Planck-like values out of permeability and permittivity

[roineust]

What happens if we make a dimensional analysis of permeability and permittivity ?

Will we get a minimum possible amount of electric and magnetic values?

 

[PeroK]

What happens if we make a dimensional analysis of permeability and permittivity ?

Will we get a minimum possible amount of electric and magnetic values?

Do such values have any meaning and use in physics? Can they tell us anything about particles? About vacuum?

The numerical values of $\epsilon_0$ and $\mu_0$ are determined by the units we use. The numbers themselves are arbitrary.

So, the answer is "no" to each of your questions.

 

[roineust]

The numerical values of $\epsilon_0$ and $\mu_0$ are determined by the units we use. The numbers themselves are arbitrary.

So, the answer is "no" to each of your questions.

I don't understand, i would appreciate if you could expand.

Why is the mathematical operation described here, not possible also for the measure types of electriciy and magnetism and the values of ϵ0 and μ0 :

 

[Ibix]

The quantities $\epsilon_0$ and $\mu_0$ are defined in terms of quantities like $c$ and $\hbar$. So you can do a similar manipulation, but you end up with nothing new. The only EM-related Planck unit I'm aware of is the Planck charge, which is about 1.875C. As we keep telling you, Planck units are not the smallest unit of anything - clearly not in this case since it's about 10²⁰ times larger than the quarks' charges.

Edit: don't trust formatting on the internet - the Planck charge is 1.875×10^-18C. The point stands, but not quite so emphatically.

 

[roineust]

As we keep telling you

What is the meaning of the Planck scale if it has any meaning at all ?

 

[Ibix]

As far as I understand it's a typical scale where you might expect classical gravity to not work correctly - where correction terms from quantum gravity might become detectable. I don't know of any such interpretation for the Planck charge, but I haven't looked into it.

Note that I've corrected the value of the Planck charge above. It's only about thirty times the quark charge.

 

[Vanadium 50]

As we keep telling you

And telling you.
And telling you.
And telling you.

 

[Vanadium 50]

What is the meaning of the Planck scale if it has any meaning at all ?

It has no meaning. The Planck resistance is 30 ohms.

 

[vanhees71]

I don't understand, i would appreciate if you could expand.

Why is the mathematical operation described here, not possible also for the measure types of electriciy and magnetism and the values of ϵ0 and μ0 :

The reason is that the SI is a very artificial system of units. It's taylored to make the life of experimental physicists and engineers easy and to provide the utmost accurate definition of a consistent set of units for all measurements right. With the revision put into force last year it's almost an ideal system using only natural constants for its definitions (the only exception is the use of $\nu_{\text{Cs}}$ to define the second, which uses a specific atom in its definition not a universal natural constant).

The reason to need $\epsilon_0$ and $\mu_0$ (which are related to the vacuum speed of light by $\mu_0 \epsilon_0=1/c^2$) is that in the SI an extra unit for electric charge, the Coulomb, or equivalently an extra unit for electric current, the Ampere, is introduced.

More natural units are the older Gaussian or Heaviside-Lorentz units, where in the Maxwell equations the one and only fundamental constant in the entire electromagnetic game is introduced, which is the vacuum-speed of light, $c$.

$\mu_0$ and $\epsilon_0$ are artificial unit-conversion constant chosen such as to make currents, voltages, etc. convenient choices of units for usual household conditions, i.e., you have simple numbers like 1 A for currents and 110 V (or 230 V) for voltages and not large powers of 10 for everyday electricians use.

Planck units are the most natural ones. Here everything is based on the fundamental constants. This cannot be done for defining the SI, which would mean to fix the value of Newton's Gravitational constant, $G$, instead of fixing the value $\nu_{\text{Cs}}$ of the hyperfine transition in Cs atoms, because $G$ cannot determined accurately enough with current technology of measurement.