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In special relativity, we write contravariant and covariant vectors respectively as $$A^\mu=(A^0, A^1, A^2, A^3), \quad A_\mu=(A_0, A_1, A_2, A_3).$$ Since $A_\mu=\eta_{\mu\nu}A^\nu$, we read off the relationship between each component, choosing the signature $(+,-,-,-$$)$: $$A_0=A^0, \quad A_1=-A^1, \quad A_2=-A^2, \quad A_3=-A^3 $$ and thus $A_\mu=(A^0, -A^1, -A^2, -A^3)$ in terms of the contravariant components. But in order to have $A^\mu A_\mu=(A^0)^2-(\mathbf A)^2$ as we should, with this notation where the metric $\eta_{\mu\nu}$ has been 'absorbed' into one of the vectors, we need to use the standard Euclidean product: $$A^\mu\eta_{\mu\nu}A^\nu=A^\mu A_\mu= (A_0, A_1, A_2, A_3)^T(A^0, -A^1, -A^2, -A^3)=(A^0)^2-(\mathbf A)^2.$$ Does this mean that going into Minkowski space correspond to staying in the Euclidean space, but inverting the spatial coordinates of one of the vectors?

ForgiveMyNoobness
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1 Answers1

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we need to use the standard Euclidean product

I wouldn't call it a Euclidean product but rather the usual Einstein convention for summation over the repeated indices:

$$A_\mu A^\mu = \sum_{\mu=0}^3A_\mu A^\mu = A_0 A^0 + A_1 A^1 + A_2 A^2 + A_3 A^3$$

This expression is the same in both Euclidean and Minkowski space. The distinction between them comes when you want to relate $A_\mu$ to $A^\mu$. As you correctly pointed out, in Minkowski this is done with the help of Minkowski metric and as a result you get

$$A_\mu A^\mu = (A^0)^2 - (\mathbf{A})^2$$

How you choose to interpret it is up to you: saying that

going into Minkowski space correspond to staying in the Euclidean space, but inverting the spatial coordinates of one of the vectors

is equivalent to just using the Minkowski metric. I am not sure that this interpretation is going to help you gain physical intuition, though. It's better to just get accustomed to using Minkowski metric.

Viking
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