# Tag Info

3

Yes it is. The volume form on any (pseudo-)Riemannian manifold $(M,g)$ of dimension $n$, where $g$ is the metric, is given in local coordinates $(x^1, \dots, x^n)$ $$\sqrt{|\det (g_{\mu\nu})|}dx^1\wedge \cdots \wedge dx^n$$ where $\det(g_{\mu\nu})$ is the determinant of the metric in these coordinates. In cartesian coordinates, the determinant of the ...

0

You could also define a LT as a transformation which transforms orthonormal basis into orthonormal basis (you have to exclude translations, since they don't belong to the Lorentz group).

2

Since a worldline along the time axis on Minkowski diagram is at rest, it is more intuitive to measure angles from that axis instead, as then 'slope' is (space)/(time), i.e., a velocity. Then we have the trigonometric relationship: $$\frac{v}{c} = \tanh\alpha$$ where Minkowski spacetime follows hyperbolic trigonometry because of the sign difference in the ...

2

Why do you stop your largest angle with ten 9s after the decimal point? If you added more of them, then you'd get a smaller bound for the velocity. And you keep adding 9s ad infinitum and you'll "eventually" reach $89.\bar{9}=90$. So eventually, you'll see that the velocity could be arbitrarily small. This just means that the worldline can be vertical... and ...

Top 50 recent answers are included