If f(r) = f(|r|) = f(r) we have Ñrf(r) = f'(r)r-1r = f'(r)r~ and Ñr2f(r) = r-1f'(r) + f"(r).
This then is the Newtonian model. The 3D gravitational force field G(r) is characterised by it's
scalar "integral" f(r) = -Ñr-1 G(r) ;
a 0-potential scalar
field which satisfies Poissons equation everywhere and provides the acceleration on a test mass at r
as GN(r)=-Ñrf(r) . We can in principle compute f(r0) as the infinite volume integral
f(r0) = gòÂ3 mr(r-r0)-½ |dr3|
The scalar field mr fully determines GN so the "flow" of matter is relevant in determining the instantanous gravitational potential only by its magnitude. It's direction effects only how GN changes over time.
Gravity is such that the force exerted on a test particle is proportional to the mass of the test particle
so that the acceleration experieneced is independent of the mass, More generally we have a force field
Fp = Ñpfp .
Under a central force with classical 0-potential f(r)=f(r) , the 2-blade angular momentum L=mrÙr' and scalar energy E=½m(r')2 + V(r) are both conserved and the radial energy equation ½m(r')2 + V(r) - ½m-1r-2L2 = E for r³0 implies that the shape of the 1D scalar effective potential f(r) - ½m-1r-2L2 = f(r) + ½m-1r-2L2 determine the possible orbits for a given L=|L2|½ and E combination. If the particle is released from rest L=0 , but otherwise the effective potential contains a term equivalent to that contributed by a repulsive inverse cube force m-1L2 r-3
Orbital Stability Indicator
Suppose a particle of inertia m is subject to a time t-independant central force
¦(r,t)r~ = ¦(r)r~.
If x=r-r0 denotes the deviation of r from cirular orbit value r0 then mx.. » (r0-13¦(r0)+¦'(r0))x to first order in r-1x where .. denotes second time derivative. For gravitationl forces ¦(r) is proportionate to the inertia m.
mr.. = ¦(r)+L2r-3 where
angular momentum L=mr2q. is
Let x=r-r0. We have mx.. = ¦(r0+x) + L2(r0+x)-3 » ¦(r0)+x¦'(r0)+½x2¦"(r0) + 3!-1¦"'(r0)x3 + ... + r0-3L2(1-3r0-1x + 6r0-2x2 - 10r0-3x3 +...
For circular orbit at r0 we require L2 = -mr03¦(r0) , w = (-r0-1m-1¦(r0))½ so
mx.. » ¦(r0)+x¦'(r0)+½x2¦"(r0) + 3!-1¦"'(r0)x3 + ... - ¦(r0)(1-3r0-1x + 6r0-2x2 - 10r0-3x3 +...)
= x¦'(r0)+½x2¦"(r0) + 3!-1¦"'(r0)x3 + ... + ¦(r0)(3r0-1x - 6r0-2x2 + 10r0-3x3 +...)
= x(¦'(r0) + 3r0-1¦(r0)) + x2( ½¦"(r0) - 6r0-2) + x3(3!-1¦"'(r0) + 10r0-3) + O(x4)
orbital stability indicator
OSI(r0) º -3r0-1¦(r0) - ¦'(r0)
= 3r0-1f'(r0) + f"(r0)
is positive we have oscillations about the
circular orbit of frequency m-½ OSI(r0)½ and so period
2pm½ OSI(r0)-½ , a proportion
of the orbit period 2p(-r0-1m-1¦(r0))-½, while if OSI(r0) is negative we have hyperbolic deviation.
We will see below that OSI(r) = 2r-1E'(r) where E(r)=f(r)+½rf'(r) is the total energy of a circular orbit of radius r.
If ¦'(r0) + 3r0-1¦(r0) vanishes, we have harmonic deviation equation
» x2( ½¦"(r0) - 6r0-2)
PSI(r0) º r0-2(6+½r02f"'(r0)) = 6r0-2 + ½f"'(r0) .
This is analytically problematic since while x" = ax2 has a particular known solution x(t)=6a-1(t+6a-1x(0)-1)-2 this is not general enough to match a given x'(0) .
The normalised orbilital stability indicator of a circular orbit of radius r0
under a central force ¦(r) £ 0 is the value
OSI(r0) º (3+r0¦(r0)-1¦'(r0))½ , where ½ denotes that for negative values we take the negative square root of the absolute value. In eliminating the magnitude of ¦ by multiplying OSI(r) by |r-1¦(r0)|½, we aquire an infinite OSI(r) when ¦(r)=0. The sign of ¦(r) has also been eliminated so OSI(r) of a repuslive central force is the OSI of the attactive force -¦(r). For nonzero ¦(r), OSI(r) is the number of orbits about 0 per peturbation oscillation.
OSI(r) provides a measure of deviation from shell condition p(t)2 = p(0)2 under acceleration p"(t) = -f'(|p(t)|)p(t)~ where |p(t)| º |p(t)2|½, assuming p(0)¿p'(0) is small. Large OSI suggests rapid oscillations across the shell with integer values suggesting petal forms. OSI<1 suggesting slowly corrected slightly noncircular orbits.
"Power law" attractive force f(r)=-ark has OSI(r) = (3+k)½
so stable circular orbits require k>-3 . Inverse square forces have unit OSI(r), so the period of the deviation
coincides with that of the orbit, stretching it into an ellipse for example.
An integer OSI(r) indicate a single repeated orbit
while rational OSI(r) pq-1 implies that after q orbits each of which contain p deviation cycles the trajectory will repeat itself.
An r-1 orbit takes ½½ orbits to complete an oscillation and, as this is irrational,
the trajectory in theory never repeats. However, as the oscillatory nature of deviations is only correct to first order in small deviations, such
multi-orbit analysis is profoundly suspect.
High OSI(r) indicate rapid oscillations about the circular orbit while high negative OSI indicate rapid hyperbolic deviations, with the caviat that ¦(r) changing sign gives infinite OSI when ¦(r)=0 corresponding to deviations from the straight line zero forces trajectory. Zero OSI indicates paraboloid deviations from u with q correspending to paraboloid deviation from u0 with q while OSI j as for ¦(r)=-arj2-3 corresponds to oscillating j times faster than the orbit.
To maximise OSI(r) º r-13f'(r) + f"(r)
we consider OSI'(r) = -3r-2f'(r) + 3r-1f"(r) + f"'(r)
which is zero when
f"'(r) = 3r-2(f'(r) - rf"(r)) whereupon
r-2(6+(3/2)(f'(r) - rf"(r)))
= (3/2)r-2(4 + f'(r) - rf"(r))
PSI'(r) = -12r-3 + ½f(4)(r)
vanishes everywhere for f(r) = 6r(ar)↓ + P4(r)
= 6(a↓r + 6r(r)↓ + P4(r)
for arbitary a and quartic polynomial P4() with
f'(r) = 6(ar)↓ + 6 + P4'(r).
f"(r) = 6r-1 + P4"(r).
f"'(r) = -6r-2 + P4"'(r).
3r0-1f'(r0) + f"(r0)
= 18r0-1(ar0)↓ + 18_roinv + 3P4"(r)
+ 6r-1 + P4"(r).
= 18r0-1(ar0)↓ + 24_roinv + 4P4"(r)
ò0r0 dr 6r(ar)↓ = 6 ò0ar0 dx a-1x (x↓) = 6a-1 [ x2(½(x)↓-¼) ]0ar0 = 6ar02(½(ar0)↓-¼) .
However, this is less intersting than PSI'(r)=0 coinciding with
OSI'(r)=0 and PSI(r)=0 .
Stability Indicators of a Klein Gordan 0-Potential
If Ñx2f(r) = af(r) then f"(r) = ±af(r) - (N-1)r-1f'(r) according as x2 is ±, and so OSI(r) = (4-N)r-1f'(r) ± af(r) and for N=4 we have OSI(r) = ±af(r) .
OSI'(r) = (±a - (4-N)r-2)f'(r) + (4-N)r-1 f"(r)
= (±a - (4-N)r-2)f'(r) + (4-N)r-1 (±af(r) - (N-1)r-1f'(r))
= (±a - (4-N)r-2 - (4-N)(N-1)r-2)f'(r) ±a(4-N)r-1f(r)
= (±a - N(4-N)r-2)f'(r) ±a(4-N)r-1f(r) .
Thus OSI(-r) is extreme when
= f'(r) = -/+a(4-N)r-1(±a - N(4-N)r-2)-1 f(r) .
Further, f"'(r) = ±af'(r) - (N-1)r-1 (±af(r) - (N-1)r-1f'(r)) + (N-1)r-2f'(r) = (±a + N(N-1)r-2)f'(r) -/+a(N-1)r-1f(r) according as x2 is ±; and so
PSI(r) = r-2(6 + ½r2f"'(r)) =
OSI'(r) = -3r-2f'(r) + 3r-1f"(r) + f"'(r) = -3r-2f'(r) + 3r-1( ±af(r)-(N-1)r-1f'(r) ) + (±a + N(N-1)r-2)f'(r) -/+a(N-1)r-1f(r)
-f'(r)r~ = -mw2r for inertial mass m>0 yields w = ±|m-1r-1f'(r)|½ Þ wr = ±m-½ |rf'(r)|½ , and so circular orbits under central force ¦(r)r~ = -f'(r)r~ have
The potential f(r)=-lrb with l>0 has f'(r)=br-1f(r) and hence squared angular momentum mlbrb+2 and kinetic energy |bf(r)|. For b<0 the toal orbit energy is thus (b+1)f(r) and so in a Coloumb potential with b=-1 all circular orbits have total energy zero, while if b=-2 all circular orbits have angular momentum (2ml)_sqdrtcnj and positve total energy -f(r).
The Coulomb Potential f(r)=-lr-1 with f'(r)=lr-2 thus has
w = ±|m-1lr-3|½ ;
2p|m½l-½ r3/2 ;
speed ±|m-1lr-1|½ ;
angular momentum ± m½r½
classical kinetic energy
½lr-1 ; and
radial acceleration -m-1lr-2.
OSI(r) = r-13f'(r) + f"(r) = -lr-3 while OSI(r) = 1.
More generally a small test particle launched from r0e1 with velocity r0w0e2 + s0e1
will seek a circular orbit having
mr02w0 = m-½r3/2f'(r)½
f(r0) + ½m(r02w02+s02) =
f(r) + ½rf'(r) .
There are two natural ways to particularise orbits under a central oscillatory potential f(r). The maxima of OSI(r) provide the stablest orbits, while orbits at r satisfying |rf'(r)| = mv2 provide the orbits having a given speed v. In particular, v=1 yeilds the lightspeed orbits with inertial mass m regarded as a coulping constant rather than a mass-energy.
The ring potential foRz(r) º 2R ò0p dq f((z2+r2+R2-2rR cosq)½) is the potential at r of a circle of radius R lieing in a plane containing 0 and r.
Setting s º (z2+r2+R2-2rR cos(q)) we have ds = 2rR sinq dq.
cos(q) = (2rR)-1(s-z2-r2-R2) so for q Î [0,p] we have
sin(q) = (2rR)-1 (4r2R2 - (s - z2-r2-R2)2)½ = (2rR)-1 (2s(z2+r2+R2) -s2 + 4r2R2 - (z2+r2+R2)2)½ = (2rR)-1 Q(s)½ where
Q(s) º -(s2 - 2s(z2+r2+R2) + (z2+ r2+R2)2 - 4r2R2 )
= -s2 + 2s(z2+r2+R2) - z4 - 2z2(r2+R2) - (r2-R2)2
=?= -(s-z2-(r-R)2)(s-z2-(r+R)2) is a quadratic in s with roots z2 + (r±R)2.
Thus foR(r) = 2r-1 òz2+(r-R)2z2+(r+R)2 ds f(s½) ( sinq)-1 = 4R ò(r-R)2 (r+R)2 ds Q(s)-½ f(s½) .
For z=0 , Q(s) = -(s2 - 2s| (r2+R2) + (r2-R2)2 ) ,
Typically used for r representing the seperation between two covalently bonded atoms, f(r) = f¥(1-(-(½kf¥-1)½(r-r0))↑)2 for dissassociation energy f¥>0, equilibrium or zero force distance r0, and bond force constant k>0.
Yellow shows potential f(r);
green is force f'(r); while red is the OSI (3+f'(r)-1rf"(r))½ with imiginary values focred negative real.
For rÎ[0,r0] we have f(r)£f0=f(0)= f¥(1-((½kf¥-1)½r0)↑)2 with f(r)³f¥ for r£ r0 - (2f¥k-1)½(2)↓.
f(r) ® f¥ from below as r®¥ and to ¥ as r®-¥ though we might typically impose r³0. We have a global minimum at r=r0 with f(r0)=0 , f'(r0)=0, and f"(r0)=k.
The comnplex Morse potential
f(r) = b2(-2(r-r0))↑ - b(1+2D)(-(r-r0))↑
= (b(-(r-r0))↑ - (½+D))2
gives real energy eigenvalues En=-(n-D)2
for integer n<D. [Saaidi]
Born Mayer Potential
The Born-Mayer potential is the exponentiion f(r) = f0 (-ar)↑ with with a>0. Since f(k) = (-a)k f(r) we have OSI(r) = a(a+3r-1) f(r) ; OSI(r) = (3+ar)½ ; and PSI(r) = 6r-2 - ½a3 f0(-ar)↑ zero when r-2(ar)↑ = 12-1a3f0-1 .
The Lennard-Jones potential is a convenient fiction used to approximate potentials that strongly repel close up but attract at a distance. f(r) = 2fr0(r0-1r)-6(1 - ½(r-1r)-6) where fr0<0 .
This is +¥ at r=0, 0 at at 26-1r0 » 1.12246r0 and ®0 as r®¥. Since f'(r) = 12fr0r-1(r0-1r)-6(1 - (r-1r)-6) , f(r) is minimised at r=r0 by f(r0) = fr0.
Tangential occlusion potential
Constructing an inverse square particle-mediated field is unrealistic because the r-1 factor representing an inverse square "dispersion" of a constant signal fails for low r . If the absorbing particle is not attracting the carriers but merely absorbing those that reach a small perimeter distance h , then we would expect the proportion of carriers absorbed to be in proportion to the area content of the spherical cap defined by the cone tangential to the h radius sphere centered on the absorber. A sphere of radius h at distance r from 0 subtends a cone of demiangle q= sin-1(r-1h) = sin-1(u) where uºhr-1 , which occludes an area 2p(1- cos(q))d2 = 2p(1-(1-u2)½)d2 of the 4pd2 surface area of a sphere of radius d at 0, rather than an area proportionate to r-2.
Attractive central force
¦(r) = (1-(hr-1)2)½ - 1 = (1-u2)½ - 1 for r³h (ie. u £1) has
¦'(r) = h2r-3(1-u2)-½ = h-1u3(1-u2)-½ = h2r-3(1+¦(r))-1 .
For r>>h we have ¦(r) » -½(hr-1)2 - 1/8(hr-1)4 - K3(hr-1)6 - K4(hr-1)8 - ... where Kk º 1*3*5*7*...*(_2k-3) / (2*4*6*8*...*(2k)) = 22-2k(2k-3)! / ((k-2)! k! ) .
For large r (so small u=hr-1), the inverse square "Coulomb" force will be massively dominant.
We find that circular orbits are stable only for u2< 3/4 (ie. r>(3-½)2h) and that the 3 + r¦'(r)¦(r)-1 indicater tends to 1 from below as r®¥ .
For a screened force
F(r) = (-lr)↑ ¦(r) we have
Z(F)(r) = F(r) + 3-1rF'(r)
= (-lr)↑ ( Z(¦)(r) - 3-1lr¦(r) ) .
Now -3-1lr¦(r) is positive
so exponentially damping ¦ makes circular orbits unstable
-1 + (1-u2)-½(1-(2/3)u2) - 3-1lr((1-u2)½-1) ³ 0 Û -1 + (1-u2)-½(1-(2/3)u2) + 3-1lr ³ 3-1lr(1-u2)½ Û -1 + 3-1lr ³ (3-1lr - 1 + (2/3)u2)(1-u2)½
Comparing this with indicater (1-3-1)rk for ¦(r)=rk we see
that for large r, circular orbits are only just stable;
verging on the the instability associated with inverse cube orbits.
The apsidal angle p(3+r¦'(r)¦(r)-1)-½ = p(3 + d2r-2(1-u2)-½ (-1 + (1-u2)½)-1 )-½ = p(3 + u2( (1-u2)½ (-1 + (1-u2)½) )-1 ) -½ = p(3 + u2( 1-u2-(1-u2)½ )-1 ) -½
When u2=½ this is p(3 + ½( ½ - (½)½ )-1 ) -½ =p(3 + (1-2½)-1 )-½ » 1.30p .
It is a rational multiple of p only when 3 + u2( 1-u2-(1-u2)½ )-1 is a squared rational number, which occurs when 1-u2-(1-u2)½ is a squared rational, .
We will here refer to V(r) = òr¥ dr ¦(r)
= r¦(r) + h sin-1(hr-1)
= r((1-(hr-1)2)½-1) + h sin-1(hr-1)
= h(u-1((1-u2)½-1) + sin-1(u))
as the tangential occlusion potential.
[ Proof : òr¥ dr ¦(r) = òu0 (-r2h-1 du) ¦(r) = h òu0 du -u-2¦(r) = h ò0u du ((1-u2)½ - 1)u-2 = h [ -(1-u2)½u-1 - sin-1(u) + u-1 ]0,u = h(u-1(1-(1-u2)½) - sin-1(u)) = -r¦(r) - h sin-1(hr-1) .]
(¶/¶r)2V + 2r-1¶V/¶r
[ Proof : ¶V/¶r = -¦(r) and (¶/¶r)2V + 2r-1¶V/¶r = = -¦'(r) + 2r-1¦(r) = -h2r-3(1-u2)-½ + 2r-1((1-u2)½-1) = r-1(1-u2)-½(-u2 + 2(1-u2) - 2(1-u2)½) = r-1(1-u2)-½(2-u2 - 2(1-u2)½) = r-1(1-u2)-½(1 - (1-u2)½)2 = -r-1(1-u2)-½¦(r)2 .]
Setting V(r) = (1-u2)a gives
so r2 ¶V/¶r = 2h2a(1-u2)(a-1)r-1
Ñ2 V(r) = r-2 (¶/¶r)(r2 ¶V/¶r) = 2r-2h2a ( (a-1)(1-u2)(a-2)2h2r-4 - (1-u2)(a-1)r-2)
= 2r-4h2a (1-u2)(a-2)( (a-1)2u2 - (1-u2) ) = 2r-4h2a (1-u2)(a-2)( (2a-1)u2 - 1 )
f'(r) = (garg-1+br-1)f(r) ;
f"(r) = (g(g-1)arg-2-br-2)f(r) + (garg-1+br-1)f'(r) = r-2(g(g-1)arg-b + (garg+b)2 )f(r) = r-2(g2a2r2g + g(g-1+2b)arg + b(b-1))f(r)
Thus f"(r)+(N-1)r-1f'(r) = (g(g-1)arg-2-br-2)f(r) + ((garg-1+br-1 + (N-1)r-1)f'(r)
= r-2(g(g-1)arg-b + (garg+b+N-1) (garg+b))f(r)
= r-2(g(g-1)arg-b + g2a2r2g + garg(2b+g+N-1) + b(b+N-1)f(r)
= r-2( g2a2r2g + garg(g+2b+N-2) + b(b+N-2))f(r)
Setting b=2-N, g=N-2 gives f"(r)+(N-1)r-1f'(r) = (N-2)2a2r2(N-3)f(r), thus for N=3 we have central f(r)=r-1(ar)↑ solving Klien Gordan equation Ñp2f =a2f.
Taking a=-m<0, b=-1, g=1, d=0 gives the
with Ñx2 (-mr)↑ r-1
= m2 (-mr)↑ r-1 .
This can be loosely regarded as the potential for a putative inverse square force (with negative potential generating an attractive force) mediated by decaying carriers, although it actually only approximates òR¥ dr r-2(-mr)↑ = (-mS)↑R-1 + m-1E(-m-1R) where E(s) º ò-¥s dt t-1(t)↑ is the integral exponential function, small for large negative s.
[ Proof : Setting s=-mr: -m ò-mR-¥ ds s-2(s)↑ = m(- [(s)↑s-1]-¥-mR + ò-¥ -mS ds s-1(s)↑ = (-mS)↑R-1 + mEi(-m-1R) .]
To explain the strong nucleonic (proton and neutron) forces,
observed to act at a range of about R=10-15 m, Yakuwa postulated massive mediator particles
of lifetime Dt = c-1R. Uncertainty principle
DEDt ³ h
then implies DEc-1R ³ h with uncertainty minimised
by DE = chR-1 » 200 MeV.
Historically first the muon, then the pion, were regarded as filling this role,
necessitating a high coupling constant of about 14, making peturbation theories
neglecting higher order couplings inapplicable for the strong nucleonic forces.
Taking a = ia gives f(r)=r-1(ari)↑ with Ñx2f = -a2f .
| f1(r) = r-1 sin(ar) = a Sin(Q)
f'(r) = -r-2 sin(ar) + ar-1 cos(ar) = ar-1( cos(Q)= Sin(Q))
f"(r) = (2r-3-a2r-1) sin(ar) - 2ar-2 cos(ar)
= ar-2(-2 cos(ar) + (2(ar)-1 - ar) sin(ar))
OSI(r) = ar-2( cos(ar) - R-a,N sin(ar))
= ar-2 ((ar)2 + (ar)-2 + 3)½ sin(ar- cot-1(R-a,N)) where R-a,N=ar+(ar)-1.
For small ar, f1(r) = a - 3!-1a3r2 + O(r-1(ar)5) is approximately harmonic.
OSI(r) is extremal when
Q - tan-1(3-1Q(3+Q2)) = kp
which is a condition of Q independant of a.
[ Proof : OSI'(r) = 2ar-3( - cos(ar) + (ar + (ar)-1) sin(ar) ) - ar-2( +a sin(ar) + (a - a-1r-2) sin(ar) + a(ar + (ar)-1) cos(ar) )
= ar-3( -2 cos(ar) + 2(ar + (ar)-1) sin(ar) - ( +ar sin(ar) + (ar - (ar)-1) sin(ar) + ar(ar + (ar)-1) cos(ar) )
= ar-3( (-2-ar(ar+(ar)-1)) cos(ar) + (2(ar + (ar)-1) - ar - (ar - (ar)-1) ) sin(ar) )
= ar-3( -(3+(ar)2) cos(ar) + 3(ar)-1 sin(ar) ) = a4Q-3( -(3+Q2) cos(Q) + 3Q-1 sin(Q) )
= a4Q-3 ( (3+Q2)2 + 9Q-2)½ sin(Q - tan-1(3-1Q(3+Q2)) .]
Define Wk to solve x - tan-1(1/3Q(3+Q2)) = kp .
E(r) = ½(a cos(ar)- Sin(ar))
f(r) = rg(arb)↓ + m has f'(r) = rg-1(g(arb)↓+b) ; f"(r) = rg-2((g-1)g(arb)↓+(2g-1)b) so that f"(r)+(N-1)r-1f'(r). = rg-2((g+N-2)g(arb)↓+(2g+N-2)b)
Setting g=2 gives f(r) = r2(arb)↓ + ½N-1(N+2)b = r2(a↓ + br↓) + ½N-1(N+2)b with f'(r) = = r(2(arb)↓+b) solving Ñx2 f(r) = ±2Nf(r) for any a,b according as x2 is ±.
OSI(r) = (4-N)r-1f'(r) ± 2Nf(r) = (4-N)(2(arb)↓+b) ± (2Nr2(arb)↓ + N(N+2)b) = 2((4-N) ± Nr2)(arb)↓.
Next : 1-Potentials