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Frame of Reference Transformation
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.
Using the Einstein convention, this can be written as
[math]\mathbf x'^{\mu}= \Lambda_{\nu}^{\mu} \mathbf x^{\nu}[/math]
If we take the 4-vector quantities to be on an infinitesimally small scale, then there exists a linear relationship between the transformation. Following the rules of partial differentiation,
[math]dt^{'} \equiv \frac{\partial t'}{\partial t} dt+\frac{\partial t'}{\partial x} dx + \frac{\partial t'}{\partial y} dy+ \frac{\partial t'}{\partial z} dz \Rightarrow dx^{'0} \equiv \frac{\partial x^{'0}}{\partial x^{0}} dx^{0}+\frac{\partial x^{'0}}{\partial x^{1}} dx^{1} + \frac{\partial x^{'0}}{\partial x^{2}} dx^{2}+ \frac{\partial x^{'0}}{\partial x^{3}} dx^{3}[/math]
[math]dx^{'} \equiv \frac{\partial t'}{\partial t} dt+\frac{\partial x'}{\partial x} dx + \frac{\partial x'}{\partial y} dy+ \frac{\partial x'}{\partial z} dz\Rightarrow dx^{'1} \equiv \frac{\partial x^{'1}}{\partial x^{0}} dx^{0}+\frac{\partial x^{'1}}{\partial x^{1}} dx^{1} + \frac{\partial x^{'1}}{\partial x^{2}} dx^{2}+ \frac{\partial x^{'1}}{\partial x^{3}} dx^{3}[/math]
[math]dy^{'} \equiv \frac{\partial y'}{\partial t} dt+\frac{\partial y'}{\partial x} dx + \frac{\partial y'}{\partial y} dy+ \frac{\partial y'}{\partial z} dz\Rightarrow dx^{'2} \equiv \frac{\partial x^{'2}}{\partial x^{0}} dx^{0}+\frac{\partial x^{'2}}{\partial x^{1}} dx^{1} + \frac{\partial x^{'2}}{\partial x^{2}} dx^{2}+ \frac{\partial x^{'2}}{\partial x^{3}} dx^{3}[/math]
[math]dz^{'} \equiv \frac{\partial z'}{\partial t} dt+\frac{\partial z'}{\partial x} dx + \frac{\partial z'}{\partial y} dy+ \frac{\partial z'}{\partial z} dz\Rightarrow dx^{'3} \equiv \frac{\partial x^{'3}}{\partial x^{0}} dx^{0}+\frac{\partial x^{'3}}{\partial x^{1}} dx^{1} + \frac{\partial x^{'3}}{\partial x^{2}} dx^{2}+ \frac{\partial x^{'3}}{\partial x^{3}} dx^{3}[/math]
Expressing this in matrix form
[math]\begin{bmatrix}
dx'^0 \\
\\
dx'^1 \\
\\
dx'^2\\
\\
dx'^3
\end{bmatrix}=
\begin{bmatrix}
\frac{\partial x^{'0}}{\partial x^0} & \frac{\partial x^{'0}}{\partial x^1} & \frac{\partial x^{'0}}{\partial x^2} & \frac{\partial x^{'0}}{\partial x^3} \\
\\
\frac{\partial x^{'1}}{\partial x^0} & \frac{\partial x^{'1}}{\partial x^1} & \frac{\partial x^{'1}}{\partial x^2} & \frac{\partial x^{'1}}{\partial x^3} \\
\\
\frac{\partial x^{'2}}{\partial x^0} & \frac{\partial x^{'2}}{\partial x^1} & \frac{\partial x^{'2}}{\partial x^2} & \frac{\partial x^{'2}}{\partial x^3} \\
\\
\frac{\partial x^{'3}}{\partial x^0} & \frac{\partial x^{'3}}{\partial x^1} & \frac{\partial x^{'3}}{\partial x^2} & \frac{\partial x^{'3}}{\partial x^3}
\end{bmatrix}
\cdot
\begin{bmatrix}
dx^0 \\
\\
dx^1 \\
\\
dx^2 \\
\\
dx^3
\end{bmatrix}[/math]
Again, using a summation over the indicies
[math]d\mathbf x^{'\mu}=\sum_{\nu=0}^3 \frac{\partial x^{'\mu}}{\partial x^{\nu}}d\mathbf x^{\nu}[/math]
Using the Einstein convention
[math]d\mathbf x^{'\mu}= \frac{\partial x^{'\mu}}{\partial x^{\nu}}d\mathbf x^{\nu}[/math]
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}^T=
\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-\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]
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