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]\displaystyle{ \mathbf{x^{\mu}},\ \mu=0,\ 1,\ 2,\ 3 }[/math], that has units of length(i.e. ct is a distance).

[math]\displaystyle{ \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]\displaystyle{ (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]\displaystyle{ (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]\displaystyle{ 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]\displaystyle{ (ds)^2\equiv d \mathbf{x_{\mu}} d \mathbf{x^{\mu}} }[/math]

[math]\displaystyle{ (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]\displaystyle{ (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]\displaystyle{ \mathbf{x_{\mu}}= \begin{bmatrix} dx_0 & -dx_1 & -dx_2 & -dx_3 \end{bmatrix} }[/math]

from the contravarient term

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

Comes from the Minkowski metric

[math]\displaystyle{ \eta_{\mu \mu}= \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 &-1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix} }[/math]

[math]\displaystyle{ 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]\displaystyle{ ds^2 \equiv \eta_{\nu}^{ \mu} \mathbf{dx^{\mu}} \mathbf{dx^{\mu}}=d\mathbf{R} \cdot d\mathbf{R} }[/math]

[math]\displaystyle{ \mathbf R \cdot \mathbf R = s = \eta_{\nu}^{ \mu} \mathbf{x^{\mu}} \mathbf{x^{\mu}} }[/math]

[math]\displaystyle{ \mathbf R_1 \cdot \mathbf R_2 = x_0^2-(x_1^2+x_2^2+x_3^2) }[/math]

Similarly, for two different 4-vectors,

[math]\displaystyle{ \mathbf R_1 \cdot \mathbf R_2 = x_{0_1}x_{0_2}-(x_{1_1}x_{1_2}+x_{2_1}x_{2_2}+x_{3_1}x_{3_2}) }[/math]

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,

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

[math]\displaystyle{ \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]\displaystyle{ \beta \equiv \frac{v}{c} }[/math]

This can be expressed in matrix form as

[math]\displaystyle{ \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]\displaystyle{ \mathbf x'^{\mu}=\sum_{\nu=0}^3 (\Lambda_{\nu}^{\mu}) \mathbf x^{\nu} }[/math]

Where [math]\displaystyle{ \Lambda }[/math] 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 [math]\displaystyle{ \Lambda=1 }[/math]. The inner product is preserved,

[math]\displaystyle{ \Lambda_{\nu}^{\mu} \eta_{\nu}^{\mu} \Lambda_{\mu}^{\nu}=\eta_{\nu}^{\mu} }[/math]

[math]\displaystyle{ \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 }[/math]

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \gamma \equiv \frac{1}{\sqrt{1-\beta^2}} }[/math]

[math]\displaystyle{ \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]\displaystyle{ \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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