Asnetcomp Tutorials: Maxwell's Equaotions and c = 1/√ε0μ0

The degree of understanding of Maxwell's Equations (there are so many arguments on so called 'Maxwell's Equations') is quite up to you (or I would say how many times you have read the textbooks on the equations does relate little with your understanding not to mention how many textbooks you purchased or downloaded but how much you thought about them does). Basically the equations are the vector field equations (directions and intensities according to him and he presented many fields). The fundamental (though subtle to me) concept of the physics field itself differs from the Maxwell's field concept because the existence of 'ether' was experimentally denied after Maxwell. One remarkable thing done by Maxwell was that he predicted the following equation and the speed of light as transverse propagation of electromagnetic wave (undulations according to him) logically by using mathematics and the reports of the experiments done by some other physicists, which was and is regarded as a big discovery (or theoretical prediction), say once per two centuries (by British standard after Newton in 1600s). I think that the above underlined part cannot be neglected after reading his some original texts.

c = \frac{1}{\sqrt{ \mu_0 \varepsilon_0}} = 2.99792458 \times 10^8 \, \mathrm{m~s}^{-1}

The discovery of this equation was introduced in his On Physical Lines of Force (published in March, 1861). It is not a waste of time to check the original works to guess what really happened in the Maxwell's brain.

Maxwell was a specialist of capacitors and inductors as well as fluid mechanism. The title of Part One of his writing published in March, 1861 "On Physical Lines of Force" is "The Theory of Molecular Vortices applied to Magnetic". Vortices (Vortex in a singular form) is the key word and it is "Curl" or "Rotation" in Vector Calculus. As we will see later without double "Curl"operation we cannot reach a wave equation including 1/√ε0μ0. Diodes and transistors, even the ideas, did not exist in his time. He used the physics and electric terms which are different from those we use now but the fundamental things are largely still valid. And his insights and understandings were more fundamental due to his knowledge of mathematics and fluid mechanism.

We saw this equation in the last post " Sphere and Inverse-square law " and μ0 is magnetic permeability in vacuum and ε0 is electric permittivity in vacuum. In terms of units, μ0 is in H (Henry)/meter while ε0 is in F (Farad)/meter. We now recall the following basic four equations.

d

Φ di(t)

v(t) = ------ = L ------- (LI =

Φ)

           dt               dt

To make some converts

v(t)           L
----- =   ----

di(t) d(t)

Unit wise

v         L
--- = ---
i          t

We can connect this with Ohm's Law

V
--- = R
I

Therefore

L = Rt

             d

Q dv(t)

i(t) = ------ = C ------- (C

V =

Q)

dt dt

To make some converts

i(t) C
----- = ----

dv(t) d(t)

Unit wise

I         C
--- = ---
V         t

We can connect this with Ohm's Law

I          1         C
--- = ---- = -----
V         R          t

Therefore

   t
C = ------
          R

Therefore

                    t
LC = Rt x ----- = t $2$
                    R

This <unit wise> equation shows a very unique relation between L, C and time.

L = Inductance in Henry and C = Capacitance in Farad. Therefore

μ0 is in H (Henry) / meter = L (in H) / meter = V t (

= Φ

) / I x meter, or L (or mμ0 ) = $Φ / I$

ε0 is in F (Farad) / meter = C (in F) / meter = I t (= Q) / V x meter, or C (or mε0) = Q / V

Since

Φ

v(t) = ------- -----> by integration with t,

Φ = vt

dt

dQ
i (t) = -------- -----> by integration with t,

Q = it

Therefore

Φ / I can be considered as

Φ / I = vt / i

Divide both side by ' t ' (or more precisely differentiate with time)

Φ / I t = v / it =

V / Q (reciprocal of Q / V)

Apart from the (profound) meanings, unit wise

μ0

= 1 /

ε 0,

μ 0 ε 0

= 1. Is this correct ? Yes, unit wise.

The unit of

μ0

is the same unit as that of

1 /

ε 0 .

while

m μ 0

= L = Φ / I = vt / i

μ 0

= vt / im

mε0 = C = Q / V = it / v

ε0 = it / vm

Therefore please note that

μ 0

and ε0 include the space dimensions, in this case 1-D, distance while the others v (V), i (I) , Q, $Φ,$ C, L do not have dimensions.

μ 0 ε 0

, also unit wise =

( vt / im) x (

it / vm) = vi

t 2 / viv m

2 =

t² /

m

2

Is this correct ? Yes, but not just unit wise but

t² /

m

2

(

t / m)

2 is the unit unit of

μ 0 ε 0 .

Please recall LC Resonant Circuit equation

___

f

= 1 /

i/L C

___ ________ ____

f

= 1 /

i/L C = 1 /

i/m μ0 m ε0 =

1 /

m i/ε0 ε0

This has a meaning since

1

m

2

------- = ------

μ 0 ε 0

t²

1

m

------- = ------ = velocity

____ t

i/ε0 ε0

And if we introduce distance or distance square, again unit wise we can get an acceleration and velocity.

Please note:

Also ＜Vdt ＞ is

Φ

(Magnetic Flux) and ＜Idt＞ is Q (Charge).

d

Φ

v = -------
dt

d

Q

i = --------
dt

vi = power

power x time = energy, so

VIt = VQ (Energy) (Q in Electric Field, Potential energy)

C (or mε) = Q / V. This simple equation means: C (Capacitor) keeps Q (Charge) against V (Voltage) and stores Energy in the form of VQ.

IVt = I

Φ (Energy) (I in Magnet Field, a kind of Dynamic energy)

L (mμ) =

Φ / I .

This simple equation means:

L (Inductor, Coil) creates and keeps

Φ

(Magnetic Fulx) against I (Electric Current) and stores Energy in the form of I

Φ

.

The key is <against>, which is similar to the Hooke's law of a spring, shows their inertial nature of C (ε0) and L (μ0), and a concept <restoring force (energy) >.

Precisely where and how to store energy in the form of VQ and I

Φ

is another story. An analogy of V to a spring is relatively easy to imagine. An analogy of I to a spring also seems to be easy when considering I (Electric Current) being a <current> or a <flow> or a <stream> of Q. But I (i) (Electric Current) is the time rate change of the amount of Q.

d

Q

i = -------
dt

Not a <current> or a <flow> or a <stream> which has a velocity (v = distance / time). Q may move so it has a velocity. But in this case we must consider distance and time. The above equation simply shows the time rate change of the amount of Q (Electric Charges). How to count a number of Q ?

Hooke's law (from wiki)

"
Hooke's law for a spring is often stated under the convention that

F

is the restoring (reaction) force exerted by the spring on whatever is pulling its free end. In that case, the equation becomes

F= -k X\,

since the direction of the restoring force is opposite to that of the displacement.

"
In terms of Energy (from wiki)

"
The potential energy

U_{{el}}(x)

stored in a spring is given by

U_{{el}}(x)={1 \over 2}kx^{2}

which comes from adding up the energy it takes to incrementally compress (or extend) a spring. That is, the integral of force over displacement.

"
(underlined by me)

This reminds us of the stored energy in a Capacitor and an Inductor.

1        2
--- Cv
2

CV = Q, so QV is energy.

1        2
---   Li
2

LI =

Φ , so

ΦI is energy.

ΦI =

QΦVI =

QΦP (VI = P, Power)

Again the basic two equations

Φ

v = -------
dt

When

Φ is constant (no change of

Φ with time), v = 0.

Q

i = --------
dt

When Q

is constant (no change of Q with time), i = 0.

To integrate the both side with ＜t＞(or 0 to ＜t＞)

vt =

Φ

it = Q

So

QV ΦI = QΦVI = QΦP (VI = P: Power) = vtitP = ptPt = energy x energy (P x time = Energy).

Above we did QV

ΦI (

x ΦI) but what is

ΦI (

QV multiplied by

ΦI) ? Does this multiplication has any meaning ? Also does the result "

energy x energy" have any meaning ? Wrong !!, maybe.

QV x

ΦI should be

QV +

ΦI (Addition).

QV seems a static energy derived from Q (fixed) multiplied by V (fixed). Verbally this static energy is proportional to the amount of Q and the strength (magnitude) of V.

On the other hand,

ΦI is an Dynamic Energy and

derived from

Φ

multiplied by I. This Dynamic energy is proportional to the amount of

Φ

and the strength (magnitude) of I. As the energy is dynamic, the amount of

Φ changes with time and

I changes with time, in which case denoted by a small letter <i>.

QV +

Φi = ItV + Vti. Is this correct ?

We must consider

Q of QV is

the amount of Q. However, QV = ItV is obtained by Q being integration with <t> by using

d

Q

i = --------
dt

QV = ItV involves time. <i> is the time rate change of Q - a derivative (rate of change, changing rate) of Q with time. When Q does not change, i = 0. Then

ItV = 0. Something wrong again.

We must consider Field. Electric Filed: E (unit: V/m)

QE = QV/m where <m> is distance. The distance from one place to another place. E means the voltage difference between this one place to this another place.

energy (work)
----------------------------- = Force
distance (displacement)

Or we can use the following empirical formulae.

a) Capacitor

Simplified structure

Dielectric is placed between two conducting plates, each of area A and with a separation of d

C = \frac{\varepsilon A}{d}

ε = C (in Farad) x d / A.

C (in F) / meter = I t (=Q) / V x meter

b) Inductor (coil)

Most simplified structure

Cylindrical air-core coil

L = \frac{1}{l} \mu_0 K N^2 A

L = inductance in henries (H)
μ₀ = permeability of free space = 4 $\pi$ × 10⁻⁷ H/m
K = Nagaoka coefficient^[14]
N = number of turns
A = area of cross-section of the coil in square metrer (m²)
l = length of coil in metres (m)

μ0 = L (in Henry) x l (m) / K (just number with no unit) x N²x A (m²)

L (in H) /meter = V t /I x meter

To put these in c = 1/√ε0μ0

c² = 1 / μ0 ε0 = 1/Vt x It / IVx m² = meter² / time²

then

c = meter / time (which is velocity)

This reminds me of another equation in which ε and μ appear - LC Resonant frequency.

\omega_0 = {1 \over \sqrt{LC}}

You can check the units and get number / time, which is frequency. Unit checking is one way to be sure of the correctness of the equation and help you to make some discovery.

We have not reached the end. Looks like a waste of time. So jumping to a conclusion without understanding to save time.

-----

From wiki (Maxwell's Equations)

In a region with no charges (

ρ = 0

) and no currents (

J = 0

), such as in a vacuum, Maxwell's equations reduce to:

\begin{align} \nabla \cdot \mathbf{E} &= 0 \quad &\nabla \times \mathbf{E} = \ -&\frac{\partial\mathbf B}{\partial t}, \\ \nabla \cdot \mathbf{B} &= 0 \quad &\nabla \times \mathbf{B} = \frac{1}{c^2} &\frac{\partial \mathbf E}{\partial t}. \end{align}

Taking the curl

(\nabla\times)

of the curl equations, and using the curl of the curl identity

\nabla\times(\nabla\times X) = \nabla(\nabla\cdot X) - \nabla 2 X

we obtain the wave equations

\frac{1}{c^2}\frac{\partial^2 \mathbf E}{\partial t^2} - \nabla^2 \mathbf E = 0\,, \quad \frac{1}{c^2}\frac{\partial^2 \mathbf B}{\partial t^2} - \nabla^2 \mathbf B = 0\,,

which identify

c = \frac{1}{\sqrt{ \mu_0 \varepsilon_0}} = 2.99792458 \times 10^8 \, \mathrm{m~s}^{-1}

with the speed of light in free space. In materials with relative permittivity

ε r

and relative permeability

μ r

, the phase velocity of light becomes

v_\text{p} = \frac{1}{\sqrt{ \mu_0\mu_\text{r} \varepsilon_0\varepsilon_\text{r} }}

which is usually less than

c

.
In addition,

E

and

B

are mutually perpendicular to each other and the direction of wave propagation, and are in phase with each other. A sinusoidal plane wave is one special solution of these equations. Maxwell's equations explain how these waves can physically propagate through space. The changing magnetic field creates a changing electric field through Faraday's law. In turn, that electric field creates a changing magnetic field through Maxwell's addition to Ampère's law. This perpetual cycle allows these waves, now known as electromagnetic radiation, to move through space at velocity

c

.

------

The point of the above explanation is ＜t

he curl of the curl identity \nabla\times(\nabla\times X) = \nabla(\nabla\cdot X) - \nabla 2 X

we obtain the wave equations.＞.

To understand this you are required to have some knowledge of and are familiarity with Vector Calculus. However, you can get some idea about this without Vector Calculus.

We have repeatedly used the following basic four equations

d

Φ di

v(t) = ------ = L ------- (LI =

Φ)

dt dt

d

Q dv

i(t) = ------ = C ------- (C

V =

Q)

dt dt

d

Φ di

di

v(t) = ------ = L ------- = mμ0 -----

dt dt dt

Now we use

dv

i(t) = C -----
dt

di d

2 v

d

2 v

= mμ0 -----   = mμ0   C ----- = mμ0 mε0 　------
             dt                      dt²dt²

d

2 v

= m²μ0 ε0 　------
dt²

This is a kind of the above curl of curl operation though we do not use 'curl' ＜place, space, or field＞ but only time derivatives.

Then

v(t)

d

2 v

---- = μ0 ε0 　------
m²dt²

And then

1 / μ0 ε0 = 1/Vt x It / IVx m² = meter² / time²

μ0 ε0= meter² / time²= velocity ²

v(t) 1

d

2 v

---- =    ----------- 　--------
m²        velocity ²dt²

        1

d

2 v

v(t)
------------ ------- -   ------ = 0
velocity ²   dt²     m²

If we change ＜v＞ to the vector field＜E(v /m)＞

        1

d

2 E

E(t)
------------ ------- -   ------ = 0
velocity ²   dt²     m²

This is the same structure of the first part of the above wave equations.

\frac{1}{c^2}\frac{\partial^2 \mathbf E}{\partial t^2} - \nabla^2 \mathbf E = 0\,, \quad \frac{1}{c^2}\frac{\partial^2 \mathbf B}{\partial t^2} - \nabla^2 \mathbf B = 0\,,

Similarly we can get the 2nd part when we take a time derivative twice of i(t).

dv

i(t) = C -------
dt

d

2 i

i (t) = m²μ0 ε0 　------
dt²

1

d

2 i

i(t)
------------ ------- -   ------ = 0
velocity ²   dt²     m²

If we change ＜i＞ to the vector field＜H (i /m)＞

Please note B = μ0H

sptt

Asnetcomp Tutorials

Monday, March 23, 2015

Maxwell's Equaotions and c = 1/√ε0μ0

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