User:First Harmonic/Sandbox

Messages, codewords, and alphabets
Alphabet A consisting of b symbols:


 * $$\mathcal{A} = \{ a_i \} $$ with $$ i \in \{ 1, 2, 3 \ ... \ b \} $$

=Lorentz transformation=

Derivation

 * Editor's Note: The following derivation is adapted from Palash B. Pal Nothing but Relativity

Two inertial observers, H and K, are moving relative to one another at a constant velocity in a two-dimensional spacetime continuum. Let v represent the velocity of K relative to H. By symmetry, then, the velocity of H relative to K is –v.


 * $$\mathbf{s} = \begin{bmatrix} t \\ x  \end{bmatrix}$$


 * $$\mathbf{s'} = \begin{bmatrix} t' \\ x'  \end{bmatrix}$$

The transformation from one frame of reference to the other is simply a function of the spacetime vector s and the relative velocity v:


 * $$\mathbf{s'} = L(\mathbf{s},v) $$

Likewise, the reverse transformation is the same function with the arguments reversed:


 * $$\mathbf{s} = L(\mathbf{s'},-v) $$

Therefore,


 * $$\mathbf{s} = L[L(\mathbf{s},v),-v] $$

Because space and time are homogeneous, we can show that the transformation L must be a linear function of s that depends only on the relative velocity v. Thus:


 * $$L(\mathbf{s},v) = \mathbf{L}(v) \cdot \mathbf{s} = \begin{bmatrix} A(v) & B(v) \\ C(v) & D(v)  \end{bmatrix} \cdot

\begin{bmatrix} t \\ x  \end{bmatrix} $$

Therefore:
 * $$\begin{bmatrix} A(-v) & B(-v) \\ C(-v) & D(-v) \end{bmatrix} \begin{bmatrix} A(v) & B(v) \\ C(v) & D(v)  \end{bmatrix}

= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

Because the positive direction of the x coordinate is arbitrary, whereas the positive direction of time t is fixed, we can show that A and D are even functions of v, whereas B and C are odd functions of v. In other words:


 * $$ \begin{matrix}

A(-v) & = & A(v) \\ B(-v) & = & -B(v) \\ C(-v) & = & -C(v) \\ D(-v) & = & D(v) \\ \end{matrix} $$

As a result, we have


 * $$\begin{bmatrix} A(v) & -B(v) \\ -C(v) & D(v) \end{bmatrix} \begin{bmatrix} A(v) & B(v) \\ C(v) & D(v)  \end{bmatrix}

= \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$$

Matrix multiplication results in four equations:


 * $$ \begin{matrix}

A^2 - BC & = & 1 \\ B(A-D) & = & 0 \\ -C(A-D) & = & 0 \\ D^2-BC & = & 1 \\ \end{matrix}$$

Unfortunately, although we have four equations in four unknowns, only two of these equations are independent. There are, however, two possible ways to proceed. The first is to assume that B = C = 0, which then implies that A = D = 1, or A = D = -1. This outcome suggests that the Lorentz transformation is nothing but the identity matrix (or the negative of the identity matrix), which is not only trivial and uninteresting, but also contrary to all known experimental evidence.

The second approach is more promising. We consolidate the four equations into two independent equations:


 * $$ \begin{matrix}

A & = & D \\ B & = & { D^2 -1 \over C } \\ \end{matrix}$$

We can derive a third independent equation by observing that the origin of

=Four-vector=

Alternative approach using complex numbers
Einstein, Lorentz, and others suggested that one way to interpret the results of special relativity is to view spacetime as a complex-valued four-dimensional vector space where the first component (index &mu; = 0) is a purely imaginary number, while the remaining three components (indices &mu; = 1, 2, 3) are purely real numbers.

World line in Minkowski space


\mathbf{x} \  = \  \{ x^{\mu} \}  \

= \begin{pmatrix} x^0 \\ x^1 \\ x^2 \\ x^3 \\ \end{pmatrix}

= \begin{pmatrix} j t \\ x/c \\ y/c \\ z/c \\ \end{pmatrix}

$$

where j is the imaginary unit and c is the speed of light in free space.

Velocity
The velocity v is defined as the rate of change in spacetime event x with respect to its time component x0:


 * $$\mathbf{v} \ = \ { \partial \mathbf{x} \over \partial x^0 } $$



= { \partial \mathbf{x} \over \partial (j t) } = { 1 \over j } { \partial \mathbf{x} \over \partial t } = -j   { \partial \mathbf{x} \over \partial t }

$$



= { -j  }  {\partial \over dt}  \begin{pmatrix} x^0 \\ x^1 \\ x^2 \\ x^3 \\ \end{pmatrix}

= { -j  }  \begin{pmatrix} j \\ v_x/c \\ v_y/c \\ v_z/c \\ \end{pmatrix}

= \begin{pmatrix} 1 \\ -j v_x/c \\ -j v_y/c \\ -j v_z/c \\ \end{pmatrix}

$$

Therefore:



\mathbf{v} \ = \  \{ v^{\mu} \} \

= \begin{pmatrix} v^0 \\ v^1 \\ v^2 \\ v^3 \\ \end{pmatrix}

= \begin{pmatrix} 1 \\ -j v_x/c \\ -j v_y/c \\ -j v_z/c \\ \end{pmatrix}

$$

First Harmonic 22:48, 28 December 2006 (UTC)