Young And On Heat

Consider Young's Modulus in the case of a sphere,  $E_s$

$F_L=E_s\cfrac{A_o}{L_o}\Delta L$

when  $L_o=1$  and  $A_o=4\pi(1)^2$

$F_L=4\pi E_s\Delta L=F_T=\sqrt{3}\rho D.\cfrac{\partial (d_s)}{\partial T}.\cfrac{1}{\alpha}$

for the case of such a sphere under high temperature.  If we have,

$E_s=\sqrt{3}E$

where  $E$ is the linear Young's Modulus and  $\sqrt{3}$  the 3D constant, then

$4\pi E\Delta L=\rho D.\cfrac{\partial (d_s)}{\partial T}.\cfrac{1}{\alpha}$

We have,

$D.\cfrac{\partial (d_s)}{\partial T}=4\pi E\cfrac{\alpha}{\rho}\Delta L$  --- (*)

where  $\Delta L$  is the change in radius of the sphere,  $\rho$ is the density of the element,  $\alpha$
 is the linear thermal coefficient, and  $E$  the Young Modulus.

If both  $D$  and  $\cfrac{\partial (d_s)}{\partial T}$  are instead independent of the specific element, then the last formula provides a simple way of finding,

$D.\cfrac{\partial (d_s)}{\partial T}$    in general.

$\cfrac{\partial (d_s)}{\partial T}$  is a function in  $T$  and  $\Delta L$ changes with  $T$.  In which case  $D$  can be found by plotting  $f(T)$  vs  $\Delta L$.  Where  $f(T)$  is a function in  $T$  and  $f(T)$  vs  $\Delta L$  is a line through the origin with gradient,

$gradient=4\pi E\cfrac{\alpha}{\rho D}$

It should be noted that expression (*) is within the frame work of a solid lattice, an array of atoms in 3D.  The limiting case of an infinitely sparse lattice will then approximate the situation of free space.

Have a nice day.