Jump to content

Planck's constant and wave impedance of vacuum


SergUpstart

Recommended Posts

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
Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.