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4-vectors
Using index notation, the time and space coordinates can be combined into a single "4-vector" [math]\mathbf{x^{\mu}},\ \mu=0,\ 1,\ 2,\ 3[/math], that has units of length(i.e. ct is a distance).
[math]\mathbf{R} \equiv
\begin{bmatrix}
x^0 \\
x^1 \\
x^2 \\
x^3
\end{bmatrix}=
\begin{bmatrix}
ct \\
x \\
y \\
z
\end{bmatrix}[/math]
We can express the space time interval using the index notation
[math](ds)^2\equiv c^2 dt^{'2}-dx^{'2}-dy^{'2}-dz^{'2}= c^2 dt^{2}-dx^2-dy^2-dz^2[/math]
[math](ds)^2\equiv (dx^0)^{'2}-(dx^1)^{'2}-(dx^2)^{'2}-(dx^3)^{'2}= (dx^0)^{2}-(dx^1)^2-(dx^2)^2-(dx^3)^2[/math]
Since [math]ds^2 [/math] is nothing more than a dot product of a vector with itself, we should expect the components of the indices to follow a similar relationship.
[math](ds)^2\equiv d \mathbf{x_{\mu}} d \mathbf{x^{\mu}}[/math]
[math](ds)^2\equiv
\begin{bmatrix}
dx_0 & -dx_1 & -dx_2 & -dx_3
\end{bmatrix} \cdot
\begin{bmatrix}
dx^0 \\
dx^1 \\
dx^2 \\
dx^3
\end{bmatrix}[/math]
[math](ds)^2 \equiv (dx^0)^{2}-(dx^1)^2-(dx^2)^2-(dx^3)^2=(dx^0)^{'2}-(dx^1)^{'2}-(dx^2)^{'2}-(dx^3)^{'2}[/math]
The change in signs in the covariant term,
[math]\mathbf{x_{\mu}}= \begin{bmatrix}
dx_0 & -dx_1 & -dx_2 & -dx_3
\end{bmatrix}[/math]
from the contravarient term
[math]\mathbf{x^{\mu}}=
\begin{bmatrix}
dx^0 \\
dx^1 \\
dx^2 \\
dx^3
\end{bmatrix}
[/math]
Comes from the Minkowski metric
[math]\eta_{\mu \mu}=
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 &-1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -1
\end{bmatrix}[/math]
[math]ds^2=
\begin{bmatrix}
dx_0 & dx_1 & dx_2 & dx_3
\end{bmatrix}\cdot
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 &-1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -1
\end{bmatrix}\cdot
\begin{bmatrix}
dx^0 \\
dx^1 \\
dx^2 \\
dx^3
\end{bmatrix}
[/math]
[math]ds^2 \equiv \eta_{\mu \nu} \mathbf{dx^{\mu}} \mathbf{dx^{\mu}}=d\mathbf{R} \cdot d\mathbf{R}[/math]
[math]\mathbf R \cdot \mathbf R = s = \eta_{\mu \mu} \mathbf{x^{\mu}} \mathbf{x^{\mu}}[/math]
[math]\mathbf R \cdot \mathbf R = x_0^2-(x_1^2+x_2^2+x_3^2)[/math]
This is useful in that it shows that the scalar product of two 4-vectors is an invariant since the time-space interval is an invariant.
Using the Lorentz transformations and the index notation,
[math]
\begin{cases}
t'=\gamma (t-vz/c^2) \\
x'=x' \\
y'=y' \\
z'=\gamma (z-vt)
\end{cases}
[/math]
[math]\begin{bmatrix}
x'^0 \\
x'^1 \\
x'^2\\
x'^3
\end{bmatrix}=
\begin{bmatrix}
\gamma (x^0-vx^3/c) \\
x^1 \\
x^2 \\
\gamma (x^3-vx^0)
\end{bmatrix}
=
\begin{bmatrix}
\gamma (x^0-\beta x^3) \\
x^1 \\
x^2 \\
\gamma (x^3-vx^0)
\end{bmatrix}[/math]
Where [math]\beta \equiv \frac{v}{c}[/math]
This can be expressed in matrix form as
[math]\begin{bmatrix}
x'^0 \\
x'^1 \\
x'^2\\
x'^3
\end{bmatrix}=
\begin{bmatrix}
\gamma & 0 & 0 & -\gamma \beta \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
-\gamma \beta & 0 & 0 & \gamma
\end{bmatrix}
\cdot
\begin{bmatrix}
x^0 \\
x^1 \\
x^2 \\
x^3
\end{bmatrix}[/math]
Letting the indices run from 0 to 3, we can write
[math]\mathbf x'^{\mu}=\sum_{\nu=0}^3 (\Lambda_{\nu}^{\mu}) \mathbf x^{\nu}[/math]
Where [math]\Lambda[/math] is the Lorentz transformation matrix for motion in the z direction.
The Lorentz transformations are also invariant in that they are just a rotation, i.e. Det [math]\Lambda=1[/math]. The inner product is preserved,
[math]\Lambda_{\nu}^{\mu} \eta_{\nu}^{\mu} \Lambda_{\mu}^{\nu}=\eta_{\nu}^{\mu}[/math]
[math]
\begin{bmatrix}
\gamma & 0 & 0 & -\gamma \beta \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
-\gamma \beta & 0 & 0 & \gamma
\end{bmatrix}\cdot
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 &-1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -1
\end{bmatrix}\cdot
\begin{bmatrix}
\gamma & 0 & 0 & -\gamma \beta \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
-\gamma \beta & 0 & 0 & \gamma
\end{bmatrix}=
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 &-1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -1
\end{bmatrix}^T[/math]
[math]
\begin{bmatrix}
\gamma^2-\beta^2 \gamma^2 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -\gamma^2+\beta^2 \gamma^2
\end{bmatrix}=
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 &-1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -1
\end{bmatrix}[/math]
[math]
\begin{bmatrix}
\gamma^2(1-\beta^2) & 0 & 0 & 0 \\
0 & -1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -\gamma^2(1-\beta^2)
\end{bmatrix}=
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 &-1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -1
\end{bmatrix}[/math]
Where [math]\gamma \equiv \frac{1}{\sqrt{1-\beta^2}}[/math]
[math]
\begin{bmatrix}
\frac{\gamma^2}{\gamma^2} & 0 & 0 & 0 \\
0 & -1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -\frac{\gamma^2}{\gamma^2}
\end{bmatrix}=
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 &-1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -1
\end{bmatrix}[/math]
[math]
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -1
\end{bmatrix}=
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 &-1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -1
\end{bmatrix}[/math]
[math]\mathbf R \cdot \mathbf R = s = \eta_{\nu}^{\mu} \mathbf{x^{\mu}} \mathbf{x^{\mu}}[/math]
Using the fact
[math]\Lambda_{\nu}^{\mu} \eta_{\nu}^{\mu} \Lambda_{\mu}^{\nu}=\eta_{\nu}^{\mu}[/math]
[math]\mathbf R \cdot \mathbf R = s =\Lambda_{\nu}^{\mu} \eta_{\nu}^{\mu} \Lambda_{\mu}^{\nu} \mathbf{x^{\mu}} \mathbf{x^{\mu}}[/math]
[math]\mathbf R \cdot \mathbf R = s =\Lambda_{\nu}^{\mu} \eta_{\nu}^{\mu} \Lambda_{\mu}^{\nu} \mathbf{x^{\mu}} \mathbf{x^{\mu}}[/math]
Using
[math] \mathbf{x^{' \mu}}\equiv \Lambda_{\mu}^{\nu} \mathbf{x^{\mu}}[/math]
[math]\mathbf R \cdot \mathbf R = x_0^2-(x_1^2+x_2^2+x_3^2)[/math]
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