I Why Does the Angle Between Coordinate Axes in Different Frames Equal atan(v/c)?

[Nugatory]

it is poosible to choose base metric is euclidean just $\vec{e_t'}=\sqrt{-1}*\vec{e_t}$ and $\vec{e_x'}=\vec{e_x}$.

No. Introducing that factor of $i$ allows many equations to take on the same form as they do for Euclidean space, but the similarity is superficial and just in the mathematical formalism. The space is still non-Euclidean in ways that cannot be avoided - for example, a straight line is not, in general, the shortest distance between two points.

There's some history here. My old copy of Goldstein and many other textbooks of that vintage used the $ict$ formalism. However, by the mid-1970's MTW had a short section stating that our old friend $ict$ was to be "put to the sword", something that had to be unlearned.

 

[Ibix]

it is poosible to choose base metric is euclidean just $\vec{e_t'}=\sqrt{-1}*\vec{e_t}$ and $\vec{e_x'}=\vec{e_x}$. Then $\vec{v_1}\cdot\vec{v_2}=|\vec{v_1}|*|\vec{v_2}|*cos(\alpha)$

then:
$\frac{e_t'}{i}\cdot e_x'=|e_t'/i|*|e_x'|*cos(\alpha_{t'-x'})$
$e_t\cdot e_x=|e_t'/i|*|e_x|*cos(\alpha_{t-x})$
$0=|e_t'/i|*|e_x|*cos(\alpha_{t-x})$
$\alpha_{t-x}=arccos(0)$
$\alpha_{t-x}=\frac{2\pi}{4}$

As @Nugatory says, not really. It's a mistake to try in my opinion. One of the key points for clear thinking is to make assumptions and differences between concepts as clear as possible. The $ict$ approach has always struck me as attempting to bury something quite subtle, and it comes back to bite when you move on to GR.

Honestly, I think the best approach is to explain that the inner product of two vectors, usually written $\vec v^T\vec v$ (in matrix notation, assuming $\vec v$ is a column vector), is more generally written as $\vec v^T\mathbf g\vec v$, where $\mathbf g$ is the metric tensor, which can be written as a square matrix. To get Euclidean geometry in Cartesian coordinates you set $\mathbf g$ to the identity matrix (which is why you don't normally see it written). To get Minkowski geometry in Einstein coordinates you make one of the diagonal elements negative (all of special relativity and our notions of cause and effect spring from that one change!). To get general relativity you let it get more complicated still.

This way you can see that the rule for taking inner products is the same in Minkowski and Euclidean spaces - but the spaces are fundamentally different in ways encoded in the metric tensor. And you are laying the groundwork for general relativity at the same time. My 2p, anyway.

 

[vanhees71]

To make the concepts even clearer, rather call $\mathbf{g}$ the components of the fundamental form, since in general it's not a metric, because it's not positive definite (but nondegenerate). Sometimes one calls it a pseudometric in this case. It's simply a non-degenerate bilinear form on a vector space.

 

[olgerm]

So on euclidean papersheet:
$\alpha_{t-t'}=arctan(v/c)$
$\alpha_{t-x}=\frac{2\pi}{4}$
$\alpha_{t-x'}=\frac{2\pi}{4}-arctan(v/c)$
$\alpha_{t'-x}=\frac{2\pi}{4}-arctan(v/c)$
$\alpha_{t'-x'}=\frac{2\pi}{4}-2*arctan(v/c)$
$\alpha_{x-x'}=arctan(v/c)$

And really in spacetime:
$\alpha_{t-t'}=arccos(\frac{\vec{e_t}\cdot \vec{e_t'}}{|\vec{e_t}|*|\vec{e_t'}|})= arccos(\frac{\vec{e_t}\cdot (\gamma*\vec{e_t}+\beta*\gamma*\vec{e_x})}{(-1)*(-1)})=arccos(\frac{1}{\sqrt{1-v^2/c^2}})=i*sgn(v)*arctanh(v/c)$
$\alpha_{t-x}=arccos(-\vec{e_x}\cdot\vec{e_t})=\frac{2\pi}{4}$
$\alpha_{t-x'}=arccos(-\vec{e_x'}\cdot\vec{e_t})=arccos(-(\gamma∗\beta∗\vec{e_t}+\gamma∗\vec{e_x})* \vec{e_t})=arccos(-\frac{v}{\sqrt{c^2−v^2}})$
$\alpha_{t'-x}=arccos(-\vec{e_x}\cdot\vec{e_t'})=arccos(-\vec{e_x}\cdot(\gamma∗\vec{e_t}+\beta∗\gamma∗\vec{e_x}))=arccos(-\frac{v}{\sqrt{c^2−v^2}})$
$\alpha_{t'-x'}arccos(-\vec{e_x'}\cdot\vec{e_t'})=\frac{2\pi}{4}$
$\alpha_{x-x'}=arccos(\frac{\vec{e_x}\cdot \vec{e_x'}}{|\vec{e_x}|*|\vec{e_x'}|})= arccos(\frac{\vec{e_x}\cdot (\beta*\gamma*\vec{e_t}+\gamma*\vec{e_x})}{(1)*(1)})=arccos(\frac{1}{\sqrt{1-v^2/c^2}})=i*sgn(v)*arctanh(v/c)$

Can you confirm these are correct?

 

[PeterDonis]

Can you confirm these are correct?

The "Euclidean papersheet" values look fine, but have no physical meaning.

The "really in spacetime" values are all nonsense except for the first one, if the first one is interpreted as the rapidity. Besides the rapidity $\alpha$, which you appear to be calling $\alpha_{t-t'}$, the only other meaningful quantities expressible in terms of trig functions (hyperbolic or otherwise) are $\gamma = \cosh (v / c)$ and $\gamma v = \sinh (v / c)$; but neither of those appear on your list.

 

[olgerm]

I had to edit $\alpha_{t-t'}$ and $\alpha_{x-x'}$.

 

[PeterDonis]

I had to edit $\alpha_{t-t'}$ and $\alpha_{x-x'}$.

That doesn't change anything I said in post #65.

At this point you have been given the correct information multiple times. There is no point in simply continuing to repeat the same corrective information if you aren't going to accept it.

Thread closed.