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Origin of the equation c^2=1/epsilon_0*mu_0 
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#1
Jan2013, 07:32 AM

P: 215

Hello,
Could someone describe the origin of the equation c^{2}=1/ε_{0}μ_{0} Is it derived from fundamental reasons? Thanks. 


#2
Jan2013, 08:07 AM

Mentor
P: 16,317

Start with [itex] \nabla \times \nabla \times E [/itex] and apply Maxwell's Equations.



#3
Jan2013, 11:15 AM

P: 215

I see, because ε_{0} and μ_{0} appear in the Maxwell's (SI) equations. Then,  how do ε_{0} and μ_{0} get to appear in the Maxwell's equations?
ε_{0} is the coefficient in the Coulomb's force (2 static charges) μ_{0} is the coefficient in the Ampere's force (2 parallel wires) I am guessing via F=qE, E gets to be expressed via ε_{0} and then ε_{0} gets to appear in Maxwell's equations because E is there. In the end, I am trying to understand how the deliberate choice (inserted "by hand", e.g. artificially) of the value of μ_{0} determines the value of ε_{0}. The shortest connection between ε_{0} and μ_{0} seems to be the relation c^{2} = 1/ε_{0}μ_{0}. But it seems that the derivation of this relation is not short, but rather long and winded. Is there a SHORTER way of seeing how the choice of μ_{0} determines the value of ε_{0}? 


#4
Jan2013, 11:45 AM

Sci Advisor
Thanks
P: 3,659

Origin of the equation c^2=1/epsilon_0*mu_0
[edit: Historically, there was a leap of imagination involved: Maxwell's equations predict electromagnetic radiation travelling at a speed given by the square root of 1/ε_{0}μ_{0}. This value is equal to the already known speed of light, leading to the suggestion that light was the electromagnetic radiation predicted by Maxwell's equations.] 


#5
Jan2013, 12:46 PM

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P: 11,743

In the case of Ampere's Law and μ_{0}, the logic is a bit convoluted, because we actually use Ampere's Law to define the ampere, by defining ##\mu_0 = 4 \pi \times 10^{7}## exactly, in SI units. But the choice isn't arbitrary, because we want to also have 1 ampere = 1 coulomb/second. 


#6
Jan2013, 02:35 PM

P: 2,251

my response to the original question is to take Maxwell's Eqs and derive the wave propagation equation (say in one dimension, to make your life easier). figure out what the speed of propagation is.



#7
Jan2113, 09:40 AM

P: 215

When one reads Wikipedia (http://en.wikipedia.org/wiki/SI_base_unit), it says plainly that the Ampere is a base unit in SI. The Coulomb is thus a derived unit.
I asked the above question in trying to understand the logical flow: Start from the Ampere, a base unit, then move on to the definition of μ_{0} >.... (missing logic here!!!) ... >then end at the definition of ε_{0} and its value. It seems the relation c^2=1/epsilon_0*mu_0 should be part of the missing logic 


#8
Jan2313, 10:31 AM

P: 215

This is the full logic as I understand it:
Decide to choose (e.g., pick) the Ampere as a base unit in SI. Then, pick the Ampere to have a value such, that it is convenient for household and engineering purposes in the measurement of electric current (no one in everyday life can use numbers like 10^{8} etc. , because most people are uncomfortable with the concept of negative numbers, let alone powers of 10). This seems to be the most "fundamental" reason for the value of μ_{0} and as we shall see, of ε_{0} also. Pick the Ampere force law as a defining equation (force between 2 parallel conductors). The Newton and the meter already have fixed definitions, thus we need to adjust the ampere to fit them in the defining equation. But we don`t want to do that, because we have already picked a householdspecific value for the ampere. Then, we insert by hand a handpicked value of a conversion factor, which we call μ_{0}, into the defining equation. We also pick a factor of 4π and insert it into the definition of μ_{0}. It will come in handy later, when we deal with spheres with surface areas of 4πr^{2}. But, recall that I=[itex]\frac{dQ}{dt}[/itex], e.g. the electrical current is a `charge flux``, much like the fluid mass flux in a pipe. Then, because time has an established unit of measurement, our handpicked definition of the Ampere imposes the unit of measurement of the electric charge. This means something has to be adjusted in the Coulomb`s law, where the electric charge is related to force and distance. Moving on to Coulomb`s law. Force, time and charge now have their defined units of measurement. To reconcile them, insert a factor of ε_{0} (and throw in a factor of 4π). Then, because the Coulomb`s law and Ampere force law determine the form of Maxwell`s equations, the new handpicked factors ε_{0} and μ_{0} get to appear in the Maxwell`s equations, where one can derive the relation c^{2}=[itex]\frac{1}{\epsilon_{0}\mu_{0}}[/itex]. 


#9
Jan2313, 11:58 AM

P: 573




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