Theory of Gyroscopic Fields.�An introduction
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Abstract
The Theory of Gyroscopic Fields is a new approach in Physics, where it is postulated the existence of a new kind of fields called “Gyroscopic Fields”, necessary in a model of Rotational Dynamics where it is not admited the existence of “action at a distance”.
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SUMMARY�Extending Fluid Mechanics for Solid-Fluid unification
ρ მv0 / მt + ρ [მv / მt + (v. grad) v] = – grad (p) + μ Δ v + ∑ Fexterior
div ρ ( v + v0 + vRT + vPT) + მρ / მt = 0 EXTENDED NAVIER-STOKES
მLR / მt = ∑ (r x FR)
curl (LR) = –2 ρ vRT – (1/vR2) მ(r x (r0 x ρ aR)) / მt
div (r x (r0 x ρ aR )) = – ρ vR2 ; div LR = 0 MAXWELL-EULER FOR ROTATION
მLP / მt + w x LP = ∑ (rP x FPT)
curl (LP) = –2 ρ vPT – (1/vP2) მ(r x (rP x ρ aP)) / მt
div (r x (rP x ρ aP)) = – ρ vP2 ; div LP = 0 MAXWELL-EULER FOR PRECESSION
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STATE OF THE ART: FLAWS (I)�Flaws in Rigid Body Dynamics
The Euler’s equations that
describe the rotational
motion of a rigid body are:
dL / dt + w x L = Σ r x F;
L = I w --> Kinetic Momentum
r x F --> Dynamic Moment
But, in this equation there is a problem. We observe that:
RIGID BODY DYNAMICS FOR ROTATION ADMITS “ACTION AT A DISTANCE”
The presence of “action at a distance” must be avoided in any rigorous theory of physics because it breaks the Principle of Causality 🡪 Something can happen without a cause
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STATE OF THE ART: FLAWS (II)�Analysis of “action at a distance”
WHEN APPLYING A DYNAMIC MOMENT M TO A BODY INITIALLY AT REST, THE ANGULAR VELOCITY w TAKES THE SAME VALUE AT ANY POINT OF THE BODY INSTANTANEOUSLY, NO MATTER HOW FAR IS THAT POINT FROM THE APPLICATION POINT OF M => EXISTENCE OF “ACTION AT A DISTANCE”
d (Iw) / dt + w x L = Σ r x F
L = Iw 🡪 Kinetic Momentum
r x F 🡪 Torque (pair of forces)
It is posible to avoid “action at a distance” if a wave equation for the propagation of angular momentum can be obtained from the model
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STATE OF THE ART: FLAWS (III)�Flaws in Classical Mechanics
According to the First Newton’ Law:
“An object at rest remains at rest, or if in motion, remains in motion at a constant velocity unless acted on by a net external force.” …But:
THERE IS A COUNTEREXAMPLE!!!
here, the net external force = 0 🡪 weight force + normal reaction = 0 whereas the translational velocity is not constant (IS NOT a straight line)
Translational velocity observed 🡪 Is a Circular trajectory with angular velocity of precession Ω
In the literature, this circular trajectory is due to the interaction between the “rolling” tip of the spinning top and the floor, but there is not a mathematical model that justifies a circular trajectory with angular velocity = angular velocity of precession Ω
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STATE OF THE ART: FLAWS (IV)�Flaws in Classical Mechanics
According to the First Newton’ Law:
“An object at rest remains at rest, or if in motion, remains in motion at a constant velocity unless acted on by a net external force.” …But:
A boomerang, thrown with intrinsic rotation, also presents a similar phenomenon: The translational velocity observed is a circular trajectory with angular velocity = angular velocity of precession Ω: another counterexample
And here, there is not a “rolling” interaction between the boomerang and the enviroment (air), but only an aerodynamic interaction that generates the lift of the object
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OBJECTIVES OF THE PRESENT APPROACH (I)
A Theory which ADMITS
“action at a distance” is about MAGIC:
A Theory which AVOIDS
“action at a distance” is about SCIENCE:
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OBJECTIVES OF THE PRESENT APPROACH (II)
Any kind of matter is only a system of particles in motion from the dynamical point of view, independently of its state of aggregation
(electrons, protons and neutrons 🡪 atoms 🡪 molecules)
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RIGID BODY DYNAMICS in Classical Mechanics
RIGID BODY DYNAMICS separate the analysis of the movement in two independent problems: the Translational problem and the Rotational problem
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RIGID BODY DYNAMICS� 1. Traslation
Translation consists in the analysis of
the Rectilinear Translation of the Centre
of Mass (G) with Angular Momentum = 0
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RIGID BODY DYNAMICS: Newton’s equations for the translation of the Rigid Body
Equations of translation of a rigid body with centre of mass G:
Newton’s second law: m aG = d mvG(t) / dt = Σ F
aG --> Acceleration of the centre of mass
vG --> Velocity of the centre of mass
F --> Force applied at the rigid body
m --> Mass of the rigid body
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RIGID BODY DYNAMICS� 2. Rotation about an axis
The Rotational problem consists in the
analysis of the gyration of the rigid
body thru an axis with instantaneous
Centre of Rotation (G) and Angular
Momentum not null
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RIGID BODY DYNAMICS: Euler‘s equations for the rotation of the Rigid Body
Equations of rotation of a rigid body about an axis thru point O:
dL/dt + w x L = Σ M0 🡪
M0 = r0 x F --> Dynamic moment = momentum of a force
w --> Angular velocity of the rigid body
L = Io w --> Kinetic momentum of the rigid body
Io --> Moment of inertia of the rigid body
These equations don’t ensure the existence of a wave equation!!!
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Maxwell‘s equations
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Fluid Mechanics (I)
Navier-Stokes (NS) equations:
THERE IS A UNIQUE VELOCITY FIELD V
We claim that an extension of the equations is necessary if we want a separated Rotational problem as in Rigid Body Dynamics
ρ 🡪 Mass density of the fluid; V 🡪 Velocity field => FLUID: MASS CURRENT = ρV
If we add a new term consisting in a vector Field
ρ მv0 / მt + ρ [მv / მt + (v. grad) v] = – grad (p) + μ Δ v
🡪 We need 3 more equations 🡪 The millenium problem must be revisited
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Fluid Mechanics (II)
for any point of the space-time (x, t): Lab reference frame at rest
🡪From this equation, it is possible to calculate the trajectory
of the material point x = x (t) given the initial values v (t=0), x (t=0)
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Unification of Fluid and Solid Dynamics (I)
has to be a Field Theory for
Continuum Mechanics in the
Eulerian description:
🡪Rigid Body Dynamics must be rewritten for this framework
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Unification of Fluid and Solid Dynamics (II)
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Solid as a continuous system of particles (I)
Classical Mechanics Unification of fluid and solid dynamics
The main difference respect to fluids: solids produce work under shear stresses 🡪 viscosity μ = 0, for solids
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Solid as a continuous system of particles (II)
SOLID LIQUID GAS
THE EQUATIONS OF DYNAMICS MUST BE COMMON FOR ALL THEM
…And the equations for a particular aggregation state is obtained when neglecting certain terms
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Extending Navier-Stokes eqs. to include solids
ρ მv0 / მt + ρ [მv / მt + (v. grad) v] = – grad (p) + μ Δ v + ∑ Fexterior
div ρ[v + v0 + (w x r) + (Ω x rP)] + მρ / მt = 0
v0 --> Translational velocity field; v --> Fluid velocity field due to molecular interaction
w x r --> Velocity field of intrinsic rotation; Ω x rP --> Velocity field of precession;
For an ideal Rigid Body: viscosity μ = 0,
but some solids present viscoelastic behaviour (μ > 0)
Extended Navier-Stokes equations:
]
ρ მv0 / მt + ρ [მv / მt + (v. grad) v] = – grad (p) + μ Δ v + ∑ Fexterior
div ρ [vT + v0 + (w x r) + (Ω x rP)] + მρ / მt = 0
Position Vector r vs Position Field – r�For (w x r):
In Classical Mechanics:
r 🡪 Position vector of a system particle P from the
instantaneous centre of rotation O
In Theory of Gyroscopic Fields:
– r 🡪 Position field of the instantaneous
centre of rotation O from a fluid particle P
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Recovering Newton’s equations of motion
For rigid bodies: v = 0 🡺 ρ მv0 / მt = ∑ Fext.
Integrating over the space occupied by
the body (Control Volume):
∫CV [ρ მv0(x, t) / მt dV] = m მv0(t) / მt = ∑ Fext.
since the mass of the rigid body is: m = ∫CV ρ dV; CV 🡪 CONTROL VOLUME
🡪 Thus, we recover Newton’s second law: m a0(t) = ∑ Fext.
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Recovering Navier-Stokes equations for fluids
For fluids:
ρ [მv / მt + (v. grad) v] = – grad (p) + μ Δ v + ρ g
div (ρv) + მρ / მt = 0;
where the angular velocities (w x r) and (Ω x r) are supposed to be contained in the fluidity field v
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Differences between NS and its extension (I)�DECOMPOSITION OF VELOCITY FIELDS
generated ONLY by the translation
of the fluid container.
curl v0 = 0 🡪 is irrotational.
v0 is also called the homogenous velocity:
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Differences between NS and its extension (II)�EXAMPLES
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Differences between NS and its extension (III)�Rotational velocities
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Analogy with Electromagnetic Fields
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The wave equation for Electromagnetism
For J2 = 0, the first equation can be written as: curl E = – μ მH / მt;
and taking the curl operator, we get;
curl (curl E) = – μ curl (მH / მt) = – μ მ(curl H) / მt = – μ მ( J1 + ε მE / მt) / მt
Knowing that J1 = σE (Ohm's law), a wave equation is reached that describes the propagation of field E:
ΔE – (1/c2) მ2E / მt2 – σε მE / მt = 0;
since με = 1/c2 and curl (curl E) = grad (div E) – ΔE
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Gyroscopic Fields�Decomposition of the Rotational velocity field
In general, the instantaneous centre of Intrinsic Rotation O does not concides with the instantaneous centre of Precession O’:
🡪Then, it is necessary to decompose the Rotational field in two independent fields: vR = w x r and vP = Ω x r, where:
with w containing two angular components: ψ and θ
with Ω containing two angular components: φ and θ’
NOTE: The theory of Rigid Body Dynamics only defines a unique angular velocity w, associated to the Euler angles: (ψ, φ, θ)
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Gyroscopic Fields: �Intrinsic Rotation
w 🡪 Angular velocity of intrinsic rotation:
It has 2 components corresponding to Euler angles
Euler Angles:
Ψ 🡪 Intrinsic Rotation
Θ 🡪 Nutation
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Concept of Intinsic Rotation
INTRINSIC ROTATION is the motion of rotation of an object thru an axis, with angular velocity w:
|w| = dΨ / dt
where Ψ is the Euler angle of rotation
It presents the property of Rotational Inertia: the object keeps rotating at the same angular velocity even after a new non-coaxial torque is applied
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Gyroscopic Fields for Intinsic Rotation (I)�Concepts of the analogy:
Gyroscopic-rotational field: Electric Field E 🡪 GR = (–1/2) r x (r0 x ρ aR)
where the INERTIAL ACCELERATION aR = w x (w x r) + (მw/მt) x r + 2 w x v0
Dynamic-moment “current”: Magnetic current J2 🡪 J2R = MR
where the DYNAMIC MOMENT MR = Σ r x FR-exterior
Kinetic-momentum field: Magnetic Induction B 🡪 KR = LR = r x (ρ vR)
where the VELOCITY OF ROTATION vR = w x r; w --> Angular velocity of rotation
r --> instantaneous centre of rotation
Mass “current”: Electric current J1 🡪 J1R = 2 ρ vRT ; r0 --> instantaneous centre of mass
vRT --> Total velocity of rotation
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Gyroscopic Fields for Intinsic Rotation (II)
The equations of dynamics for intrinsic rotation are:
where
GR --> Gyroscopic-rotation field; J2R --> Dynamic Moment “current”
KR --> Kinetic moment field; J1R --> Mass “current”
ρ --> Mass density; c’ --> Gyro-kinetic rotational velocity
vR --> module of velocity of intrinsic rotation;
These equations resemble the Maxwell’s equations
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div GR = – ρ vR2 (1) curl GR = J2R + (vR2/c’2) მKR / მt (2)
div KR = 0 (3) curl KR = J1R – (4/vR2) მGR / მt (4)
Gyroscopic Fields for Intinsic Rotation (III)
Expanding terms:
curl ((– ½) r x (r0 x ρ aR)) = r x FR + (vR2/c’2) მ(LR) / მt
curl (LR) = – 2 ρ vRT – [4/vR2] მ((– ½) r x (r0 x ρ aR)) / მt
where:
aR = w x (w x r) + (მw/მt) x r + 2 w x v 🡪 Inertial Acceleration
LR = r x (ρ vR) 🡪 Kinetic Momentum
r x FR 🡪 Dynamic Moment;
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Gyroscopic Fields for Intinsic Rotation (IV)
Expanding the Gyroscopic-rotation field term curl (GR) using curl (r x A) = –2A, we get:
curl (GR) = curl ((–1/2)r x (r0 x ρ aR)) = r0 x ρ aR = r0 x ρ (მvR/მt + w x vR)
= მ(r0 x ρ vR)/მt – (მr0 /მt x ρ vR) + r0 x ρ (w x vR);
მr0 /მt = v0 🡪 (მr0 /მt x ρ vR) = (v0 x ρ vR); 🡪 If v0 = 0 🡪 Substituting in equation (2), we obtain:
[1– (vR2/c’2)] მLR / მt = Σ (r x FR) – r0 x ρ (w x vR)
The last term can be expanded using the vector expression a x (b x c) = (a.c) b – (a.b) :
r0 x ρ (w x vR) = (r0 . vR) w – (r0 . w) vR
Then, if (r0 . w) = 0 and ignoring the factor [1– (vR2/c’2)], the resultant equation is:
მLR / მt = Σ (r x FR) – (r0 . vR) w
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Analysis of WAKE TURBULENCE
მLR / მt = Σ (r x FR) – r0 x ρ (w x vR)
Expanding the last term:
მLP / მt = Σ (r x FR) – (r0 . (w x r )) w
Turbulence’s eddies at the extrados of an airplane wing
Here, we postulate that the term – (r0 . (w x r )) w corresponds to the field of eddies that appear in “turbulence at “large scale separation” in a airplane wing
where r0 . (w x r ) = 0 in the laminar regime and r0 . (w x r ) > 0 in the turbulent regime
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Gyroscopic Fields for Intinsic Rotation (V)
Ignoring the last term, we have obtained an equation very similar to the Euler equation (dL/dt = Σ r x F), but only for the intrinsic rotational motion:
[1– (vR2/c’2)] მLR / მt = Σ (r x FR)
where the subindex R means that the corresponding field only applies for the intrinsic rotational motion
We observe a relativistic correction [1– (vR2/c’2)] that can be ingnored if we asume that vR << c’, leaving :
მLR / მt = Σ (r x FR);
🡪 Conservation of Intrinsic Rotation’s angular momentum:
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Gyroscopic Fields for Intinsic Rotation (VI)
From equation (4) we obtain a kinematic description of the velocity field:
curl (LR) = curl (ρ r x (w x r)) = – 2 ρ w x r,
If we define J1R = – 2 ρ vRT, then,
through substitution in (4), we obtain:
– 2 ρ w x r = – 2 ρ vRT – (4/vR2) მGR / მt
If we ignore the transient term მGR / მt, we recover the kinematic definition of the rotational velocity vR described in classical dynamics:
vR = w x r
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Gyroscopic Fields for Intinsic Rotation (VII)
Conclusion
In the permanent regime, we have recovered the classical description of Classical Mechanics:
მLR / მt = Σ r x FR ;
vR = w x r
but only for Intinsic Rotation
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Wave equation for Gyroscopic Rotation Fields
Taking the curl operator to (2) and using (4), after some algebra:
Δ GR – (1/c’2) მ2GR / მt2 = FRext – (vR2/c’2) მ(2ρvTR) / მt;
here we cannot define something analogous to Ohm's law (E = σ J) that would link the field GR with the velocity field vTR, however the following can be fulfilled:
FRext – (vR2/c’2) მ(2ρvTR) / მt = (2/r2) GR + f(r, t),
giving the following Helmholtz wave equation:
Δ GR – (1/c’2) მ2GR / მt2 – (2/r2) GR = f(r, t)
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Applications of the Wave equation (I)
If f(r, t) = 0 🡪 Δ GR – (1/c’2) მ2GR / მt2 – (2/r2) GR = 0
If Δ GR << – (1/c’2) მ2GR / მt2 – (2/r2) GR 🡪 Then, Δ GR can be ignored:
DRAGONEHEAD CYCLONE 🡪 (1/c’2) მ2GR / მt2 = – (2/r2) GR
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Applications of the Wave equation (II)
Mathematical model of the DRAGONEHEAD CYCLONE
If GR = (– 1/2) w2r2r 🡪 d2r / dt2 = – μ r / r4,
μ = 2 c’2 r2; h --> Specific Angular Momentum
Solution 🡪 Cotes’s Spiral, for the case μ < h2 🡪 r = BA sec(kθ + ϵ) + C,
k2 = 1 − μ / h2
Cotes’s Spiral 🡪
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Gyroscopic Fields�Precession
Ω 🡪 Angular velocity of precession:
It has 2 components corresponding to Euler angles
Euler Angles:
φ 🡪 Precession
Θ’ 🡪 Nutation
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Coupling of Translation and Precession
AN AXIOM OF THE PRESENT THEORY:
There is a coupling between the velocity of translation v0 that appears in the extended Navier-Stokes equation:
ρ მv0 / მt + ρ [მv / მt + (v. grad) v] = – grad (p) + μ Δ v + ∑ Fexterior
and the classical definition of a rotational velocity Ω x rP. Then, the new definition for the velocity of precession is:
vP = v0 + Ω x rP ; Ω --> Angular velocity of precession
rP --> instantaneous centre of precession
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Proof of the coupling from the flight of the boomerang
Coupling of translational and precession velocity 🡪 vPT = v0 + Ω x rP
where vPT is the total velocity of precession.
This coupling can be proven experimentally from the flight of the boomerang:
A boomerang is an example of gyroscopic precession coupled to translation
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Proof from the dancing spinning top
Coupling of translational and precession velocity 🡪 vPT = v0 + Ω x rP
where vPT is the total velocity of precession.
This coupling can be seen experimentally from the dancing spinning top:
The tip of the spinning top performs a circular translational trajectory with angular velocity equal to the angular velocity of precession Ω
A dancing spinning top is an example of gyroscopic precession coupled to translation
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Gyroscopic Fields for Precession (I)�Concepts of the analogy with EM:
Gyroscopic-precession field: Electric Field E 🡪 GP = (–1/2) rP x [(r x ρ aP) + (w x LP)]
where the INERTIAL ACCELERATION aP = მv0 / მt + Ω x (Ω x rP) + (მΩ / მt) x rP + 2 Ω x v0
Dynamic-moment “current” of precession : Magnetic current J2 🡪 J2P = MP
where the DYNAMIC MOMENT MP = Σ (rP x FP); FP --> External Force, non-coaxial to FR
Kinetic-moment field of precession: Magnetic Induction B 🡪 KP = – LP = – rP x (ρ vP)
where the VELOCITY OF PRECESSION vP = v0 + Ω x rP;
Mass “current” : Electric current J1 🡪 J1P = 2 ρ vPT; vPT --> Total velocity of precession
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Gyroscopic Fields for Precession (II)
The equations of dynamics for precession are:
where
GP --> Gyroscopic-precession field; J2P --> Dynamic Moment “current”
KP --> Kinetic moment field; J1P --> Mass “current”
ρ --> Mass density;
vP --> velocity of precession; cP’--> Gyro-kinetic precession-velocity
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div GP = – ρ vP2 (5) curl GP = J2P + (vP2/cP’2) მKP / მt (6)
div KP = 0 (7) curl KP = J1P – (4/vP2) მGP / მt (8)
Gyroscopic Fields for Precession (III)
Simplifying:
curl ((–½) rP x (r x ρ aP)) = rP x FP + (vP2/c’P2) მ(LP) / მt
curl (LP) = – 2 ρ vPT – [4/vP2] მ((–½) rP x (r x ρ aP)) / მt
where:
aP = მv0/მt + Ω x (Ω x rP) + (მΩ/მt) x rP + 2 Ω x v0 🡪 Inertial Acceleration:
LP = rP x (ρ vP) 🡪 Kinetic Momentum
rP x FP 🡪 Dynamic Moment;
w x LP 🡪 Gyroscopic couple
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Gyroscopic Fields for Precession (IV)�EXPLAINING TURBULENCE
Expanding curl (GP) using: curl (r x A) = –2A, we get:
curl (GP) = curl ((–1/2)rP x (r x ρ aP)) = r x ρ aP = r x ρ (მvP/მt + Ω x vP)
= მ(rP x ρ vP)/მt – (მrP/მt x ρ vP) + rP x ρ (Ω x vP).
მrP/მt = vP🡪 (მrP/მt x ρ vP) = 0; Substituting in equation (3), we obtain:
[1– (vR2/c’2)] [მLP / მt + w x LP] = Σ (rP x FP-ext) – rP x ρ (Ω x vP)
The last term can be expanded using the vector expression a x (b x c) = (a.c) b – (a.b) :
rP x ρ (Ω x vP) = (r . vP) Ω – (r . Ω) vP ; but vP = v0 + Ω x r which gives r . vP = r . v0, and r . Ω = 0.
Then, the resultant equation is:
[1– (vP2/cP’2)] [მLP / მt + w x LP] = Σ rP x FP – (r . v0) Ω 🡪 WAKE TURBULENCE
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Analysis of WAKE TURBULENCE
[1– (vP2/cP’2)] [მLP / მt + w x LP] = Σ (rP x FP) – (r . v0) Ω
For fluids, precession corresponds to the transversal
flow which is perpendicular to the usual velocity flow.
Simplifying: მLP / მt + w x LP = Σ (rP x FP) – (r . v0) Ω
Turbulence’s eddies at the extrados of an airplane wing:
Here, we postulate that the term – (r . v0) Ω corresponds to the field of eddies that appear in turbulence of “Vortex Shedding” type in a airplane wing
where r . v0 = 0 in the laminar regime and r . v0 > 0 in the turbulent regime
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Gyroscopic Fields for Precession (V)
Then, we have obtained an equation very similar to the Euler’s equations:
(dL/dt + w x L = Σ r x F), but only for the precession’s motion:
[1– (vP2/cP’2)] [მLP / მt + w x LP] = rP x (Σ FP – ρ (Ω x vP))
where the subindex P means that the corresponding field only applies for the motion of precession
We observe a relativistic correction: [1– (vP2/c’2)] that can be ignored if vP << cP’, leaving:
where FPint. = – ρ (Ω x vP)
of Precession;
🡪მLP / მt + w x LP = Σ rP x FPext.; for solids (FPint. = 0)
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Gyroscopic Fields for Precession (VI)
From equation (8) we obtain a kinematic description of the velocity field:
curl (LP) = curl (rP x ρ (v0 + Ω x rP)) = – 2 ρ (v0 + Ω x rP) = – 2 ρ vP,
If we define J1P = –2 ρ vPT, then, through substitution in (8), we obtain:
–2 ρ (v0 + Ω x rP) = –2 ρ vPT – (4/vP2) მGP / მt
If we ignore the transient term მGP / მt, we get the kinematic definition of the velocity of precession vP as the coupling of the translational velocity v0 and Ω x rP:
vP = v0 + Ω x rP 🡨 Example: Dancing Spinning top
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Gyroscopic Fields for Precession (VII)
Conclusion
In the permanent regime, we have recovered the classical description of Classical Mechanics:
მLP / მt + w x LP = Σ rP x FP; FP = FP-ext + FP-int
but only for Precession, and we have a new definition of the velocity of precession as:
🡨 Dancing
vP = v0 + Ω x rP Spinning top
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Gyroscopic Fields for Intrinsic Rotation + Precession
We have seen that we have recovered:
მLR / მt = Σ r x FR; Conservation law for Intrinsic Rotation:
and:
მLP / მt = Σ rP x FP; Conservation law for Precession:
The sum gives: მLR / მt + მLP / მt + w x LP = Σ r x FR + Σ rP x FP
Then🡪 მL / მt + w x LP = Σ r x FR + Σ rP x FP
🡪 Conservation of Total Angular Momentum
Where L = LR + LP🡪 TOTAL ANGULAR MOMENTUM
But: Σ r x FR + Σ rP x FP is not Σ r x FTOTAL
🡪 Conservation of Total Angular Momentum is different than in Classical Mechanics
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Wave equation for Precession Fields
Taking the curl operator to (6) and using (8), after some algebra:
Δ GP – (1/c’P2) მ2GP / მt2 = – 2FP – (vP2/c’P2) მ(ρ vTP) / მt forced vibrations
of the square plate
here we cannot define something analogous to Ohm's law (E = σ J) that would link the field GP with the velocity field vTP, however the following can be fulfilled:
– 2FP – (vP2/c’P2) მ(ρ vTP) / მt = (2/rP2) GP + f(r, t),
giving the following Helmholtz wave equation: 🡪Solution for seismic waves of Helmoltz equation
Δ GP – (1/c’P2) მ2GP / მt2 – (2/rP2) GP = f(r, t)
Applications of the Wave equation
Δ GP – (1/c’P2) მ2GP / მt2 – (2/rP2) GP = f(r, t)
If we ignore the third term when (1/c’P2) მ2GP / მt2 >> (2/rP2) GP, then:
Δ GP – (1/c’P2) მ2GP / მt2 = f(r, t);
If f(r, t) = 0 🡪 Δ GP – (1/c’P2) მ2GP / მt2 = 0
When a secondary flow is generated.
For example: The “tea leaf paradox”:
NOTE: The equivalent of the precession movement for rigid bodies is called ‘Secondary flow‘ or ‘ transversal flow’ for fluids
DYNAMIC EQUILIBRIUM FOR FLUIDS (I)
No One Can Explain Why Planes Stay in the Air
Scientific American - FEBRUARY 1, 2020.
https://www.scientificamerican.com/video/no-one-can-explain-why-planes-stay-in-the-air/
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DYNAMIC EQUILIBRIUM FOR FLUIDS (II)�Two theories have been proposed to explain what keeps a plane in the air:
the area of higher speed and lower pressure
on the top of the wing:
p + 1/2 v2 + ρgh = Constant
________________________
of action and reaction 🡪 Explains lift
as an upward push on the wing from
the moving air bellow:
ACTION = REACTION
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DYNAMIC EQUILIBRIUM FOR FLUIDS (III)�Bernoulli’s theorem flaws
It is supposed that if two particles
are initially close each other, they
will reach the trailing end of the wing
at the same time:
But, …
this is not true;
the air across the
top moves even
faster than its
“paired” particle:
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DYNAMIC EQUILIBRIUM FOR FLUIDS (IV)�Bernoulli’s theorem flaws
Doesn’t explain why planes can
fly inverted or with flat wings:
Doesn’t fully address lower-pressure zone above wing:
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DYNAMIC EQUILIBRIUM FOR FLUIDS (V)�Newton’s third law flaws
Air below wing is pushed down, resulting in
an equal and opposite force: lift
But, …
doesn’t fully address
lower-pressure
zone above wing:
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DYNAMIC EQUILIBRIUM FOR FLUIDS (VI)�New ideas about lift (according to Scientific American)
Co-dependency of lift’s FOUR elements:
1.- Downward turning air
2.- Speeding up of air flow
3.- Low pressure
4.- High pressure
But,…
this doesn’t explain
how the reduced pressure
up top got there initially
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DYNAMIC EQUILIBRIUM FOR FLUIDS (VII)�New ideas about lift (according to Scientific American)
How low pressure forms above the wing:
The air above the wing momentarily
flows straight back (A), forming a void
or vaccum. This will then strongly pull
the air back down (B), filling in and thus
eliminating most -but not all- of the vaccum.
Just enough vaccuum remains to pull the air
into the curved path that follows the wing (C):
(Known as the “Coandă effect” in the literature)
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DYNAMIC EQUILIBRIUM FOR FLUIDS (VIII)�“Coandă effect”
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DYNAMIC EQUILIBRIUM FOR FLUIDS (IX)�New ideas about lift, based on Gyroscopic Fields
According to the Theory of Gyroscopic Fields, the equations of dynamics for fluid flows can be written as:
div ((– ½) r x (r0 x ρ aR)) = – ρ vR2 (1)
div (LR) = 0 (3)
curl ((– ½) r x (r0 x ρ aR)) = Σ (r x FR) + (vR2/c’2) მ(LR) / მt (2)
curl (LR) = – 2 ρ vRT – (4/vR2) მ((– ½) r x (r0 x ρ aR)) / მt (4)
where aR is the inertial acceleration:
aR = მvR/მt + w x (w x r) + (მw/მt) x r + 2 w x v
We have ignored the term (vR2/c’2) მLR / მt for being small (vR << c’)
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DYNAMIC EQUILIBRIUM FOR FLUIDS (X)�New ideas about lift, based on Gyroscopic Fields
The DYNAMIC EQUILIBRIUM introduced by the Theory of Gyroscopic Fields, when the unique force is the gradient of pressure + viscosity force: ΣFR = (grad (p) + μ Δ v), is given by the following equations:
curl ((– ½) r x (r0 x ρ aR)) = r x (grad (p) + μ Δ v) (1) 🡪 (2)
curl (LR) = – 2 ρ vRT – (4/vR2) მ((– ½) r x (r0 x ρ aR)) / მt (2) 🡪 (3)
div ρ (v + v0 + vRT + vPT) + მρ / მt = 0 (3) 🡪 (4)
ρ მv0 / მt + ρ [მv / მt + (v. grad) v] = – grad (p) + μ Δ v + ∑ Fexterior (4) 🡪 (5)
Then, the dynamic equilibrium can be described as:
(1) A dynamic moment generated by a gradient of pressure r x grad (p) 🡪
(2) INDUCES 🡪 A rotation of the field of inertial accelerations ρ aR of the flow 🡪 A variation of LR
(3) GENERATES 🡪 A rotation of the field of kinetic momentum LR of the flow
(4) INDUCES 🡪 A new profile of velocities of rotation vRT 🡪 A new profile of flow velocities v + v0
(5) GENERATES 🡪 A new pressure’s distribution p 🡪 A new grad (p) 🡪 (2) …
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DYNAMIC EQUILIBRIUM FOR FLUIDS (XI)�New ideas about lift, based on Gyroscopic Fields
Representation of
the DYNAMIC
EQUILIBRIUM
introduced by the
Theory of
Gyroscopic Fields:
(explains the
“Coandă effect”)
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THEORY OF GYROSCOPIC FIELDS FOR SOLIDS�Application to the PRECESSION of the spinning top
According to the Theory of Gyroscopic Fields, the equations of dynamics of PRECESSION for solid objects can be written as:
div ((– ½) r x ( rP x ρ aP)) = – ρ vP2 (1)
div LP = 0 (3)
curl ((– ½) r x ( rP x ρ aP)) = Σ rP x FP + (vP2/c’2) მLP / მt (2)
curl LP = – 2 ρ vPT – [4/vP2] მ((– ½) r x ( rP x ρ aP)) / მt (4)
Where aP is the inertial acceleration:
aP = მv0/მt + Ω x (Ω x rP) + (მΩ/მt) x rP + 2Ω x v0 = a0 + Ω2 rP + (მΩ /მt) x rP + 2Ω x v0
and LP is the kinetic momentum:
LP = rP x ρ vP
It can be proven that: curl ((– ½) r x (rP x ρ aP)) = მLP/მt
DYNAMIC EQUILIBRIUM FOR SOLIDS (I)�New ideas introduced by Gyroscopic Fields
The DYNAMIC EQUILIBRIUM introduced by the Theory of Gyroscopic Fields, when
initially the torque is Σ (r x F) = rP x FP, is given by the following equations:
მLP / მt + ρrP2 w x Ω = rP x FP (1) 🡪 (2)
curl LP = – 2 ρ vPT – [4/vP2] მ((– ½) rP x ( r x ρ aP)) / მt; (2) 🡪 (3) 🡪 (4) 🡪 (5)
Where we have ignored the term (vP2/c’2) მLP / მt, for being small (vP << c’);
Then, the dynamic equilibrium can be described as:
(1) Initially, Ω = 0 🡪 w x Ω = 0, then, the dynamic moment is generated by the couple of forces rP x FP 🡪
(2) INDUCES 🡪 A variation მLP / მt 🡪 A new kinetic momentum of precession LP = rP x vP
(3) GENERATES 🡪 A rotation of the field LP of the object
(4) INDUCES 🡪 A new profile of velocities of precession vPT
(5) GENERATES 🡪 A new kinetic momentum LP = rP x vPT 🡪 A gyroscopic couple: w x LP 🡪 (2) …
DYNAMIC EQUILIBRIUM FOR SOLIDS (II)�New ideas introduced by Gyroscopic Fields
Initial velocity profile (w x r) of the
Spinning Top generated by intrinsic rotation:
Velocity profile (Ω x rP) for the precession
of the Spinning Top produced by a pair of
forces generated by weight + reaction (rP x FP):
DYNAMIC EQUILIBRIUM FOR SOLIDS (III)�New ideas introduced by Gyroscopic Fields
Initial velocity profile (w x r):
From (4): curl LP = – 2ρ vPT – (4/vP2) მ((– ½) rP x ( r x ρ aP)) / მt;
where: LP = rP x ρ vP 🡪 curl LP = curl (rP x ρ vP) = – 2 ρ vP;
მ(r x ( rP x ρ aP)) / მt = მ[(r . ρ aP) rP – (r . rP) ρ aP] / მt;
= [(მ(ρ r . aP) / მt) rP + (r . aP) (ρ მrP/მt) – მ(r . rP)/მt ρ Ω2 rP – (r . rP) ρ მaP/მt]
= [rrP cos(ϴ) Ω2) ρ vP – (rrP cos(ϴ))) ρ მaP/მt]; ϴ = angle(r, rP)
🡪 ρ vPT = ρ vP0 – (1/vP2) [ (rrP cos(ϴ)) ρ (Ω2 vP – მaP/მt)]; vP0 🡪 Initial value of vP
In the permanent regime: (მaP /მt) = 0 🡪 ρ vPT = ρ vP0 – (1/vP2) [rrP cos(ϴ) ρ Ω2 vP ]
If (v0 = 0), then: vP2 = Ω2rP2 🡪 ρ vPT = ρ vP0 – ρ cos(ϴ) vP; where ϴ = angle(r, rP) and |r| = |rP|;
DYNAMIC EQUILIBRIUM FOR SOLIDS (IV)�New ideas introduced by Gyroscopic Fields
vPT ≈ vP0 – cos(ϴ) vP; vP0🡪 Initial value of vP
vP = v0 + Ω x rP New velocity profile ~ cos(ϴ):
From (2): curl ((– ½) rP x (r x ρ aP)) = Σ rP x FP; where the term: (vP2/c’2) მLP/მt is ignored
= rP x ρ aP0 + rP x [ρ|w|sin(ϴ) vP] – rP x [ρ cos(ϴ) aP]; aP0🡪 Initial value of aP
But wsin(ϴ) is perpendicular to (r x vP):
= r P x ρ aP0 + w x (rP x ρ vP) – rP x [ρ cos(ϴ) aP]
🡪 r P x ρ aP0 + w x LP = rP x [ρ cos(ϴ) aP] + Σ rP x FP = rP x ρ aPF + Σ rP x FP
If aP0 = მ(rP x ρ vP)/მt and FP = 0 🡪 მLP/მt + w x LP = Σ rP x ρ aPF; aPF🡪 Final value of aP
DYNAMIC EQUILIBRIUM FOR SOLIDS (V)�New ideas introduced by Gyroscopic Fields
aPF = aP0 + |w|sin(ϴ) vP – rP x [ρ cos(ϴ) aP]
New acceleration profile ~ sin(ϴ):
🡪 Correspond to dynamic moments:
rP x aPD = rP x [|w|sin(ϴ) vP]; Dynamic Interaction
rP x aPF = rP x [ρ cos(ϴ) aP] = rP x FP; Final Interaction
Because wsin(ϴ) is perpendicular to (r x vP) 🡪 rP x aPD = w x (rP x vP) = w x LP
🡪 LP = rP x ρ vP = rP x (v0 + Ω x rP) = ρ rP2 Ω + rP x v0
If v0 = 0 then 🡪 w x LP = ρrP2 w x Ω 🡪 GYROSCOPIC COUPLE;
🡪 Where the pair Σ rP x FP has been substituted by the Dynamic Interaction rP x ρ aPF
DYNAMIC EQUILIBRIUM FOR SOLIDS (VI)�New ideas introduced by Gyroscopic Fields
Representation of
the DYNAMIC
EQUILIBRIUM
introduced by the
Theory of
Gyroscopic Fields:
მLP / მt = rP x FP
– 2 ρ vP0 = – 2 ρ vP – (4/vP2) მGP / მt
DYNAMIC EQUILIBRIUM FOR SOLIDS (VII)�New ideas introduced by Gyroscopic Fields
The final result of this dynamic equilibrium is the well-known motion of precession, where the angular velocity vector Ω is in a vertical position and the spinning top precesses around this Ω vector:
მ(LR + LP)/მt = τnon-coaxial
τ 🡪 Pair of forces
generated by
weight + normal reaction
F 🡪 Force of gravity =
spinning top weight
DYNAMIC EQUILIBRIUM FOR SOLIDS (VIII)�New ideas introduced by TID & Gyroscopic Fields
where α = Ωt; Ω = |Ω|
ρ v0 (v0 . rP) + ρ (v0 . rP) (Ω x rP) – ρ [rP . (Ω x v0) + Ω2rP2]rP – ρ rP2 (Ω x v0) = 0
Ignoring some terms (in red), which must be null in the permanent regime:
v0 (v0 . rP) – rP2 (Ω x v0) = 0
This is the equation of the orbit of an object with translational velocity v0 being attracted by an apparent (not real) central force (– Ω x v0)
DYNAMIC EQUILIBRIUM FOR SOLIDS (IX)�NEW DEFINITION OF EULER ANGLES
For the total Rigid Body: w and Ω are independent vectors
მ [I1 (მψ/მt ) + I2 (მ(θ’+θ)/მt) + I3 (მφ/მt)] /მt = τnon-coaxial
მ(θ+θ’)/მt🡪 Nutation
w = (0, w2, w3) = (0 , მθ’/მt, მψ/მt);
Ω = (Ω1, Ω2, 0) = (მφ/მt , მθ /მt, 0);
Euler-Maxwell Euler equations for rotation
equations for Non-Linear:
rotation. Linear:
I2 (მw2 /მt) = τ2-coaxial
I3 (მw3/მt) = τ3-coaxial
I1 (მΩ1/მt) = τ1
I2 (მ(Ω2+w2)/მt) = τ2
DYNAMIC EQUILIBRIUM FOR SOLIDS (X)�Application to the Spinning Top
Euler-Maxwell equations for the rotation of the Spinning Top
For the Spinning Top: τ = rP x FP 🡪 τ1 = rP FP sin(θ+θ’+ π/2) = – mgh cos(θ+θ’)
🡪 τ2 = 0 🡪 I2 (მΩ2/მt) + I2 (მw2/მt) = 0
Where:
FP = mg is the external force of gravity which generates the non-coaxial pair of forces
|rP| = hcos(θ+θ’) is the distance from the centre of precession P to the centre of gravity
I2 (მw2 /მt) = τ2-coaxial
I3 (მw3/მt) = τ3-coaxial
I1 (მΩ1/მt) = – mgh cos(θ+θ’)
I2 (მ(Ω2+w2)/მt) = 0
DYNAMIC EQUILIBRIUM FOR SOLIDS (XI)�Application: The Spinning Top
We have obtained: I2 (მ(Ω2+w2)/მt) = 0 🡪 I2 (მΩ2/მt) + I2 (მw2/მt) = 0
From a system of equations where we have introduced one more unknown w2, so we need one more equation.
Now, let’s suppose that the missing equation takes the following form:
I2 (მw2/მt) = – (I3 – I1) w3Ω1
Then, we get something similar to the Euler equations:
I1 (მΩ1/მt) = – mgh cos(θ+θ’)
I2 (მΩ2/მt) – (I3 – I1) w3Ω1 = 0
I3 (მw3/მt) = τ3-coaxial
The only difference with the Euler equations for the symmetric Spinning Top is at the first line:
I1 (მΩ1/მt) + (I3 – I2) w3Ω2 = τ1
But this can be explained if the nutation component Ω2 is null in the permanent regime
DYNAMIC EQUILIBRIUM FOR SOLIDS (XI)�The Spinning Top: Lagrangian approach
GP = grad (ϕ) + მAP / მt; KP = curl AP;
ϕ = rP . (– ½) [r x (rP x (Ω x ρ v0)) – vR x LP]; vR = მr/მt
AP = (– ½) r x LP vP = v0 + Ω x rP
LP = rP x ρ vP
🡪 grad (ϕ) = (– ½) [r x (rP x (Ω x ρ v0)) – vR x LP]; rP🡪 Instantaneous centre of precession
= (– ½) ρ r x [(rP x (მv0/მt + Ω x (Ω x rP) + (მΩ/მt) x rP + 2(Ω x ρ v0)))]
= (– ½) ρ [r x (rP x aP)]
🡪KP = curl AP = LP
DYNAMIC EQUILIBRIUM FOR SOLIDS (XII)�The Spinning Top: Lagrangian approach
L = L2/2ρ + AP . vP + ϕ = ½ ρ (v0 + Ω x rP)2 + AP . vP + ϕ
= ½ ρ (v0 + Ω x rP)2 + (– ½) (r x LP) . (v0 + Ω x rP)
+ rP . (– ½) [r x (rP x (Ω x ρ v0)) – vR x LP]
= ½ ρ [v02 + 2 v0 . (Ω x rP) + (Ω x rP)2 – (r x LP) . v0 – (r x LP) . (Ω x rP)]
– rP . [(r . (Ω x ρ v0)) rP – (r . rP) (Ω x ρ v0) + vR x LP]
If (v0 = 0) 🡪
L = q [½ ρ (rP x (Ω x rP))2 – ½ ρ (r x (rP x ρ (Ω x rP))) . (Ω x rP) – [(w x r) x (rP x ρ (Ω x rP))] . rP]
= q [½ ρ (rP x (Ω x rP))2 – ½ ρ (r . rP) (Ω x rP)2 – ½ ρ[(w x r) x ((rP2 Ω ) – (rP . Ω) rP)] . rP]
= ½ ρ vP2 – ½ ρ cos(ϴ) vP2 + ½ ρ[(Ω . r) (w . rP)]; Gyroscopic Charge q = 1/rP2
= ½ ρ vP2 – ½ ρ cos(ϴ) vP2 + ½ ρ[rP x |w|Ω] . rP cos(ϴ); rP cos(ϴ) = r
🡪 L = ½ ρ [vP2 – cos (ϴ) vP2 + (rP x (w x Ω)) . r]; GYROSCOPIC CHARGE: q = 1/rP2
Conclusions
We have got an extended system of Partial Differential Equations to describe the real behaviour of solid objects and fluids when they perform rotational and translational motions, appart from the fluidity phenomenon described in Fluid Mechanics.
We have discarded “vorticity” as the source of angular momentum
We claim that this new approach constitutes a new paradigm for CLASSICAL MECHANICS that could be called MAXWELLIAN MECHANICS
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Conclusions: System of equations (PDE’s)
ρ მv0 / მt + ρ [მv / მt + (v. grad) v] = – grad (p) + μ Δ v + ∑ Fexterior
div ρ ( v + v0 + vRT + vPT) + მρ / მt = 0 EXTENDED NAVIER-STOKES
მLR / მt = ∑ (r x FR)
curl (LR) = – 2 ρ vRT – (4/vR2) მ(r x (r0 x ρ aR)) / მt
div (GR) = – ρ vR2 ; div (LR) = 0 MAXWELL-EULER FOR ROTATION
მLP / მt + w x LP = ∑ (rP x FPT)
curl (LP) = – 2 ρ vPT – (4/vP2) მ(rP x (r x ρ aP)) / მt
div (GP) = – ρ vP2 ; div ( LP) = 0 MAXWELL-EULER FOR PRECESSION
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FIN
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