# Planck's constant and wave impedance of vacuum

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I noticed that the wave resistance of the vacuum is in some sense equivalent to the Planck constant

where e is the elementary charge, alpha is the fine structure constant, and Z0 is the vacuum wave resistance

What does this mean?

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What do you mean by “wave resistance”?

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2 minutes ago, swansont said:

What do you mean by “wave resistance”?

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1 hour ago, SergUpstart said:

What does this mean?

Nothing. You're going in circles and then playing a naming game. The definition of the fine structure constant, $$\alpha$$ in SI units is,

$\alpha=\frac{e^{2}}{4\pi\epsilon_{0}\hbar c}$

$$e$$, $$\hbar$$, and $$c$$ are measured quantities.

$$\epsilon_{0}$$ is a convention.

$$\mu_{0}$$ is another convention coming from convention $$\epsilon_{0}$$, and measured $$c$$,

$\left(\epsilon_{0}\mu_{0}\right)^{-1}=c^{2}$

Then you play with the identity,

$\hbar=\sqrt{\epsilon_{0}\mu_{0}}\frac{e^{2}}{4\pi\epsilon_{0}\alpha}=\sqrt{\frac{\mu_{0}}{\epsilon_{0}}}\frac{e^{2}}{4\pi\alpha}$

And then you name,

$Z_{0}=\sqrt{\frac{\mu_{0}}{\epsilon_{0}}}$

the impedance of the vacuum. You have no operational definition for that as an impedance.

Edited by joigus

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IOW, Z depends on Planck’s constant because it depends on c, and you rewrote it in terms of the fine structure constant, which also depends on c.

But this is little like a fraction you haven’t simplified.

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