Chapter 32, Problem 57GP

Solution

(a)

To show: In the unified grad theory the unification distance of ${10}^{-31}\text{m}$ is equivalent to ${10}^{16}\text{GeV}$ energy.

Given data:

The unification distance in grand field theory, $\Delta x={10}^{-32}\text{m}$

The Planck's constant, $h=6.626\times {10}^{-34}\text{J}\cdot \text{s}$

The speed of the light in vacuum, $c=3\times {10}^8\text{m}/\text{s}$

Formula used:

The relation between the energy $E$ and momentum $p$ of a particle is

  $\begin{array}{l}E=pc\\ \Delta E=\Delta p\times c\\ \therefore \Delta p=\frac{\Delta E}{c}\mathrm{...}(1)\end{array}$

From Heisenberg's uncertainty relation, we have

  $\Delta x\Delta p\ge \frac{h}{2\pi }$

  $\Delta x\Delta p\approx \frac{h}{2\pi }\mathrm{...}(2)$

From equation (1) and (2)

  $\Delta E\approx \frac{hc}{2\pi \Delta x}\mathrm{...}(3)$

Calculation:

Substituting value in the equation (3)

. $\begin{array}{c}\Delta E\approx \frac{hc}{2\pi \Delta x}\\ =\frac{\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3.00\times {10}^8\text{m}/\text{s}\right)}{2\pi \left({10}^{-32}\text{m}\right)}\frac{\left(1\text{GeV}/{10}^9\text{eV}\right)}{\left(1.60\times {10}^{-19}\text{J}/\text{eV}\right)}\\ =2\times {10}^{16}\text{GeV}\end{array}$

Next, we use de Broglie's wavelength formula. We take the de Broglie wavelength as the unification distance.

  $\lambda =\frac{h}{p}=\frac{h}{Ec}$

After rearranging the above equation

  $\begin{array}{c}E=\frac{hc}{\lambda }\\ =\frac{\left(6.63\times {10}^{-34}\text{J}\cdot \text{s}\right)\left(3.00\times {10}^8\text{m}/\text{s}\right)}{\left({10}^{-32}\text{m}\right)}\frac{\left(1\text{GeV}/{10}^9\text{eV}\right)}{\left(1.60\times {10}^{-19}\text{J}/\text{eV}\right)}\\ =12.4\times {10}^{16}\text{GeV}\end{array}$

Conclusion:

Both energies are reasonably close to ${10}^{16}\text{GeV}.$

To find: The temperature corresponding to ${10}^{16}\text{GeV}$

  ${10}^{29}\text{K}$

Given data:

Energy, $E={10}^{16}\text{GeV}\text{=}{\text{10}}^{25}\text{eV}$

Boltzmann's constant $k=1.38\times {10}^{-23}\text{J}/\text{K}$

Formula used:

The corresponding temperature of the energy $E$ is calculated by the following formula.

  $E=\frac{3}{2}kT\mathrm{....}(1)$

Where, $k=$ Boltzmann's constant

Calculation:

After rearranging the energy equation

  $\begin{array}{l}T=\frac{2E}{3k}\\ =\frac{2\left({10}^{25}\text{eV}\right)\left(1.6\times {10}^{-19}\text{J}/\text{eV}\right)}{3\left(1.38\times {10}^{-23}\text{J}/\text{K}\right)}\\ =7.7\times {10}^{28}\text{K}\\ \approx {10}^{29}\text{K}\end{array}$

Conclusion:

The temperature corresponding to ${10}^{16}\text{GeV}$ is ${10}^{29}\text{K}$