Frame of Reference Transformation

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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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