While playing with Planck constants I noticed the following:

 

The electric constant can be figured from: ((Tp^2 * Qp^2) / (Mp * Lp^3)) * (1/(4*pi))

Where,

Tp = Planck Time , Qp = Planck Charge, Mp = Planck Mass, Lp = Planck Length

 

And the magnetic constant can be determined with:

((Mp * Lp) / (Qp^2)) * (4*pi)

 

 

Anyone know why the "4 pi" pops up like this?

Edited March 7, 200916 yr by gre

Artifact of SI units, probably. The electrostatic force constant in Coulomb's law is [math]\frac{1}{4\pi\epsilon_0}[/math], but that's because of how the Coulomb is defined. If you go to the cgs system, charge is defined differently and the constant 1.

Well.. What does the "4 pi" represent in Coulomb's force equation??

I don't know that it "represents" anything. If the Coulomb were defined differently (or, technically, the Ampere, since that's the base unit) then the constant would be different.

 

One ampere is defined to be the constant current which will produce an attractive force of 2×10–7 newton per metre of length between two straight, parallel conductors of infinite length and negligible circular cross section placed one metre apart in free space.

http://en.wikipedia.org/wiki/Ampere

The expression of [math]4\pi[/math] usually indicates that an integration over three-space has taken place, and this is the solid angle. Swansont, interesting to read this surprising definition. Surprising to me because I assume such essences.

Edited March 8, 200916 yr by Norman Albers

The expression of [math]4\pi[/math] usually indicates that an integration over three-space has taken place, and this is the solid angle. Swansont, interesting to read this surprising definition. Surprising to me because I assume such essences.

 

Yes, [math]4\pi[/math] can mean such an integration has taken place. But does it have a physical significance? It may just mean you defined something as a surface element and then integrated over the whole surface. You have arbitrarily-defined terms, and you can put the constants anywhere. What would happen if we redefined the Coulomb (assuming for the moment that it was the SI base unit) to be [math]2\sqrt{\pi}[/math] larger? The [math]4\pi[/math] in the SI version of Coulomb's law goes away, but nothing physical has changed. Some other units may end up having to be redefined as a result, but since they are arbitrary, why does it matter?

Generally, I agree, Swansont. I suspect I came on something more profound in seeing that the Compton wavelength is [math] 4\pi[/math] greater than the GR expression for the geometric angular momentum radius. Over good beers, we might agree that ultimately we will see deeper unity.

Yes, [math]4\pi[/math] can mean such an integration has taken place. But does it have a physical significance? It may just mean you defined something as a surface element and then integrated over the whole surface. You have arbitrarily-defined terms, and you can put the constants anywhere. What would happen if we redefined the Coulomb (assuming for the moment that it was the SI base unit) to be [math]2\sqrt{\pi}[/math] larger? The [math]4\pi[/math] in the SI version of Coulomb's law goes away, but nothing physical has changed. Some other units may end up having to be redefined as a result, but since they are arbitrary, why does it matter?

 

 

 

4 pi just looks like a conversion factor going from a spherical surface area of a field, to cubical surface area.

 

 

Since coulombs constant is:

 

1 / (4*pi * (Pt^2 * Qp^2 / Pm * Pl^3 * 4 * pi)

 

After the 4pi(s) cancel out. Wouldn't you then be working with a cubic surface area or volume?

Edited March 10, 200916 yr by gre