Show That alpha:beta:gamma=1:2:3??

Consider a cube of length l, dt is the change in temp, dl is the change in length, ds is the change in area, dv is the change in volume.

Linear coefficient of expansion Alpha- $\alpha$
 coefficient of area expansion Beta- $\beta$
coefficient of volume expansion Gama- $\gamma$

Relation between  $\alpha$ and $\beta$ :-

From thermal expansion of solids,

dl= $\alpha$ldt.(1)

ds=$\beta$sdt..(2)

ds=new area - original area

=(l+dl)²-l²

=l²+2ldl+(dl)²-l²

=2ldl( dl is very small and the square of it will be very very small so it can be neglected)

=2l $\alpha$dtl( from (1))

=2l² $\alpha$dt

=2s $\alpha$dt..(3)

From (2) and (3)

$\beta$sdt=2s $\alpha$dt

$\beta$=2 $\alpha$

hence, $\alpha = \frac{\beta }{2}$

Relation between a and $\gamma$ :-

dl= $\alpha$ldt.(1)

dv=$\gamma$vdt.(2)

dv=new volume-original volume

=(l+dl)³-l³

=l³+3(l)²dl+3l(dl)²+(dl)³-l³

=3l²dl (dl2 and dl3 can be neglected)

Substituting dl = $\alpha$ldt

=3l²2 $\alpha$ldt

=3l³ $\alpha$dt

Since l³ = v
=3v $\alpha$dt..(3)

From (2) and (3)

$\gamma$vdt=3v $\alpha$dt

$\gamma$=3 $\alpha$

$\alpha$=$\gamma$/3

SO $\alpha$=$\beta$/2= γ/3