Machian space quanta

1.4 MB·

by  David W. Essex

Machian space quanta
David W. Essex∗
Clare College, Trinity Lane, Cambridge, CB2 1TL, England
(Dated: February 4, 2015)
A new model for space and matter is obtained by joining every pair of point charges in the
observableuniversebyanetherealstring. Positivegravitationalpotentialenergyineachstringgives
an attractive gravitational force due to the action of an energy conservation constraint. Newton’s
laws of motion are derived and inertia is explained in accordance with Mach’s principle. The
Machian string model gives a surprisingly simple way to understand the expansion history of the
5 Universe. The decelerating expansion in the radiation era and the matter era is explained without
1 using General Relativity and the transition from deceleration to acceleration is explained without
0 theneedtointroduceaseparate‘darkenergy’component. TheinteractionbetweenMachianstrings
2 gives a physical model for modified Newtonian dynamics (MOND) and is therefore an alternative
to ‘dark matter’.
n
a “Iconsideritentirelypossiblethatphysicscannotbebaseduponthefieldconcept,thatisoncontinuous
J structures. Thennothingwillremainofmywholecastleintheair,includingthetheoryofgravitation,but

5 alsonothingoftherestofcontemporaryphysics.”(AlbertEinstein)
2
PACSnumbers:

]
h
p I. MOTIVATION given elementary particle can be imagined as a quantum
- of space with the size of the observable universe and will
n
e The purpose of the present paper is to suggest a new be referred to as a Machian space quantum. The strings
g modelforthestructureofelementaryparticlesandanew containenergy,andhencemass,butarenotcharged. The
s. modelforspaceandtoofferanewapproachtotheprob- strings associated with the space quanta of two nearby
c lems of ‘dark matter’ and ‘dark energy’. particles of mass m1 and m2 are illustrated in Figure 1.
i The strings connecting two such masses will be referred
s One of the basic properties of elementary particles is
y that they have spin one-half. Quantum mechanically, a toasdirectstringsandstringsconnectedtodistantmat-
h ter will be referred to as Machian strings.
spinone-halfparticleendsupinadifferentstatewhenro-
p
tatedthroughanangleof2π andonlyreturnstoitsorig-
[
inal state when rotated through an angle of 4π. There is

1 no analogue of such behaviour for a classical point par-
v
ticle but a mechanical model for spin one-half can be
0
constructed when the particle is connected to its sur-
6
roundings, for example by someone holding onto it or
6
0 by strings connecting it to a rigid external frame [1, 2].
0 Suppose that the strings are initially not entangled. A
. rotation of the particle through an angle of 4π results in
2
0 a configuration that is equivalent to the initial configu-
5 ration, in the sense that the strings can be disentangled,
1 but a rotation through an angle of 2π results in a config-
: urationthatisnotequivalenttotheinitialconfiguration.
v
i In searching for a new model for space it is natural to FIG. 1: Schematic diagram of the direct strings and the
X reconsider Mach’s principle, which was never fully incor- Machian strings in the Machian space quanta of two masses
r poratedintoGeneralRelativity[3–5]. Insteadofdefining m1 and m2. The direct strings join the centres of the space
a quanta associated with the two masses and the Machian
inertial frames in terms of a local absolute space, Mach
strings connect the centres with the centres of all the other
arguedthattheinertialforcesexperiencedbyanacceler-
space quanta in the observable universe.
atedparticleareduetosomesortofinteractionwiththe
distantmatterintheUniverse. Perhapsaphysicalmodel
for space can be constructed by defining space itself in The idea that an elementary particle could have the
terms of distant matter [6]. size of the observable universe appears, at first sight, to
InthemodelofMachianspacequanta,themechanical be completely contrary to experience. Indeed, high en-
modelforspinone-halfandMach’sprincipleareincorpo- ergyscatteringexperimentshavefailedtofindanyspatial
ratedbypostulatingthatanelementaryparticleconsists extentatallinthechargedistributionofeitherquarksor
of a set of ethereal strings connecting a charged centre electrons. However, experiments cannot probe an entire
to the centre of every other elementary particle in the spacequantumbecausetheonlypartofaspacequantum
observable universe. The set of strings associated with a that can interact directly with another space quantum
is the centre. The only conclusion that may therefore
be drawn from scattering experiments is that the charge
distributionatthecentreofaspacequantumispointlike.
∗Electronicaddress: [email protected] The gravitational interaction between the two masses
2
m and m illustrated in Figure 1 is defined by speci- product is h. The period is carried by the waves
1 2
fying the total energy, Gm m /r, in the direct strings without change, but the energy spreads out in
1 2
joiningthem,wherer isthedistancebetweenthecentres an ever-widening circle. Eight years and nine
ofthetwomasses. Toafirstapproximation, thegravita- monthsaftertheemissionthewave-frontisdueto
reachtheearth. Afewminutesbeforethearrival
tional masses m and m are simply the total masses in
1 2
some person takes it into his head to go out and
the associated space quanta and it makes no difference
admire the glories of the heavens and - in short
at all whether all the mass of a space quantum is at the
- to stick his eye in the way. The light waves
centreorwhetherthemassisdistributedthroughoutthe
when they started out could have had no notion
strings. For the further development of the theory, how-
ofwhattheyweregoingtohit;foralltheyknew
ever,thefactthatthemassisdistributedthroughoutthe they would be bound on a journey through end-
strings is very significant. The fundamental variables in lessspace,asmostoftheircolleagueswere. Their
theproblemarethestringsthemselves,ratherthanposi- energywouldseemtobedissipatedbeyondrecov-
tions and velocities of the particle centres. The energies eryoverasphereof50billionmiles’radius. And
in the strings may be calculated explicitly, including the yetifthatenergyisevertoentermatteragain,if
dependence on the velocities of the particles, and it is itistoworkthosechemicalchangesintheretina
which give rise to the sensation of light, it must
then possible to derive the properties of particle inertia
enter as a single quantum of action h [8].
andkineticenergyfromfirstprinciplesinsteadofputting
them in by hand. When the contribution to the energies How can a photon possibly manage to keep itself all to-
ofthestringsfromtheexpansionoftheUniverseistaken gether as a single tiny quantum when it is so incredibly
into account, the string model is found to give an accel- spreadoutacrosstheUniverse? Theanswer, perhaps, is
erating expansion of the Universe without the need to that a photon is also a Machian space quantum. A pho-
introduce any mysterious ‘dark energy’. tonspacequantumstillhasacentre,definedasthepoint
Itmightbethought,atfirstsight,thatMachianstrings of intersection of all the photon strings, but the centre
give a very simple solution to the problem of ‘dark mat- has no mass and no charge.
ter’. Itiswellknownthatflatgalaxyrotationcurvescan Machian space quanta may also help to understand
be accounted for by assuming that galaxies lie within the invariance of the speed of light. Special Relativity
a dark matter halo whose density at a distance r from is based on the postulate that the speed of a light wave,
the centre of the galaxy is proportional to 1/r2 [7]. For c, is independent of the motion of the light source rel-
strings with uniform mass per unit length, the mass in ative to the observer. Is there any way to understand
the strings enclosed within a spherical shell of thickness the invariance of c with some sort of physical model or
dr at any radius is proportional to dr and it follows that must the postulate simply be accepted? In the old ether
the density in the strings is proportional to 1/r2, which model, the speed of light was assumed to be equal to c
appears to be exactly what is required. This observa- in the rest frame of an ‘ether’ that pervades all space.
tionwas,infact,theoriginalmotivationfortheMachian The invariance of the speed of light waves with respect
string model. Unfortunately the idea doesn’t work be- tomotionofthesourceisthenobvious,justasthespeed
cause the only gravitational interaction between space of sound waves in air and water waves in water are also
quanta is the interaction due to the energies in their independent of the motion of the source. However, the
common direct strings. Unlike a conventional mass dis- invariance of c with respect to motion of the observer
tribution,themasselementswithinaMachianstringare was much harder to explain and led to the conclusion
invisibletootherspacequantaanddonot, therefore, act that motion through the ether somehow causes the time
as additional point sources for the gravitational field. In dilation of clocks and the length contraction of metre
any case, such a model would give an additional gravi- sticks. The Machian string model offers an alternative
tational acceleration proportional to M/r, whereas the point of view because the strings of the observer space
additionalaccelerationrequiredtoaccountforgalaxyro- quantum define an ether that is always at rest relative
√
tationcurvesisproportionalto M/r. Adifferentmodel to the observer. Suppose the speed c is a characteristic
for‘darkmatter’,basedonacontactinteractionbetween of the interaction between a photon space quantum and
thestringsofdifferentspacequanta,isintroducedinSec- the space quantum of an observer. The invariance of c
tion V. withrespecttomotionoftheobserverthenfollowsauto-
Whereas the quantisation of matter is automatically matically because, when the observer starts moving, the
present at the classical level a photon is, by definition, a space quantum of the observer moves with it.
quantum mechanical object. The strange quantum me- Although the centre of a space quantum can be any-
chanical behaviour of the photon was described by Ed- where in the continuous three-dimensional coordinate
dington in 1928 as follows. space, physical space is defined as the space occupied by
the strings and is therefore quantised. The space quan-
The pursuit of the quantum leads to many sur- tum of an elementary particle is an extended object and
prises; but probably none is more outrageous to
Quantum Mechanics also requires that every elementary
our preconceptions than the regathering of light
particle is associated with an extended object, namely
and other radiant energy into h-units, when all
thewavefunction. PerhapsSchrodinger’s3N-dimensional
the classical pictures show it to be dispersing
configurationspaceshouldnotberegardedasanabstract
more and more. Consider the light waves which
areemittedasaresultofasingleemissiononthe mathematicalspacebutshouldinsteadbeidentifiedwith
star Sirius. These bear away a certain amount thephysicalspacedefinedbythesuperpositionofNspace
ofenergyendowedwithacertainperiod,andthe quanta.
3
The inertial mass constraint (IMC) for space quanta B. Newtonian gravity and the constant of
is introduced in Section II and, in Section III, the IMC gravitation
is used to give a physical model for Newtonian gravity.
Whenrelativisticeffectsareincludedandtheexperimen-
In the Machian string model, Newton’s point masses
tal constraints are applied, the equations of motion are
andnegativegravitationalpotentialenergiesarebothre-
found to be the same as in General Relativity. Inertia
placed by positive gravitational potential energy in the
and kinetic energy are explained in terms of the energies
strings. In the simplest model of Machian space quanta,
in the Machian strings and a Machian formula is given
the direct strings in Figure 1 joining the centres of the
for G, the constant of gravitation. Possible solutions to
two masses m and m , a distance r apart, have en-
1 2
the dark energy and dark matter problems are given in
ergy+Gm m /r,whereGistheconstantofgravitation.
1 2
Sections IV and V, respectively.
Due to the action of the IMC, the total energy at the
centre and in the Machian strings connected to m is
1
m c2−Gm m /r,andsimilarlyform . Thetotalenergy
1 1 2 2
in the centres and all the strings of m and m is there-
1 2
II. THE INERTIAL MASS CONSTRAINT (IMC) fore m c2+m c2−Gm m /r, so the interaction energy
1 2 1 2
is the same as in Newtonian gravity. The idea that the
gravitational mass energy is positive, with a compensat-
Inertialmassisdefinedasthetotalenergyinthespace
ing reduction in the energy of each mass, was suggested
quantum of a massive particle in its instantaneous rest
many years ago [11, 12].
frameandtheinertialmassconstraint(IMC)statesthat
the inertial mass of every massive particle is constant. There is a well-known Machian relation for G [5],
The IMC allows an attractive gravitational force to be namely
obtained with positive potential energies in the strings,
as explained in Section III. The IMC is also used in Sec- (cid:16)M (cid:17)−1
tionIVon‘darkenergy’andSectionVon‘darkmatter’. G ∼ c2 U , (1)
R
U
where M and R are the mass and radius of the ob-
U U
servable universe, respectively. In the string model, the
III. A MACHIAN MODEL FOR GRAVITY
relation (1) follows directly from the fact that the iner-
tial mass, mc2, is the same order of magnitude as the
The equations of motion for a Machian model of grav- total gravitational potential energy in the strings. Since
ity are derived in the Appendix. The model is a direct the average length of a string is of order RU, the total
interaction theory, as opposed to a field theory, and is gravitationalpotentialenergyinthestringsconnectinga
definedbytheenergiesinthestringsconnectingallpairs mass m to distant matter of total mass MU is of order
ofparticlesintheobservableuniverse. Theenergyinthe GmMU/RU so mc2 ∼GmMU/RU.
strings has the form of positive Newtonian gravitational
potentialenergy,withadditionaltermsdependingonthe
velocities of the particles.
C. Inertial and gravitational mass
The inertial mass in the space quantum of the ith par-
A. Inertia and Newton’s laws of motion
ticle,m ,isdefinedasthemassatthecentreofthespace
i
quantum plus the mass in all the strings connected to it.
According to Newton’s laws of motion, a free particle The gravitational mass of the ith particle, m˜i, is defined
in an ‘inertial frame’ moves with uniform velocity and a asthemassatthecentreplusthemassinthe(verylarge)
force ma is required to give a mass m an acceleration a. subsetofstringsconnectedtodistantmatteronly. When
In General Relativity, acceleration is defined relative to all the space quanta are superposed, each pair of space
the local ‘spacetime’ around the particle and it has been quanta has two strings connecting their centres. The to-
suggested that an interaction with the quantum vacuum tal energy in the two strings is Gm˜im˜j/rij so the energy
is responsible for the existence of inertial forces [9, 10]. ineachofthetwostringsisGm˜im˜j/2rij. Sincethediffer-
In the Machian string model, the energy of a particle ence between inertial mass and gravitational mass is due
is not concentrated at a point but is instead distributed to the mass in the strings connected to nearby matter,
throughouttheMachianstrings. Inertialframesareiden- itfollowsthatinertialandgravitationalmassarerelated
tified with frames moving at uniform velocity relative by the equation
to distant matter, in accordance with Mach’s principle.
Appendix B shows that the kinetic energy of a parti- m c2 =m˜ c2 + (cid:88) Gm˜im˜j , (2)
cle is part of the gravitational interaction energy in the i i 2r
ij
Machian strings. Appendix D shows that distant matter j∈Sn,j(cid:54)=i
exerts a force −ma on a particle that has acceleration a
relative to distant matter, so that an external force ma whereS denotesthesetofnearbyparticlesinthesystem
n
is needed to maintain the acceleration. under consideration.
4
D. Experimental constraints IV. THE EXPANSION OF THE UNIVERSE
AND ‘DARK ENERGY’
1. The two body and three body problems
At first sight, the Machian relation (1) implies that G
Relativistic corrections for the motion of two gravitat- is a functionoftime since M and R vary with time as
U U
ing masses and the acceleration of the Earth and Moon the Universe expands. It is shown in Appendix H, how-
towards the Sun in the three-body Earth-Moon-Sun sys- ever, that the relation (1) is not only consistent with the
tem are calculated in Appendix E. After applying the expansion history of the Universe but it can, in fact, be
experimental constraints, the equations of motion are used to calculate the expansion history of the Universe.
found to be the same as in General Relativity. In the
corresponding Machian string model, approximately 8% Oncosmologicaltimescales,theexpansionoftheUni-
ofthemassofaspacequantumisintheMachianstrings verse has a significant effect on the number of Machian
and the remaining 92% of the mass is at the centre. The strings in the Universe and on the energy in each string.
effect of a gravitational field on the precession of a gyro- In all space quanta, the number of Machian strings in-
scope is considered in Appendix F and the usual results creases with time because the number of particles in the
for the rates of geodetic and gravitomagnetic precession observableuniverseincreasesastheUniverseexpands. In
are recovered. addition,themassassociatedwiththepotentialenergyin
eachstringhasakineticenergyassociatedwithitsreces-
sionalvelocity. Intheearlyuniverse,itturnsoutthatthe
kinetic energy of the potential energy is small compared
2. Gravitational redshift and the propagation of light withthepotentialenergyso,astheUniverseexpands,the
totalenergyineveryMachianstringdecreases. TheIMC
requires the effects of the increasing number of strings
Gravitationalredshiftandthepropagationoflightina
and the decreasing energy of each string to balance and
gravitational field are discussed in Appendix G. Photons
the result is that the expansion of the Universe deceler-
propagate with constant energy in a gravitational field
ates. The expansion rates in the radiation era and the
because a massive particle propagates with constant en-
matter era are the same as in the conventional ΛCDM
ergy and a photon may be thought of as a massive par-
model.
ticle with vanishingly small rest mass. From the relation
E = hf, it follows that the frequency of a photon prop-
agating in a gravitational field is constant. To explain
the gravitational redshift, consider an atom of mass m
at rest at a distance r from a mass M. The energy in
the strings joining the two masses is GMm/r and the
IMC implies that the energy at the centre of the atom is
mc2−GMm/r, so the atomic energy levels are reduced
by a factor of 1−GM/rc2. The frequency of a photon
corresponding to a given atomic transition is therefore
lowerwhenthephotonisemittedinagravitationalfield.
The frequency remains unchanged as the photon propa-
gatesoutofthegravitationalfield,sothephotonemerges
with a lower frequency and therefore appears to be red-
shifted. AppendixGalsoshowsthatthespeedoflightis
reduced in a gravitational field and Fermat’s principle is
used to calculate the deflection of a light beam and the
time delay of radar signals.
FIG. 2: Scale factor, R(t), in the string model (red) and in
theΛCDMmodel(green)asafunctionofthetime,t. Inboth
3. Gravitational radiation models, the present time is at R = 1. The two scale factors
are almost indistinguishable experimentally, as discussed in
Appendix H.
ThecalculationsinAppendixIshowthattheMachian
stringsaroundamassivebodyaredeformedbythepres-
ence of a nearby mass. When two masses orbit one an-
other, the deformation in the strings varies periodically In the later universe, the kinetic energy in each string
intimeandwavesintheMachianstringsareexpected. A becomes larger than the potential energy and the IMC
formulafortherateatwhichenergyisradiatedisderived then requiresan accelerating expansion tokeepthe total
inAppendixJforthecaseofcircularorbits. Comparison kinetic energy in the space quantum of a massive parti-
with the experimentally verified formula for the rate of cle constant. The time evolution of the scale factor in
emission of gravitational radiation in General Relativity the string model calculated in Appendix H is shown in
suggests that gravitational radiation may be attributed Figure 2 and compared to the time evolution in the con-
to the generation of gravitational waves in the strings. ventional ΛCDM model.
5
V. GALAXY ROTATION CURVES AND ‘DARK A plot of the Newtonian gravitational acceleration and
MATTER’ the additional acceleration calculated in Appendix I is
shown in Figure 3, plotted in units of a . The total ac-
0
It is known that Newton’s law of gravity, when ap-
plied to the observed mass distribution, does not give
the correct dynamics on the scale of galaxies and galaxy
clusters. Either there is a very large amount of still un-
seen dark matter or Newton’s law is incorrect. In mod-
ified Newtonian dynamics (MOND) [13], the Newtonian
1/r2 force law is replaced by a 1/r force law when the
Newtonian acceleration is less than a characteristic ac-
celeration, a . It has been shown [14] that MOND gives
0
averygoodfittotheobservedgalaxyrotationcurvesfor
a ≈1.2×10−10m/s2.
0
In the Machian string model, an acceleration scale of
ordera isanaturalconsequenceofacontactinteraction
0
between Machian strings. Consider a test mass in the
vicinity of a larger mass M. In addition to the Machian FIG.3: TheNewtonianacceleration(redcurve)andthead-
strings of M, the test mass is immersed within the back- ditional acceleration due to the interaction between Machian
ground space defined by the Machian strings joining all strings (green curve) around a point mass M in the Machian
stringmodelforthemodelconsideredinAppendixI.Theac-
the pairs of particles in the Universe. It is therefore rea-
celerationisplottedinunitsoftheMONDaccelerationscale,
sonabletosupposethattheeffectoftheMachianstrings
a , and the radial distance r is plotted in units of the radius
of M on the Machian strings of the test mass depends 0
R , defined as the radius at which the Newtonian gravita-
g
on the ratio of the density of the Machian strings of M
tional acceleration is equal to a , so the MOND acceleration
0
to the density of background strings. In Appendix I, it is equal to 1/r2 for r<1 and 1/r for r>1.
is shown that the ratio of densities is equivalent to the
ratio of the Newtonian gravitational acceleration to a .
0 celeration in the string model is seen to be very close to
Physically, the additional acceleration is due to the the MOND acceleration when the acceleration is much
forces exerted on the centre of the test mass by the less than a , so the string model is expected to give a
0
Machian strings connected to it. Due to the action of good fit to the observed galaxy rotation curves.
the IMC, the force between every pair of space quanta is
attractive so the strings are in tension. In the absence of
the mass M, the distribution of Machian strings around VI. SUMMARY
the test mass would be spherically symmetrical and all
the tension forces on the centre would cancel out. The
In the Machian string model, direct strings give New-
interaction with the Machian strings of M causes the
tonian gravity, Machian strings give inertia and the ex-
distribution of Machian strings around the test mass to
pansion of the Universe and dark matter effects are due
become asymmetrical, so the forces exerted on the cen-
to the interaction between Machian strings.
tres no longer cancel out and the test mass experiences
Aformulaforthegravitationalpotentialenergyinthe
an additional acceleration.
strings is derived from a relativistic string action in Ap-
To estimate the magnitude of the additional accelera- pendix A. The usual formula for the Newtonian kinetic
tion,considerthetotalenergyintheMachianstringscon- energy of a massive particle can be derived by a suitable
nectedtoatestmassm. AsnotedinSectionIIID,about choice of the fraction, κ, of the mass of a space quan-
8% of the rest mass, 0.08mc2, is in the Machian strings. tum that lies in the strings. It is shown in Appendix B
Dividing the total energy in the Machian strings by the that the relativistic correction to the kinetic energy and
average length of a Machian string, RU/2, gives a force all the experimental results for the relativistic two-body
of about 0.16mc2/RU. In the limiting case of maximum problem, the Nordtvedt effect and gyroscope precession
asymmetry, with all the Machian strings pointing in the are obtained if the free parameters a , a , b and b are
1 2 1 2
same direction at the centre, an additional acceleration suitablychosen. Thetheorygivesthestandardresultsfor
of 0.16c2/RU would be expected. The acceleration a0 is gravitationalredshift,lightbendingandradartimedelay
approximately0.4c2/RU,whereRU =3.6×1026misthe andtheelectromagneticinteractionbetweenchargedpar-
radius of the observable universe found in Appendix H, ticlescanalsobeincludedintheformalism, asexplained
so the maximum additional acceleration is about 0.4a0. in Appendix G. A physical model for the observed rate
In contrast to MOND, which simply postulates a ofemissionofgravitationalradiationfrombinarypulsars
change in Newton’s law, the additional acceleration in is discussed in Appendix J.
thestringmodelisaddedtotheNewtonianacceleration. The expansion history of the universe can be calcu-
Thus,althoughtheadditionalaccelerationissignificantly lated by considering the total energy in all the Machian
lessthana ,itisstillsufficienttoexplaintheMONDphe- strings of a space quantum. The usual results for the
0
nomenology. Aspecificmodelfortheinteractionbetween decelerating expansion of the Universe in the radiation
MachianstringsisconsideredinAppendixIandthemax- and matter eras are obtained if the mass fraction κ is
imum additional acceleration is found to be about 0.2a . assumed to be constant. By including the effect of the
0
6
HubbleflowontheenergyofaMachianstring,agoodfit It is desired to find the Lagrangian L in a general
totheexperimentaldataisobtainedwithoutintroducing frame, S, in which the velocities of the ith and jth par-
‘dark energy’. ticles are v and v , respectively. In the special case
i j
The interaction between Machian strings gives effects B(s)=sA(s),theactionisinvariantunderreparameter-
usually attributed to ‘dark matter’. The strength of the isationoftheτ coordinateoftheformτ →τ(cid:48)(τ,σ)andτ
interaction depends on the density of space defined by can then be chosen to be the time coordinate in the gen-
the Machian strings. Since the ratio of the density of eral frame S as well as in S [17]. More generally, the
cm
space around a mass to the background density of space Lagrangian must first be evaluated in S , using (A5),
cm
is equivalent to the ratio of the Newtonian gravitational and a Lorentz transformation must then be applied to
accelerationtotheaccelerationscalea ,theMONDphe- find the Lagrangian in S.
0
nomenology is explained. In the Machian string model, the energy in a string of
length r is proportional to 1/r, so n = −1. Consider a
straight string in the centre of mass frame. Then x(cid:48) =ξξξ,
VII. ACKNOWLEDGEMENTS whereξξξistheinstantaneousseparationofthetwocentres
in S , so (∂Xµ/∂τ)(∂X /∂σ) = −ξξξ.x˙, (∂Xµ/∂σ)2 =
cm µ
Iwouldliketothankbothmyparentsforalltheirhelp −ξ2 and s = 1−x˙2/c2. It is convenient to rewrite the
and support over the years. They gave me the time and Lagrangian density (A2) in the form
space in which to work and the courage to think along
different lines. The financial support of Trinity College, (cid:104)g(z)(cid:16)∂Xµ∂Xµ(cid:17)2 (cid:16)∂Xµ(cid:17)2(cid:105)−1/2
L = f(z) − (A6)
Cambridge, is also gratefully acknowledged. The pro- c2 ∂τ ∂σ ∂σ
gramming language used for the numerical calculations f(z)(cid:104) (ξ(cid:98)ξξ.x˙)2(cid:105)−1/2
= 1+g(z) , (A7)
is available online [15].
ξ c2
where f(z) and g(z) are arbitrary functions of z =x˙2/c2
Appendix A: Lorentz invariance and the Lagrangian
defined by A(s)=g(z)/f(z)2 and B(s)=1/f(z)2 andξ(cid:98)ξξ
for a string
denotes the unit vector along ξξξ. Expanding f and g in
powers of z, the Lagrangian (A4) has the form
Let Xµ(τ,σ) be the spacetime coordinates of a classi-
cal string joining the centres of the ith and jth particles, Γ(cid:90) 1
where τ is a time-like coordinate and σ ∈ [0,1] parame- Lcm = ξ dσ(1+a1z+a2z2+...)
terises the position along the string [16]. An action for 0
the string may be written in the form (cid:104) (ξ(cid:98)ξξ.x˙)2(cid:105)−1/2
× 1+(b +b z+...) , (A8)
1 2 c2
(cid:90)
S = L dτdσ, (A1)
forsomeconstantsΓ,a ,a ,b andb . Let(ct,r)denote
1 2 1 2
the coordinates in S. To lowest order, the relative posi-
wheretheLagrangiandensityLissomescalarfunctionof tions and velocities of the ends of the string are Lorentz
thefour-vectors∂Xµ/∂τ and∂Xµ/∂σ. Themostgeneral invariant, so ξ(cid:98)ξξ may be replaced by (cid:98)rij, the unit vector
form of a Lagrangian density with dimensions rn is
along the string in S, and the velocity of the string as a
function of σ may be written as x˙ =(1/2−σ)v . Then,
(cid:104)A(s)(cid:16)∂Xµ∂X (cid:17)2 (cid:16)∂Xµ(cid:17)2(cid:105)n/2 ij
L = µ −B(s) , (A2) to O(v2/c2), (A8) gives
c2 ∂τ ∂σ ∂σ
where A(s) and B(s) are arbitrary functions of s = L = Γ(cid:104)1+ a1 vi2j − b1 ((cid:98)rij.vij)2 +...(cid:105). (A9)
(∂Xµ/∂τ)2/c2. The Nambu-Goto Lagrangian [16], for cm ξ 12 c2 24 c2
example, has A = 1, B = s and n = 1. Let the centre
If S moves with velocity w relative to S then dτ =
ofmassframeofthestring, S , bedefinedastheframe cm
cm
dt/γ(w). From (A3), L=L /γ(w), so
in which the velocities of the ends of the string are equal cm
andopposite. Iftheparameterτ ischosentobethetime
Γ (cid:104) a v2 b (r .v )2 (cid:105)
coordinate in Scm then L = 1+ 1 ij − 1 (cid:98)ij ij +... .(A10)
γ(w)ξ 12 c2 24 c2
(cid:90)
S = L dτ, (A3)
cm The quantity 1/γ(w)ξ must now be rewritten in terms
of the positions and velocities at a given time in frame
where the Lagrangian in Scm is given by S. The instantaneous separation ξξξ of the two centres in
frameS canberelatedtotheinstantaneousseparation
(cid:90) cm
Lcm = L dσ. (A4) rij inframeS byamodificationofthemethodin[18]that
is symmetrical with respect to the two centres. Consider
thetwoeventsAandC inFig4, whicharesimultaneous
If x denotes the spatial coordinates in S then Xµ =
cm in S at time τ and separated by the distance ξ. In
(cτ,x) and cm A
S, the events A and C are not simultaneous. A line
∂Xµ ∂Xµ of simultaneity in S is shown in Fig 4 connecting the
= (c,x˙) and = (0,x(cid:48)). (A5) events B and D. The events B and D are defined to be
∂τ ∂σ
7
For interactions with distant matter, the centre of mass
frame of m and particles of distant matter is effectively
i
the rest frame of the distant matter. The calculation of
relativistic corrections to kinetic energy in Appendix B
requires O(v4/c4) terms proportional to v4/c4. The ve-
i
locityofm inS isv /(1−v .v /c2),buttheO(v2/c2)
i cm ij i j
correction to v does not lead to any terms in the La-
ij
grangianproportionaltov4/c4andisthereforeneglected.
i
The velocity of the string as a function of σ may there-
fore be written as x˙ = σv . Similarly, it is sufficient to
ij
use the approximation ξ(cid:98)ξξ =(cid:98)rij in the evaluation of (A8)
and expansion to O(v4/c4) then gives, instead of (A10),
Γ (cid:104) a v2 b (r .v )2 a v4
L = 1+ 1 ij − 1 (cid:98)ij ij + 2 ij
γ(w)ξ 3 c2 6 c2 5 c4
FIG. 4: Diagram showing the worldlines of the ith and jth 3b2(r .v )4 b +a b (v )2(r .v )2(cid:105)
particles in the centre of mass frame, S . The line AC is + 1 (cid:98)ij ij − 2 1 1 ij (cid:98)ij ij .
cm 40 c4 10 c4
a line of simultaneity in S and the line BD is a line of
cm (A17)
simultaneity in the frame S.
Evaluating the prefactor (A14) with w=v instead of
j
symmetrical relative to the line of simultaneity in Scm, w=(vi+vj)/2 gives, instead of (A15),
so that τ −τ =τ −τ . The Lorentz transformations
B A A D
relatingthepositionsofeventsBandDinthetwoframes
are 1 1 (cid:104) v2 (r .v )2 (cid:105)
= 1− j − (cid:98)ij j +... , (A18)
γ(w)ξ r 2c2 2c2
x = r +(γ−1)w(w.r )−γwt (A11) ij
B B (cid:98) (cid:98) B B
and x = r +(γ−1)w(w.r )−γwt . (A12)
D D (cid:98) (cid:98) D D
Since the velocities of the centres are equal and opposite with no O(v4/c4) terms proportional to vi4/c4. Substi-
in S and τ −τ = −(τ −τ ), it follows that x − tuting into (A17) then gives, instead of (A16),
cm B A D C B
x =x −x . Then
A D C
ξξξ = xA−xC = xB −xD = rij +(γ−1)w(cid:98)(w(cid:98).rij) L = Γ (cid:104)1− vj2 − ((cid:98)rij.vj)2 + a1vi2j − b1((cid:98)rij.vij)2
⇒ ξξξ =r + w(w.rij) +... rij 2c2 2c2 3 c2 6 c2
ij 2c2 a v4 3b2(r .v )4 b +a b (v )2(r .v )2(cid:105)
(r .w)2 + 2 ij + 1 (cid:98)ij ij − 2 1 1 ij (cid:98)ij ij .
⇒ ξ2 = r2 + ij +... . (A13) 5 c4 40 c4 10 c4
ij c2
(A19)
Forinteractionswithnearbymatter,thecentreofmass
velocity is w=(v +v )/2 so
i j
1 (cid:16) w2 (cid:17)(cid:104) (r .w)2(cid:105)−1/2
= 1− +... r2 + ij (A14)
γ(w)ξ 2c2 ij c2
1 (cid:104) w2 (r .w)2 (cid:105) Appendix B: The total Lagrangian and experimental
= 1− − (cid:98)ij +... constraints
r 2c2 2c2
ij
1 (cid:104) v2 v .v (r .v )2
= 1− ij − i j − (cid:98)ij ij TodefinethetotalLagrangianforasystemofparticles
r 8c2 2c2 8c2
ij
interactingwitheachotherandwiththedistantuniverse,
(r .v )(r .v ) (cid:105)
− (cid:98)ij i (cid:98)ij j +... (A15) it is necessary to define the gravitational mass of the ith
2c2 particle, m˜ , for each interaction. For interactions be-
i
tweena particlein thesystem and adistant particle, the
and the Lagrangian (A10) becomes
gravitational mass of each particle is equal to the iner-
Γ (cid:104) (cid:16)a 1(cid:17)v2 v .v tial mass, mi. For interactions between two particles in
L = 1+ 1 − ij − i j thesystem,thegravitationalmassesaredefinedbyequa-
r 12 8 c2 2c2
ij tion (2). The total Lagrangian for a system of nearby
−(cid:16)b1 + 1(cid:17)((cid:98)rij.vij)2 − ((cid:98)rij.vi)((cid:98)rij.vj) +...(cid:105). particles interacting with distant matter, using (A9) for
24 8 c2 2c2 interactions between pairs of particles within the system
(A16) and (A16) for interactions with distant matter, is there-
8
fore If the constant κ satisfies the relation
L = (cid:88) Gm˜rim˜j(cid:104)1+(cid:16)1a21 − 18(cid:17)vci22j − v2ic.v2j κ3(cid:16)a1− b61(cid:17) = 21 (B7)
ij
i,j∈Sn,i<j
and the parameters a and b in (B4) are such that
(cid:16)b 1(cid:17)(r .v )2 (r .v )(r .v ) (cid:105) 2 2
− 1 + (cid:98)ij ij − (cid:98)ij i (cid:98)ij j +...
24 8 c2 2c2 1
κδ = (B8)
+ (cid:88) Gmimj(cid:104)1− vj2 − ((cid:98)rij.vj)2 8
r 2c2 2c2 then,afterdroppingthefirsttwotermsin(B6),whichdo
ij
i∈Sn,j∈Sd notcontributetotheequationsofmotion,theLagrangian
+ a1vi2j − b1((cid:98)rij.vij)2 + a2vi4j + 3b21((cid:98)rij.vij)4 has the form
3 c2 6 c2 5 c4 40 c4
(cid:88) (cid:104)1 1 v4 (cid:105)
b +a b (v )2(r .v )2 (cid:105) L = m v2 + m id
− 2 1 1 ij (cid:98)ij ij +... , (B1) 2 i id 8 i c2
10 c4 i∈Sn
whereSndenotesthesetofnearbyparticlesinthesystem + (cid:88) Gmimj(cid:104)1+β(vi2+vj2) +γvi.vj
andSddenotesthesetofparticlesinthedistantuniverse. rij c2 c2
Let the gravitational potential due to distant matter, κ, i,j∈Sn,i<j
be defined by +µ((cid:98)rij.vi)2+((cid:98)rij.vj)2 +ν((cid:98)rij.vi)((cid:98)rij.vj)(cid:105)
c2 c2
(cid:88) Grijmc2j = κ, (B2) + (cid:15) (cid:88) G2mimjmk + ... , (B9)
j∈Sd rijrjkc2
where r is the distance from a distant particle of mass i,j,k∈Sn,j(cid:54)=i,k(cid:54)=j
ij
m to some nearby mass m , and suppose that, in frame where the parameters β, γ, µ, ν and (cid:15) are given by
j i
S, all distant matter particles have velocity v . The av-
d a 1 a 1 b 1
erage values of (r .n)2 and (r .n)4 when summed over β = 1 − , γ = − 1 − , µ = − 1 − ,
(cid:98)ij (cid:98)ij 12 8 6 4 24 8
distantmatterare1/3and1/5,respectively,foranyfixed
b 1 1
unit vector n, so (B1) becomes ν = 1 − and (cid:15) = − . (B10)
12 4 2
L = (cid:88) κm c2(cid:104)1− 2vd2 + 1(cid:16)a − b1(cid:17)vi2d +δvi4d ...(cid:105) The derivation of the Lagrangian (B9) shows that the
i 3c2 3 1 6 c2 c4 kineticenergyofamassisduetothevelocity-dependence
i∈Sn
of the energy in the Machian strings connected to it.
+ (cid:88) Gm˜im˜j(cid:104)1+(cid:16)a1 − 1(cid:17)vi2j − vi.vj The equations of motion are presented in Appendix C
r 12 8 c2 2c2
ij and relativistic effects are calculated in Appendix E.
i,j∈Sn,i<j
The experimental constraints (E8) and (E9) from the
(cid:16)b 1(cid:17)(r .v )2 (r .v )(r .v ) (cid:105)
− 1 + (cid:98)ij ij − (cid:98)ij i (cid:98)ij j +... , two-body problem and (E31), (E32) and (E33) from the
24 8 c2 2c2
Nordtvedt effect have the unique solution
(B3)
3 7 1 1
where β = , γ = − , µ = 0, ν = − and (cid:15) = − .
2 2 2 2
δ = 1(cid:16)a + 3b21 − b2+a1b1(cid:17). (B4) (B11)
5 2 40 6
With the parameters in (B11) and in the special case
From equation (2), the gravitational mass m˜ is given in
i where the reference frame is at rest relative to distant
terms of the inertial mass m by
i matter, the Lagrangian simplifies to
m˜ic2 = mic2 − (cid:88) Gm2˜rim˜j L = (cid:88) (cid:104)1m v2+ 1m vi4(cid:105)
j∈Sn,j(cid:54)=i ij 2 i i 8 ic2
= mic2 − (cid:88) Gm2riimj j + ... . (B5) i∈+Sn (cid:88) Gmimj(cid:104)1+ 3(vi2+vj2) − 7vi.vj
j∈Sn,j(cid:54)=i rij 2c2 2c2
i,j∈Sn,i<j
Substitution into (B3) gives
(r .v )(r .v )(cid:105)
− (cid:98)ij i (cid:98)ij j
L = (cid:88) κm c2(cid:104)1− 2vd2 + 1(cid:16)a − b1(cid:17)vi2d +δvi4d +...(cid:105) 2c2
i∈Sn i 3c2 3 1 6 c2 c4 − (cid:88) G22rmirmjcm2k + ... , (B12)
+ (cid:88) Gmimj(cid:104)1+(cid:16)a1 − 1(cid:17)vi2j − vi.vj i,j,k∈Sn,j(cid:54)=i,k(cid:54)=j ij jk
rij 12 8 c2 2c2 whichisthesameastheLagrangianderivedbyEinstein,
i,j∈Sn,i<j
Infeld and Hoffmann [19]. The Lagrangian (A8) describ-
(cid:16)b 1(cid:17)(r .v )2 (r .v )(r .v ) (cid:105)
− 1 + (cid:98)ij ij − (cid:98)ij i (cid:98)ij j +... ing the motion of the string in its centre of mass frame
24 8 c2 2c2
is therefore consistent with experiment if
− (cid:88) G2mimjmk + ... . (B6) 39
2r r c2 a = and b = −3. (B13)
i,j,k∈Sn,j(cid:54)=i,k(cid:54)=j ij jk 1 2 1
9
Equation (B7) then gives Equation (D1) defines the total force acting on the mass
m in an accelerated reference frame. When m = 0,
3 1 2
κ = (B14) the additional force that has to be added to the right
40
hand side to give ¨r = 0 is m a. A force F = m a is
1 1 1
and (B8) requires a and b to satisfy thereforeneededtomaintainanaccelerationarelativeto
2 2
distant matter, which is Newton’s second law of motion.
b 251
a − 2 = − . (B15) Newton’s first law of motion, that a particle moves with
2 6 120 ¨r = 0 in an inertial frame when there are no external
1
Equation (B14) implies that the total energy in the forces, corresponds to the special case a=0.
Machian strings of a space quantum is
(cid:88) Gmimj = 3 m c2, (B16) Appendix E: Relativistic gravity
r 40 i
ij
j∈Sd
1. The two body problem
so that approximately 8% of the energy of a space quan-
tum is in the Machian strings and the remaining 92% is
Consider the motion of two masses, m and m , in an
1 2
at the centre.
inertial frame. Setting ¨r = 0 and v = 0, the equation
d d
of motion (C2) for m is, to first order in v2/c2,
1
Appendix C: Equations of motion 1(cid:16) v2 v .¨r (cid:17)
¨r + ¨r 1 +2v 1 1
1 2 1c2 1 c2
givVeasrtiahteioenquoaftitohneoLfamgoratinogniafnor(mB9i,) with respect to ri = −Grm22(cid:98)r(cid:104)1+β(v12c+2v22) +γv1c.2v2
∂L d(cid:16)∂L(cid:17) (r.v )2+(r.v )2 (r.v )(r.v )(cid:105)
− = 0. (C1) + 3µ (cid:98) 1 (cid:98) 2 +3ν (cid:98) 1 (cid:98) 2
∂r dt ∂v c2 c2
i i
Gm (cid:104) (r.v )v +(r.v )v
For the mass m , the equation of motion is + 2 2µ (cid:98) 1 1 (cid:98) 2 2
1
rc2 r
(cid:40)
− (cid:88) Gm1mir 1+β(v12+vi2) +γv1.vi + ν((cid:98)r.v2)v1+((cid:98)r.v1)v2(cid:105)
r 3 1i c2 c2 r
1i
i∈Sn (cid:41) + Gm2(r.v)(cid:104)2βv +γv +6µ(r.v )r+3ν(r.v )r(cid:105)
(r .v )2+(r .v )2 (r .v )(r .v ) r3c2 1 2 (cid:98) 1 (cid:98) (cid:98) 2 (cid:98)
+ 3µ (cid:98)1i 1 (cid:98)1i i +3ν (cid:98)1i 1 (cid:98)1i i
c2 c2 −Gm2(cid:104)2β¨r +γ¨r +2µ(v.v1)r+2µ((cid:98)r.v1)v
rc2 1 2 r (cid:98) r
+ (cid:88) Gm1mi(cid:104)2µ((cid:98)r1i.v1)v1+((cid:98)r1i.vi)vi (r.v ) (v.v ) (cid:105)
r c2 r + 2µ(r.¨r )r+ν (cid:98) 2 v+ν i r+ν(r.¨r )r
i∈Sn 1i 1i (cid:98) 1 (cid:98) r r (cid:98) (cid:98) 2 (cid:98)
(r .v )v +(r .v )v (cid:105) 2(cid:15)G2
+ ν (cid:98)1i i 1 (cid:98)1i 1 i − (m m +m2)r. (E1)
r r3c2 1 2 2 (cid:98)
1i
 
+ ∂  (cid:15) (cid:88) G2mimjmk −AfGtemr sru/brs3tiatnudtin¨rg=thGemNerw/tro3n,iaeqnuaaptipornox(Eim1a)trieodnusc¨re1s t=o
∂r r r c2 2 2 1
1 ij jk
i,j,k∈Sn,j(cid:54)=i,k(cid:54)=j
(cid:40) ¨r1 = A1(cid:98)r + B1v1 + B2v2, (E2)
d (cid:16) v2 (cid:17)
− m 1+ 1d v with
dt 1 2c2 1d Gm G2m (cid:104) (cid:105)
A = − 2 + 2 (2β+2µ)m −(γ+ν)m
+ (cid:88) Gm1mi(cid:104)2βv +γv +2µ(r .v )r 1 r2 r3c2 2 1
i∈Sn r1ic2 1 i (cid:98)1i 1 (cid:98)1i + Gr2mc22(cid:104)−β(v12+v22)−γv1.v2
(cid:41)
(cid:105) − 3µ(cid:8)(r.v )2+(r.v )2(cid:9)+6µ(r.v)(r.v )−2µv.v
+ ν(r .v )r = 0. (C2) (cid:98) 1 (cid:98) 2 (cid:98) (cid:98) 1 1
(cid:98)1i i (cid:98)1i
(cid:105)
− 3ν(r.v )(r.v )−ν(v.v )+3ν(r.v)(r.v )
(cid:98) 1 (cid:98) 2 2 (cid:98) (cid:98) 2
Gm 2(cid:15)G2
Appendix D: Inertia and Newton’s laws of motion + 2r2c22v12− r3c2 (m1m2+m22),
Gm (cid:104) (cid:105)
B = 2 r.v +2βr.v
Suppose that the masses m1 and m2 are in a reference 1 r2c2 (cid:98) 1 (cid:98)
frame accelerating with respect to distant matter with Gm (cid:104) (cid:105)
acceleration a. In the accelerated frame, distant matter and B2 = r2c22 (ν+2µ)((cid:98)r.v1+(cid:98)r.v2)+γ(cid:98)r.v .
has an acceleration −a, so¨r =−a. Equation (C2) then
d (E3)
has the form
Gm m The equation of motion for r in the centre of mass
m1¨r1 =−m1a− r13 2r + ... . (D1) frame may be found by subtracting the corresponding
10
equation for m and substituting v = m v/M and the acceleration of a composite body towards the Sun is
2 1 2
v =−m v/M, where M =m +m , which gives independent of its gravitational self-interaction energy.
2 1 1 2
Foracompositebodyconsistingoftwopointmasses, m
i
¨r = A(cid:98)r + B((cid:98)r.v)v, (E4) and mj, separated by a distance of rij, the acceleration
GM G2 (cid:104) of the centre of mass towards a massive external body
with A = − r2 + r3c2 (2β+2µ−2(cid:15))(m21+m22) must therefore be independent of Gmimj/rij.
(cid:105)
Consider the motion of the two masses m and m in
−2(γ+ν+2(cid:15))m m i j
1 2
thepresenceofanexternalmassm . Theaccelerationof
e
G (cid:104) the centre of mass is given by
+ {3µ(m2+m2)−3νm m }(r.v)2
Mr2c2 1 2 1 2 (cid:98)
(cid:26)(cid:16) 1(cid:17)
− β+2µ− 2 (m21+m22) a = mi¨ri+mj¨rj
M
−(cid:16)2ν+γ− 12(cid:17)m1m2(cid:27)v2(cid:105) = mMi¨ri + (cid:110)i↔j(cid:111), (E10)
G (cid:104)
and B = (1+2β)(m2+m2)
Mr2c2 1 2 where M = mi +mj. The acceleration ¨ri of mi in the
(cid:105) presenceofm andm maybefoundbyextendingequa-
j e
− (1+2γ)m m . (E5)
1 2 tion(E1)tothecaseoftwonearbymasses. Therequired
generalisation of (E1) may be written in the form
Introducing polar coordinates for r, the angular com-
ponent of (E4) may be integrated to give r2θ˙ =
hexp[(cid:82)rBdr(cid:48)], for some constant h, and the radial com- ¨r = − 1 (v2¨r +v .¨r v )
ponent becomes i 2c2 i i i i i
Gm (cid:104) (cid:105)
− j 2β¨r +γ¨r +2µ(r .¨r )r +ν(r .¨r )r
d2u + u = − A e−2(cid:82)rBdr(cid:48) rijc2 i j (cid:98)ij i (cid:98)ij (cid:98)ij j (cid:98)ij
dθ2 h2u2 Gm (cid:104) (cid:105)
= GM + G2u(cid:104)(2+2β−2µ+2(cid:15))(m2+m2) − riec2e 2β¨ri+γ¨re+2µ((cid:98)rie.¨ri)(cid:98)rie+ν((cid:98)rie.¨re)(cid:98)rie
h2 h2c2 (cid:105)1 2 −2(cid:15)G2(cid:110)(cid:104)mimj +m2j + mjme(cid:16) 1 + 1 (cid:17)(cid:105)r
− (2+2γ−2ν−4(cid:15))m1m2 c2 ri4j rij3 rie rje ij
3G[νm m −µ(m2+m2)] (cid:104)m m +m2 m m (cid:16) 1 1 (cid:17)(cid:105) (cid:105)(cid:111)
+ 1 2 1 2 r˙2 + i e e + j e + r
Mh2c2 r4 r r r ie
ie ie3 ij je
G (cid:104)(cid:16) 1(cid:17)
+ β+2µ− (m2+m2) + Xij + Xie, (E11)
Mh2c2 2 1 2
(cid:16) 1(cid:17) (cid:105)
− γ+2ν− 2 m1m2 v2, (E6) where Xij denotes the remaining acceleration-
independent terms of the right hand side of (E1)
whereu=1/r. Afterusingtheenergyconservationequa- corresponding to the interaction of m and m , i.e.
i j
tion v2 =2GMu+2EM/m m to substitute for v2, the
1 2
rateofperiastronprecessionisfoundfromthecoefficient
Gm (cid:104) (v2+v2) v .v
of u on the right hand side to be X = − jr 1+β i j +γ i j
ij r3 ij c2 c2
ij
πG2(cid:104)
∆θ = h2c2 (1+4β+2µ+2(cid:15))(m21+m22) + 3µ((cid:98)rij.vi)2+((cid:98)rij.vj)2 +3ν((cid:98)rij.vi)((cid:98)rij.vj)(cid:105)
c2 c2
(cid:105)
− (1+4γ+2ν−4(cid:15))m m (E7) Gm (cid:104) (r .v )v +(r .v )v
1 2 + j 2µ (cid:98)ij i i (cid:98)ij j j
r c2 r
ij
per orbit. The prediction of General Relativity is ∆θ =
(r .v )v +(r .v )v (cid:105)
6πG2M2/h2c2,soagreementwiththeexperimentaldata + ν (cid:98)ij j i (cid:98)ij i j
r
for the precession of the perihelion of Mercury and the ij
Gm (cid:104)
periastron of a binary pulsar requires + j(r .v ) 2βv +γv +6µ(r .v )r
r3c2 ij ij i j (cid:98)ij i (cid:98)ij
ij
4β + 2µ + 2(cid:15)= 5 (E8) (cid:105)
+ 3ν(r .v )r
and 4γ + 2ν − 4(cid:15)= −13. (E9) (cid:98)ij j (cid:98)ij
Gm (cid:104) (v .v ) (r .v )
− j 2µ ij i r +2µ (cid:98)ij i v
r c2 r (cid:98)ij r ij
ij ij ij
2. The Nordtvedt effect
(r .v ) (v .v ) (cid:105)
+ ν (cid:98)ij j v +ν ij j r . (E12)
r ij r (cid:98)ij
The absence of a Nordtvedt effect in the Earth-Moon- ij ij
Sun system [20] implies that the accelerations of the
Earth and Moon towards the Sun are the same, so that The acceleration-dependent terms in (E11) may be eval-