Grok Generated LaTex for Fine-structure Constant Equations

### Given Equations

We start with the following system of equations:

1. **\(\alpha^2 = \pi \cdot r_p \cdot R_\infty \cdot \frac{R_H}{R_\infty - R_H}\)**  

   - This equation relates \(\alpha^2\) to the variables \(r_p\), \(R_\infty\), and \(R_H\).

2. **\(r_p = \frac{2}{\pi R_H} \cdot \left( \frac{v_u}{c} \right)^2\)**  

   - This equation expresses \(r_p\) in terms of \(R_H\), \(v_u\), and \(c\).

The goal is to solve for \(\alpha\) in terms of \(R_\infty\), \(R_H\), \(v_u\), and \(c\) by eliminating \(r_p\).

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### Solution Process

To find \(\alpha\), we substitute the expression for \(r_p\) from the second equation into the first equation and simplify step-by-step.

#### Step 1: Substitution

- Substitute \(r_p = \frac{2}{\pi R_H} \cdot \left( \frac{v_u}{c} \right)^2\) into the first equation:  

  \(\alpha^2 = \pi \cdot \left( \frac{2}{\pi R_H} \cdot \left( \frac{v_u}{c} \right)^2 \right) \cdot R_\infty \cdot \frac{R_H}{R_\infty - R_H}\)

#### Step 2: Simplification

- **Cancel the \(\pi\) terms:**  

  The \(\pi\) in the numerator and denominator cancels out:  

  \(\alpha^2 = \frac{2}{R_H} \cdot \left( \frac{v_u}{c} \right)^2 \cdot R_\infty \cdot \frac{R_H}{R_\infty - R_H}\)

- **Simplify the \(R_H\) terms:**  

  The \(R_H\) in the numerator and denominator simplifies:  

  \(\frac{2}{R_H} \cdot \frac{R_H}{R_\infty - R_H} = \frac{2}{R_\infty - R_H}\)  

  Thus:  

  \(\alpha^2 = \frac{2}{R_\infty - R_H} \cdot \left( \frac{v_u}{c} \right)^2 \cdot R_\infty\)

- **Rearrange:**  

  Rewrite the expression for clarity:  

  \(\alpha^2 = \frac{2 R_\infty}{R_\infty - R_H} \cdot \left( \frac{v_u}{c} \right)^2\)

#### Step 3: Solve for \(\alpha\)

- Take the square root of both sides:  

  \(\alpha = \sqrt{ \frac{2 R_\infty}{R_\infty - R_H} \cdot \left( \frac{v_u}{c} \right)^2 }\)

- **Apply the square root property:**  

  Since \(\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}\) (for positive \(a\) and \(b\)):  

  \(\alpha = \sqrt{ \frac{2 R_\infty}{R_\infty - R_H} } \cdot \sqrt{ \left( \frac{v_u}{c} \right)^2 }\)

- **Simplify:**  

  The square root of the squared term simplifies:  

  \(\alpha = \sqrt{ \frac{2 R_\infty}{R_\infty - R_H} } \cdot \frac{v_u}{c}\)

- **Alternative form:**  

  Rearrange for conciseness:  

  \(\alpha = \frac{v_u}{c} \cdot \sqrt{ \frac{2 R_\infty}{R_\infty - R_H} }\)

This form is clear and expresses \(\alpha\) in terms of the required variables.

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### Final Answer

The solution for \(\alpha\) is:  

\(\alpha = \frac{v_u}{c} \sqrt{\frac{2 R_\infty}{R_\infty - R_H}}\)

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### Notes

- The expression assumes \(R_\infty - R_H > 0\) to ensure the denominator is positive and the square root is defined.  

- All variables (\(R_\infty\), \(R_H\), \(v_u\), \(c\)) are treated as positive quantities for simplicity, consistent with typical mathematical or physical contexts.

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