Question

If velocity of light c, Planck’s constant $h$ and gravitational constant $G$ are taken as fundamental quantities then express mass, length and time in terms of dimensions of these quantities.

Answer 1

Step 1: Find the dimension of mass $\left(M\right)$ in terms of $c,h,$and$G$.
As we know that, dimensions of

$\left[h\right]=\left[M{L}^2{T}^{-1}\right]$
$\left[c\right]=\left[L{T}^{-1}\right]$
$\left[G\right]=\left[M^{-1}{L}^3{T}^{-2}\right]$

By using principle of homogeneity,

Let $M\alpha c^a{h}^b{G}^c$

Then, $M=k{c}^a{h}^b{G}^c\dots \dots ..\left(i\right)$
where, k is a dimensionless constant of proportionality. by putting the values of $c,h$ and$G$in equation (i)

$\left[M{L}^0{T}^0\right]$

$=[L{T}^{-1}{]}^a\times [M{L}^2{T}^{-1}{]}^b\times [M^{-1}{L}^3{T}^{P-2}{]}^c$

Comparing powers of same terms on both sides, we get
$b-c=1\dots \dots \dots \left(ii\right)$
$a+2b+3c=0\dots \dots \dots .\left(iii\right)$
$-a-b-2c=0\dots \dots \dots \left(iv\right)$

Adding Equations (ii), (iii) and (iv), we get

$2b=1\Rightarrow b=\frac{1}{2}$

Substituting value of b in Eq. (ii), we get $c=\frac{-1}{2}$

From Eq. (iv) $a=-b-2c$
Substituting values of b and c, we get

$a=\frac{-1}{2}-2\left(\frac{-1}{2}\right)=\frac{1}{2}$

Putting values of a,b and c in Eq. (i), we get

$M=k{c}^{\frac{1}{2}}{h}^{\frac{1}{2}}{G}^{\frac{-1}{2}}\Rightarrow M=k\sqrt{\frac{ch}{G}}$

Step 2: Find the dimension of length

(L) in terms of $c,h,$and$G.$

Let $L\alpha c^a{h}^b{G}^c$
Then,$L=k{c}^a{h}^b{G}^c\dots \dots \dots .\left(v\right)$
where k is a dimensionless constant.
Substituting dimensions of each term in Eq. (v), we get

$\left[M^{0}L{T}^0\right]=[L{T}^{-1}{]}^a\times [M{L}^2{T}^{-1}{]}^b\times \left[M^{-1}{L}^3{T}^{-2}{]}^c\right[M^{0}L{T}^0]=[M^{b-c}{L}^{a+2b+3c}{T}^{-a-b-2c}]$

On comparing powers of same terms, we get

$b-c=0\dots \dots ..\left(vi\right)$

$a+2b+3c=1\dots \dots \dots \left(vii\right)$

$-a-b-2c=0\dots \dots \left(viii\right)$

Adding equation (vi), (vii) and (viii), we get

$2b=1\Rightarrow b=\frac{1}{2}$
Substituting value of b in Eq. (vi), we get ,
$c=\frac{1}{2}$

From Eq. (viii), $a=-b-2c$
Substituting values of b and c, we get
$a=\frac{-1}{2}-2\left(\frac{1}{2}\right)=\frac{-3}{2}$
Putting values of a,b and c in Eq. (v), we get

$L=k{c}^{\frac{-3}{2}}{h}^{\frac{1}{2}}{G}^{\frac{1}{2}}=$ $L=k\sqrt{\frac{hG}{c^3}}$

Step 3: Find the dimension of Time (T) in terms of $c,h,$and$G$.
Let $T\alpha c^a{h}^b{G}^c$

Then,$T=c^a{h}^b{G}^c\dots \dots ..\left(ix\right)$
where, k is a dimensionless constant.

Substituting dimensions of each term in Eq. (ix), we get

$\left[M^0{L}^0{T}^1\right]=[L{T}^{-1}{]}^a\times [M{L}^2{T}^{-1}{]}^b\times \left[M^{-1}{L}^3{T}^{-2}{]}^c\right[M^0{L}^0{T}^1]=[M^{b-c}{L}^{a+2b+3c}{T}^{-a-b-2c}]$

On comparing powers of same terms, we get
$b-c=0\cdots \dots ..\left(x\right)$

$a+2b+3c=0\cdots \dots ..\left(xi\right)$

$-a-b-2c=1\cdots \dots .\left(xii\right)$

Adding Equation (x), (xi) and (xii), we get

$2b=1\Rightarrow b=\frac{1}{2}$
Substituting the value of b in Eq. (x), we get

$c=b=\frac{1}{2}$
From Eq. (xii),
$a=-b-2c-1$
Substituting values of b and c, we get

$a=\frac{-1}{2}-2\left(\frac{1}{2}\right)-1=\frac{-5}{2}$

Putting values of a,b and c in Eq. (ix), we get

$T=k{c}^{\frac{-5}{2}}{h}^{\frac{1}{2}}{G}^{\frac{1}{2}}$

$T=k\sqrt{\frac{hG}{c^5}}$

Final Answer:$L=k\sqrt{\frac{hG}{c^3}},M=k\sqrt{\frac{ch}{G}},T=k\sqrt{\frac{hG}{c^5}}$