4-vectors

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4-vectors

Using index notation, the time and space coordinates can be combined into a single "4-vector" $\mathbf{x^{\mu}},\ \mu=0,\ 1,\ 2,\ 3$, that has units of length(i.e. ct is a distance).

$\mathbf{R} \equiv \begin{bmatrix} x^0 \\ x^1 \\ x^2 \\ x^3 \end{bmatrix}= \begin{bmatrix} ct \\ x \\ y \\ z \end{bmatrix}$

We can express the space time interval using the index notation

$(ds)^2\equiv c^2 dt^{'2}-dx^{'2}-dy^{'2}-dz^{'2}= c^2 dt^{2}-dx^2-dy^2-dz^2$

$(ds)^2\equiv (dx^0)^{'2}-(dx^1)^{'2}-(dx^2)^{'2}-(dx^3)^{'2}= (dx^0)^{2}-(dx^1)^2-(dx^2)^2-(dx^3)^2$

Since $ds^2$ 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.

$(ds)^2\equiv d \mathbf{x_{\mu}} d \mathbf{x^{\mu}}$

$(ds)^2\equiv \begin{bmatrix} dx_0 & -dx_1 & -dx_2 & -dx_3 \end{bmatrix} \cdot \begin{bmatrix} dx^0 \\ dx^1 \\ dx^2 \\ dx^3 \end{bmatrix}$

$(ds)^2 \equiv (dx^0)^{2}-(dx^1)^2-(dx^2)^2-(dx^3)^2=(dx^0)^{'2}-(dx^1)^{'2}-(dx^2)^{'2}-(dx^3)^{'2}$

The change in signs in the covariant term,

$\mathbf{x_{\mu}}= \begin{bmatrix} dx_0 & -dx_1 & -dx_2 & -dx_3 \end{bmatrix}$

from the contravarient term

$\mathbf{x^{\mu}}= \begin{bmatrix} dx^0 \\ dx^1 \\ dx^2 \\ dx^3 \end{bmatrix}$

Comes from the Minkowski metric

$\eta_{\mu \mu}= \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 &-1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix}$

$ds^2= \begin{bmatrix} dx_0 & dx_1 & dx_2 & dx_3 \end{bmatrix}\cdot \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 &-1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix}\cdot \begin{bmatrix} dx^0 \\ dx^1 \\ dx^2 \\ dx^3 \end{bmatrix}$

$ds^2 \equiv \eta_{\mu \nu} \mathbf{dx^{\mu}} \mathbf{dx^{\mu}}=d\mathbf{R} \cdot d\mathbf{R}$

$\mathbf R \cdot \mathbf R = s = \eta_{\mu \mu} \mathbf{x^{\mu}} \mathbf{x^{\mu}}$

$\mathbf R \cdot \mathbf R = x_0^2-(x_1^2+x_2^2+x_3^2)$

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,

$\begin{cases} t'=\gamma (t-vz/c^2) \\ x'=x' \\ y'=y' \\ z'=\gamma (z-vt) \end{cases}$

$\begin{bmatrix} x'^0 \\ x'^1 \\ x'^2\\ x'^3 \end{bmatrix}= \begin{bmatrix} \gamma (x^0-vx^3/c) \\ x^1 \\ x^2 \\ \gamma (x^3-vx^0) \end{bmatrix} = \begin{bmatrix} \gamma (x^0-\beta x^3) \\ x^1 \\ x^2 \\ \gamma (x^3-vx^0) \end{bmatrix}$

Where $\beta \equiv \frac{v}{c}$

This can be expressed in matrix form as

$\begin{bmatrix} x'^0 \\ x'^1 \\ x'^2\\ x'^3 \end{bmatrix}= \begin{bmatrix} \gamma & 0 & 0 & -\gamma \beta \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -\gamma \beta & 0 & 0 & \gamma \end{bmatrix} \cdot \begin{bmatrix} x^0 \\ x^1 \\ x^2 \\ x^3 \end{bmatrix}$

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

$\mathbf x'^{\mu}=\sum_{\nu=0}^3 (\Lambda_{\nu}^{\mu}) \mathbf x^{\nu}$

Where $\Lambda$ 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 $\Lambda=1$. The inner product is preserved,

$\Lambda_{\nu}^{\mu} \eta_{\nu}^{\mu} \Lambda_{\mu}^{\nu}=\eta_{\nu}^{\mu}$

$\begin{bmatrix} \gamma & 0 & 0 & -\gamma \beta \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -\gamma \beta & 0 & 0 & \gamma \end{bmatrix}\cdot \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 &-1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix}\cdot \begin{bmatrix} \gamma & 0 & 0 & -\gamma \beta \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -\gamma \beta & 0 & 0 & \gamma \end{bmatrix}= \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 &-1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix}^T$

$\begin{bmatrix} \gamma^2-\beta^2 \gamma^2 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -\gamma^2+\beta^2 \gamma^2 \end{bmatrix}= \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 &-1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix}$

$\begin{bmatrix} \gamma^2(1-\beta^2) & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -\gamma^2(1-\beta^2) \end{bmatrix}= \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 &-1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix}$

Where $\gamma \equiv \frac{1}{\sqrt{1-\beta^2}}$

$\begin{bmatrix} \frac{\gamma^2}{\gamma^2} & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -\frac{\gamma^2}{\gamma^2} \end{bmatrix}= \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 &-1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix}$

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

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