Linear and Non Linear Effects on the Newtonian Gravitational Constant as deduced from the Torsion Balance

0.16 MB·English

by  Michele Rossi

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International JournalofModernPhysicsA
(cid:13)c WorldScientificPublishingCompany
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a Linear and Non Linear Effects on the Newtonian Gravitational
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Constant as deduced from the Torsion Balance
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]
h
M.Rossi
p
- Dipartimento di Matematica,
s Universit`a degli Studi di Torino
s
Via Carlo Alberto 10,
a
10123 Torino, Italy
l
c
.
s L.Zaninetti
c
Dipartimento di Fisica Generale,
i
s Universit`a degli Studi di Torino
y
Via Pietro Giuria 1,
h I-10125 Torino, Italy
p phone [phone], fax [phone]
[

1 ReceivedDayMonthYear
v RevisedDayMonthYear
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TheNewtoniangravitationalconstanthasstill150partspermillionofuncertainty.This
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paperexaminesthelinearandnonlinearequationsgoverningtherotationaldynamicsof
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the torsion gravitational balance. A nonlinear effect modifyingthe oscillation periodof
.
1 thetorsiongravitationalbalanceiscarefullyexplored.
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8 Keywords: Experimentalstudiesofgravity;Determinationoffundamentalconstants;
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PACSnumbers:04.80.-y;06.20.Jr;
:
v
i
X
1. Introduction
r
a After many years of measurements, begun by H. Cavendish, the Newtonian gravi-
1
tational constant value, said G , is still affected by a large error . The CODATA
recommends G=(6.6742±0.001)×10−11 m3 , meaning a relative standarduncer-
kgs2
2
tainty of 150 parts per million (in the following ppm) . This value has had recent
3
confirmation by means, on one hand, of a super-conducting gravimeter and, on
4
the other hand, of a careful analysis of the possible beam balance nonlinearity .
The originalCavendish method of measure, employing the torsionbalance, still re-
5
veals a large discrepancy from the recommended value: about 500 ppm . This is
probably due to imperfections of the crystalline structure of the torsion fibre 6,7,8.
Moreoverrecentstudiespointoutthattheperiodofatorsionpendulummightvary
9
under disturbances ofenvironmentalnoise factors,see see . Other authorssuggest
a possible deviation from Newton’s law specified as an additional contribution of
1
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2 Rossi Zaninetti
Torsion Balance
r
ϕ
distant scale
luminous source
Fig.1. Schematicviewofthetorsionalbalance
10
Yukawapotentialtype .Thispaperfirstanalysesthe linearandnonlinearequa-
tions governing the torsional balance rotational dynamics (Section 2). By means
of a gravitational torsion balance, same values of G are obtained and summarised
(Section 3). The oscillation period’s variation, due to a non linear effect, is then
discussed (Section 4).
2. The basic equations
The form of Newton’s law of gravitation is
mM
F =G , (1)
r2
where G is the gravitational constant , M the great mass , m the small mass and
r their relative distance. The Leybold balance represents a widespread instrument
to determine the constant G, see Figure 1, and is constructed with the following
components.
(1) Afreely oscillatinghorizontalbar,oflength 2d,holdingtwo smallleadballs of
massmasinFigure1supportedbyatorsionfibrethathasatorsionalconstant
τ .
(2) Twolargerballs ofmass M thatcanbe positionednextto the smallballs as in
Figure 1. The center of mass of the two m and M are supposed to be all on a
plane perpendicular to the fibre.
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Torsion Balance 3
Forces
u =r+s
1
GF
1
u ϕ
α
GF
2
Fig.2. TopviewoftheCavendishbalance
(3) A luminous source directed toward the center of mass of the bar where is re-
flected by a mirror.
(4) A scale at distance l where the reflected light beam is measured.
Thus the equilibrium position about which the pendulum oscillates is different for
thetwopositionsanditisthisdifferencewhichweusetodetermineG.TheFigure2
reports a plot of the motion. The moment of inertia of the bar,I is
I ∼2md2 . (2)
The fundamental equation of rotational dynamics is
..
I ϕ=M +M +M , (3)
g v t
where
M =−β ϕ. , (4)
v
M =−τϕ , (5)
t
here β is the coefficient of viscosity of air , ϕ the angle between bar and bar itself
whenthetorqueiszero.Thisangleis measuredinthe anticlockwisedirection.The
term M represents the torque of the gravitational forces. From equation (1) , we
g
obtain
M =2dGF(ϕ) , (6)
g
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4 Rossi Zaninetti
whereF isafunctionoftheangleϕ.ThelawofdependenceofF withϕiscomplex
and will here be analysed. When the motion starts the resulting force is
F(ϕ)=F (ϕ)−F (ϕ) , (7)
1 2
where
cosϕ
F (ϕ)=mM , (8)
1 u2
1
cos(π −(ϕ+α)) sin(ϕ+α)
F (ϕ)=mM 2 =mM , (9)
2 u2 u2
now α:=arcsin(r−s) (where r e s are defined as in Figure 1). With our data , see
u
Section 3 the maximum angular excursion of the angle ϕ is
x −x
max min
∆=arcsin( ) . (10)
2l
The angle ϕ has a low values, see Table 1, and a Taylor series expansion that keep
termstoorderϕ2 willbeadopted.Thismeanstoforgetquantitieslessthan310−5.
The series representation gives
u1 =p(r+s)2+(d−dcosϕ)2 ∼r(1+ϕ(2rd + dr22ϕ+ 4dr22ϕ3))12 . (11)
Developing the last term with a Maclaurin series we obtain
ϕ 2d d2 d2
u ∼r(1− ( + ϕ)− ϕ2+O(ϕ3))∼r+dϕ . (12)
1 2 r r2 2r2
As a consequence
1− ϕ2
F ∼mM 2 , (13)
1 (r+dϕ)2
and expanding the denominator we obtain
(r+dϕ)−2 =(r−2−2 d ϕ+3d2ϕ2+O(cid:0)ϕ3(cid:1)) , (14)
r3 r4
that means
mM 2d 3d2 1
F ∼ (1− ϕ+( − )ϕ2) . (15)
1 r2 r r2 2
An expression for F can be obtained from equation (9)
2
sinϕcosα+cosϕsinα r+dϕ− rϕ2
F =mM ∼mM 2 . (16)
2 u2 (d2+r2−2rdϕ−d2ϕ2)32
On Taylor expanding the denominator
(4d2+r2−2rdϕ−d2ϕ2)−32 =(4d2+r2)−32(1+ 3dr ϕ+3d2(4d2+5r2)ϕ2+O(ϕ3)) ,
4d2+r2 2(4d2+r2)2
(17)
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Torsion Balance 5
and therefore
r+ 4d(d2+r2)ϕ+ r(20d4+13d2r2−r4)ϕ2
F ∼ 4d2+r2 2(4d2+r2)2 . (18)
2 (4d2+r2)23
Now F from equation (7) can be expressed like a Taylor expansion truncated at
O(ϕ3)
F ∼A +A ϕ+A ϕ2 , (19)
0 1 2
where
1 r
A :=mM( − ) , (20)
0 r2 (4d2+r2)23
d 2d(d2+r2)
A =−2mM( + ) , (21)
1 r3 (4d2+r2)25
mMd2 mM mMr(cid:0)4d4−10r2d2+r4(cid:1)
A =3 −1/2 +1/2 . (22)
2 r4 r2 (4d2+r2)7/2
Three methods that allow to obtain an expression for G in terms of measurable
quantities are now introduced. Further on the well known formula for G extracted
from the Leybold manual is reviewed.
2.1. Averaged G
LetF(ϕ)=F for allthe experience ;in firstapproximationwe mayassume thatF
is given by equation (19) to first order
F =A +A ϕ , (23)
0 1
where ϕ is the averageof the values that ϕ assumes between the first position, P ,
1
and the last position ,P∞, of the balance ; A0,A1 are given by equations ( 20) and
(21) . The differential equation that describes the motion is
.. .
I ϕ+β ϕ+τϕ=2dGF , (24)
and it’s solution is
2dGF
ϕ(t)=ce−δtcos(ωt+φ)+ , (25)
τ
where c represents the amplitude and
β
δ :=− , (26)
2I
p4Iτ −β2
ω := . (27)
2I
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6 Rossi Zaninetti
The angle ϕ∞, that represents the bar position at the end of the phenomena P∞
can be determined as follows
ϕ∞ ∼ x∞−xo = x∞−x1 , (28)
2l 4l
and should be the same as predicted by the theory
2dGF
lim ϕ(t)= ; (29)
t→+∞ τ
therefore
τϕ∞
G= . (30)
2dF
Inorderto continuea value forτ shouldbe derived.This canbe obtainedfromthe
period of oscillation of the bar
2π 4πI
T = = . (31)
ω p4Iτ −β2
We continueby identifying T withthe empiricalvalueT.We continue onassuming
that β2 is small ; therefore from equations (2), (30) and (23), the following is
4I
obtained
2π2Iϕ∞
G= . (32)
2
d(A +A ϕ)T
0 1
2.2. G with air viscosity
From formula (4) is possible to deduce the viscosity of the air once the coefficient
of damping δ is known, see Section 3 on data analysis. From formula (30) and (26)
we should add to the value of G reported in equation (32)
β2ϕ∞
G := , (33)
β
8IdF
obtaining
2π2Iϕ∞ β2ϕ∞
G= + . (34)
d(A0+A1ϕ)T2 8Id(A0+A1ϕ)
2.3. G to the first order
Let assume that F(ϕ) is not constant, we can assume at the order O(ϕ2) with the
aid of formula (19)
F ∼A +A ϕ , (35)
0 1
where A e A are defined in equations (20) and (21) respectively. In this case the
0 1
law of motion is still equation (3)
I ϕ.. +β ϕ. +(τ −2dGA )ϕ=2dGA , (36)
1 0
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Torsion Balance 7
and the solution is
2dGA
ϕ(t)=ce−δtcos(ω′t+φ)+ 0 , (37)
τ −2dGA
1
′
where the angular velocity ω has now the following expression
′ p4I(τ −2dGA1)−β2
ω := . (38)
2I
As a consequence
4π2I β2
τ = +2dGA + , (39)
T2 1 4I
2dGA
0
ϕ∞ =t→li+m∞ϕ(t)= τ −2dGA1 , (40)
and therefore
τϕ∞
G= . (41)
2d(A0+A1ϕ∞)
Once equation (39) is substituted in this relationship we obtain
2π2Iϕ∞ β2ϕ∞
G= + . (42)
dA0T2 8dIA0
2.4. G from Leybold manual
The deduction of G through the Leybold torsional balance is widely known , see
11
. We simply report the final expression
π2b2d∆S
G= ×(1+β) , (43)
MT2l
where
b3
β = . (44)
(b2+4d2)pb2+4d2
The meaning of the symbols is
• b:Distance between center of the great mass and small mass
• ∆ S:Total deflection of the light spot
• d:The length of the lever arm
• l:Distance between mirror and screen
• M: Great mass
• T Period of the oscillations
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8 Rossi Zaninetti
parameter value unit
M (1.5 ± 10−3) Kg
m (1.5·10−2 ± 10−3) Kg
r (4.65·10−2 ± 10−3) m
d (5.0·10−2 ± 10−3) m
l (5.475 ± 10−3) m
ϕ (-6.715·10−3 ± 1.3·10−5) rad
β ( 1.432·10−7 ± 1.11·10−8) kg·m2
s
parameter value unit
A (8.188·10−2 ±1.11·10−4) m
0
A ( 0.1470 ± 2.85·10−4) m
1
T (552.98 ± 0.16) s
τ (1047.0 ± 4.8) s
method equation value uncertainty [ppm] accuracy %]
G averaged (32) (6.67 ± 0.34)·10−11 m3 52433 0.0161
kg·s2
G with air viscosity (34) (6.72 ± 0.35)·10−11 m3 52433 0.69
kg·s2
G to the first order (42) (6.80 ± 0.34)·10−11 m3 49989 1.92
kg·s2
G from Leybold manual (43) (6.71 ± 0.33) ·10−11 m3 49600 0.64
kg·s2
3. Analysis of the data
The physical parameters as well their uncertainties are reported in Table 1. The
data were analysed through the following fitting function
2πt t
y(t)=A +A cos( )exp(− ) . (45)
0 1
T τ
The data has been processed through the Levenberg–Marquardtmethod ( subrou-
12
tine MRQMIN in ) in order to find the parameters A ,A ,T and τ. The results
0 1
are reported in Table 2 together with the derived quantities.
The value of G can be derived coupling the basic parameters of the torsion
balance,seeTable1,andthe measuredparametersofthe dampedoscillations,see
Table 2.Table 3 reportsthe four valuesofGhere consideredwiththe uncertainties
expressed in absolute value and in ppm ; the precision of the measure in respect
of the so called ”true” value is also reported. A considerable source of error is the
uncertainty in the determination of the span between the two spheres that in our
case is ≈ 10−3m . Adopting a rotating gauge method 13 the uncertainty in the
determination of the span between the two spheres is ≈ 0.5×10−6m; this is the
way to lower the uncertainty in Table 3.
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Torsion Balance 9
4. Non linear effects in the vacuum
By starting from the equation of rotational dynamics up to the second order
I ϕ.. +β ϕ. +(τ −2dGA )ϕ−2dGA =2dGA ϕ2 , (46)
1 0 2
the case of β=0 is analysed ,
I ϕ.. +(τ −2dGA )ϕ−2dGA =2dGA ϕ2 , (47)
1 0 2
that corresponds to perform the experiment in the vacuum. On dropping the con-
stant term and dividing by I we obtain
ϕ.. +(τ −2dGA1)ϕ= 2dGA2ϕ2 . (48)
I I
On imposing
(τ −2dGA )
ω2 = 1 , (49)
0 I
the nonlinear ordinary differential equation , in the following ODE , has the form
ϕ.. +ω2ϕ= 2dGA2ϕ2 . (50)
0 I
On adopting the transformation T =t∗ω the nonlinear ODE is
0
ϕ.. +ϕ−ǫϕ2 =0 , (51)
where
2dGA
2
ǫ= . (52)
(τ −2dGA )
1
The solution of equation (51) is reported in the Appendix Appendix A and in our
case ǫ=0.0187.We now have a measured period ,T , that is equalised to the non
MS
linear value , T . The period of the linear case , T , can be written as
NL L
T =T =1.00014T , (53)
NL MS L
and therefore
T
MS
T = . (54)
L
1.00014
InthevariousformulaeofGwithoutdamping,forexampleequations(32)and(43),
the periods T and T are raised to the square
L MS
T2
T2 = MS , (55)
L 1.00029
and in the denominator, making the non linear G greater than the linear G
NL L
G =1.00029G . (56)
NL L
The value of this correction , δG, can be evaluated as a difference between 1 and
the multiplicative factor of G
L
δG=(1.00029−1)×6.6742×10−11m3kg−1s−2 =0.19×10−13m3kg−1s−2 . (57)
The official error on G is 0.1×10−13 m3kg−1s−2 and therefore the nonlinear cor-
rection can be expressed as the double of the official error on G.
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10 Rossi Zaninetti
Appendix A. The eardrum equation
The equation
x¨+x+ǫx2 =0 , (A.1)
14
is wellknownunder the name ”eardrumequation”.It canbe solved,see ,trans-
forming it in
d2
Ω2 X(T)+X(T)+ǫ (X(T))2 =0 , (A.2)
dT2
and adopting the method of Poisson that imposes the following solution to X
x(T)=x0 (T)+x1(T)ǫ+x2(T)ǫ2 , (A.3)
and to Ω
Ω=1+ω1 ǫ+ω2ǫ2 . (A.4)
The computer algebra system (CAS) gives
ω =0 ω :=−5/12 . (A.5)
1 2
Acknowledgements
We thank Richard Enns who has provided us the Maple routine Example 04-S08
14
extracted from .
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