Question

The rest mass of the observable universe is 1.5 × 1053 kg as measured in the Earth frame. If βc is the maximum possible speed of an electron in this frame, find 1 − β. Ignore the effects of general relativity.

Answer 1

1.84 × 10-167

Answer 2

To solve this problem, we need to relate the rest mass of the observable universe to the maximum possible speed of an electron. The phrase "maximum possible speed of an electron in this frame" implies that the electron has acquired an enormous amount of energy, potentially equivalent to the entire rest mass energy of the observable universe.

1. Relativistic Energy:
   According to special relativity, the total relativistic energy ($E$) of a particle with rest mass ($m_0$) moving at a speed $v$ is given by:
   $E = \gamma m_0 c^2$
   where $\gamma = \frac{1}{\sqrt{1 - \beta^2}}$ is the Lorentz factor, $\beta = \frac{v}{c}$, and $c$ is the speed of light.

2. Energy Equivalence:
   The rest mass energy of the observable universe ($M_{universe}$) is given by $E_{universe} = M_{universe} c^2$.
   If the electron attains its "maximum possible speed" by absorbing all the energy from the universe, then its total relativistic energy will be equal to the total rest mass energy of the universe:
   $E_{electron} = E_{universe}$
   $\gamma m_e c^2 = M_{universe} c^2$
   where $m_e$ is the rest mass of the electron.

3. Calculate the Lorentz Factor ($\gamma$):
   From the energy equivalence, we can find $\gamma$:
   $\gamma = \frac{M_{universe}}{m_e}$

Given:
Rest mass of the observable universe, $M_{universe} = 1.5 \times 10^{53} \text{ kg}$
Rest mass of an electron, $m_e = 9.109 \times 10^{-31} \text{ kg}$

$\gamma = \frac{1.5 \times 10^{53} \text{ kg}}{9.109 \times 10^{-31} \text{ kg}}$
$\gamma \approx 0.16467 \times 10^{84}$
$\gamma \approx 1.6467 \times 10^{83}$

4. Calculate $1 - \beta$:
   We know that $\gamma = \frac{1}{\sqrt{1 - \beta^2}}$.
   Squaring both sides:
   $\gamma^2 = \frac{1}{1 - \beta^2}$
   Rearranging to find $1 - \beta^2$:
   $1 - \beta^2 = \frac{1}{\gamma^2}$
   We can factor the left side:
   $(1 - \beta)(1 + \beta) = \frac{1}{\gamma^2}$

Since $\gamma$ is extremely large ($1.6467 \times 10^{83}$), $\beta$ must be very, very close to 1.
Therefore, $1 + \beta \approx 1 + 1 = 2$.

Substituting this approximation:
$(1 - \beta) \times 2 \approx \frac{1}{\gamma^2}$
$1 - \beta \approx \frac{1}{2\gamma^2}$

Now, substitute the value of $\gamma$:
$1 - \beta \approx \frac{1}{2 \times (1.6467 \times 10^{83})^2}$
$1 - \beta \approx \frac{1}{2 \times (1.6467)^2 \times (10^{83})^2}$
$1 - \beta \approx \frac{1}{2 \times 2.7115 \times 10^{166}}$
$1 - \beta \approx \frac{1}{5.423 \times 10^{166}}$
$1 - \beta \approx 0.18439 \times 10^{-166}$
$1 - \beta \approx 1.844 \times 10^{-167}$

The value of $1 - \beta$ is extremely small, indicating that the electron's speed is incredibly close to the speed of light.

Explanation of the solution:

The problem implies that the electron achieves its maximum possible speed by acquiring energy equivalent to the total rest mass energy of the observable universe. Using Einstein's mass-energy equivalence, $E = Mc^2$, we equate the total relativistic energy of the electron ($\gamma m_e c^2$) to the universe's rest mass energy ($M_{universe} c^2$). This allows us to calculate the Lorentz factor, $\gamma = M_{universe}/m_e$. Since $\gamma$ is extremely large, the electron's speed ($v = \beta c$) is very close to $c$. We use the relativistic formula $\gamma = 1/\sqrt{1-\beta^2}$ and the approximation $1-\beta \approx 1/(2\gamma^2)$ for $\beta \approx 1$ to find the value of $1-\beta$.