# Redshift Academy

Wolfram Alpha:

Last modified: January 26, 2018
```Lorentz Transformation of the Electromagnetic Field
---------------------------------------------------

An observer measures a charge at rest in a frame F.  He
will see a static electric field but no current because
the charge is not moving. Since there is no cuurent there
is no magnetic field.  Another observer in a frame F'
moving at a velocity, v, w.r.t. F, however, will see an
electric current and hence a magnetic field.

Consider the electromagnetic tensor:

-                     -
|   0   Ex/c  Ey/c  Ez/c |
Fμν = | -Ex/c  0     Bz   -By  |
| -Ey/c -Bz    0     Bx  |
| -Ez/c  By   -Bx    0   |
-                     -

Which is contructed as follows:

Fμν = ∂νAμ - ∂μAν

Where,

Aμ = (φ,-Ax,-Ay,-Az)

= (A0,-A1,-A2,-A3)

Therefore,

F01 = ∂1A0 - ∂0A1

= -∂Ax/∂t - ∂φ/∂x

= -Ex

F23 = ∂2A3 - ∂3A2

= -∂Az/∂y + ∂Ay/∂z

= (∇ x A)x

= Bx

Now add a boost:

-          -
|  γ -βγ 0 0 |
Λμν = Λ = ΛT = | -βγ γ  0 0 |
|  0  0  1 0 |
|  0  0  0 1 |
-

Therefore, we get:

Fμν' = ΣΣΛμσΛμρFσρ (see note on Lorentz Group)
σρ

Where σ amd ρ run from 0 to 3.

For F01 we get:

F01' = ΣΣΛ0σΛ1ρFσρ = Ex'/c

= Λ00Λ11F01 + Λ01Λ10F10  (all other terms are 0)

= γ2Ex/c - β2γ2Ex/c

= (1 - β2)γ2Ex/c

= Ex/c

We get the same answer using the matrix form:

F' = ΛTFΛ

-         -  -          -  -         -
|  p -q 0 0 ||  0  a  b c ||  p -q 0 0 |
= | -q  p 0 0 || -a  0  d e || -q  p 0 0 |
|  0  0 1 0 || -b -d  0 f ||  0  0 1 0 |
|  0  0 0 1 || -c  e -f 0 ||  0  0 0 1 |
-         -  -          -  -         -

-                                          -
|     0       ap2 - aq2   bp - dq    cp + eq |
= | aq2 - ap2       0       dp - bq   -cq - ep |
|  dq - bp    bq - dp        0          f    |
| -cp - eq    cq + ep       -f          0    |
-                                          -

Therefore,

F01' = ap2 - aq2 = γ2Ex/c - γ2β2Ex = Ex/c

If we repeat the process for all of the components of
F we get:

Bx' = Bx

By' = γ(By + (v/c2)Ez)

Bz' = γ(Bz - (v/c2)Ey)

Ex' = Ex

Ey' = γ(Ey - vBz)

Ez' = γ(Ez + vBy)

What appears to be a magnetic field to one observer looks
like an electric field to another, and vice versa.

Maxwell's Equations
-------------------

Maxwell's equation can be written as:

μ0jμ = ∂μFμν

and,

∂μFμν = 0

μ0j0 = ∂F00/∂t + ∂F10/∂x + ∂F20/∂y + ∂F30/∂z

= (1/c)(0 - ∂Ex/∂x - ∂Ey/∂y - ∂Ey/∂z)

= -∇.E  ... Gauss's Law

μ0j1 = ∂F01/∂t + ∂F11/∂x + ∂F21/∂y + ∂F31/∂z

= (1/c)∂Ex/∂t + 0 - ∂Bz/∂y + ∂By/∂z

μ0jx = [(1/c)∂E/∂t - ∇ x B]x  ... Ampere's Law

If we make the substitution E/c -> B and B -> -E/c we get
the DUAL ELECTROMAGNETIC TENSOR, Gμν.  Note this operation
leaves the 6 boost equations shown above unchanged.  When
we do this and repeat the above procedure for ∂μFμν = 0
we get:

0 = ∂G00/∂t + ∂G10/∂x + ∂G20/∂y + ∂G30/∂z

= 0 + ∂Bx/∂x + ∂By/∂y + ∂By/∂z

= ∇.B

And,

0 = ∂G01/∂t + ∂G11/∂x + ∂G21/∂y + ∂G31/∂z

= ∂Bx/∂t + 0 + (1/c)∂Ez/∂y - (1/c)∂Ey/∂z

= [∇ x E + ∂B/∂t]x  ... Faraday's Law

Alternative Derivation
----------------------

Aμ = (-φ/c,A)  ... the vector potential

jμ = (cρ,j)  ... the 4 current.

μ0jμ = ∂μFμν

= ∂μ(∂μAν - ∂νAμ)

= ∂μ∂μAν - ∂μ∂νAμ

j0 = (∂0∂0A0 - ∂0∂0A0) + (∂1∂1A0 - ∂1∂0A1) + (∂2∂2A0 - ∂2∂0A2)
+ (∂3∂3A0 - ∂3∂0A3)

= 0 + ∂1∂1A0 + ∂2∂2A0 + ∂3∂3A0 - ∂1∂0A1 - ∂2∂0A2 - ∂3∂0A3

= -∂2φ/∂x2 - ∂2φ/∂y2 - ∂2φ/∂z2 - ∂/∂t(∂A1/∂x) - ∂/∂t(∂A2/∂y)
- ∂/∂t(∂A3/∂z)

= -∂2φ/∂x2 - ∂2φ/∂y2 - ∂2φ/∂z2 - ∂/∂t(∂A1/∂x + ∂A2/∂y + ∂A3/∂z)

ρ/ε0 = -∇2φ - ∂/∂t(∇.A)

This is equivalent to:

∇ x E = -∂B/∂t

Proof:

The fields E and B are expressed in terms of the potentials
as:

B = ∇ x A and E = -∇φ - ∂A/∂t

Therefore,

∇ x E = -∂B/∂t

= -∂(∇ x A)/∂t

= -∂(∇ x A)/∂t

∇ x E + ∂(∇ x A)/∂t = 0

∇ x (E + ∂A/∂t) = 0

Now, ∇ x ∇φ = 0 (curl of the gradient of a scalar = 0).
Therefore, by comparison:

∇φ = (E + ∂A/∂t)

∇.(∇φ) = ∇.E - ∇.(∂A/∂t)

∇2φ = -ρ/ε0 - ∂(∇.A)/∂t

Which is what we had before.

Lorentz Invariance
------------------

∂μFμν = μ0jν and ∂μFμν = 0 arn not Lorentz invariant
because they have a dangling index, ν, (i.e. they are
not scalars like FμνFμν or xμxμ).  However, we will show
that the equations derived from these equations are, in
fact, Lorentz invariant.

-           -  -   -     -           -
|  γ  -γβ 0 0 || -cρ |   | -γcρ - γβjx |
| -γβ  γ  0 0 ||  jx | = |  γjx + γβcρ |
|  0   0  1 0 ||  jy |   |      jy     |
|  0   0  0 1 ||  jx |   |      jz     |
-           -  -   -     -           -

-           -  -    -     -            -
|  γ  -γβ 0 0 || -φ/c |   | -γφ/c - γβAx |
| -γβ  γ  0 0 ||   Ax | = |  γAx + γβφ/c |
|  0   0  1 0 ||   Ay |   |      Ay      |
|  0   0  0 1 ||   Az |   |      Az      |
-           -  -    -     -            -

Length contraction has an effect on charge density and
current density, and time dilation has an effect on the
rate of flow of charge.  Therefore, the charge and current
distributions must transform in a related way under a
boost.

Consider a boost in the x directions.  The y component is:

∂Ex'/∂z' - ∂Ez'/∂x' = -∂By'/∂t

Substituting Ez' = γ(Ez + vBy) and By' = γ(By + (v/c2)Ez)
The equation becomes:

∂Ex'/∂z' - γ∂(Ez + vBy)/∂x' = -γ∂(By + (v/c2)Ez)/∂t'

∂Ex'/∂z' - γ∂Ez/∂x' + γv∂By/∂x' = -γ∂By/∂t' - (v/c2)γ∂Ez/∂t'

Rearranging we get:

∂Ex'/∂z' + (v/c2)γ∂Ez/∂t' - γ∂Ez/∂x' = vγ∂By/∂x' - γ∂By/∂t'

... Equation 1.

In order to go any further it is necessary to see how
the derivatives of the E and B components transform in
the direction of the boost (x).  To do this we make use
of the chain rule.

∂/∂x = (∂/∂x')(∂x'/∂x) + (∂/∂t')(∂t'/∂x)

and

∂/∂t = (∂/∂x')(∂x'/∂t) + (∂/∂t')(∂t'/∂t)

Also;

∂/∂t' = (∂/∂x)(∂x/∂t') + (∂/∂t)(∂t/∂t')

The transformations in the y and z directions are simply:

∂/∂y = ∂/∂y'

∂/∂z = ∂/∂z'

For a boost in the x direction the cordinates transform
as:

x' = γ(x - vt)

x = γ(x' - v't') ≡  γ(x' + vt')

y' = y

z' = z

t' = γ(t - (v/c2)x)

t = γ(t' - (v'/c2)x') ≡ γ(t' + (v/c2)x')

Therefore,

∂x'/∂x = γ

∂t'/∂t = γ

∂t/∂t' = γ

∂x'/∂t = -γv

∂x/∂t' = γv

∂t'/∂x = -γv/c2

Therefore,

∂Ez/∂x = (∂Ez/∂x')(∂x'/∂x) + (∂Ez/∂t')(∂t'/∂x)

= (∂Ez/∂x')(γ) + (∂Ez/∂t')(-γv/c2)

= γ∂Ez/∂x' - (γv/c2)∂Ez/∂t'

∂By/∂t = (∂By/∂x')(∂x'/∂t) + (∂By/∂t')(∂t'/∂t)

= (∂By/∂x')(-γv) + (∂By/∂t')(γ)

= γ∂By/∂t' - γv∂By/∂x'

Armed with this information we can rewrite equation 1
as:

∂Ex/∂z - ∂Ez/∂x = -∂By/∂t

Or,

(∇ x E)y = -∂By/∂t  Q.E.D

The x component is a little simpler to calculate:

∂Ez'/∂y' - ∂Ey'/∂z' = -∂Bx'/∂t'

Again, Ey' = γ(Ey - vBz) and Ez' = γ(Ez + vBy). Therefore,
the equation becomes:

(γ∂Ez/∂y' + γv∂By/∂y') - (γ∂Ey/∂z' - γv∂Bz/∂z') = -∂Bx'/∂t'

Using the chain rule from above, and noting that
Bx' = Bx, the RHS becomes:

∂Bx'/∂t' = (γ∂Bx/∂t + γv∂Bx/∂x)

Therefore,

(γ∂Ez/∂y + γv∂By/∂y) - (γ∂Ey/∂z - γv∂Bz/∂z) = -(γ∂Bx/∂t + γv∂Bx/∂x)

Note:  The derivatives for the y and z directions do
not transform in accordance with the chain rule as
described for the x direction.

Now ∇.B = 0 meaning that ∂Bx/∂x = ∂By/∂y = ∂Bz/∂z = 0.
Therefore,

γ∂Ez/∂y - γ∂Ey/∂z = -γ∂Bx/∂t

Or,

(∇ x E)x = -∂Bx/∂t  Q.E.D

Using similar techniques we can demonstrate that the other
Maxwell equations are also Lorentz invariant.
```