Chapter 17, Problem 17.21E

Interpretation Introduction

(a)

Interpretation:

The expression for $\Phi$ explicitly is to be stated. The given expressions are to be verified.

Concept introduction:

A mathematical function represents the way in which its value depends on some variable. A function can contain some variables and constants. The term multiplied by a variable is called its coefficient. When derivative of a function is taken with respect to a variable, then the terms that only contain other variables are considered as zero.

Answer to Problem 17.21E

The expression for $\Phi$ explicitly is represented as,

$\Phi =C_1{\xi }_1+C_2{\xi }_2+C_3{\xi }_3$

The given expressions are verified as shown below.

Explanation of Solution

The given function $\varphi$ can be expressed as,

$\Phi ={\displaystyle \sum _{i=1}^3{C}_i{\xi }_i}$ …(1)

Where,

• ${\xi }_i$ represents the $i^{\text{th}}$ variable

• $C_i$ represents the value of the coefficient multiplying variable ${\xi }_i$.

The expanded form of the equation (1) is represented as,

$\Phi =C_1{\xi }_1+C_2{\xi }_2+C_3{\xi }_3$ …(2)

Derivate equation (2) with respect to ${\xi }_1$.

$\begin{array}{rll}\frac{\delta \Phi }{\delta {\xi }_1}& =& \frac{\delta }{\delta {\xi }_1}\left(C_1{\xi }_1+C_2{\xi }_2+C_3{\xi }_3\right)\\ & =& \frac{\delta }{\delta {\xi }_1}\left(C_1{\xi }_1\right)+\frac{\delta }{\delta {\xi }_1}\left(C_2{\xi }_2\right)+\frac{\delta }{\delta {\xi }_1}\left(C_3{\xi }_3\right)\end{array}$

The term ${\xi }_2$, and ${\xi }_3$ are constant with respect to ${\xi }_1$. Therefore, their derivative with respect to ${\xi }_1$ will be zero.

$\begin{array}{rll}\frac{\delta \Phi }{\delta {\xi }_1}& =& C_1+0+0\\ & =& C_1\end{array}$

Derivate equation (2) with respect to ${\xi }_2$.

$\begin{array}{rll}\frac{\delta \Phi }{\delta {\xi }_2}& =& \frac{\delta }{\delta {\xi }_2}\left(C_1{\xi }_1+C_2{\xi }_2+C_3{\xi }_3\right)\\ & =& \frac{\delta }{\delta {\xi }_2}\left(C_1{\xi }_1\right)+\frac{\delta }{\delta {\xi }_2}\left(C_2{\xi }_2\right)+\frac{\delta }{\delta {\xi }_2}\left(C_3{\xi }_3\right)\end{array}$

The term ${\xi }_1$, and ${\xi }_3$ are constant with respect to ${\xi }_2$. Therefore, their derivative with respect to ${\xi }_1$ will be zero.

$\begin{array}{rll}\frac{\delta \Phi }{\delta {\xi }_2}& =& 0+C_2+0\\ & =& C_2\end{array}$

Derivate equation (2) with respect to ${\xi }_3$.

$\begin{array}{rll}\frac{\delta \Phi }{\delta {\xi }_3}& =& \frac{\delta }{\delta {\xi }_3}\left(C_1{\xi }_1+C_2{\xi }_2+C_3{\xi }_3\right)\\ & =& \frac{\delta }{\delta {\xi }_3}\left(C_1{\xi }_1\right)+\frac{\delta }{\delta {\xi }_3}\left(C_2{\xi }_2\right)+\frac{\delta }{\delta {\xi }_3}\left(C_3{\xi }_3\right)\end{array}$

The term ${\xi }_2$, and ${\xi }_1$ are constants with respect to ${\xi }_3$. Therefore, their derivative with respect to ${\xi }_1$ will be zero.

$\begin{array}{rll}\frac{\delta \Phi }{\delta {\xi }_3}& =& 0+0+C_3\\ & =& C_3\end{array}$

Therefore, the given expressions are verified.

Conclusion

The expression for $\Phi$ explicitly is represented as,

$\Phi =C_1{\xi }_1+C_2{\xi }_2+C_3{\xi }_3$

The given expressions are verified.

Interpretation Introduction

(b)

Interpretation:

The general expression for the derivative of $\Phi$ with respect to ${\xi }_i$ is to be stated. The corresponding expression is to be compared with the equation $17.15$. The explanation for the derivation of the equation $17.15$ from the preceding equation is to be stated.

Concept introduction:

A mathematical function represents the way in which its value depends on some variable. A function can contain some variables and constants. The term multiplied by a variable is called its coefficient. When derivative of a function is taken with respect to a variable, then the terms that only contain other variables are considered as zero.

Answer to Problem 17.21E

A general expression for the derivative of $\Phi$ with respect to ${\xi }_i$ is represented as,

$\frac{\delta \Phi }{\delta {\xi }_1}=C_i$

Where,

• $i$ represents a number that can be either $1,2$, or $3$.

The general expression for the derivative of $\Phi$ with respect to ${\xi }_i$ and the given equation $17.15$ are similar.

In the given expression it was assumed that for a large system $N_i=N_j$ and $g_j$ is rewritten as $g_i$. After the assumption, the equation was derived to get the equation $17.15$.

Explanation of Solution

The given function $\varphi$ can be expressed as,

$\Phi ={\displaystyle \sum _{i=1}^3{C}_i{\xi }_i}$ …(1)

Where,

• ${\xi }_i$ represents the $i^{\text{th}}$ variable

• $C_i$ represents the value of the coefficient multiplying variable ${\xi }_i$.

A general expression for the derivative of $\Phi$ with respect to ${\xi }_i$ is represented as,

$\frac{\delta \Phi }{\delta {\xi }_1}=C_i$

Where,

• $i$ represents a number that can be either $1,2$, or $3$.

The given equation $17.15$ is shown below as,

$\ln \frac{g_i}{N_i}-1+\alpha -\beta {\in }_i=0i=1,2,3,\mathrm{...}$ …(3)

Where,

• $\alpha$and $\beta$ represents the constant.

• $N_i$ represent $i^{\text{th}}$ a number of the particle.

• $N_j$ represent $j^{\text{th}}$ a number of the particle.

• $g_j$ represents degeneracy for $j^{\text{th}}$ the number.

• $g_i$ represents degeneracy for $i^{\text{th}}$ the number.

The expression from which equation $17.15$ is derived is shown below as,

$\frac{\partial }{\partial N_j}\left[\left(N\ln N+{\displaystyle \sum _j{N}_j\ln \frac{g_j}{N_j}}\right)+\alpha \cdot {\displaystyle \sum _i{N}_i}-\beta \cdot {\displaystyle \sum _i{N}_i}\cdot {\in }_i\right]=0$ …(4)

For a large number, it is assumed that $N_i=N_j$ and $g_j$ is rewritten as $g_i$.

$\frac{\partial }{\partial N_i}\left(N\ln N\right)+\frac{\partial }{\partial N_i}\left({\displaystyle \sum _i{N}_i\ln \frac{g_i}{N_i}}\right)+\frac{\partial }{\partial N_i}\left(\alpha \cdot {\displaystyle \sum _i{N}_i}\right)-\frac{\partial }{\partial N_i}\left(\beta \cdot {\displaystyle \sum _i{N}_i}\cdot {\in }_i\right)=0$

The term $N\ln N$ is constant with respect to $N_i$ therefore, its derivative with respect to $\partial N_i$ will be zero.

$\begin{array}{rll}0+\left(1\right)\ln \frac{g_i}{N_i}+N_i\left(-\frac{1}{N_i}\right)+\alpha \left(1\right)-\beta \left(1\right){\in }_i& =& 0\\ \ln \frac{g_i}{N_i}-1+\alpha -\beta {\in }_i& =& 0\end{array}$

The general expression for the derivative of $\Phi$ with respect to ${\xi }_i$ and the given equation $17.15$ are similar. In both cases, the derivative of a function that contains summation is expressed by the coefficient multiplying variable.

Conclusion

A general expression for the derivative of $\Phi$ with respect to ${\xi }_i$ is represented as,

$\frac{\delta \Phi }{\delta {\xi }_1}=C_i$

Where,

• $i$ represents a number that can be either $1,2$, or $3$.

The general expression for the derivative of $\Phi$ with respect to ${\xi }_i$ and the given equation $17.15$ are similar.

In the given expression it was assumed that for a large system $N_i=N_j$ and $g_j$ is rewritten as $g_i$. After the assumption, the equation was derived to get the equation $17.15$.