

The idea is that if matter has a net charge, there is an electrostatic repulsion that opposes gravitational attraction. For expansion to start, the repulsive force must exceed the attractive gravitational force.

Method: Force Balance at a Test Particle

We consider a test hydrogen atom at a distance $r$ from the centre of a uniform, static spherical universe of constant number density $N$.

Steps:

1. Gravitational force on the test atom (mass $m_p$):
   • Mass inside radius $r$: $M(r) = \frac{4}{3} \pi r^3 N m_p$
   • Gravitational force (inward):

2. Electrostatic force on the test atom (net charge on proton $e_p = -(1+y)e$, electron charge $= -e$):
   • Net charge on each hydrogen atom: $q = e_p + (-e) = -(1+y)e - e = -(2+y)e$
   • Charge inside radius $r$: $Q(r) = \frac{4}{3} \pi r^3 N q$
   • Electrostatic force on the test atom (charge $q$):

3. Condition for expansion: The net outward force must be positive:

Substituting $q = -(2+y)e$:

4. Cancel common factors ($\frac{4}{3} N r$):

5. Solve for critical $y$ (when equality holds): $(2+y)^2 = \frac{4\pi\varepsilon_0 G m_p^2}{e^2}$ …

Common Mistakes & How to Avoid Them

Mistake 1: Forgetting that expansion requires a net repulsive force

Many students jump straight into equating forces without first identifying the physical condition for "expansion may start."

How to avoid:

• Expansion starts when the net gravitational + electrostatic force on a test particle becomes repulsive (outward).
• For a hydrogen atom at the edge of a spherical universe of radius $R$, the inward gravitational force must be balanced or overcome by the outward electrostatic repulsion.
• Write the condition:

where outward is positive.

Mistake 2: Using the wrong sign for the proton charge

Students often take $e_p = -(1+y)e$ and forget that $y$ can be positive or negative, leading to sign errors in the Coulomb force.

How to avoid:

• Write explicitly:
  • Electron charge: $q_e = -e$
  • Proton charge: $q_p = -(1+y)e$
• The net charge of a hydrogen atom is:

• This net charge determines the electrostatic force between atoms.

Mistake 3: Incorrectly applying Gauss's law for the electric field

Students sometimes treat the charge distribution as point charges rather than a continuous uniform distribution.

How to avoid:

• For a uniform spherical distribution of net charge density $\rho_q$, the electric field at distance $r$ from the centre is:

where $Q_{\text{enc}} = \frac{4}{3}\pi r^3 \rho_q$.

• Similarly, gravitational field uses mass density $\rho_m$:

(negative sign for inward direction).

Mistake 4: Confusing number density with mass/charge density

Given $N$ = number density of hydrogen atoms, students sometimes forget to multiply by mass or charge per atom.

How to avoid:

• Mass density: $\rho_m = N \cdot m_H$ (where $m_H \approx m_p$)
• Charge density: $\rho_q = N \cdot q_{\text{atom}} = N \cdot [-(2+y)e]$

Mistake 5: Forgetting that both forces act on the same test particle

When writing the net force on a hydrogen atom at distance $r$, both gravitational and electrostatic forces must be considered simultaneously.

How to avoid:

• Net force on a test hydrogen atom at distance $r$:

• Substitute $E(r)$ and $g(r)$ from Gauss's law, then set $F_{\text{net}} = 0$ for critical $y$.

Mistake 6: Algebraic slip when solving for critical $y$

After substituting, students often make sign errors in the equation:

How to avoid:

• Cancel common factors ($\frac{4}{3}\pi r$) carefully.
• Write the simplified equation: $\frac{q_{\text{atom}} \rho_q}{3\varepsilon_0} - G m_H \rho_m = 0$ …