91影视

Theexpression for the average energyE of a system of harmonic oscillators is given as follows,

$\stackrel{炉}{E}=\frac{\mathrm{鈩�}{蝇}_0}{e^{\mathrm{鈩�}{蝇}_0/k_{B}T}-1}$

The expression for the firstorder Taylor expansion of the function$e^x$is given by:

$e^x鈮�1+x$

To carry out the approximation of E in the limit $k_{B}T>>\mathrm{鈩�}{蝇}_0$, recall that the first-order Taylor expansion of the function$e^x$ that is a reasonable approximation for$x<<1$

Now, consider the exponential in the expression for E which is$e^{\mathrm{鈩�}{蝇}_0/k_{B}T}$ . Consider$x=\mathrm{鈩�}{蝇}_0/k_{B}T$ , and $x<<1$then the expression becomes as follows,

$\mathrm{鈩�}{蝇}_0<<k_{B}T$

which is the assumption needed. So, the Taylor鈥檚 expansion becomes as follows,

$e^{\mathrm{鈩�}{蝇}_0/k_{B}T}鈮�1+\frac{\mathrm{鈩�}{蝇}_0}{k_{B}T}$

Substitute the above value in the expression for the average energy E,.

$\begin{array}{rcl}\stackrel{炉}{E}& 鈮�& \frac{\mathrm{鈩�}{蝇}_0}{\left(1+\frac{\mathrm{鈩�}{蝇}_0}{k_{B}T}\right)-1}\\ & 鈮�& \frac{\mathrm{鈩�}{蝇}_0{k}_{B}T}{\mathrm{鈩�}{蝇}_0}\\ & 鈮�& k_{B}T\\ & & \end{array}$

Thus, the given expression is verified.