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The fine structure constant $\alpha$ actually is speed parameter of electron $\beta_e$, moving around proton in hydrogen atom.

$v_n=\frac {\alpha_c}{n}=c\frac {\alpha}{n}=c\frac {\beta_e}{n}$

How can it be a constant!?

photo atom invariance is not a synonym for being a constant

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The fine structure constant is a dimensionless number $\alpha^{-1} \approx 137$ (see en.wikipedia.org/wiki/Fine-structure_constant). Could you clarify your confusion in light of this? Perhaps you're referring to some calculation you've seen in the Bohr model of Hydrogen? –  joshphysics Feb 20 '13 at 0:53

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The fine structure constant effectively measures the strength of the electromagnetic interaction between unit charges - governed by the constant $e^2/4\pi\epsilon_0$ - in quantum mechanical, relativistic terms. It is the ratio of the electromagnetic interaction's (dimensional) coupling constant $e^2/4\pi\epsilon_0$ to the equivalent constant with the same dimensions in relativistic quantum mechanics, which turns out to be $\hbar c$. Thus, $$\alpha=\frac{e^2/4\pi\epsilon_0}{\hbar c}.$$

The speed you are talking about, $\alpha c=\frac1\hbar\frac{e^2}{4\pi\epsilon_0}$, is the only speed you can obtain from quantum electrostatics. Since the hydrogen atom is the simplest quantum electrostatic problem, it is no surprise that its characteristic speeds are simple multiples of $\alpha c$.

You should also note that velocities inside the hydrogen atom are ill-defined (there's a broad-ish distribution over different speeds, over a range about the size of $\alpha c$) and that the picture you drew (the "solar system" atom) is only valid as a very crude picture of what goes on inside it. A more accurate picture of a hydrogen atom in its ground state is the top left: enter image description here

Finally, I'm answering because other lazy people might come back with the same question, but by rights you should get double downvotes for not bothering to look it up first.

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Well, the quantum mechanically correct ground state is in fact static and spherically symmetric. Truly, there is nothing orbiting anything else. Also note that the probability distribution although negligible, is in fact nonzero everywhere; i.e., the electron cloud occupies all of space... If this is not confusing... Probably nothing is. –  safkan Dec 18 at 14:18

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