Minkowski and Euclidean 4 vectors

Question

I am reading these lecture notes. On page 114 they define Minkowskian space time coordinates:

$$\mathcal{X}=(\mathcal{X}^0,\mathcal{X}^1,\mathcal{X}^2,\mathcal{X}^3)=(t,x^i),$$

where $x^1=x,x^2=y,x^3=z$. They also define momenta:

$$\mathcal{K}=(\mathcal{K}^0,\mathcal{K}^1,\mathcal{K}^2,\mathcal{K}^3)=(k^0,k^i)$$

where $k^1=k_x,k^2=k_y,k^3=k_z$.

The Euclidean counterparts are defined as:

$$X=(X^0,X^1,X^2,X^3)=(\tau,x^i),$$

with $\tau=i t$

$$K=(K^0,K^1,K^2,K^3)=(k_n,k_i),$$

where $k_n=-ik^0$.

My first question is why don't we define $K=(k_n,k^i)$ ?

Next they calculatet in the notes $X\cdot K$. When I plug in all definitions I get:

$$X\cdot K=X_0K^0+X_iK^i=\tau k_n-X^iK^i=\tau k_n -x^i k_i=\tau k_n +x^i k^i=t k^0+\vec{k}\cdot\vec{x}$$

where $\vec{x}=(x^1,x^2,x^3)$ and $\vec{k}=(k^1,k^2,k^3)$ and I have used $\eta_{\mu\nu}={\rm diag}(1,-1,-1,-1)$.

Second question: In the lecture notes that I have linked they get a slightly different result:

$$X\cdot K=\tau k_n -\vec{k}\cdot\vec{x}=t k^0 -\vec{k}\cdot\vec{x}.$$

But this doesn't make sense because it is the same as $\mathcal{X}\cdot\mathcal{K}$ ???

Answer 1

You're making an error when using the Minkowski metric on the Euclidean components $X$ and $K$, which is why you get a different result. The result from the lecture notes is indeed correct and rightly in agreement with $\mathcal{X}\cdot\mathcal{K}$.

This convention of describing Minkowski space with a Euclidean metric and imaginary zero-component of vectors was more popular around the first half of the last century, and is now, I think, widely considered bad practice.