Science Forums

# A quantum picture of torsion

## Recommended Posts

Posted (edited)

Sorry I haven't been around, if you noticed, OceanBreeze fancied abusing his powers, spoke to me like ****, couldn't handle getting a bit back, including the fact he was doubly ashamed when citations where thrown his way when he said I couldn't. I have respect for the site, but not when it comes to him. Anyway, thanks to him, my month exile allowed me to write up a theory for torsion in some linearized gravitoelectromagnetic theory which includes a core aspect of torsion dynamics in spin and orbit coupling. My preliminary work interested Ruth Kastner (look her up) enough that I've been told not to drop my investigations.

Torsion in particle dynamics

The torsion has been proposed in literature to be related to the frequency of the spin

ω = -1/2 Ω = −(e/2mc)B

Where I gather the last expression under inspection from manipulation of dimensional analysis. It still remains though that torsion and spin while related, are still different dynamics. The gravitational field Γ is related to torsion as:

∇x Γ = - 1/2c² ⋅∂Ω/∂t

The magnetic field associated to spin and orbit is:

B = 1/2emc² ⋅ 1/r ⋅ ∂U/∂r ⋅ J
= (1/mc² ⋅ ∂U/∂t) J/2e

(where J is the total angular momentum)

The Josephson constant appears J/2e from the definitions and I identify torsion here as:

Ω = 1/2mc² ⋅ (∂U/∂t)

A gravielectro field would be related as

E = 1/e ⋅ ∂U/∂t = c²/e ⋅ ∂m/∂t

And from Sciama units I identify:

a/G = 1/r ⋅ ∂m/∂r = ω²r/G

When putting a particle or sphere into motion by spinning it, it experiences a Lorentz force If the magnetic field has origins in the gravitational field, it couples equally to the Gravimagnetic field and is associated with torsion

B x v = 1/2emc² ⋅ 1/r ⋅ ∂U/∂r ⋅ Jv
= (1/mc² ⋅ ∂U/∂t) Jv/2e

Since Jv ~ e we can state roughly that

B x v ~ (1/2mc² ⋅ ∂U/∂t)

It's approximate because in exact units, Jv is really twice the value of e. An exact equality would be

B x v = (e/2mc² ⋅ ∂U/∂t)

Some more results Bohr obtained two major objects of importance, the Bohr radius and the Bohr inverse mass. He derived the inverse mass from the known classical laws

1/m≡(4π²Be²)/h

is:

B = 1/2emc² ⋅ 1/r ⋅ ∂U/∂r ⋅ J

=1/me ⋅ (Φ/c²) ⋅ ∂v/∂t J

= 1/me ⋅ (a/G) J

= 1/me ⋅ ω²r/G ⋅ J

=1/me ⋅ m/r² ⋅ J

Where

1/G = Φ/c²

a/G = 1/r ⋅ ∂m/∂r = ω²r/G

1/e⋅ ∂U/∂r = c²/e⋅ ∂m/∂r

In CGS units. A gravielectro field now defined as

E = 1/2emc² ⋅ 1/r ⋅ ∂U/∂r ⋅ Jc = 1/2e ⋅ ∂U/∂r

Plugging in now the inverse Bohr mass we get

E = 4π²eB/2mc⋅ 1/r ⋅ ∂U/∂r ⋅ J/h

Here we can identify an important term arising as the magnetic dipole:

μ = eJ/2mc

So we can write

μ(S) = -g μ J/h

This now gives

E = -4π²μB⋅ 1/r ⋅ ∂U/∂r ⋅ J/h

Where

H = μB

Is an interaction energy.
The gravielectro potential is
From Sciama theory we have

E = - ∇φ - 1/c (∂v/∂r) ~ Φ/c² (∂v/∂r)

E = ∇x A = 0

The total electric field is

Φ = ∫ ρ/r ⋅ c²t²

And if density is uniform

Φ ~ -2πρc²t²

And by symmetry the vector potential is

A = 0

So the total gravielectro field is,

E = m/r² + 1/c² (Φ + φ)a

Rewriting the spin orbit equation in terms of the Sciama gravielectro field we get

B = 1/2emc² ⋅ 1/r ⋅ ∂U/∂r ⋅ J

The electric field encoded in this is

E = 1/e⋅ ∂U/∂r

B = 1/2mc² ⋅ 1/r ⋅ Φ/c² (∂v/∂r) ⋅ J
= -1/2mc² ⋅ 1/r ⋅ (∇φ - 1/c (∂v/∂r))⋅J

With a total Gravimagnetic field as

B = 1/2mc² ⋅ 1/r ⋅ ∂U/∂r J
= 1/2mc² ⋅ 1/r ⋅(m/r² + 1/c² [Φ + φ]a) ⋅ J

a =  Ω x v

So that we get after plugging it in

B = 1/2mc² ⋅ 1/r ⋅(m/r² + 1/c² [Φ + φ]Ω x v) ⋅ J

Need to knows!

We can define an acceleration as

a = ω x v

And

v = ω x r = ωr sinθ

Under this interpretation, the acceleration can be attributed to the centrifugal pseudo force

a = ω x (ω x r) = ω²r sinθ

And we recognise the orthogonality of ω⋅r =0 from the triple cross product rule

a x (b x c) = b(a⋅b) - c(a⋅b)

Mass to charge ratio has the same dimensions as angular momentum to magnetic moment,

L/μ = m/e

The angular momentum is

L = mωR^2

Derivation:

L = r × p
L = r × mv
L = mr × (ω × r)
L = mω(r · r) − mr(r · ω)
L = mr²ω − 0
L = mr²ω

Edited by Dubbelosix

• Replies 53
• Created

#### Popular Days

I changed the result from the first part of the investigation so that the units of torsion is satisfied in the second part. I was using two different ideas.

##### Share on other sites

Posted (edited)

In the modified units, we retain the usual standard units,

B x v ~ (1/2mc² ⋅ ∂U/∂t)

Which means we now identify more acutely a torsion with dimensions of inverse time, just like the second part of my investigation, so everything remains consistent. This means a second velocity cross would produce an acceleration.

Edited by Dubbelosix
##### Share on other sites

Posted (edited)

In previous derivations, when we take the triple cross product involving two terms of angular frequency with a radius, we find the result of orthogonality satisfying  (r · ω) = 0, and with that of the terms of what happens physically speaking when you take the velocity cross product of

v x (B x v)

Then

· v = 0

(iff) the velocity vectors are perpendicular to each other, so this is another little "need to know."

Edited by Dubbelosix
##### Share on other sites

This model to finish for now, does rely on one specific difference to some earlier works from different authors who had identified a torsion directly as Ω = B, but I can argue, I think better than those models, that the Gravimagnetic field, isn't directly related to torsion, since velocity must be included so that torsion arises from a spinning top in a Gravimagnetic field. It's more acceptable then in the case of treating the two as closely related topics instead of a direct equality.

##### Share on other sites

On 6/7/2021 at 5:38 PM, Dubbelosix said:

Where I gather the last expression under inspection from manipulation of dimensional analysis.

Yeah I read to the bottom earlier and I don't see anything more than two D equivalencies

##### Share on other sites

6 hours ago, JeffreysTubes8 said:

Yeah I read to the bottom earlier and I don't see anything more than two D equivalencies

I'm talking about the remodelling for

Since Jv ~ e we can state roughly that

B x v ~ (1/2mc² ⋅ ∂U/∂t)

##### Share on other sites

Josephson uses branes in his math but you haven't touched those and he uses 4D digital analysis. Nothing here is algorithmic in your op.

And furthermore, inflation can't be caused by linear spin because it's observation in the earlier than universe flung in every direction and, defiant to the inverse square law, it ignores conservation of momentum apparent in torsion and it moves FTL and you can't explain that with Newtonian bodies.. especially not with planes

##### Share on other sites

Posted (edited)

Exe claims require evidence.

You have failed to provide that extraordinary evidence.

Edited by JeffreysTubes8
##### Share on other sites

Posted (edited)

Why are you on about branes? Why would torsion speak about branes, you clearly have no idea what you're talking about anyway. Ridiculous jargon reminds me of another here who keeps making various new accounts in hope that his dialogue would go unnoticed, revealing a massive crank.

You're not qualified in any way either to say I've provides no evidence, I've done my bit, the onus is now on others to prove me wrong. Not with words. But with experiment and math.

Edited by Dubbelosix
##### Share on other sites

Don't bother replying by the way,you're even talking about inflation in a topic unrelated. I don't have time for people like you.

##### Share on other sites

Pot. Kettle black

##### Share on other sites

Posted (edited)
2 hours ago, Dubbelosix said:

the onus is now on others to prove me wrong. Not with words. But with experiment and math.

Wrong the burden of proof is on the one making claims such as linear spin being related to expansion 🤣😂in your other thread so you're clearly a crackpot

2 hours ago, Dubbelosix said:

Why would torsion speak about branes, you clearly have no idea what you're talking about anyway.

Wrong again, you can't have torsion without depth

Torsion-

• MATHEMATICS
the extent to which a curve departs from being planar.

A brane is like the plane in which the values youre inputting for spin go

Only you're excluding key geometric values for that

Edited by JeffreysTubes8
##### Share on other sites

Btw I do know what I'm talking about, while you scour the web for incomplete material to integrate using whatever tools you know, I started dimensional analysis from scratch for my tedious work which gives fresh physics definitions of these quantum interactions arising from pure arithmetic.

##### Share on other sites

No you really don't, it's a mish-mash of buzzwords with no relevance to what is being discussed in the Op.

##### Share on other sites

Again it's not a buzzword. You need it to express torsion, I literally gave you it's definition it's not a buzzword

##### Share on other sites

You really weren't. You started talking about branes and inflation. Take a walk. A long one and stop spamming my thread. You're a poster who used to do exactly the same thing before, just under a different guise. You have no knowledge worth participating here.

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

×   Pasted as rich text.   Paste as plain text instead

Only 75 emoji are allowed.

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.