Chapter A.6, Problem 24P

(a)

To determine

To Write: The formula for the total zero-point energy of all the modes of the field inside the box in terms of a triple integral over the mode numbers in the x, y, and z direction.

Answer to Problem 24P

  $\frac{E}{V}=\frac{\hslash }{2{\pi }^2{c}^2}{\displaystyle {\int }_0^{\infty }{\omega }^{3}d\omega }$

Explanation of Solution

Formula used:

Total energy within the box can be written as:

  $\begin{array}{c}E=2{\displaystyle \sum _{n_x}{\displaystyle \sum _{n{y}_{}}{\displaystyle \sum _{n{z}_{}}\epsilon \left(\overrightarrow{n}\right)}}}\\ =2{\displaystyle \iiint \epsilon \left(\overrightarrow{n}\right)d{n}_{x}d{n}_{y}d{n}_z}\end{array}$

Where, $n$ is positive integer$n_x,n_y,n_z$ is a triplet of positive number.

Factor of $2$ is for the two-spin orientation for each $\left(\overrightarrow{n}\right)$

Calculation:

Wavelength can be given as:

  ${\lambda }_n=\frac{2L}{n}$

So, speed of light:

  $\begin{array}{l}c=\frac{\omega }{2\pi }{\lambda }_n\\ c=\frac{\omega }{2\pi }\frac{2L}{n}\\ C=\frac{\omega L}{n\pi }\\ n=\frac{\omega L}{c\pi }\mathrm{...}\left(1\right)\end{array}$

  

So, $E=2{\displaystyle {\int }_0^{n_{\max }}dn}{\displaystyle {\int }_0^{\pi }d\theta }{\displaystyle {\int }_0^{\pi /2}d\varphi n^2\sin \theta \epsilon \left(n\right)}$

From equation (1)

  $\begin{array}{l}E=2{\displaystyle \int \frac{L}{c\pi }}d\omega {\displaystyle {\int }_0^{\pi }d\theta }{\displaystyle {\int }_0^{{\omega }_{\max }}{\omega }^{3}d\omega }\\ E=2\left(\frac{L}{c\pi }\right){\left(\frac{L}{c\pi }\right)}^2\left(\frac{\pi }{2}\right)\hslash {\displaystyle {\int }_0^{\infty }{\omega }^{3}d\omega }\\ E=\frac{L^3\hslash }{2c^2{\pi }^2}{\displaystyle {\int }_0^{\infty }{\omega }^{3}d\omega }\\ \frac{E}{L^3}=\frac{\hslash }{2c^2{\pi }^2}{\displaystyle {\int }_0^{\infty }{\omega }^{3}d\omega }\\ \frac{E}{V}=\frac{\hslash }{2c^2{\pi }^2}{\displaystyle {\int }_0^{\infty }{\omega }^{3}d\omega }\end{array}$

Conclusion:

The formula for the total zero-point energy of all the modes of the field inside the box is

   $\frac{E}{V}=\frac{\hslash }{2c^2{\pi }^2}{\displaystyle {\int }_0^{\infty }{\omega }^{3}d\omega }$ .

(b)

To determine

To Show: The Planck length indeed has units of length.

To Calculate:Numerical value of Planck length.

Answer to Problem 24P

  $1.61\times {10}^{-35}\text{m}$

Explanation of Solution

Given:

Planck Length = $\sqrt{\frac{G\hslash }{c^3}}$

Formula used:

Dimension formula of$\begin{array}{l}G=\left[M^{-1}\right]\left[L^3\right]\left[T^{-2}\right]\\ \hslash =\left[M\right]\left[L^2\right]\left[T^{-1}\right]\\ c=\left[M^0\right]\left[L^1\right]\left[T^{-1}\right]\end{array}$

Calculation:

Dimension formula for given expression can be calculated as:

  $\begin{array}{l}\sqrt{\frac{G\hslash }{c^3}}=\sqrt{\frac{\left(\left[M^{-1}\right]\left[L^3\right]\left[T^{-2}\right]\right)\left(\left[M\right]\left[L^2\right]\left[T^{-1}\right]\right)}{\left[M^0\right]\left[L^3\right]\left[T^{-3}\right]}}\\ \sqrt{\frac{G\hslash }{c^3}}=\left({\left[L^2\right]}^{1/2}\right)\\ \sqrt{\frac{G\hslash }{c^3}}=\left[L\right]\end{array}$

Planck length can be calculated as:

  $\begin{array}{l}\sqrt{\frac{G\hslash }{c^3}}=\sqrt{\frac{6.674\times {10}^{-11}\times 1.053\times {10}^{-34}}{27\times {10}^{24}}}\\ \sqrt{\frac{G\hslash }{c^3}}=\sqrt{2.6028\times {10}^{-70}}\\ \sqrt{\frac{G\hslash }{c^3}}=1.61\times {10}^{-35}\text{m}\end{array}$

Conclusion:

It is proved that the Planck length indeed has units of length and its magnitude is $1.61\times {10}^{-35}\text{m}$ .

(c)

To determine

To Estimate: The energy per unit volume in empty state and equivalent mass density.

Answer to Problem 24P

  $2.52\times {10}^{123}{\text{ J/m}}^3$

  $2.8\times {10}^{106}{\text{ kg/m}}^3$

Explanation of Solution

Formula used:

  $\frac{E}{V}=\frac{\hslash }{2{\pi }^2{c}^2}{\displaystyle {\int }_0^{\infty }{\omega }^{3}d\omega }$

Where, $\hslash$ is the reduced Planck’s constant$c$ is the speed of light$\omega$ is the angular velocity

Calculation:

  $\begin{array}{l}\frac{E}{V}=\frac{\hslash }{2{\pi }^2{c}^2}{\displaystyle {\int }_0^{\infty }{\omega }^{3}d\omega }\\ \frac{E}{V}=\frac{\hslash }{2{\pi }^2{c}^2}\times \frac{{\omega }_{\max }^4}{4}\\ \frac{E}{V}=\frac{\hslash {\omega }_{\max }^4}{8{\pi }^2{c}^2}\end{array}$

But,

  $\begin{array}{l}c=\frac{\omega }{2\pi }\lambda \\ \lambda =L_p\end{array}$

Therefore,

  $\begin{array}{l}\frac{E}{V}=\frac{\hslash }{8{\pi }^2{c}^2}\times {\left(\frac{2\pi c}{L_p}\right)}^4\\ \frac{E}{V}=\frac{2\hslash {\pi }^2{c}^2}{L_p^4}\\ \frac{E}{V}=\frac{2\times 1.051\times {10}^{-34}\times {\left(3\times {10}^8\right)}^2{\left(3.14\right)}^2}{{\left(1.65\times {10}^{-35}\right)}^4}\\ \frac{E}{V}=2.52\times {10}^{123}{\text{ J/m}}^3\end{array}$

Now, mass density can be calculated as$\begin{array}{l}m=\frac{\frac{E}{V}}{c^2}\\ m=\frac{2.52\times {10}^{123}}{{\left(3\times {10}^8\right)}^2}\\ m=2.8\times {10}^{106}{\text{ kg/m}}^3\end{array}$

It is much greater than the average mass density of ordinary matter in the universe.

Conclusion:

Average mass density is $2.8\times {10}^{106}{\text{ kg/m}}^3$ .