If f(r) satisfies f"(r) - (N-1)r-3f(r) + (N-1)r-1f'(r)
= lf(r) then f(|x|)x~ satisfies Ñx2 f(|x|)x~ =
l f(|x|)x~ .
Newton Potential
To understand relativistic gravitation, it is helpful to understand the model of which it is essentially
a refinement, Newtonian gravitation and, in particular, gravitational potential theory.
In the Newtonian model, we can represent gravity by a 1-vector "acceleration" field GN(p)
[ where p is a 4D Euclidean spacetime coordinate ] representating the instantaneous acceleration exerted on a
small test mass placed at event p.
Time is global in Newtonian physics so for a given t, we can regard G as a function
of 3D spacial position r. GN is assumed
to be irrotational (ÑrÙGN(r)) = 0 and so expressable as the gradient of a
scalar field -f(r) known as the gravitational 0-potential.
The minus sign is incorporated so that the force ¦(r) = -Ñrf(r) tends to
move matter from high to lower potential areas .
We can add an arbitary point-independant
scalar constant to such a field without effecting its gradient and we typically seek to choose this constant so that
f(r) ® 0 as |r|®¥ (in any direction).
A single point mass m at the origin generating an attarctive gravitational field
GN(r) = - gmr~(r-2)
can be regarded as generating a 0-potential f(r) = -gm(r2)-½ where point-independant scalar g is
the universal gravitational constant.
Let V be an arbitary volume (3-curve) having boundary surface (2-curve) dV.
By Gauss's law,
the integral of GN(r) over dV
is zero unless 0 Î V in which case it is -4pgm (regardless of the geommetry of
V) . In general, if scalar mr is the local "mass density" at r, assumed to be a continuous function of r, we have
òdV GN(r)¿dr2
= -4pgòV mr dr3
Now
òdV GN(r)¿dr2
= òV (Ñ¿GN(r))|dr3|
by the divergence theorem so for arbitary small V we have
òV (Ñ¿GN(r)-4pmr)|dr3| = 0
whence ѿGN(r) = -4pgmr and so
Ñr2f(r) = Ñr¿(Ñrf(r))
= -4pgmr
which (for scalar f(r)) is known as Poisson's equation , or Laplace's equation
if mr=0 .
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 .
Effective Potential
Under a central force with classical 0-potential f(r)=f(r) ,
the 2-blade angular momentum L=mrÙr' and scalar energy E=½m(rdt)2 + V(r) are both conserved
and the radial energy equation
½m(rdt)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
conserved.
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)
Thus if
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
(-3-r0¦'(r0)¦(r0)-1)-½
of the orbit period 2p(-r0-1m-1¦(r0))-½, while if OSI(r0) is negative we have hyperbolic deviation.
If ¦'(r0) + 3r0-1¦(r0) vanishes, we have harmonic deviation equation
mx"
» x2( ½¦"(r0) - 6r0-2)
= -PSI(r0)x2
where
PSI(r0) º
r0-2(6+½r02f"'(r0))
.
This is analytically problematic since while x" = ax2 has a ready particular solution
x(t)=6a-1(t+6a-1x(0)-1)-2 this is not general enough to match a given xdt(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) .
Circular Orbits
-f'(r)r~ = -mw2r where m represents a positive inertial mass 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 ;
period
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
and
f(r0) + ½m(r02w02+s02) =
f(r) + ½rf'(r) .
Quantised Orbits
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.
with a particular speed
v orbits which we might
Ring Potential
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 ) ,
Stream Function
Setting scalar stream function yp º
òC dV2 V1
- òC dV1 V2 for any path C from 0 to p
yields
V1 = Ðe2yp ; V2 = - Ðe1yp , ie.
Vp = Ñp×(ype3)
= (ÑPÙ(ype3))e123-1 .
If yp is any ananalytic scalar field over Â2
satisfying this then its analyticity provides incompressability condition Ðe1V1 + Ðe2V2 = 0, ie.
ÑPVp=0 and also Ñp2yp = 0.
Hence Vp = (ÑP(ype3))e123-1 = -ÑPype12 .
Also note that ÐVp yp = (Vp¿Ñp)yp = 0 so yp is constant when following the flow.
Velocity Potential
For a steady irrotational incompressible flow with ÑpÙVp = 0 we can constuct a 0-potential fp = òC d1p¿Vp for any path C from 0 to p so that Vp=Ñpfp , ie. Vi = Ðeifp . If the flow is conserved we also have Ñp¿Vp=0 so ÑpVp = Ñp2fp = 0.
Complex Potential
The Complex 0-potential fp º fp + iyp
satisfies ÑP2fp = 0 and it is natural to here associate i with e123.
It is defined for P=x Î ÂN0=Â2
but by defining z=x1+ix2 where p=x1e1+x2e2 we can regard
fp = fz,t) as a t-dependant analytic function mapping C®C with
dfz,t)/dz =
df/dx1 + idy/dx1 =
V1 - iV2 and
|d_cvphiz/dz|+ = |Vp| .
The complex potential thus fully embodies an incompressible irrotational 2D flow.
Morse Potential
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
- (½+D)2
gives real energy eigenvalues En=-(n-D)2
for integer n<D. [Saaidi]
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(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
whenever
-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
= -r-1(1-u2)-½¦(r)2
[ 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
¶V/¶r
= a(1-u2)(a-1)2h2r-3
so r2 ¶V/¶r = 2h2a(1-u2)(a-1)r-1
whence
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 )
Next : 1-Potentials