Expression substitution

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

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}]


• How does that work? Don't get it Mar 30, 2023 at 0:23
• In fact, this is not practical，We usually define a variable with = because we need to use that variable later Mar 30, 2023 at 1:45
• No problem, use: var= alpha /. Solve[...][[1]] You should try to learn MAthematica. Mar 30, 2023 at 7:39
• @DanielHuber Can you explain your domain parameter {e}? Mar 30, 2023 at 15:31
• 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}]` Mar 30, 2023 at 19:34