Planck's constant and wave impedance of vacuum

[SergUpstart]

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?

 

[swansont]

What do you mean by “wave resistance”?

 

[SergUpstart]

2 minutes ago, swansont said:

What do you mean by “wave resistance”?

 

https://en.wikipedia.org/wiki/Impedance_of_free_space

[joigus]

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 November 21, 2020 by joigus

[swansont]

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.