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

We have to apply the principle of homogeneity to solve this problem. Principle of homogeneity states that in a correct equation, the dimensions of each term added or subtracted must be same, i.e., dimensions of LHS and RHS should be equal,
We know that, dimensions of
$\left[h\right]=\left[M{L}^2{T}^{-1}\right],\left[c\right]=[L{T}^{-1}{]}^a,[G]=[M^{-1}{L}^3{T}^{-2}]$

i) Let, $m\infty c^x{h}^y{G}^z$

$\Rightarrow m=l{c}^a{h}^b{G}^0$...........(i)

where, $k$ is a dimensionless constant of proportionality.
Substituing dimensions of each term in eq.(i), we get

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

On comparing powers of same terms on both sides , we get
$b-c=1$................(ii)
$a+2b+3c=0$.................(iii)
$-a-b-2c=0$.............(iv)
Adding eq (ii),(iii) and (iv) we get
$2b=1\Rightarrow 2b=1\Rightarrow b=\frac{1}{2}$
Substituting value of b in eq.ii.
$c=-\frac{1}{2}$
From eq.iv
$a=-b-2c$
Substituting values of b and c,we get
$a=-\frac{1}{2}-2(-\frac{1}{2})=\frac{1}{2}$
Putting values of a ,b and c in eq.(i), we get
$m=k{c}^{1/2}{h}^{1/2}{G}^{1/2}=k\sqrt{\frac{ch}{G}}$

ii) Let $L\infty c^a{h}^b{G}^c$

$\Rightarrow L=k{c}^a{h}^b{G}^0$,............(v)

where k is a dimensionless constant.
Substituting dimensions of each term in eq.v, we get

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

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

Comparing powers of same terms on both sides, we get
b-c=1........{vi}
$a+2b+3c=1$...............(vii)

$-a-b-2c=0$...............(viii)
Adding eqs $vi,vii$ and $viii$, we get

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

Substituting value of $b$ in eq $vi$, we get

$c=\frac{1}{2}$

From eq.$viii,a=-b-2c$

Substituting values of $a,b,c$ in eq v,we get

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

iii) Let $T\infty c^a{h}^b{G}^c$

$\Rightarrow =T=c^a{h}^{b}G6c$........(ix)

where k is a dimensionless constant.
Substituting dimensionless of each term in eq.

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

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

On comparing powers of same terms, we get
$b-c=\mathrm{0................}x$
$a+2b+3c=\mathrm{1.............}xi$
$-a-b-2c=1$

Adding eq. $x,xixii,$ 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 vales 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

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