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5 votes

Complex coordinates $ds^2 = dzdz̄$ in 2d

It helps to notationally distinguish between skew-symmetric exterior products and ordinary tensor products: $$ dz \wedge d\bar z= (dx+idy)\wedge (dx-idy) = -2i dx\wedge dy $$ for the volume element, ...
mike stone's user avatar
  • 53.8k
3 votes

How do we interpret measurements of Mercury's position?

This is a good question. The 3D space part of a 3+1D flat Minkowski spacetime is a Euclidean space. We can make an identification when it is so. You are correct that the coördinates that we are using ...
naturallyInconsistent's user avatar
2 votes

Generating function condition not satisfied?

OP's condition (3) is not there/not necessary. Instead a type-2 generating function $F_2(q,P,t)$ satisfies the following $2n+1$ conditions: $$ p_j~=~\frac{\partial F_2}{\partial q^j},\qquad Q^j~=~\...
Qmechanic's user avatar
  • 204k
2 votes

Understanding differentials in an equation in general relativity

This can be made rigorous, but this is one of the instances in which the notation is chosen so that the calculations get intuitive. $ds^2$ is just a bookkeeping notation to express the metric tensor, ...
Níckolas Alves's user avatar
2 votes

Complex coordinates $ds^2 = dzdz̄$ in 2d

Consider $z=x+i y$ and $\bar z=x-iy$ ($x$ and $y$ are real). It is easy to show that: $$ dz\land d\bar z=(dx+i\,dy)\land(dx-i\,dy) =dx\land dx+i \, dy\land dx -i \, dx\land dy-dy\land dy. $$ However, ...
Quillo's user avatar
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1 vote

Does quantum entanglement arise from perpendicular time vectors?

First, what you are proposing as a “perpendicular” time axis is a new dimension, making 5 rather than the traditional 4. Considering how well general relativity does with four (and not five), that’s a ...
DrChinese's user avatar
  • 1,193
1 vote

Complex coordinates $ds^2 = dzdz̄$ in 2d

The other answers are perfectly fine, but I think they might be a bit too advanced for what you're looking for (if I could read your misunderstanding correctly). The difficulty you're having is with ...
Níckolas Alves's user avatar
1 vote

Schroedinger equation with value of wave function in polar coordinates?

Schrödinger equation written as a product of a real modulus $R$ and a unitary phase $U$ $$\Psi(t,x)= R(t,x) U(t,x)$$ with $ R>0, U^* U = 1$, such that for a scalar wave function $U=e^{i\theta}$ ...
Roland F's user avatar
1 vote

Whether nonlinear coordinate transformations are symmetries of flat spacetime

"The laws of physics must have the same form under transformations from one inertial frame to another inertial frame". … nonlinear transformations from Cartesian to curvilinear coordinate ...
Dale's user avatar
  • 102k
1 vote

Whether nonlinear coordinate transformations are symmetries of flat spacetime

Poincaire invariance in special relativity is a global symmetry that relates one physical state to another, different physical state. By Noether's theorem, the existence of this symmetry implies the ...
Andrew's user avatar
  • 49.6k
1 vote

Whether nonlinear coordinate transformations are symmetries of flat spacetime

My one cent. A rotating frame or accelerating frame or other non-inertial frame is not inertial frame of reference. On the other hand, merely a different choice of coordinates description (eg ...
Nikos M.'s user avatar
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