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=Frame of Reference Transformation=
Using the Lorentz transformations and the index notation,

<center><math>
\begin{cases}
t'=\gamma (t-vz/c^2) \\

x'=x' \\

y'=y' \\

z'=\gamma (z-vt)
\end{cases}
</math></center>


<center><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></center>



<center>Where <math>\beta \equiv \frac{v}{c}</math></center>

This can be expressed in matrix form as

<center><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></center>


Letting the indices run from 0 to 3, we can write

<center><math>\mathbf x'^{\mu}=\sum_{\nu=0}^3 (\Lambda_{\nu}^{\mu}) \mathbf x^{\nu}</math></center>


<center>Where <math>\Lambda</math> is the Lorentz transformation matrix for motion in the z direction.</center>


Using the Einstein convention, this can be written as

<center><math>\mathbf x'^{\mu}= \Lambda_{\nu}^{\mu} \mathbf x^{\nu}</math></center>

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,


<center><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></center>


<center><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></center>


<center><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></center>


<center><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></center>



Expressing this in matrix form

<center><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></center>


Again, using a summation over the indicies

<center><math>d\mathbf x^{'\mu}=\sum_{\nu=0}^3 \frac{\partial x^{'\mu}}{\partial x^{\nu}}d\mathbf x^{\nu}</math></center>


Using the Einstein convention

<center><math>d\mathbf x^{'\mu}= \frac{\partial x^{'\mu}}{\partial x^{\nu}}d\mathbf x^{\nu}</math></center>


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,



<center><math>\Lambda_{\nu}^{\mu} \eta_{\nu}^{\mu} \Lambda_{\mu}^{\nu}=\eta_{\nu}^{\mu}</math></center>


<center><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></center>



<center><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></center>



<center><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></center>



<center>Where <math>\gamma \equiv \frac{1}{\sqrt{1-\beta^2}}</math></center>


<center><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></center>




<center><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></center>


<center><math>\underline{\textbf{Navigation}}</math>

[[4-momenta|<math>\vartriangleleft </math>]]
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</center>