4-vectors

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4-vectors

Using index notation, the time and space coordinates can be combined into a single "4-vector" [math]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 x_{\mu} x^{\nu}[/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]x_{\mu}= \begin{bmatrix} dx_0 & -dx_1 & -dx_2 & -dx_3 \end{bmatrix}[/math]

from the contravarient term

[math]x^{\nu}= \begin{bmatrix} dx^0 \\ dx^1 \\ dx^2 \\ dx^3 \end{bmatrix} [/math]

Comes from the Minkowski metric

[math]\eta_{\mu \nu}= \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=\Eta_{\mu \nu} x^{\mu} x^{\nu}[/math]

This is useful in that it shows that the "dot 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]x'^{\mu}=\sum_{\nu=0}^3 (\Lambda_{\nu}^{\mu})x^{\nu}[/math]

Where [math]\Lambda[/math] is the Lorentz transformation matrix for motion in the z direction.

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