# Redshift Academy

Wolfram Alpha:

Last modified: April 25, 2022 ✓
```Lorentz Transform
-----------------

The Lorentz transformation converts between two
different observers' measurements of space and
time, where one observer is moving a CONSTANT
velocity with respect to the other.  Coordinate
frames that are moving at a constant velocity
with respect to one another are called INERTIAL
reference frames.  Special Relativity does not
apply to the situation where the frames are
accelerating with respect to each other.  This
is the subject of General Relativity.

In classical physics, 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 SR
this is only a good approximation at speeds small
compared to the speed of light.  At higher speeds
the result is not just an offsetting of the x
coordinates but times, lengths and masses are also
effected.

Consider 2 cartesian coordinate systems (t,x,y,z)
and (t',x',y',z') such that the x and x' axes are
coincident and the other 2 axes are parallel.
Let (t,x,y,z) be a fixed reference frame and let
(t',x',y',z') be a reference frame moving relative
to the fixed frame.  The reference frames coincide
at t = t' = 0.  For a frame boosted along the x-
axis).

Time Dilation
-------------

The Lorentz transformation for time is:

t - vx/c2
t' = ------------
√(1 - v2/c2)

= γ(t - vx/c2)

Consider a time interval:

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

If the time measurements in the moving frame are
made at the same location, the expression reduces
to:

t12' = γt12

Example:

Consider 2 people - Bob and Alice.  Bob's frame
is stationary and Alice's frame is moving with
v = 0.5c, therefore γ = 1.154.  Alice measures
1s in her frame (t' = 1) but when Bob looks at
her clock he sees.

t = t'/γ = 1/1.154 = 0.866s

Likewise, if Bob measures 1s in his frame (t = 1)
then Alice will observe Bob's time to be:

t' = γt = 1.154s

Therefore, a clock in an inertial reference frame
moving relative to an observer in a stationary
reference frame appears to run more slowly.

Length Contraction
------------------

The Lorentz Transformation for spacial coordinates
is:

x' = ct' = γ(ct - vx/c)

= γ(x - vt)

Consider a length measurement:

x1' - x2' = γ{x1 - vt1 - x2 + vt2}

If the length measurements in the moving frame
are made at the same time, the expression reduces
to:

x12' = γx12

Example:

Again, consider Bob and Alice under the same
conditions. Therefore,

x = x'/γ = 0.866m

x' = γx = 1.154m

Therefore, a meter stick in an inertial reference
frame moving relative to an observer in a stationary
reference frame appears to be shorter.

Proper Time and Proper Length
-----------------------------

The proper time, τ, and proper length, L0, is
defined as the time and length that is always
measured in the observer’s own frame regardless
of whether that frame is moving or not.  This
is because an observer is always at rest in his
own reference frame and cannot be moving relative
to himself.  Thus a clock or meter stick moving
along a world line will show the proper time to the
observer carrying the clock or meter stick.

Velocity Addition
-----------------

Consider the inverse transformations:

x = γ(x' + vt')

t = γ(t' + vx'/c2)

w = x/t = γ(x' + vt')/γ(t' + vx'/c2)

= (x' + vt')/(t' + vx'/c2)

÷ RHS by t'. Let u = x'/t':

w = (u + v)/(1 + vu)

For small v and/or u this becomes the standard
Newtonian formula for relative velocity.

The Lorentz Transformation in 4-vector Form
-------------------------------------------

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     |
-          -

Likewise, 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.

Energy and Momentum
-------------------

Energy and momentum can be combined into a single
4-vector, (E/c,px,py,pz).  Therefore,

-           -  -    -
|   γ -βγ 0 0 || E/c  |
| -βγ  γ  0 0 || px   |
|   0  0  1 0 || py   |
|   0  0  0 1 || pz   |
-           -  -    -

-            -
|  γE/c - βγpx |
= | -βγE/c + γpx |
|      py      |
|      pz      |
-            -

Mass (Energy) Contraction
-------------------------

E'/c = γE/c - βγpx

= γ(E/c - vpx/c)

∴ E' = γ(E - vpx)

∴ m'c2 = γ(mc2 - vpx)

∴ m' = γ(m - vpx/c2)

For a particle at rest px = 0.  Therefore,

m = γm0

Note:  There is another derivation of this equation
that relies on analyzing the elastic collision of
2 masses in the moving frame.  This derivation uses
the formula for relativistic velocity addition
derived above in conjunction with the conservation
of linear momentum law.

-----> v
t'
|
|      u    u
|    o ->  <- o  Before
|    1        2
|     <- oo ->   After
|     -u    u
|
-------------- x'

u1 = (u + v)/(1 + uv/c2)  u2 = (-u + v)/(1 - uv/c2)

m1u1 + m2u2 = (m1 + m2)v

After a considerable amount of math one arrives at:

m1/m2 = √(1 - u22/c2)/√(1 - u12/c2)

If u2 = 0 then m2 = m0 and we get:

m1 = m0/√(1 - u12/c2)

If we then let u1 = we get:

m = m0/√(1 - v2/c2)

Momentum Contraction
--------------------

px' = -βγE/c + γpx

= γ(px - Ev/c2)

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 moving 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
reference frames.  We start with,

x2 - c2t2 = x'2 - c2t'2

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 transformations 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 gives:

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
--------------------

In addition to boosts we can also have rotations about
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
field about the z axis.

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

```