Wolfram Alpha:

```Lorentz Transform
-----------------

The Lorentz transformation converts between two different
observers' measurements of space and time, where one
observer is in constant motion with respect to the other.
In classical physics (Galilean relativity), the only
conversion believed necessary was x' = x - vt, describing
how the origin of one observer's coordinate system slides
through space with respect to the other's, at speed v and
along the x-axis of each frame. According to special
relativity, this is only a good approximation at speeds
small compared to the speed of light, and in general the
result is not just an offsetting of the x coordinates;
lengths and times are distorted as well.

Consider 2 cartesian coordinate systems (x,y,z) and
(x',y',z') such that the x and x' axes are coincident
and the other 2 axes are parallel (i.e. a boost along
the x-axis).   The Lorentz transformation for time is:

t - vx/c2
t' = --------
√1-v2/c2

Consider a time interval

t1' - t2' = γ{t1 - vx1/c2 - t2 + vx2/c2}

The time measurements in the moving frame are made at
the same location so the expression reduces to:

t12' = γt12

= γτ

A clock in motion slows down as observed by a
stationary observer in an inertial reference frame.

t12' = time interval we observe in the other
reference frame τ = time observed in observers own
frame of reference ('Proper Time') γ = 1/√(1 - v2/c2)

The Lorentz Transformation for spacial coordinates is:

x - vt
x' = ------
√1-v2/c2

y' = y

z' = z

Consider a length measurement

x1' - x2' = γ{x1 - vt1 - x2 + vt2}

The length measurements in the moving frame are made at
the same time so the expression reduces to:

x12' = γx12

which can also be written

x = γx0

where x0 is the 'Proper Length'

The Lorentz Transform can also be written as a matrix
(see the note on the Lorentz Group for more details).
For a boost of a contravariant vector in the x-direction
we get:

-           -  -  -
|   γ -βγ 0 0 || ct |
| -βγ  γ  0 0 ||  x | where β = v/c
|   0  0  1 0 ||  y |
|   0  0  0 1 ||  z |
-           -  -  -

-          -
|  γct - βγx |
= | -βγct + γx |
|      y     |
|      z     |
-          -

For a boost of a covariant vector in the x-direction
we get:

-         -  -  -
|  γ βγ 0 0 ||-ct |
| βγ  γ 0 0 ||  x |
|  0  0 1 0 ||  y |
|  0  0 0 1 ||  z |
-         -  -  -

-         -
| γβx - γct |
= | γx - βγct |
|     y     |
|     z     |
-         -

Note:  The boost matrix for a covariant vector is
the inverse of the boost for a contravariant vector.
(see notes on the Lorentz Group).

Therefore, the product of the 2 resulting vectors is
equal to:

-                           -  -         -
| (γct - βγx) (-βγct + γx) y z|| γβx - γct |
-                           - | γx - βγct |
|      y    |
|      z    |
-         -

= (γct - βγx)(γβx - γct) + (-βγct + γx)(γx - βγct) + y2 + z2

= γ2(ct - βx)(βx - ct) + γ2(-βct + x)(x - βct) + y2 + z2

= γ2[(ct - βx)(βx - ct) + (-βct + x)(x - βct)] + y2 + z2

= γ2[(-c2t2 - β2x2 + 2ctβx) + (-2βctx + x2 + β2c2t2)] + y2 + z2

= γ2[-c2t2 - β2x2 + x2 + β2c2t2] + y2 + z2

= γ2[-c2t2(1 - β2) + x2(1 - β2] + y2 + z2

= γ2(1 - β2)[-c2t2 + x2] + y2 + z2

= -c2t2 + x2 + y2 + z2

Note: the above matrix will be different for a boost
in either the y or z direction.

Note:  Proper time can be a little confusing.  It is
the time measured by an observer in his own frame of
reference regardless of whether that frame is moving
or not.  Thus a clock moving along a world line will
show the proper time to the observer carrying the
clock.  Likewise for the proper length.

Rapidity and Boosts
-------------------

The Lorentz transformations for boosts can also be
derived in a way that resembles circular rotations
in 3d space using the hyperbolic functions.  For the
moment consider rotations in circular space:

A rotation of x and y by θ around the z axis is
given by:

x' = xcosθ + ysinθ
y' = -xsinθ + ycosθ
z' = z
t' = t

The length of the vector, R, has to be the same in
both frames.  Therefore,

x2 + y2 = x'2 + y'2

The rotation can also be written as the matrix.
_            _
| cosθ sinθ 0 0|
R  = |-sinθ cosθ 0 0|
|  0     0  1 0|
|  0     0  0 1|
-            -

Boosts
------

Consider a boost along the x coordinate.

t     t'
|     |  / x = vt
|     | /
|     |/  o A
|     +----- x'
|    /
|   /
|  /
| /
------------------- x

The origin of the prime frame is moving on the line
x = vt.  The point A in the prime frame is at (x',t').
Therefore:

x' = x - vt

t = t'

But this doesn't work for a light ray x = ±ct because

x' = ct - vt = (c - v)t

and

x' = -ct - vt = -(c + v)t

violates the idea that the laws of physics are the
same in all inertial reference frames.

To compensate for the fact that the speed of light
in free space has to be the same value in all inertial

x2 - c2t2 = x'2 - c2t'2

Note:  Although not discussed in this particular note,
each side of the equation is the 4-vector formalism
xμxμ for 2 dimensions where xμ = x - ct and xμ = x + ct

This is the equation of a hyperbola.  Instead, of
using circular geometry using sines and cosines we
now need to switch to the hyperbolic functions sinh
and cosh.

The tranformations that satisfy x2 - c2t2 = x'2 - c2t'2
are:

x' = xcoshζ - ctsinhζ
t' = -(x/c)sinhζ + tcoshζ
y' = y
z' = z

In the prime frame x' = x - vt We can write:

x - vt = xcoshζ - ctsinhζ

∴ x - xcoshζ =  -ctsinhζ + vt

∴ x(1 - coshζ) =  t(v - csinhζ)

∴ v(1 - coshζ) =  (v - csinhζ)

∴ vcoshζ = csinhζ)

∴ v = ctanhζ

or

tanhζ = v/c

tanh2ζ = v2/c2

Now,

tanh2ζ = 1 - sech2ζ

∴ sech2 = 1 - v2/c2

∴ 1/cosh2ζ = 1 - v2/c2

∴ coshζ = 1/√(1 - v2/c2)

Now,

sinhζ = tanhζcoshζ

= (v/c)/√(1 - v2/c2)

Substituting into the original coordinate transformation
formulas give.

x' = x/√(1 - v2/c2) - (v/c)ct/√(1 - v2/c2)

= (x - vt)/√(1 - v2/c2)

ζ is the RAPIDITY defined as tanh-1β where β = v/c.  ζ
is the hyperbolic angle.  It is analagous to but does
not have the dimensions of velocity.

In summary, the connections between γ, β and ζ are:

β = v/c = tanhζ
γ = 1/√(1 - v2/c2/) = coshζ
βγ = (v/c)/√(1 - v2/c2) = sinhζ

Rotational Transform
--------------------

the 3 different spatial axes.  In this case the Lorentz
transformation matrix is:

-       -
| 1 0 0 0 |
| 0 x x x |  where the x's represent a 3D rotation matrix.
| 0 x x x |
| 0 x x x |
-       -

In general there can be combination of boosts and
rotations.

Active versus Passive Transformations
-------------------------------------

An active transformation is a transformation which
actually changes the physical position of a point.
A passive transformation is a change in the coordinate
system in which the object is described.

Lorentz Transformation of Fields
--------------------------------

For a basic scalar field:  φ(x) -> φ'(x) = φ(Λ-1x)

The inverse appears in the argument because we want to
express φ'(x) in terms of the 'untransformed' field,
φ.  In other words, the transformed field evaluated
at the transformed point gives the same value as the
original field evaluated at the point before it was
transformed.  For example consider a rotation, R, of a

z
|
|
|
|     (1,0,0)
--------o---- x
/        A
/
B o (0,1,0)
/
y

The point A goes to B.  So from B's perspective, A looks
like R-1(0,1,0) = (1,0,0).  The definition of a Lorentz
invariant theory is that if φ(x) solves the equations of
motion then φ(Λ-1x) also solves the equations of motion.

The derivative of φ(Λ-1x) can be found as follows:

Let yν = (Λ-1)νμxμ

We can use the Chain rule to write:

(∂'φ(xμ)/∂xμ) = (∂yν)/∂xμ)(∂φ(yν)/∂yν)

Now,

∂yν/∂xμ = ∂(Λ-1x)νμ/∂xμ = (Λ-1)νμ

So we end up with:

(∂φ'(xμ)/∂xμ) = (Λ-1)νμ(∂φ(yν)/∂yν)

≡ (Λ-1)νμ∂νφ(y)

Therefore, the derivative transforms like a vector which
is not surprising since derivative of a scalar field is
a vector.

We can find the second derivative as follows:

(∂φ(x))2 = ∂μφ(x)∂νφ(x)ημν

= (Λ-1)ρμ∂ρφ(y)(Λ-1)σν∂σφ(y)ημν

= ∂ρφ(y)∂σφ(y)ηρσ since ΛρμΛσνημν = ηρσ

= (∂φ(y))2

We now show that the Klein-Gordon, Dirac, and Weyl
equations are Lorentz invariance.

Klein-Gordon:  (∂2 + m2)φ = 0

(∂2 + m2)φ'(x) -> ((Λ-1)νμ∂ν(Λ-1)νμ∂ν + m2)φ(Λ-1x)

= (gνσ∂ν∂σ + m2)φ(Λ-1x)

= (∂2 + m2)φ(Λ-1x)

= 0

Dirac:  (iγμ∂μ - m)ψ = 0

In the case of the Dirac field the spinor also carries
an orientation that can be rotated or boosted in
spacetime. Therefore, ψα = S[Λ]αβψβ(Λ-1x)

(iγμ∂μ - m)ψ(x) -> (iγμ(Λ-1)νμ∂ν - m)S[Λ]ψ(Λ-1x)

Multiply from the right by S[Λ]S[Λ]-1 = I to get:

= S[Λ]S[Λ]-1(iγμ(Λ-1)νμ∂ν - m)S[Λ]ψ(Λ-1x)

= S[Λ](iS[Λ]-1γμS[Λ](Λ-1)νμ∂ν - m)S[Λ]ψ(Λ-1x)

= S[Λ](i(Λ)μσγσ(Λ-1)νμ∂ν - m)S[Λ]ψ(Λ-1x)

= S[Λ]{iγν∂ν - m)ψ(Λ-1x)

= 0

Weyl:  iσμ∂μψL = 0

iσμ∂μψL -> (iσμ(Λ-1)νμ∂ν)S[Λ]ψL(Λ-1x)

Using the same process as above we get:

= S[Λ]{iσν∂ν)ψL(Λ-1x)

= 0

```