[math]\underline{\textbf{Navigation}}[/math]
[math]\vartriangleleft [/math]
[math]\triangle [/math]
[math]\vartriangleright [/math]
4-vectors
Using index notation, the time and space coordinates can be combined into a single "4-vector" [math]\mathbf{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. Following the rules of matrix multiplication, the dot product of two 4-vectors should follow the form:
[math](ds)^2\equiv d \mathbf{x_{\mu}} d \mathbf{x^{\mu}}[/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]
This gives the desired results as expected.
[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]\mathbf{x_{\mu}}= \begin{bmatrix}
dx_0 & -dx_1 & -dx_2 & -dx_3
\end{bmatrix}[/math]
from the contravarient term
[math]\mathbf{x^{\mu}}=
\begin{bmatrix}
dx^0 \\
dx^1 \\
dx^2 \\
dx^3
\end{bmatrix}
[/math]
Comes from the Minkowski metric
[math]\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]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 \equiv \eta_{\nu}^{ \mu} \mathbf{dx^{\mu}} \mathbf{dx^{\mu}}=d\mathbf{R} \cdot d\mathbf{R}[/math]
[math]\mathbf R \cdot \mathbf R = s = \eta_{\nu}^{ \mu} \mathbf{x^{\mu}} \mathbf{x^{\mu}}[/math]
[math]\mathbf R_1 \cdot \mathbf R^1 = x_0^2-(x_1^2+x_2^2+x_3^2)[/math]
Similarly, for two different 4-vectors,
[math]\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.
[math]\underline{\textbf{Navigation}}[/math]
[math]\vartriangleleft [/math]
[math]\triangle [/math]
[math]\vartriangleright [/math]