Talk:Spin tensor

Rank 2 or 3?

In section "Noether currents", you indicate the angular momentum tensor once as tensor of rank 3, $M_{y}^{\alpha \mu \nu }$ or $M_{0}^{\alpha \mu \nu }$, and once as tensor of rank 2, $M_{0}^{\mu \nu }$ or $M^{\mu \nu }$. So, what is the rank now? --195.138.38.10 (talk) 17:01, 8 September 2011 (UTC)

The moment/angular momentum tensor $M^{\mu \nu }$, given as the integral itself, is a rank 2 tensor, and provides the Noether charges associated respectively with boosts (for $\mu \nu =01,02,03$) and spatial rotations (for $\mu \nu =23,31,12$). The integrand $M_{0}^{\alpha \mu \nu }$, which is the spin tensor itself - is technically a rank 3 tensor density (rather than a tensor) that provides the local current density for $M^{\mu \nu }$. The tensor associated with this is rank 3.

It's customary to use German script for densities, so that the integrand would actually be denoted ${\mathfrak {M}}_{0}^{\alpha \mu \nu }$, while the corresponding tensor would be $M_{0}^{\alpha \mu \nu }$. The components of both coincide only for Cartesian (and Minkowski) coordinates, otherwise they differ by the factor which appears in the integral measure; e.g. the measure in Cartesian coordinates $dx\wedge dy\wedge dz$ becomes $r^{2}sin\,\theta \,dr\wedge d\theta \wedge d\phi$ in spherical coordinates, so that ${\mathfrak {M}}_{0}$ = $r^{2}sin\,\theta \,M_{0}$, when expressed in spherical coordinates. The confusion between tensors and densities is prevalent throughout the Physics literature and is a contributing factor to other mistakes further down the line (e.g. the frequent citing in the literature of the wrong power of c for the coupling coefficient in the Einstein field equation).

Stress tensor being non symmetric?

Uhm....(scratching head) ...since when are Hermertian operators non symmetric?

What they call continuity here is a misnomer. When you summate the derivatives for the 4 momentum components in a given vector in some direction in the context of the stress energy tensor, they add to 0.

It's symmetric as when you swap indices, you get the negative result. The diagonal line is kinda like an identity matrix; the difference for vector component x in the x direction is 0. It's a derivative of itself.

On either side of that diagonal, you've got shear and the other stress. As 4 momentum moves forward you have shear as space time is moving in some direction paralell to an axis. Force over a distance that's distributed is shear.

The perpendicular vector in the stress energy tensor denotes the twisting or torsion of things. The 4 momentum is moving perpendicular to the other axis, and it's rotational. So it's moving in an opposing direction and traverse to the other.

That's not continuity. That's equilibrium to account for all energy in that system. It's called law of conservation of energy at an instant of observed time. That tensor is in fact symmetric. So that aspect is wrong. Period.

Continuity...try the unit circle. Radius r is 1 for all phase angles theta over the interval -pi to pi. In cartesian coordinates, the distance from the circle itself from its origin is 1-unless you've put a scalar to it, but that's why it's a scalar

Why are we integrating with respect to x 3 times over a 4 momentum vector? You've got x y and z; space time is your 4th variable. So let's integrate over x alone 3 times and screw all else.

Why do we have 2 stress tensors in the definition? If you've got 2 different systems, that's one thing. Or is someone confused by applying momentum as 3 separate parameters plus time there be it observed or relativistic, square that and summate it. Wtf? That makes no sense at all and doesn't tell me squat about spin!

If I've got a car on a car carrier moving forward along some mountain road, that defines the spin of the wheels of the car being carried? That's about what this entire thing looks like as it's relating things to which don't relate in a way that makes any physical sense at all! 24.74.113.253 (talk) 00:30, 27 March 2022 (UTC)

Re: rank 2 or 3

Are you getting spin tensor and spin from angular 4 momebtum mixed up? I believe what you're refering to is the spin of a particle with respect to its angular 4 momentum if im not mistaken.

The spin tensor itself describes a particle that has 0 angular momentum in 4 space; it simply spins about its own axis-unless I'm getting that confused with the 0 spin tensor. (Rubbing face)

I think what you're referring to is a vector between 2 points: xy, yz, zx which describes the 4 angle momentum of a particle in motion. If there's an additional intrinsic spin within the coil or some like condition, that's a different story. :-s 24.74.113.253 (talk) 00:42, 27 March 2022 (UTC)