Dot Product of Four Vectors

We are going to use the metric signature (-,+,+,+).

The dot product of two four-vectors, $(\vec{A})$ and $(\vec{B})$ , can be expressed as:

$\vec{A} \cdot \vec{B} = -A^t B^t + A^x B^x + A^y B^y + A^z B^z$

Here, $(A^t, A^x, A^y, A^z)$ represent the components of vector $(\vec{A})$ and $(B^t, B^x, B^y, B^z)$ represent the components of vector $(\vec{B})$ .

Discussions on this topic will probably mention a Minkowski metric. We used…

$\eta_{\mu\nu} = \text{diag}(-1, 1, 1, 1)$

There is a matrix that corresponds to the Minkowski metric…

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

You may find it named by three different names:

Minkowski metric matrix
Minkowski spacetime matrix
metric tensor matrix

We should now take…

$\vec{A} \cdot \vec{B} = -A^t B^t + A^x B^x + A^y B^y + A^z B^z$

…and consider what happens if we input the same vector twice…

$\vec{A} \cdot \vec{A} = -A^t A^t + A^x A^x + A^y A^y + A^z A^z$

You are more likely to see the above typed as…

$\vec{A}^2 = -(A^t)^2 + (A^x)^2 + (A^y)^2 + (A^z)^2$

…and there is a very good chance that they will use {0,1,2,3} instead of {t,x,y,z} and if so you get…

$\vec{A}^2 = -(A^0)^2 + (A^1)^2 + (A^2)^2 + (A^3)^2$

Finally, there may be some confusion because you might see author after author telling what answer you get from $\vec{A} \cdot \vec{A}$ and then someone says that we can’t do that math unless we have the metric in it…

$\mathbf{A} \cdot \mathbf{B} = g_{\mu\nu} A^\nu B^\mu$

What happens is shown below:

$\begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{bmatrix} a^1 \\ a^2 \\ a^3 \\ a^4 \end{bmatrix} \begin{bmatrix} b^1 \\ b^2 \\ b^3 \\ b^4 \end{bmatrix}$