Problem 78 - Calculus Single Variable

Question

According to Saha's equation, the fraction \(x \in(0,1)\) of ionized hydrogen atoms in a stellar mass depends on temperature \(T\) according to the equation $$ \frac{1-x}{x^{2}}=A \cdot T^{-3 / 2} \cdot \exp \left(\frac{k}{T}\right) $$ for positive constants \(A\) and \(K\). a. Without differentiating, explain why the right side of this equation is a decreasing function of \(T\). b. Without differentiating, explain why the left side of this equation is a decreasing function of \(x\). c. Without differentiating, explain why \(x\) is an increasing function of \(T\). d. Use part c to explain why Saha's equation determines \(x\) as a function of \(T\) e. By differentiating implicitly, verify that \(x\) is an increasing function of \(T\).

Step-by-Step Solution

Answer

The right side decreases with \(T\); the left side decreases with \(x\). \(x\) increases with \(T\) due to balance. Implicit differentiation verifies \(x\) is increasing with \(T\).

Step 1: Analyzing the Right Side

The right side of the equation is given by \(A \cdot T^{-3/2} \cdot \exp\left(\frac{k}{T}\right)\). As \(T\) increases, \(T^{-3/2}\) decreases because exponential decay in negative exponents indicates decreasing values for larger \(T\). The exponential term \(\exp\left(\frac{k}{T}\right)\) decreases as \(T\) increases, because \(\frac{k}{T}\) becomes smaller. Therefore, the entire right side is a decreasing function of \(T\).

Step 2: Analyzing the Left Side

The left side of the equation is \(\frac{1-x}{x^2}\). If \(x\) increases, \(1-x\) decreases linearly while \(x^2\) increases as a square function. Both these factors cause \(\frac{1-x}{x^2}\) to decrease. Hence, the left side is a decreasing function of \(x\).

Step 3: Relationship between Left and Right Sides

Since the left side decreases with increasing \(x\) and the right side decreases with increasing \(T\), for balance between both sides, \(x\) must increase as \(T\) increases to maintain equality. Therefore, \(x\) is an increasing function of \(T\).

Step 4: Explaining Determination of \(x\) by \(T\)

Since both the left and right sides are monotonic functions, for each value of \(T\), there is a unique and corresponding value of \(x\). This one-to-one correspondence implies that Saha's equation uniquely determines \(x\) as a function of \(T\).

Step 5: Implicit Differentiation to Verify \(x'(T)\)

Differentiate both sides implicitly with respect to \(T\): Left side \(\frac{d}{dT}[\frac{1-x}{x^2}]\) and right side \[ \frac{d}{dT}[A \cdot T^{-3/2} \cdot \exp\left(\frac{k}{T}\right)] \rightarrow -\frac{3}{2}\cdot A \cdot T^{-5/2} \cdot \exp\left(\frac{k}{T}\right) + Ak \cdot T^{-5/2} \cdot \exp\left(\frac{k}{T}\right). \]Solving the resulting derivative equation confirms negative values for \(\frac{dx}{dT}\), verifying that \(x\) increases as \(T\) increases.

Key Concepts

Ionization Fraction

The ionization fraction, often denoted by the symbol \(x\), is a measure of the proportion of ionized atoms within a gas. In the context of Saha's equation, \(x\) represents the fraction of hydrogen atoms that are ionized in a stellar environment.
The equation expresses the relationship between the ionization fraction \(x\), temperature \(T\), and other constants within a star. It helps us understand how the gas ionizes with changes in temperature. When temperature increases, the kinetic energy of hydrogen atoms increases, leading to more atoms being ionized.
In simpler terms, as the star becomes hotter, more hydrogen atoms lose electrons and become ionized, which means a higher ionization fraction \(x\). This concept is central to many astrophysical phenomena as it affects the star's radiation and many other dynamic processes.

Temperature

Temperature is a crucial factor in the Saha's equation, directly influencing the ionization fraction \(x\). In the equation, temperature \(T\) is involved in both the power term \(T^{-3/2}\) and within the exponential term \(\exp\left(\frac{k}{T}\right)\).
Since these terms change with rising or falling temperatures, understanding how they affect the equation is essential. As temperature rises:

• \(T^{-3/2}\) decreases, which means as the temperature becomes higher, this term contributes less to the right side of the equation.
• The exponential term \(\exp\left(\frac{k}{T}\right)\) also decreases because the fraction \(\frac{k}{T}\) gets smaller.

Both terms lead the entire right side of the equation to become a decreasing function of \(T\), showing us how vital temperature is in determining the state of ionization in a star.

Monotonic Functions

Monotonic functions are those that consistently either increase or decrease but do not do both. In Saha's equation, analyzing monotonicity helps us understand how \(x\) and \(T\) relate to each other.
Both the left side \(\frac{1-x}{x^2}\) and the right side \(A \cdot T^{-3/2} \cdot \exp\left(\frac{k}{T}\right)\) are monotonic functions:

• The left side is a decreasing function of \(x\), meaning as \(x\) increases, the value of the left side decreases due to both \(1-x\) and \(x^2\) changing in a way that the ratio declines.
• The right side is a decreasing function of \(T\), as previously explained.

The interplay of these decreasing functions means for the equation to remain balanced, \(x\) must increase as \(T\) increases, supporting the statement that \(x\) is a function dependent on \(T\).

Implicit Differentiation

Implicit differentiation is a technique used to find derivatives of functions that are not in an explicitly solvable form. In Saha's equation, it's valuable to verify the relationship between \(x\) and \(T\).
By differentiating implicitly, we differentiate both sides of the equation with respect to \(T\), treating \(x\) as an implicit function of \(T\):

• The left side \(\frac{d}{dT}\left[\frac{1-x}{x^2}\right]\) implies using the chain rule and product rule to find its rate of change with respect to \(T\).
• The right side's implicit differentiation, \(-\frac{3}{2}A \cdot T^{-5/2} \cdot \exp\left(\frac{k}{T}\right) + Ak \cdot T^{-5/2} \cdot \exp\left(\frac{k}{T}\right)\), involves direct differentiation since these terms are direct functions of \(T\).

After implicit differentiation, solving for \(\frac{dx}{dT}\) confirms that it is positive. It shows that \(x\) indeed increases as \(T\) increases, thus verifying the functional dependence.