Physics question about a particle

[Boof-head]

What am I?

 

In a vector coordinate system with respect to a ground-state with the x axis always horizontal from left to right (a right hand perspective of spin-momentum).

The one at the origin represents the L-frame. The C-frame represents the system for the spin vector. The x coordinate in this case is along the x vector so your right thumb orients spin.

 

In addition there are three internal angles (θ, φ, Ψ) that describe the orientation of vectors; θ is the angle made with the x direction; φ is the angle the spin axis makes with the xy plane 'into the distance'; this angle determines the amount of 'top' or 'bottom' spin and transverse spin. Ψ is the angle between the spin axis and the x spin axis.

 

Ψ is the case where the spin axis would be pointing forward or behind the yz plane, measured through the vertex in the C-frame. We can simplify the system by forcing Ψ to be always perpendicular to the x axis, so that it will be restricted to 90°.

 

Fg is the gauge force, and g a scalar, FD is a viscosity, FL is a linear force, and FC is all the forces that pertain to the object in motion. Think of FC as a remainder for additional interactions due to angular momentum interactions, vortex shedding, or distortions in the inner or outer surface of "particle p" that could be added in. We will neglect these forces initially and assume it isn't sufficiently distorted (i.e. is smooth) that chaotic effects can be ignored during its travel or "flight".

 

We now write the force equation:

 

F(tot) = Fg + FD + FL + FC

 

spin momentum is conserved as: [math] \frac {\omega_z} {\omega} \,= \theta.[/math]

 

P.S. hint(1): usually the gauge means the orientation wrt the L-frame; in this case the z axis points along the yz plane. The gauge is assumed to be toward the ground (state).

 

Am i: a proton, paintball, bouncing bomb, lawn bowl, Higgs boson, or all the above, a quasi-particle?

[Boof-head]

Imagine a quasiparticle is 'stuck' in a tube, a cylinder with smooth sides. These are 'frictional' or exert a drag on the particle q. You stand the tube on its end with q exactly 1/2 the distance from either open end of the tube.

 

You drop some mass into the tube so q is 'freed' and travels down the tube with a velocity, v.

 

This models a particle interaction, or "a frozen particle gains mass and momentum". You have created a universe with a particle 'interaction'.

[Erasmus00]

In a vector coordinate system with respect to a ground-state with the x axis always horizontal from left to right (a right hand perspective of spin-momentum).

 

This doesn't make sense- why does placing the x axis have any bearing on spin or momentum? What is spin-momentum? Angular-momentum?

 

The one at the origin represents the L-frame. The C-frame represents the system for the spin vector. The x coordinate in this case is along the x vector so your right thumb orients spin.

 

The what at the origin? Whats a C-frame? What is an L-frame?

 

In addition there are three internal angles (θ, φ, Ψ) that describe the orientation of vectors; θ is the angle made with the x direction; φ is the angle the spin axis makes with the xy plane 'into the distance'; this angle determines the amount of 'top' or 'bottom' spin and transverse spin. Ψ is the angle between the spin axis and the x spin axis.

 

What is "top" or "bottom" spin? Why do we need three angles in a 3-d coordinate system? Shouldn't it be 2?

 

Fg is the gauge force, and g a scalar, FD is a viscosity, FL is a linear force, and FC is all the forces that pertain to the object in motion. Think of FC as a remainder for additional interactions due to angular momentum interactions, vortex shedding, or distortions in the inner or outer surface of "particle p" that could be added in. We will neglect these forces initially and assume it isn't sufficiently distorted (i.e. is smooth) that chaotic effects can be ignored during its travel or "flight".

 

What is causing these forces? What is a gauge force? Why is g a scalar? What is a linear force? Why are we using forces at all?

 

 

spin momentum is conserved as: [math] \frac {\omega_z} {\omega} \,= \theta.[/math]

 

What does that even mean? Your equation seems non-sensical.

 

You seem to leave the reader with the impression you've strung a bunch of a poorly remembered or understood concepts into a conceptual mess.

[Boof-head]

"spin" isn't hard to visualise; if you have a coin and access to a tabletop, stand the coin on its edge and flick one side; the coin should spin on the tabletop: thus, spin.

 

You require a spin measurement; so that spin is conserved, given a z-spin the xy spin momentum is available, since you know there are 3 spin components.

 

L-frames, C-frames are well-known tools in inertial systems, velocity/momentum equations of groups of particles; it's a very common formalism.

Have you done any Physics?

 

There is sufficient information in the post to construct a Lagrangian ansatz, and derive general equations of transfer for spin-momentum and linear velocity. Forces are the formulation, to illustrate "what is a force"; in this case there are a total of 4; one is the gauge of the field, or g.

 

You may have heard of the "force of gravity", or perhaps the "magnetic force"..?

 

An L-frame - C-frame:

 

This diagram has two frames, S and S'. S is equivalent to "the L-frame", so the C-frame is S'. If the L-frame is a tabletop, the coin is in the C-frame spinning 'on' the L-frame.

[Erasmus00]

You require a spin measurement; so that spin is conserved, given a z-spin the xy spin momentum is available, since you know there are 3 spin components.

 

Again, what is "spin momentum"? Do you, perhaps, mean "angular momentum?"

 

L-frames, C-frames are well-known tools in inertial systems, velocity/momentum equations of groups of particles; it's a very common formalism.

Have you done any Physics?

 

I have done quite a bit of physics, and taught at university level. I have never encountered the phrase "L-frame and C-frame." The nearest I can figure, is you mean a "lab frame" and a "center of mass frame," which might have been abbreviated with L and C in a class, but its not a common convention. But the rest of your description appears to be gibberish.

 

There is sufficient information in the post to construct a Lagrangian ansatz, and derive general equations of transfer for spin-momentum and linear velocity. Forces are the formulation, to illustrate "what is a force"; in this case there are a total of 4; one is the gauge of the field, or g.

 

Do you know what a gauge is? Saying the "gauge of the field" is a force doesn't make any sense. And again, I'm also certain you mean "angular momentum" rather than "spin momentum."

[Boof-head]

"The C-frame of reference coincides with the center of mass of the system of particles. The L-frame of reference is fixed relative to the observer" - undergrad physics text (1980)

 

Spin in QM = angular momentum; a spinning coin or weight on a string (torsion pendulum) has angular momentum, because it's spinning. Have you not seen either or something?

 

Sure you teach physics... obviously.

 

Do you know what a gauge is? Saying the "gauge of the field" is a force doesn't make any sense. And again, I'm also certain you mean "angular momentum" rather than "spin momentum."

A gauge is a measurement; railway tracks have a gauge - a standard width. SWG is a gauge of steel wire diameter, etc.

 

I didn't say the gauge of the field is a force, Fg is the force. Do you mark student's work at all? Do you read it first before you do? I'm skeptical of your claims about your knowledge of physics, at this point. [sorry, I do imply it's a force - after the first version in the OP - please read this again - with "one is the gauge", which should say "one is the gauge force". Apologies for the word that was missing. But a physicist who understood "gauge" would know there was either a missing word, the one I have now added, or there was an inconsistency. Not assume it was a load of rubbish]

 

Perhaps, given the obvious perplexity, assume I'm dealing with mostly high-school, or grade-school level at this site? Or just not bother?

 

Bear in mind, this is for my own benefit - I am trying to understand certain things by seeing if I can explain them. Usually I encounter 'mods' and other objectors; this is a challenge since I then need to put aside assumptions, like thinking a physicist will know what the term "spin-momentum" means, or "L-frame".

 

That is, in my university, a physicist would understand "Fg is the gauge force, and g a scalar"

 

P.S. "spin-momentum" = "angular-momentum", in QM; why doesn't this equality exist in inertial spin-momentum, is rotation not momentum (angular)?

[Boof-head]

See how moderators with a sense of 'scientific importance', manage to import insults to one's intelligence?

you've strung a bunch of a poorly remembered or understood concepts into a conceptual mess.

...the rest of your description appears to be gibberish.

Which leads me to the question - since this moderator has decided something, does that mean everyone else reading this will do the same?

(am I wasting my time...)

[Boof-head]

So, you assume that angular momentum (spin) is conserved, and you can set z spin (angle) to 1.

So then you measure the spin along xy, with "one z unit" of spin as above.

 

You also assume that the particle is smooth (Newtonian) and FC can be ignored; there are no deformations in the particle itself, and it 'stays together' or has a constant diameter d.

So if d exists p exists.

 

So that leaves the scalar g force Fg, FL and FD. We know how to write a general Lagrangian to account for all three. The particle has a velocity v, because of three forces and conserves angular momentum as spin, so also has linear momentum as a distance s(d). There is a Cartesian R3 space with xy,z; xy in the projective plane, z is the azimuthal spin angle measure.

[Ben]

Though I freely confess to knowing very little physics, I am pretty sure that, in quantum mechanics, spin and angular momentum are different beasts entirely.

 

Angular momentum is easy; it simply assigns a value to a (usually) massive object at constant speed in an (open or closed) arc. So let's think about "spin". As far as I understand it, spin (in the QM sense) refers to symmetries under some sort of group operation. (If I am wrong, I am happy to be corrected).

 

So specifically, the symmetry described by the group [math]SO(3)[/math] is such that, for any point [math]p[/math] one will have that [math]p+2\pi =p[/math]. Any point particle that can be so described is said to have integer spin

 

Now it is not hard to show that the symmetry group [math]SU(2)[/math] is a double cover for [math]SO(3)[/math] which leads (indirectly) to the conclusion that this group group describes symmetries of the form [math]p+4\pi = p[/math]. Any point particle so described is accordingly assigned a 1/2 integer spin.

 

Easy to see this has nothing to do with angular momentum, right?

 

I confess I didn't understand the rest of your post(s), but that's probably just me...

[Boof-head]

Total angular momentum is all the spin in the system - the L-frame is fixed locally and the gauge is constant, but the particle might have zero spin at v = 0;

Or, it might be a magnet, or a fluid-filled elastic container, or the fluid might be elastic and the container rigid.

 

The particle experiences 4 forces; initially ignore FC the force contribution from local deformations in or on p at the C-vertex as it translates with velocity v along x.

 

Fg is scalar, FL is linear and FD is drag from the medium it travels through; if it can spin (sorry, rotate) in three directions then spin is spherical, it is independent of the geometry of p.

 

If p is not smooth and deforms then drag goes up; so Fg,FL are linear. FD depends on the medium and p's geometry. FC is the remaining force pertaining to p.

 

If p is filled with a fluid containing ferrite beads, then p is accelerated down a magnetic track, say, so it arrives at a 'launchpad' with velocity v, then is launched at an angle with a boosted v through an electric field transverse to x the direction of motion in the C-frame..?

 

Or a 'simpler' particle p: your thumb is the L-frame and a coin is p; describe Fg,Fl, then FD (FC i guess will be insignificant); are all rotations conserved?

[Ben]

Total angular momentum is all the spin in the system

I am pretty sure this is not true - see my earlier post

the L-frame is fixed locally and the gauge is constant, but the particle might have zero spin at v = 0

I'm not sure what an "L-frame" is, but spin, as I (perhaps wrongly) described it, has nothing - absolutely nothing - to do with velocity. Why would a particle with, say integer spin, have a different spin if it were to be at rest. Which also begs the question; at rest relative to what, exactly?

 

Fg is scalar

Pardon? I seem to recall that earlier (3 days ago, 3:32 pm GMT) you were at great pains to emphasize that Fg is a force. How can a force be scalar?

spin is spherical, it is independent of the geometry of p.

Spin is spherical? Now what does that mean? Nothing as far as my understanding goes. And I had assumed, obviously incorrectly, that p was a point particle (hint: no other objects posses this property). Thus p has no geometry, it is just a point

 

If p is not smooth

Careful there. "Smooth" has a rather precise meaning in math, and also, I suspect, in math as applied to physics. Try not to add your own confusion to ours by abusing terminology.

 

Though I doubt you know what you are talking about, I am happy to be proved wrong

[Boof-head]

g is the gauge in both frames; the particle is at rest initially and might have an inner spin momentum.

This is because it might have some kind of inner structure (imagine a fluid that's inside a spherical object).

Fg is the gauge force, g is scalar; FL is linear momentum, etc.

Spin is conserved; if L is a center of rotation so that the C-frame is 'spinning around L', and the particle has a surface which is/is not spinning, and the fluid inside the surface is/is not spinning, describe all forces and momentum; ignore FC, any chaotic behavior, and treat the system as having 3 initial components Fg,Fl,FD.

Assume v = 0 at initial time t = 0. The particle's spin axis has an outer product which is the surface of p., the spin component w is conserved along z, xy as above. There is a spin measurement available as a ratio of total spin in C. Any rotations around L are contributions to total angular momentum, which includes total particle spin angular momentum.

 

So, you have an idea I don't know what I'm talking about? I have an idea I know what you're talking about, and I have an idea I know what I'm talking about.

 

Therefore if talking() then if know() then if about() then if(what) then

if know(talking()) then if about(know(talking)) then {know(talking(about))}

if know(what) then if {know(talking(about))) then {know(what(about))}

if not know(what) then {}

 

If sphere() then if rotate() then if spin() then

if rotate(sphere) then {sphere(spin)}

if x then if y then if z then

if sphere(spin()) then {spin(x,y,z)}

if sphere(rotate(spin()) then {rotate(sphere)}

if not sphere() then {spin(x,y,z)}

if not spin then {spin(0,0,0)}

if not rotate then if not spin() then {}

if not sphere(rotate(),spin()) then {}

 

/lambda fall-through

 

if L(x,y,z) then if C(x,y,z) then

if equal(L,C) then {rotate(L(0,0,0)}

if spin(x,y,z) then {rotate(C(x,y,z)}

...

[Boof-head]

Quantum coin:

 

if table(xy,z) then if coin(xy,z) then

if table(coin(xy,z)) then {equal(table(xy),coin(xy))}

if rotate(coin(z),table()) then {rotate(coin(0,z))}

if rotate(coin(z),table(xy)) then {rotate(coin(xy,z)}

if rotate(coin(xy,table()) then {rotate(coin(xy,z))}

if spin(coin(z)) then {rotate(coin(xy,z)}

if rotate table(xy,0) then {rotate(coin(xy,z)}

if rotate table(0,z) then {rotate(coin(xy,z)}

if rotate() then {}

 

p.s. Note how there's no "mass", just (xy,z), a table (L) and a coin ©.

This coin has a strange quantum, since if you rotate the z axis, at coin(xy,z), then coin(0,z) implies xy is zero for quantum particle: "coin"; it has no "diameter", so what is wrong with this quantum, should it have a xy too, and how to account for it if it does?

 

hint(1): see what QM says about particle spin in quanta with no xy. }I don't think there can be any such thing, however quanta have spin(0,1) along x,y,z at all times.

[Boof-head]

Who let the cubes out?

 

Cube particle

1) rotate 1x(cube) or a 1-cube so it has a single upper vertex in graph G; the upper 3-vertex is z, and has 3 colors; the lower vertex z' has 3 colors.

 

2) embed 2-slice cube group in n-slice group at z; rotate outer slices to polynomial (color map) in space + time.

 

3) find nullspace wedge product in rotation group of polytope; measure curvature of color polynomial (time,space) -> bifurcated conic over cube = punctured nullspace.

 

4) derive spectral triple G(H), A(D), where G is cycle group of 2-slice polytope; A is commutative algebra associated to spin manifold M, D is operators on M(A); degree zero is distance measure in M(G). Triple is (A,H,D)

 

5) prove that embedded 2-slice in N-slice over N-cube is infinite-recursive and apex of cone is at infinity. Derive equations for (time,space) as color polynomial in time/nullspace. Map to {(3,2),(2,3)}-torus.

[Erasmus00]

Fg is the gauge force, g is scalar; FL is linear momentum, etc.

 

Linear momentum can't also be a force. Further, the sentence Fg is a gauge force, g is a scalar makes no sense. You particle has spin, so it is not a scalar field, and g cannot be a force carrying particle, because a gauge field needs to be Lorentz vector, not a scalar. Your phrase is nonsense.

 

Spin is conserved; if L is a center of rotation so that the C-frame is 'spinning around L',...

 

Spin won't be conserved if the center of mass is moving around the observer, only spin plus angular momentum will be. FURTHER, L and C frame is NOT a common notation. It is a shorthand that most textbooks do not use. Please use lab, and center of mass.

 

If sphere() then if rotate() then if spin() then

if rotate(sphere) then {sphere(spin)}

if x then if y then if z then

if sphere(spin()) then {spin(x,y,z)}

if sphere(rotate(spin()) then {rotate(sphere)}

if not sphere() then {spin(x,y,z)}

if not spin then {spin(0,0,0)}

if not rotate then if not spin() then {}

if not sphere(rotate(),spin()) then {}

 

/lambda fall-through

 

if L(x,y,z) then if C(x,y,z) then

if equal(L,C) then {rotate(L(0,0,0)}

if spin(x,y,z) then {rotate(C(x,y,z)}

 

You have not defined any of the functions your pseudocode is using.

[Boof-head]

FL = force [due to] linear momentum = momentum due to Fg; you're the sort of pedantic person who 'likes' to correct mistakes aren't you? Three missing words and you couldn't see them?

g can't be a particle? I guess not, but then, g is the gauge, the particle in the spin frame (of reference) sees g. The particle isn't g; where did you get that from?

 

Do you know what gravity is, what g is, the constant that mass sees in the inertial frame of this planet?

You have constructed a nonsensical proposition, from invalid predicates. Gravity on a table top looks exactly like gravity on the surface of any body with mass. There is a force called "gravity", G is a universal bound or limit, a constant, g is a scalar gauge on table top T, in the L-frame (momentum => L-frame)

L and C frame is NOT a common notation. It is a shorthand that most textbooks do not use.

So what? is that a really big problem for a conservative thinker?:confused:

"A frame" in physics generally implies two frames, a fixed frame, a moving frame. If there's a moving frame what's 'moving'?

 

You have not defined any of the functions your pseudocode is using.

You want me to define table(), coin(), rotate() ...??

Well, you see, I figured most 'ordinary' people who can read, would understand 'table', 'coin', 'rotate' and so on; are you seriously being dismissive because I didn't define 'rotate'?

 

Your post doesn't make much sense, => your post is rubbish aimed in a sniping, pedantic way, you are being dismissive because you assume I am making mistakes; it appears to be a result of a poor education or something (imho), or of someone who simply can't see past their own cloistered opinion of every idea they've ever seen. Maybe you should watch some TV, and figure out how to get the colors to conserve spin; while your on it, see if you can rotate the frame (walk past the TV).

 

I think you're asleep, mate.

 

p.s. do you know spectral theory, do you know what a spectral triple is, or a bundle map?

[Jay-qu]

Boof-head you are being rude and in general pretty much living up to your user name. Your pseudo code makes no sense to me either, you either have to explain it a bit better or define what these functions are that you are using - table() as a function could mean a lot of things..

 

I suggest you cool off your posting style or risk being infracted...again