Gravitational rotation of polarization: Clarifying the gauge dependence and prediction for double pulsar

0.44 MB·

by  Ue-Li Pen

Gravitational rotation of polarization:
Clarifying the gauge dependence and prediction for double pulsar
Ue-Li Pen,1,2,3,4,∗ Xin Wang,1,† and I-Sheng Yang1,4,‡
1Canadian Institute of Theoretical Astrophysics, 60 St George St, Toronto, ON M5S 3H8, Canada.
2Canadian Institute for Advanced Research, CIFAR program in Gravitation and Cosmology.
3Dunlap Institute for Astronomy & Astrophysics, University of Toronto,
AB 120-50 St. George Street, Toronto, ON M5S 3H4, Canada.
4Perimeter Institute of Theoretical Physics, 31 Caroline Street North, Waterloo, ON N2L 2Y5, Canada.
Fromthebasicconceptsofgeneralrelativity,weinvestigatetherotationofthepolarizationangle
by a moving gravitational lens. Particularly, we clarify the existing confusion in the literature by
showingandexplainingwhysuchrotationmustexplicitlydependontherelativemotionbetweenthe
observerandthelens. WeupdatethepredictionofsucheffectonthedoublepulsarPSRJ0737-3039
and estimate a rotation angle of ∼10−7rad. Despite its tiny signal, this is 10 orders of magnitude
larger than the previous prediction by Ruggiero and Tartaglia [1], which apparently was misguided
7
by the confusion in the literature.
1
0
2 PACSnumbers:

n
a I. INTRODUCTION mentsofitsrotation. Thissystemhasanalmostedge-on
J
orbitwhichallowstheimpactparametertobeverysmall

8 Einstein’s theory of General Relativity has been the at the superior conjunction. Finally, its compact orbit,
2
dominant theory of gravity for a century. Many of its withatwo-hourperiod,meansalargevelocity. Onemain

] signature outcomes, such as light-bending, orbital pre- pointofthispaperistoshowthatonecanexpecttohave
c cession,andgravitationalwaves,havebeenconfirmedby ∆φ ∼ 10−7 from double pulsar, which might be observ-
q precision tests in astronomy [2–5]. One of the last pend- able given a dedicated observation campaign.
-
r ing tests is the gravitational rotation of the polarization The gravitational rotation of polarization from this
g angle. Since gravity directly affects the spacetime geom- doublepulsarsystemwaspreviouslystudiedin[1]. They
[
etry, a gravitational lens not only can bend the light ray, however derived a much smaller number which is incor-

1 it may also rotate the polarization. Such gravitational rect. In fact, various theoretical derivations of this rota-
v rotation of polarization has not been measured so far. 1 tion of polarization have probably brought more confu-
3 One obvious reason is that it is usually very small. The sion than clarity since its first appearance in [7]. It was
4
rotation angle of a linearly polarized light ray, caused by summarized in [8] that three different values of ∆φ can
2
agravitationallens,issuppressedbytwosmallnumbers. be derived from existing literature for seemingly iden-
8
tical physical situations. Many authors disagreed on
0
1. ∆φ≈ 4GM ·v . (1) “whether there is a nonzero rotation in Schwarzschild
r metric”, while none of them correctly pointed out that
0
it is not even a well-defined question to ask. Although
7 Here M is the mass of the lens, r is the impact
Eq. (1) has been derived by some authors, such as in
1
parameter—theshortestdistancewhenthelightraypass
: [6,9],theydidnotexplicitlyexplainwhyitisthecorrect
v near the lens, and v is the velocity of the lens. Un-
result.
i lessthelightgoesthroughsomewherecomparabletothe
X Inthispaper,wewillresolvetheconfusionsbyderiving
Schwarzschildradius, thefirst factorissmall. Unless the
r Eq.(1)fromtheverybasicconceptofgeneralrelativity—
a velocity is almost relativistic, the second factor is small. parallel transports. It turns out that ∆φ, despite being
Luckily,analmostedge-on,compactpulsarbinarysys-
a number, is not a gauge-invariant scalar. It is actually
tem,suchasthedoublepulsarPSRJ0737-3039,canbea
anSO(2)projectionofanSO(3,1)tensor. TheSO(3,1)
verystrongcandidatetomeasuresucheffect. Pulsarsig-
rotation is a gauge-invariant property associated with a
nalsareoftenhighlypolarized,allowingprecisemeasure-
lightraythatstartsandendsfarawayfromthelens,but
∆φ depends on which SO(2) we project to. Therefore
it is natural for ∆φ to be gauge-dependent. More pre-
cisely, the rotation of polarization depends on the frame
∗Electronicaddress: [email protected]
of the observer. Such observer/gauge-dependence was
†Electronicaddress: [email protected]
‡Electronicaddress: [email protected] thesourceofconfusions. Forexample,onecannotsimply
1 The E-mode and B-mode in CMB are defined as the relative ask whether there are rotations in Schwarzschild metric
angle between polarization and gradient. The observed leading without specifying who the observer is. For an observer
order effect is a consequence of a rotated gradient but a fixed
at rest, there is indeed zero rotation; for a moving ob-
polarization, thus it does not count. The actual rotation of po-
server, however, the rotation will be nonzero.
larizationcontributestosub-leadingeffects, whichareanalyzed
in[6],anditisunclearwhetherwewilleventuallyobservethem. We also identify the “correct” gauge to compute such
2
rotation, in which the answer will agree with the actual First of all, parallel transport is not limited to null
observation of double-pulsar signal from the earth. It rays. Onevectoreµatapointcanbeparalleltransported
involves comparing the polarization of two signals, and alonganypathbyanintegralsimilartoEq.(3). Afterwe
botharemeasuredintherestframeoftheearth. Thusit parallel transport from A to B along a null ray, we can
isonlynaturalthatwecalculateintheearth(observer’s) again parallel transport the resulting vector in B along
frame. The v in Eq. (1) is the relative velocity between another curve back to A. If we compare the final vector
the observer and the lens. back at A with the initial vector, the result is a loop
In Sec.II, we will provide an operational definition integral.
of polarization rotation from the basic principles in
(cid:90) B (cid:90) A
general relativity and explain the inevitable observer- ∆ec = kˆaebΓc dl+ pˆaebΓc dl . (4)
0 ab 0 ab
dependence. In Sec.III, we will derive Eq. (1) in the A B
smallrotationlimitforonespin-less,point-masslens. We
Since now we are comparing two vectors on the tangent
thengeneralizeitintomultiplelensesevenwithspins. In
space of one point, it is mathematically meaningful. It
Sec.IV, we will describe the observable effects on double
is also gauge invariant. The gauge freedom allows us to
pulsar. In Sec.V, we conclude with a discussion of an
change the value of Γc locally but not globally, as there
appropriate observation campaign of future detection. ab
will be constraints. Parallel transport in any close loop
is indeed one of those constraints. Its answer carries the
gauge-invariant information about spacetime curvature
II. DEFINITION FROM SCRATCH enclosed by such loop, and it must be gauge invariant.
Eq. (3) is just providing a convenient way to evaluate
A. Parallel Transport around a Loop such gauge-invariant loop integral. Assume that we are
in asymptotic Minkowski space, and both points A and
Intuitively,onecanimaginethepolarizationasavector B are in the asymptotic region. We can then choose this
attached to a light ray that is spacelike and orthogonal path from B back to A to stay in the asymptotic region.
to the direction of propagation. Let k be the null vector IntheasymptoticMinkowskigaugethatΓc →0asymp-
ab
of the light ray and e be a polarization vector, a parallel totically, the segment of integral through the asymptotic
transportofeshouldbevalidinthegeometricopticlimit, region contributes nothing. Therefore, the line integral,
i.e. Eq. (3), in the asymptotic Minkowski gauge, gives ex-
actly the gauge-invariant answer of the loop integral in
ka∇ ec =ka∂ ec+kaebΓc =0 . (2) Eq. (4).
a a ab
Furthermore, when there are multiple matter sources
When the rotation is small, using the Born approxima- in asymptotic Minkowski space, there is a well-defined,
tion, we can integrate along a light ray from point A to common asymptotic Minkowski gauge. That simply
point B to get the change of polarization vector. meansΓc isonlynonzeronearsources,decaysawayfrom
ab
individualsourceslikeinaSchwarzschildmetric,andthe
(cid:90) B contributionfromeachsourcesuperimposesintheregion
∆ec = kˆaebΓc dl , (3)
0 ab far away from all sources. Computing a gauge-invariant
A
loop integral in this gauge allows us to identify the con-
where e is the original vector and ∆e is the change. tributions to rotation from individual sources, since only
0
Straightforwardly, ∆φ ≡ |∆e|/|e| could serve as a def- the segments of integral near sources have nonzero con-
inition of how much a polarization vector has been ro- tributions.
tated. This however, cannot be the full story. That is An actual observation, as we depict that in Fig.1, is
becauseEq.(3)literallycomparestwovectorsonthetan- closely related to a loop integral. What we have is a
gent spaces of two different points,. Such comparison is source(pulsar)whichconstantlyemitsafixed(albeitun-
mathematicallymeaningless. Thetwovectorsmustbein known)polarization. Wemeasurethepolarizationduring
thesametangentspacetoprovideameaningfulrotation. a usual time, which is a light ray from A1 to B1. And
Another way to state the same problem is that the value thenwecompareitwiththepolarizationmeasuredwhen
of connections, Γc , depends on the coordinate choice. itsbinarycompanionpassesveryclosetothelineofsight,
ab
By choosing a different gauge, one can change the value which is another light ray from A2 to B2. We take the
of the line integral of Eq. (3) to any value. difference between these two measurements, which is a
Many literature used Eq. (3) and computed its value loop integral if we add two extra time-like segments.
in one very natural gauge, for example [6, 9]. We
(cid:90) B2 (cid:90) B1
can probably call that the asymptotic Minkowski gauge ∆ec = ec| −ec| = kˆaebΓc dl+ pˆaebΓc dl
or the Schwarzschild gauge, in which Γcab falls of to B2 B1 A2 0 ab B2 0 ab
zero asymptotically away from matter sources as they (cid:90) A1 (cid:90) A2
would in the Schwarzschild coordinate. It turns out that + kˆaeb0Γcab dl+ pˆaeb0Γcabdl . (5)
Eq.(3)insuchgaugehappenstogivethecorrect,gauge- B1 A1
independent answer. Here we will explain why. The two time-like segments, A A and B B , and the
1 2 1 2
3
gauge-invariant rotation. In the asymptotic Minkowski
∆φ
B
2 gauge, such rotation can be calculated along a light ray.
This however, does not resolve all the confusion in the
rotation of polarization. The next problem is which vec-
k’
µ tor do we rotate? Polarization lives in a 2-dimensional
plane, which cannot be uniquely determined by a light
pulsar v (out of the plane) ray. Thus there is no unambiguous answer to the ques-
tionof“howmuchrotationofpolarizationhasalightray
B
1 gone through.” We need both the light ray and the ob-
earth server’s 4-velocity to determine the 2-dimensional plane
k
µ on which the polarization lives. Therefore, “how much
A2 rotation” must have an observer dependent answer.
Mathematically, we can see that Eq. (3) is the leading
order effect of a small rotation matrix.
ec =ec +∆ec =Λc eb =(gc +∆c )eb , (6)
0 b 0 b b 0
A
1 where
FIG.1: Paralleltransportalongaloopcontains4segments: (cid:90) B
∆ = kˆaΓd g dl . (7)
bent light ray (solid black), worldline of the pulsar (thick, cb ab cd
green,left),worldlineoftheobserver(thick,green,right),and A
anunbentlightray(dashed,bottom). ItleadstoanSO(3,1) By definition of a rotation matrix, ∆ has to be anti-
cb
rotation of the tangent space, and contains the information
symmetric, which can be verified explicitly.
of both deflection of light (from k to k(cid:48)) and the rotation
µ µ
of polarization, ∆φ. In practice, we can measure this effect (cid:90) 1(cid:90)
by comparing the pulsar signal when the companion neutron ∆ = kbΓd g dλ = kb(∂ g +∂ g −∂ g )dλ
ac ab cd 2 a bc b ac c ab
star (blue dot) passes through the line-of-site (solid line) to
the same signal in other times (dotted line). = 1(cid:90) kb(∂ g −∂ g )dλ . (8)
2 a bc c ab
Note that we have to drop the boundary term in the
light ray A B , are all far away from the companion.
1 1 above integral, which is allowed because this is effec-
TheyallcontributenothingintheasymptoticMinkowski
tively a loop integral as we explained in the previous
gauge. Thustheaboveloopintegralcanbecalculatedin
section. This demonstrates that there is actually a full
theasymptoticMinkowskigaugeasonlythelineintegral
SO(3,1) rotation, Λa ≈ (ga +∆a ), that is associated
A B . Such line integral actually only has contribution b b b
2 2 with a loop, therefore a light ray which starts and ends
near the companion, thus it is indeed the rotation of po-
in asymptotic Minkowski region.
larization caused by the passage of the companion. 2
This SO(3,1) rotation is the gauge-invariant property
associated with the light ray, but it does not yet deter-
minetherotationofpolarization,whichisonlyanSO(2).
B. Observer Dependence It also contains extra information such as the deflection
of the light ray itself. Particularly, one needs to spec-
ify a two-dimensional plane of polarization to determine
In the previous section, we explained that a given vec-
which SO(2) to project to. For any observer, the po-
tor, parallel-transported along a loop, goes through a
larization vectors are orthogonal to both the incoming
light ray and its own worldline. Thus a projection to the
co-dimension-twosurfaceorthogonaltothelightrayand
the observer 4-velocity is the desired SO(2) rotation of
2 Thissimplifiedstoryistruewhenboththeemitterandreceiver
are light so we can ignore their contribution to Γc . In reality, polarization. Therefore, it is natural and necessary that
ab
points A1, A2 are near the pulsar, and B1, B2 are on earth, polarizationrotationdependsontheobserverframe. This
so neither is in the asymptotic region. Thus the time-like seg- also explains the v dependence in Eq. (1), which has to
ments can be nonzero, and the null segments will have extra
be the velocity of the lens in the observer’s frame.
contributionneartheendpoints. Nevertheless,theextracontri-
One last possible confusion is why such dependence is
butionstothenullsegmentswillcanceleachother. Thetime-like
contributions has nothing to do with the companion, and they on the observer instead of the source, since they seem to
aredegeneratewithanintrinsicvariationofpulsarsignalorthe play equivalent roles in the integral of Eq. (3). Again,
telescopereceivingfunction. Wecansimplyobserveandfitsuch we remind the readers that the apparent line integral in
behaviour when the companion is not passing through the line-
Eq. (3) is a convenient illusion. The physically mean-
of-sight,andsubtractitfromthedata. Thustreatingthemasin
ingful rotation is defined by a loop, where one compares
the asymptotic region is a simple way to show the contribution
fromthecompanionwithoutlossofgenerality. a vector to its parallel transported outcome after going
4
around the loop. Thus there does exist one unique point form, expanded to the leading order of (M/r).
at which the rotation is defined, which is where the two
(cid:18) (cid:19)
polarizations are being compared. In practice, we will g dxadxb = − 1− 2M dt2 (9)
have no idea about the actual polarization when the sig- ab r
nlaarliziastieomnsittweedraetcetihveedpounlsaera.rthA,llsowtehiksnioswthaerefrathmeepwoe- + (cid:18)1+ 2M(cid:19)(cid:0)dx2+dy2+dz2(cid:1),
r
have to choose.
where r2 = x2 +y2 +z2, and the Newton constant G
is conveniently set to 1. Instead of studying an arbi-
trary light ray in the above coordinate, we will shift and
C. Beyond Born Approximation boost this metric such that the lens has arbitrary po-
sition and velocity, and the relevant light ray is aways
x = t. In principle, we need six parameters, i.e.
The actual effect we will calculate in the rest of this
(x ,y ,z ,v ,v ,v ). By applying symmetries, we can
0 0 0 x y z
paperwillbequitesmall,soBornapproximationisjusti-
furthersimplifytheproblemsothateventuallyonlythree
fied. Nevertheless, the above abstract explanation must
will be needed.
stillbetruebeyondtheBornapproximation,andwewill
First we use shift symmetries in x and t to set x =0,
0
spend this subsection to demonstrate that.
which simply means that t = 0 is defined as the time
It is straightforward to actually solve the parallel when the light ray is closest to the lens. Next, we set
transport equation, Eq. (2), instead of using the Born- v =0,soinsteadoflettingthelenstohaveanx-velocity,
x
approximation integral in Eq. (3). A loop-parallel trans- the asymptotic observer who measures the polarization
port back to the same point is obviously an SO(3,1) ro- will have a nonzero x-velocity. This changes nothing be-
tation. So one can see that up to getting the SO(3,1) cause the light ray is in the x direction, kµ =(1,1,0,0).
rotation, everything we said in the previous section di- Independent of what x-velocity the observer has, the
rectly generalizes beyond Born approximation. The only plane orthogonal to both the light ray and the observer
question is that we have used the unique kµ to deter- willbethey-z plane. Thuswearealwayscalculatingthe
mine the SO(2) projection in the Born approximation. rotation of polarization on the y-z plane. Finally, using
Now the direction of light is also deflected significantly, rotational symmetry on the y-z plan, we can set v =0,
z
kµ → k(cid:48)µ. Do we still have an unambiguous way to de- leaving the remaining three parameters to be v =v, y
y 0
termine which SO(2) to project into? and z .
0
Employingthesesymmetriessignificantlysimplifiesthe
The answer is yes, and this is how we do it. First of
problem, and we illustrate the final situation in Fig.2
all, the observer’s 4-velocity reduces SO(3,1) down to
Since the lens is the centre of the coordinate in Eq. (10),
SO(3). The light rays, before and after the deflection, k
andk(cid:48), arealsoreduceddowntotwospacelikevectorsin weneedtoapplytheappropriatecoordinatetransforma-
theobserver’sframe,κandκ(cid:48). Aslongasκ(cid:54)=−κ(cid:48),there tion to accommodate our symmetry choice.
is a unique, minimal SO(3) rotation that aligns them. 3
γ =(1−v2)−1/2 , t→γ(t−vy) ,
Aligning κ and κ(cid:48) also merges their polarization planes,
in which an SO(2) rotation is uniquely defined. Thus, y →γ(y−vt−y0) , z →(z−z0) . (10)
one can see that even beyond Born approximation, the
The resulting metric becomes
rotation of polarization is still a well-defined, unambigu-
ous, observer-dependent SO(2) projection of a covariant (cid:20) (cid:21)
2M
SO(3,1) tensor. g dxadxb = − 1−γ2(1+v2) dt2
ab r
(cid:20) 2M(cid:21) 8Mvγ2
+ 1+γ2(1+v2) dy2− dtdy
r r
(cid:18) (cid:19)
III. EXPLICIT CALCULATION + 1+ 2M (dx2+dz2) , (11)
r
A. Point Mass (cid:112)
where r = γ2(y−vt−y )2+x2+(z−z )2.
0 0
While calculating the connections,
We will treat the gravitational lens as a point mass
andmodelitwithaSchwarzschildmetricintheisotropic Γc = gcd (∂ g +∂ g −∂ g ) , (12)
ab 2 a bd b ad d ab
wecantreatthefirstgcdastheflatmetricηcdsinceweare
onlykeepingtheleadingorderresult. Thisappliestoany
3 Note that there are many rotations which can align them, but
gab that is not hit by a derivative in the calculation, e.g.
there is a unique minimal rotation, that is rotating along the
directionorthogonaltobothofthem,(κ×κ(cid:48)). theoneinEq.(8). Wealsoassumethatboththenullray
5
z Furthermore, let us imagine that there is a continuous
source of signals. The light along the x direction con-
tinues to shine while the lens is moving in its constant
velocity in the y direction. The light ray which is clos-
M v
est to the lens will get the maximum rotation. In other
z words, we get maximum rotation when the lens’ veloc-
0 ity is orthogonal to its distance to the light way. This
corresponds to y =0 in the above calculation.
0
4M
∆φ = v . (15)
Max z
0
This maximum rotation along a source trajectory is the
promised result in Eq. (1).
∆φ
B. General Case
y The above point-mass calculation assumes that it car-
y ries no spin. Many papers employed a Kerr metric in-
0
stead to calculate how the angular momentum from the
spin also contributes to the rotation of polarization. In
thelimitofsmallrotations,wecaninsteadgeneralizethe
aboveresultwithoutexplicitlystartingfromaKerrmet-
FIG.2: Usingallsymmetries,wecanreduceanycalculation
ric. That is because Eq. (11) allows superposition when
of polarization-rotation from a spin-less, point-mass lens to
alllensesarenotmovingtoofastandnottooclosetothe
thispicture. Thereddotattheoriginisthelightraytraveling
inthexdirection. M isthemassofthelens. v isitsvelocity light ray. At the leading order (ignoring sub-leading ve-
(onlyinydirection). (y ,z )isthelocationofthelensrelative locityterms),themetricofmultiplemovingpointmasses
0 0
to the light ray. ∆φ is the rotation of polarization, which is are given by
drawn in the appropriate direction in the picture. One can
(cid:32) (cid:33)
visualize it as being “dragged” by the motion of the lens. (cid:88)m(n)
g dxadxb = − 1−2 dt2 (16)
ab r(n)
n
(cid:32) (cid:33)
directionandthepolarizationdirectionareonlychanged (cid:88)m(n)
+ 1+2 (dx2+dy2+dz2)
by a small amount, i.e. the Born approximation. Thus r(n)
we can compute ∆ by Eq. (8) along the undeflected n
light ray x=t. Manayb components of g are zero due to (cid:88)m(n)(cid:18) (cid:19)
ab − 8 v(n)dx+v(n)dy+v(n)dz ,
our choice of symmetries, so it is straightforward to see r(n) x y z
i
that
(cid:20)(cid:16) (cid:17)2
1(cid:90) ∞ r(n) = x−x(0n)−vx(n)t (17)
∆ = −∆ = dt∂ g
zy yz 2 −∞ z ty (cid:16) (cid:17)2 (cid:16) (cid:17)(cid:21)1/2
= −2Mvγ2z (cid:90) ∞ dt .(13) + y−y0(n)−vy(n)t + z−z0(n)−vz(n)t
0 −∞ [γ2(vt+y0)2+t2+z02]3/2 Their contributions to the total rotation also superim-
pose linearly. If we further assume that the masses are
For v (cid:28)1, we can perform the integral and keep only
distributed in a small enough region such that their lo-
the leading order value to get
cationsstaythesameduringthepassageofthelightray,
4Mvz 4|J | 4M|k·(r ×v)| the answer is very simple.
∆φ≡|∆ |≈ 0 = x = 0 .
yz y2+z2 r2 r2 (cid:12) (cid:12)
0 0 0 0 (cid:12) (cid:12)
Here J is the lens’ angular momentum with respec(t1t4o) ∆φ=4(cid:12)(cid:12)(cid:12)(cid:88)m(n)v(cid:16)y(n)z0((cid:17)n2)−v(cid:16)z(n)y0((cid:17)n2)(cid:12)(cid:12)(cid:12) . (18)
x=0,i.e. thepointwherethelightraywasclosesttothe (cid:12)(cid:12) n y0(n) + z0(n) (cid:12)(cid:12)
lens. The direction of the light ray ka determines which
componentofJ wecareabout,whichisthex-component Since their velocities are small, they are roughly in the
in our symmetry choice. The final, generalized format of samelocationafterthelightraygoesthroughallofthem,
Eq. (14) should be straightforward from our symmetry thus x(n) do not matter at all.
0
choice. A more explicit calculation in [9] led to the same Undertheseassumptions,Eq.(18)providesthegeneral
result. answer to any mass and velocity distribution. By the
6
neutron stars are slightly larger. A typical neutron star
is slightly smaller than 10 times its own Schwarzschild
radius, so we take the radius of the lens neutron star to
be 30km. If it is a slow pulsar, we take the spin period
to be about 1s. If the spin aligns with the line of sight,
we estimate its contribution in Eq. (20) as
30km/1s
|S |∼3km× ×30km∼104m2 .
x c
WehaveusedbothcandGtomakethisquantitytohave
theunitoflength2,whichmakesiteasiertocalculatethe
unitless ∆φ. Similarly, assume the binary orbit is 109m,
velocity is about 0.1% speed of light, and the orbital tilt
is 2 degree, so the impact parameter is roughly 109 ×
2π/180 ≈ 3.5×107m, then the orbital contribution is
roughly
|J |≈3km×10−3×109×2π/180≈107m2 . (21)
x
In this case, the spin contribution is negligible.
If the lens is a recycled (fast) neutron star, its the
periodwouldbe∼1ms, anditsspinangularmomentum
isincreasedbyafactorof∼1000. Ifitisalsoalignedwith
thelineofsight,thespincontributionwillbecomparable
to the orbital contribution. Fortunately, fast pulsars are
FIG. 3: The angle of polarization rotation for binary pul-
sar PSR J0737-3039, the signals are shifted to peak at π/2 usually spun-up by accretion from the companion, so its
only for aesthetic reason. Top: Rotation angle ∆φ of mil- spin is usually aligned with the orbital plane. In our
lisecondpulsarA,thegreyareahighlightstheregionblocked case, it means that the spin is almost perpendicular to
by eclipse. Bottom: The inclination dependence on the peak the line-of-sight, so its contribution is again negligible.
rotation, assuming all other orbital parameters unchanged. Therefore, in either case, we can estimate the maxi-
The dashed line indicates the amplitude of real signal. mal rotation from the orbital contribution only, which is
about
107m2
uniquenesstheorem,theeffectfromaKerrmetricofmass ∆φ≈4 ≈3∗10−8 . (22)
M and spin S can be mimicked by a two-particle system [109m×2π/180]2
at the leading order.
The actual value for double pulsar is slightly larger. In
Fig. (3), we calculated the signal ∆φ using the orbital
v(1) = v(2) =0 , v(1) =v−δv , v(2) =v+δv ,
z z y y informationofdoublepulsarsystemPSRJ0737-3039[3].
y(1) = y(2) =y , z(1) =z −d , z(2) =z +d , We ignore the spin contribution since they are negligible
0 0 0 0 0 0 0
as we explained. We can see that during a rotation pe-
m(1) = m(2) =M/2 , S =M(δv)d . (19)
riod, we can expect a maximal rotation of polarization
(∆φ)atabout10−7rad. Thishappenswhenthecompan-
Taking d→0 while holding S fixed, we get
ion(lens)isalmostinfrontofthepulsar. Itiswell-known
(cid:12) (cid:12)
∆φ=4(cid:12)(cid:12)(cid:12)Myv2z+0+z2S(cid:12)(cid:12)(cid:12)=4|Jxr+2Sx| . (20) tmhaayt awtotrhryistmhaotmweentc,atnhneoret wacitlluaallslyosbeeeatnhiescmlipasxei,msaolornoe-
0 0 0
tation. We specifically zoomed in and blacked-out the
The spin of the point mass contributes in exactly the eclipsed. We found that the peak of the ∆φ curve is
samewayasits“orbital”angularmomentumaroundthe significantly wider than the eclipse duration. Thus for
light ray. Intriguingly, although [9] agrees with Eq. (14), a (relatively) long duration before and after the eclipse,
theyclaimedthatthespindoesnotcontributeatall. We we can still observe ∆φ ∼ 10−7rad. We can also see
cannot see any physical reason for such statement since thatalthoughthedouble-pulsarisalreadyquiteedge-on,
Eq. (20) seems to be the most natural result. one can hope to get luckier and discover another system
whose inclination angle is even closer to 90 degree. The
resulting rotation of polarization can be even larger.
IV. EXAMPLE: DOUBLE PULSAR
Before putting in the actual numbers, let us first give V. OBSERVATIONAL CONSIDERATIONS
a rough estimation on the maximal rotation we can
get from the double pulsar system. Recall that the Such a small swing in polarization angle will be chal-
Schwarzschildradiusofthesunisroughly3km,andthese lenging to detect. For parameters of PSR J0737-3039
7
with a polarized fraction of about 50 % [10], raw ther- tems may be discovered, for example pulsar-black hole
mal sensitivity requires a signal-to-noise of at least 107 systems. Our estimates indicates that this effect is in
in polarization. For a pulsar self-noise dominated tele- principle observable in the foreseeable future. The grav-
scope such as FAST [11] or SKA [12] with a band width itational Faraday effect can be added to the wish-list of
of ∼1 GHz, it requires 106 seconds of on target on-pulse pulsar tests of general relativity.
integration to achieve a 5−σ detection. This is a sub-
stantial commitment of resources. On top of that, there
arepracticalitiesonemustconsidercarefully. Firstofall,
any given object can be only seen for a limited about of
Acknowledgments
time each day. This observable duration for FAST could
be short at the low declination of PSR J0737-3039, so
SKA is likely the more suitable facility. At a duty cycle WethankJohnAntoniadisandMichaelKramerforin-
of about 20%, this requires a few years of observations formative discussions. We also thank the hospitality of
with the full phase 2 telescopes. During superior con- Max-Planck Institute for Radio Astronomy during the
junction, the companion’s magnetosphere partially ab- Scintillometry Workshop. This work is supported by
sorbs the pulsar, and plasma faraday effects may also the Canadian Government through the Canadian Insti-
complicate the analysis. 4 tuteforAdvanceResearchandIndustryCanada, andby
Despite the substantial efforts required, the pulsar is Province of Ontario through the Ministry of Research
likelytobemonitoredextensivelyforotherreasons. Thus and Innovation. The Dunlap Institute is funded through
a detection may be eventually achieved over decades of an endowment established by the David Dunlap family
SKA operations. Alternatively, other more optimal sys- and the University of Toronto.
[1] M.L.RuggieroandA.Tartaglia,Mon.Not.Roy.Astron. [7] G. V. Skrotskii, Soviet Physics Doklady 2, 226 (1957).
Soc. 374, 847 (2007), astro-ph/0609712. [8] A.Brodutch,T.F.Demarie,andD.R.Terno,Phys.Rev.
[2] F. W. Dyson, A. S. Eddington, and C. Davidson, Philo- D84, 104043 (2011), 1108.0973.
sophicalTransactionsoftheRoyalSocietyofLondonSe- [9] S. Kopeikin and B. Mashhoon, Phys. Rev. D65, 064025
ries A 220, 291 (1920). (2002), gr-qc/0110101.
[3] M. Kramer, I. H. Stairs, R. N. Manchester, M. A. [10] P. Demorest, R. Ramachandran, D. C. Backer, S. M.
McLaughlin, A. G. Lyne, R. D. Ferdman, M. Bur- Ransom, V. Kaspi, J. Arons, and A. Spitkovsky, ApJ
gay, D. R. Lorimer, A. Possenti, N. D’Amico, 615, L137 (2004), astro-ph/0402025.
et al., Science 314, 97 (2006), ISSN 0036-8075, [11] C. J. Jin, R. D. Nan, and H. Q. Gan, Proceedings of
http://science.sciencemag.org/content/314/5796/97.full.pdf, the International Astronomical Union 3, 178 (2007),
URL http://science.sciencemag.org/content/314/ URL https://www.cambridge.org/core/article/
5796/97. div-class-title-fast-telescope-and-its-possible-contribution-to-high-precision-astrometry-div/
[4] J. M. Weisberg and J. H. Taylor, ASP Conf. Ser. 328, 1F91AA86ABBD28527A4887B59562C38C.
25 (2005), astro-ph/0407149. [12] Smits, R., Kramer, M., Stappers, B., Lorimer, D. R.,
[5] B. P. Abbott et al. (Virgo, LIGO Scientific), Phys. Rev. Cordes, J., and Faulkner, A., Astronomy and Astro-
Lett. 116, 061102 (2016), 1602.03837. physics 493, 1161 (2009).
[6] L. Dai, Phys. Rev. Lett. 112, 041303 (2014), 1311.3662.