I konw: $$ a_0=\frac{4 \pi \epsilon_0 \hbar^2}{m_e e^2} $$ and $$ \alpha = \frac{m_{\mathrm{e}}^2 e^4}{18 \pi^3 \epsilon_0^2 \hbar^4} $$

So we can use $a_0$ for $\alpha$,like this: $$ \alpha=\frac{m_{\mathrm{e}}^2 e^4}{18 \pi^3 \epsilon_0^2 \hbar^4}=\frac{16}{18 \pi} \cdot \frac{m_{\mathrm{e}}^2 e^4}{16 \pi^2 \epsilon_0^2 \hbar^4}=\frac{8}{9 \pi} \cdot \frac{1}{a_0^2} $$

alpha = (e^4 me^2)/(18 eps^2 h^4 \[Pi]^3);
a0 = (4 Pi eps h^2)/(me e^2);

I've tried a few things and failed. What should I do


1 Answer 1

eq1 = alpha == (e^4 me^2)/(18 eps^2 h^4 \[Pi]^3);
eq2 = a0 == (4 Pi eps h^2)/(me e^2);

Solve[{eq1, eq2}, alpha, {e}]

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  • $\begingroup$ How does that work? Don't get it $\endgroup$ Mar 30, 2023 at 0:23
  • $\begingroup$ In fact, this is not practical,We usually define a variable with = because we need to use that variable later $\endgroup$ Mar 30, 2023 at 1:45
  • 1
    $\begingroup$ No problem, use: var= alpha /. Solve[...][[1]] You should try to learn MAthematica. $\endgroup$ Mar 30, 2023 at 7:39
  • $\begingroup$ @DanielHuber Can you explain your domain parameter {e}? $\endgroup$
    – Edmund
    Mar 30, 2023 at 15:31
  • $\begingroup$ Solve has a third argument that is no more documented, similar to Reduce (see. e.g.: reference.wolfram.com/legacy/v7/ref/Solve.html or mathematica.stackexchange.com/questions/83902/…). This argument specifies variables to eliminate. Why it is no more documented is beyond me. Here is a simple example: Solve[{a == b, b == c}, a, {b}]` $\endgroup$ Mar 30, 2023 at 19:34

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