Chapter 1, Problem 91P

(a)

To determine

The combination of constants that has dimension of time.

Answer to Problem 91P

The combination of constants that has dimension of time is $\underset{\_}{\sqrt{\frac{hG}{c}}}$.

Explanation of Solution

Write an expression for Plank’s time using the combinations of $c$, $G$, and $h$

$t_{planck}=c^p{G}^q{h}^r$ (I)

Here, $c$ is the speed of light, $G$ is the gravitational constant, $h$ is the Planck’s constant, $t_{\text{planck}}$ is the Planck’s time

$p,q,r$ are integers. Rewrite the equation (I) in terms of units

${\text{s}}^1={\left(\frac{\text{m}}{\text{s}}\right)}^p{\left(\frac{{\text{m}}^3}{{\text{kg}}^1\cdot {\text{s}}^2}\right)}^q{\left(\frac{{\text{kg}}^1{\text{m}}^2}{{\text{s}}^1}\right)}^r$ (II)

Conclusion:

The power of base units should be same in both sides of equation (II). Consider the units separately.

For seconds,

$1=-p-2q-r$ (III)

For meters,

$0=p+3q+2r$ (IV)

For kilogram,

$\begin{array}{c}0=0-q+r\\ q=r\end{array}$ (V)

Substitute equation (V) in (III)

$\begin{array}{l}1=-p-2q-q\\ 1=-p-3q\end{array}$ (VI)

Substitute equation (V) in (IV)

$\begin{array}{c}0=p+3q+2q\\ 0=p+5q\\ p=-5q\end{array}$ (VII)

Substitute $-5q$ for $p$ in equation (VI)

$\begin{array}{c}1=-\left(-5q\right)-3q\\ q=\frac{1}{2}\end{array}$

Substitute $\frac{1}{2}$ for $q$ in equation (V)

$p=-\frac{5}{2}$

From equation (V)

$q=r=\frac{1}{2}$

Thus, according to the values of $p,q,\text{and }r$ the only possible combination that has unit of time is

$t_{\text{planck}}=\sqrt{\frac{hG}{c}}$

Therefore, the combination of constants that has dimension of time is $\underset{\_}{\sqrt{\frac{hG}{c}}}$.

(b)

To determine

The Planck’s time in seconds.

Answer to Problem 91P

The Planck’s time in seconds is $\underset{\_}{t_{\text{planck}}=1.3\times {10}^{-43}\text{s}}$.

Explanation of Solution

Write the expression for Planck’s time

$t_{\text{planck}}=\sqrt{\frac{hG}{c}}$

Here, $c$ is the speed of light, $G$ is the gravitational constant, $h$ is the Planck’s constant, $t_{\text{planck}}$ is the Planck’s time

Conclusion:

Substitute $6.6\times {10}^{-34}\text{kg}\cdot {\text{m}}^2\cdot {\text{s}}^{-1}$ for $h$, $6.7\times {10}^{-11}{\text{m}}^3\cdot {\text{kg}}^{-1}\cdot {\text{s}}^{-2}$ for $G$ , and $3.0\times {10}^8\text{m}/\text{s}$ in above equation

$\begin{array}{c}{t}_{\text{planck}}=\sqrt{\frac{6.6\times {10}^{-34}\text{kg}\cdot {\text{m}}^2\cdot {\text{s}}^{-1}\times 6.7\times {10}^{-11}{\text{m}}^3\cdot {\text{kg}}^{-1}\cdot {\text{s}}^{-2}}{3.0\times {10}^8\text{m}/\text{s}}}\\ =1.3\times {10}^{-43}\text{s}\end{array}$

Therefore, the Planck’s time in seconds is $\underset{\_}{t_{\text{planck}}=1.3\times {10}^{-43}\text{s}}$.