grav radiation project

8/3/2019 Grav Radiation Project

1/21

Gravitational Radiation from the PSR 1913+16

Binary Pulsar System

Final Project for ASTR G6005, May 2011

James Mossman

I. Introduction

This report will discuss the effects of gravitational radiation on the PSR 1913+16 Binary Pulsar System. This

system, located about 10,000 light years from us, consists of two stars circling around each once every 7 hours, 45

minutes and 6.9816132 seconds. The precision of this measurement allows for a number of tests of Einsteins

theory of General Relativity -- including its prediction of gravitational radiation.

The binary system consists of a rapidly spinning neutron star (a pulsar) and a companion, also believed to be a

neutron star. The pulsar emits a pulse of electromagnetic radiation in our direction every 59 milliseconds --indicating that it is rotating around its axis 17 times a second -- and acts as a very stable clock. The system was

discovered in 1974 by Russell Hulse and Joseph Taylor, who used the arrival time of these pulses to make very

accurate measurements of the motion of the pulsar and its companion.

As will be shown below, a system emits gravitational radiation, and hence loses energy, by a factor proportional to

the square of the third time derivative of its quadrupole moment. The binary pulsar system has just such a

moment. If a binary star system loses orbital energy its two stars should slowly begin spiraling inward towards

each other. This will cause a shortening of its orbital period. The energy loss rate predicted by General Relativity

is in very close agreement with the measured value, as shown in the following graph of periastron advance from

Joseph Taylors 1993 Nobel Lecture [1]. The predicted parabolic curve is due to energy losses from gravitational

radiation, relative to a constant period orbit.


2/21

General Relativity also predicts that as the orbit of the binary system decays it should evolve from an elliptical

shape to a circular form. Ultimately, gravitational radiation will cause the binary system to collapse within a

calculable lifetime.

The report will begin in Section II with an overview of the theory of gravitational radiation. Much of the material

is based on discussions found in the following texts:

Relativity, Astrophysics and Cosmology by Radoje Belusevic [2]

Relativity, Gravitation and Cosmology by Ta-Pei Cheng [3], and

The Classical Theory of Fields by Landau and Lifshitz [4]

The report attempts to fill in the blanks in the development of the theory by showing all the steps involved in

deriving the various relationships.

Section III deals with calculations specific to the PSR 1913+16 system. First the calculations of energy loss and

angular momentum loss will be developed for the case of a circular orbit. These will then be generalized to an

elliptical orbit with eccentricity e. The theoretical rate of orbital period decay for our system will be found to beabout 7 10-5 seconds per year. The agreement between theory and observation turns out to be accurate to about

one part in a thousand -- a strong testament to the validity of the theory of general relativity.

The relationship between orbit size and eccentricity will be explored next. Finally, the lifetime, or decay time due

to gravitational radiation, of a binary system will be discussed. It turns out that the PSR 1913+16 system will

decay due to gravitational radiation in about 3 108 years.

Grav Radiation Project.nb | 2


3/21

II. Overview of the Theory of Gravitational Radiation

The Linearized Einstein Field Equations:

Assume a small linear perturbation, hn, on a flat Minkowski background metric, hmn

gmn = hmn + hn with hn`

1 (1)

Then the related Christoffel symbol, since hmn,l = 0, is

Gamr =1

2hasIhsm,p + hsr,m - hmr,sM

And the Ricci tensor, to first order in h, is

Rmn = (Gma,ua - Gmn,a

a ) =1

2[!n IhaaIham,a + haa,m - hma,aMM - !a IhaaIham,n + han,m - hmn,aMME

= 12

(h,mn - hm,naa - hn,ma

a + !l!lhmn) with h = haa (2)

The curvature scalar is

R = Rmm =1

2I2 h, mm - 2ham,ma ) = h - hmn, mn with = !l!l (3)

SubstitutingRmn and R into the Einstein fieldequations, Gmn = Rmn - nmn R/2 = kTmn gives

2kTmn = (h,mn - hm,naa - hn,ma

a + hmn) - hmn (h - hab

, ab) (4)

For later simplicity, define hmn as a traceless version ofhmn ==> hmn = hmn - hmnh

2 Then (5)

h = hmn Ihmn - hmn h2

) = h - 2h = - h and

hmn = hmn + hmnh

2= hmn - hmn

h

2

Substitute this into the expression for Tmn in (4)

2kTmn = - h,mn

- ham,na + h

am

h,na /2 - han,ma + h

an

h,ma /2 + ( hmn - hmnh

2) +

hmn(h + h

ab

,ab - h

ab

h ,ab /2)

= - h,mn

- ham,na + h,mn/2 - h

an,ma + h,mn/2 + hmn - hmn

h

2+ hmnh +hmn h

ab,ab - hmn

h

2

Tmn =1

2 k(hmn - h

am,na - h

an,ma +hmn h

ab,ab) This is the field equation for the linear perturbation. (6)



4/21

Gauge Freedom:

The basic defining equation, gmn = hmn + hn with hn`1, does not specify a unique reference frame.Consider a transformation to a new reference frame noted byxm' where

xm' =xm + cm(x) Here the cm are four small functions of x such that !m cn ` 1, and

! xm'

!xs = d

ms + !s c

m and the inverse! x

m

!xs' = d

ms - !s c

m to first order (7)

If these are applied to the metric tensor, then

ga' b' =! x

m

!xa'

! xn

!xb'

gmn = ( dma - !a c

m )( dnb - !b cn) (hmn + hn )

= dmadnb (hmn + hn ) - d

ma !b c

nhmn - dnb!a c

mhmn to first order

= gab - !b ca - !a cb

= hab + ha' b' Where

ha' b' = hab - !b ca - !a cb This is the general gauge transformation of the linear gravitational field,

similar to the electromagnetic field transformation. (8)

Since ga' b' was defined in a proper tensorial way it will leave all tensor equations invariant. Hence specifying

ha' b' via theabove equation will leave all results invariant as well Hat least to first orderL.

The idea is to now choose a form ofha' b' to make the field equations (6) simpler. If we set

ha' b' = hab - !b ca - !a cb and recall that hmn = hmn - h/2 hmn then

hm' n' = hm' n' - ha'a'/2 hm' n' = hmn - !n cm - !m cn - h

a'a'/2 hm' n' = hmn - !n cm - !m cn+ ( h

aa - h

a'a') hmn2

and

haa - ha'a' = h

ab hab - habIhab - !a cb - !b caM = 2 !a ca hence

hm' n' = hmn - !n cm - !m cn+ hmn!a ca is the gauge transformation that we want for hmn. (9)

In fact, if we choose cm = !uhmn

then

!nhm' n'

= !nhmn

- cm = 0 ==> !n I!mxm'!nxn'hmn L = 0 ==> !nhmn = 0 to first order using (7)

!nhmn

= 0 is the Lorentz gauge condition, which will simplify the field equations. (10)



5/21

The Gravitational Wave Equation

In the Lorentz gauge using (5), !nhmn = hmn!nh/2 = !mh/2, then using this in (2) and (3) forRmn andR gives

Rmn =1

2(h,mn - hm,na

a - hn,maa + hmn) =

1

2(h,mn -

1

2h,mn -

1

2h,mn + hmn) =

1

2hmn

R = h - hmn, mn = h -1

2!m!mh =

h

2

Thus Gmn =1

2( hmn -

1

2hmnh) =

1

2hmn using (5) and finally using the Einstein field equation

hmn = 2 kTmn This is the gravitational wave equation in the Lorentz gauge. (11)

The vacuum solution to (11) can be expressed as a superposition of plane waves

hmn = emn ka x

a

where the emn are components of a constant symmetric (0,2) polarization tensor (12)

Substituting (12) into (11) with Tmn = 0 implies:

km km = 0 ==> ifkm = (w/c, k) then HwcL2 = k2 ==> this dispersion relationship implies that gravita

tional waves move at the speed of light.

The Lorentz gauge condition !nhmn

= 0 implies that emn kv = 0 which means that the wave vector kn is orthogonal to

the polarization.

Now the gauge condition !nhmn

= 0 or cm

= 0 still leaves degrees of freedom in the cm

which may be specified tofurther simplify emn. It can be shown (see [2]) that the further conditions:

emm = 0 (traceless) and em0 = 0 (transverse) can be derived from a subset of the allowed cm under the

Lorentz gauge.

Along with the symmetry requirement this gives, for a wave propagating in the z direction with km = (k,0,0,k)

eTTmn =

0 0 0 0

0 h+ h 0

0 h -h+ 00 0 0 0

This is the Transverse-Traceless or TT gauge polarization tensor. (13)

Note that since the trace hTT

= 0 then hmn = hmn Hsee H5LL and wecan dropthebar over h in this gauge.

A more illuminating way of writing the plane wave solution is to explicitly break out the two polarization states,

h+ and h for a wave traveling in the z direction:



6/21

hmn = Ih+ e+mn + hemn L ka xa where

e+mn =

0 0 0 0

0 1 0 0

0 0 -1 00 0 0 0

and emn =

0 0 0 0

0 0 1 0

0 1 0 00 0 0 0

Energy Density of Linearized Gravitational Waves

In order to calculate the energy density of gravitational waves we must recognize that the gravity waves them-

selves carry energy and thus produce additional small amounts of curvature in background spacetime. Thus

instead of (1) we should write

gmn = gHbL + hmn where g

HbL = hmn + OIh2M is the new adjusted background metric

We can also decomposeRmn as

Rmn =RHbLmn +R

(1)mn + R

(2)mn + ....... whereR

HnLmn is of order n in hmn

Now in the vacuum, with Tmn = 0, the Einstein equation givesRmn = 0. This must be true separately for all powers

ofRHnLmn . Thus

RH1Lmn = 0 and RHbLmn +R

H2Lmn = 0 since both of these terms are of order h

2. (14)

Call the energy-momentum tensor of the gravity wave tmn . Then tmn andRHbLmn are related through the Einstein

equation:

RHbLmn -1

2hmnR

HbL = ktmn Using (14) we can solve for tmn in terms ofRH2Lmn and use k= -

8G

c4

tmn =c4

8 G(RH2Lmn -

1

2hmn R

H2L )

A subtlety which must be recognized is that it is possible to always transform to a frame of reference where at a

single point spacetime is flat and hence Rmn and tmn vanish. Thus one cannot not talk about gravitational energy

density at a single point in spacetime. Instead we must look at an average over a spatial volume which is greater

than the relevant gravitational wavelength to obtain:

tmn =c4

8 G[YRH2Lmn] - 1

2hmn YRH2L]] This is the energy momentum tensor for the gravity wave. (15)

Our goal is to find t00, the energy density. For a simple plane wave moving in the z direction, using (12) and (13)

we have:



7/21

hmn =

0 0 0 0

0 h+ h

0

0 h -h

+ 0

0 0 0 0

with h+ = h+ cos[w (t - z/c)] and h

= h cos[w (t - z/c)] (16)

The related metric for the h+ polarization state is to first order:

gmn = hmn + hmn =

-1 0 0 0

0 1 + h+ cos@w Ht-zcLD 0 00 0 1 - h+ cos@w Ht-zcLD 00 0 0 1

and the inverse gmn =

-1 0 0 0

0 1 - h+ cos@w Ht-zcLD 0 00 0 1 + h+ cos@w Ht-z cLD 00 0 0 1

(17)

Using this metric we may now calculate the Christoffel symbols Gamn. For example:

G110 =1

2g11Hg11,0 + g10,1 - g10,1L = 1

2( !0h

+ - h

+!0h

+)

As shown in [3], the other non-zero terms are:

G110 = G101 = G

011 =

1

2( !0h

+ - h

+!0h

+) and G

113 = G

131 = -G

311 = -

1

2( !0h

+ - h

+!0h

+)

NowRmnrs isof the form !G - !G + GG-GG andfrom

H15

Ltmn includes only time-averaged second order terms

in hmn. The second order condition implies that the h+!0h

+ only enter into the !G terms, producing elements "X!

sin[2w (t-z/c)]\"Xcos[2w (t-z/c)]\ = 0. Hence there are no !G contributions to the Ricci tensor. Only the GGcontribution enters into the Ricci tensor and only via the !0h

+ element ofG. Using

YRH2Lmn] = XGaalGlmn - GamlGlan\ then for example,

RH2L00 = GaalG

l00 - G

a0 lG

la0 = 0 - G

10 lG

l10 - G

a01G

1a0 = -2G

101G

110 = -

1

2I!0h+M2

Similarly

RH2L33 = -1

2I!0h+M2 and RH2L11 = RH2L22 = 0

And the curvature scalar is:

RH2L = hmnRH2Lmn = -RH2L

00 +RH2L

11 + RH2L

22 +RH2L

33 = 0

Now if we include the same contribution from the h polarization state, and take the positive value for energy



8/21

density we find using (15) that

t00 =c4

16GZI!0h+M2 + I!0hM2^ (18)

Writing h

+ as h

TT

11 = - h

TT

22 and h

as h

TT

21 = h

TT

12 and note that h

TT

3 j= 0. Also define h

TT

ij =!th

TT

ij so thatZI!0h+M2 + I!0hM2^ = 12 c2

Xh TTijh TTij\. Then the flux (f), defined here as energy per unit area per unit time, isequal to ct00, and can be written as:

f =c3

32 GXh TTijh TTij\ This is the flux (per unit time) of gravitational radiation energy. (19)

Emission of Gravitational Radiation

Following the same procedure as in electromagnetism when a source exists, the solution to (11):

hmn = -16 G

c4

Tmn (in the TT gauge ) is

hmn(t, x) = -4 G

c4

TmnHtR, x'Lx-x' 3 x' where tR is the time at the source, or the retarded time,

tR = t - x-x/cTo simplify the solution, we assume we are in the radiation zone far away from the source. Then we can replace

x-x ==> x = r and tR = t - x-x/c ==> tR = t - r/c Then

hmn(t, x) - 4Gc4 r TmnHtR, x'L 3 x' (20)We now want to simplify (20) so that it can be used to evaluate the flux via (19). Recall that energy-momentum

conservation requires that !mTmn = 0. Setting n = 0, differentiating by !0 one more time and expanding we obtain

!2T

00

c2!t

2 = -!2T

i0

c !t!xi = -

!2T

i0

c !xi!t

Now if we use the continuity equation again with n = i we have !0T0 i + !jT

ji = 0 and then

!

2

T

00

c2!t

2 = !2

T

ij

!xi!x

j Now multiply both sides byxkxland integrate over the source volume:

!2

c2!t

2 T00xkxl3x = !2T

ij

!xi!x

j xkxl3x =

!Tij

!xj x

kxl -+

- 2 !Tij

!xj x

kdli 3x

= - 2 !Tij

!xj x

kdli 3x = 2Tij dkjdli 3x = 2Tkl 3x (21)



9/21

The surface terms in the above integration by parts vanish since the source Tij is assumed to extend to only finite

range. Now combining (20) and (21) we find that:

hij(t, x) = -2 G

c4

r

!2

c2!t

2 T00xixj3x

If we assume the energy density is related only to the mass, then T00 = rc2 (i.e. non - relativistic case for the

source). The second mass moment is IijHx, tRL = rHx, tRL xi xj3x , then

hij(t, x) = -2 G

c4

rI..

ijHx, tRL The linear gravitational wave is proportional to the second time derivative of the

second mass moment - or the quadrupole mass moment. (22)

As we are working in the transverse and traceless (TT) gauge, we define Qij as a traceless version ofIij

Qij = 3Iij - dijIkk where Ikk is the trace ofIij (23)

Clearly Qjj = 0, hence Qij is traceless. Writing (22) in terms ofQij we obtain:

hij = -2 G

3 c4

rH Qij.. + dij !

2

!t2

Ikk ) (24)

Total Power Radiated by a Gravitational Source

In order to calculate the Energy flux we see from (18) and (19) that we need:

I!th+M2

+ I!thM2

= H!th11L2 + H!th12L2 We want this in terms ofQijWe now do some algebra. Since h11 = - h22 then (h11 - h22L2 = h112 + h222 - 2h11h22 = 4 h112 then

H!th+L2 + H!thL2 = 14H!th11 - !th22L2 + H!th12L2 Now use (24) with hij = hij in the TTgauge

!th12 = -2 G

3 c4

rQ...

12 and !th11 - !th22 = -2 G

3 c4

r(Q...

11 - Q...

22) so we have

H!th+L2 + H!thL2 = 4 G2

9 c8

r2

BJQ... 12N2 + 14JQ... 11- Q

...

22N2F

Using (18) for the flux (per unit time) = ct00

fz =G

36 c51

r2 ZJQ

...

12N2 + 14JQ... 11- Q

...

22N2^ This is the flux per unit time in the z direction. (25)

This expression for the flux in the z direction will need to be generalized to any direction, so that we can then



10/21

integrate it over all directions to find the total flux. This will take some algebra.

Since Qij is traceless Q33 = - Q11 - Q22 and Qij = Qji from the symmetrical definition ofIij. Then

1

4H2 QijQij - 4 Qi3 Qi3 + Q33 Q33 L = 1

4(2 (Qi1Qi1 + Qi2Qi2 + Qi3Qi3) - 4 Qi3 Qi3 + Q33 Q33 L

=1

4(2 (Qi1Qi1 + Qi2Qi2 - Qi3Qi3) + Q33 Q33 L

=1

4(2 IQ112 + 2 Q122 + Q222) - Q332M

=1

4(2 IQ112 + 2 Q122 + Q222) - HQ11 + Q22L2M

=1

4(Q11

2 + Q222 + 4Q12

2 - 2 Q11 Q22)

= Q122 +

1

4HQ11 - Q22L2

Therefore

fz =G

36 c51

r2

1

4Z2 Q... ijQ

...

ij - 4 Q...

i3 Q...

i3 + Q...

33 Q...

33^ (26)

If we would like the flux in an arbitrary direction n`

= Hn1, n2, n3 L which is quadratic in Q...

ij then the most general

form of (26) is:

fn =G

36 c51

r2 Z aJQ

...

ik ni nkN2 + bJQ...

ijQ...

jk ni nkN + gJQ...

ijQ...

ij N (27)

Now we must solve for a, b and gmatching (27) to (26) for the case that n`

= H0, 0, 1 L. By inspection we see that

a =1

4 b = -1 g=

1

2which gives

fn

=G

36 c5

1

r2

Z1

4 JQ...

ikninkN2

-

JQ...

ijQ...

jkninkN

+1

2 JQ...

ijQ...

ij N

Define Intensity (I) as energy flux area per unit time = -E

t. Therefore the infinitesimal amount of Intensity

per unit time, In, flowing through the infinitesimal surface area element r2W in the direction n

`is

In =G

36 c5Z 14JQ... ik ni nkN2-JQ

...

ijQ...

jk ni nkN + 12JQ... ijQ

...

ij N W This is the differential Intensity in directionn^

through solid angle W. (28)

The total Intensity per unit time (or power) is found by integrating (28) over the solid angle W = sinqqf. Toperform the integration over all directions of a sphere we make use of:

W = 4nknl W = 4 3 dklni njnknl W = 415 Idij dkl + dkj dil + dik djl) (29)



11/21

The above integration results can be shown explicitly by using spherical polar coordinates with

nx = sinqcosf , ny = sinqsinf and nz = cosf. For example:

nx ny W = sin3 q sinf cosf q f = 0 and

nx nx W = sin3 q cos2 f q f = 4

3

Using (28) and (29) the total power radiated is then:

I =G

36 c5 Z1

4JQ... ik ni nkN2 - JQ

...

ij Q...

jk ni nkN + 12 JQ...

ij Q...

ij N W

=G

36 c54Y 1

60(dij dkl + dkj dil + dik djl)Q

...

ijQ...

kl - Q...

ijQ...

jk I dik3 M +1

2JQ... ijQ

...

ij N^

=G

9c5 ZQ

...

ij Q...

ij ^@ 260 - 13 + 12 ] using the traceless and symmetry properties ofQij

I =G

45 c5i,jZQ

...

ij Q...

ij ^ This is the total power radiated by a gravitational source with quadrupole Qij. (30)

This expression compares with -E

t= I =

1

720 c5i,jZQ

...

ij Q...

ij ^ in the electromagnetic field case using Heavi-side-Lorentz units.

Angular Momentum and Gravitational Radiation

Just as a gravitational system loses energy via radiation it also loses angular momentum, L. Now L = S (r p)assuming a group of discrete particles, therefore

L

= S ( r p) , since r

p = 0. HenceL

=S (r fgr), where fgr is the damping force from gravitational

radiation. We actually are interested in the time weighted average loss of momentum, hence we need:

XL \ = SXr fgr\ ===> XL a\ = SYeabgrb fg\ (31)

To find fgr we note that Egr = Sfgr r hence

Z

Egr

t

^=

Xfgr v

\= -

G

45c5

ZQ...

ij Q...

ij

^ using (30) And (32)

ZQ... ij Q...

ij ^ = kQ...

ij Q...

ij t then integrate by parts twice

= k Q..

ijQ...

ij -+

- kQ.. ij Q....

ij t = kQ ij QH5Lij t where the surface terms vanish at

ZQ... ij Q...

ij ^ = YQ ij QH5Lij] with QH5L representing the fifth time derivative (33)



12/21

Using the definitions ofQij in (23) andIij above, we have

Q

ij = @m H3ri vj + 3vi rj - 2 dij r v)] (34)

Then using (32) - (34) we have

Xfgr v\ = - G45 c5

XmH3ri vj + 3vi rj- 2 dij r v)QH5Lij \

This implies (using Qjj = 0) that

(fgrMj = -G

45 c5mI3 ri + 3 rj)QH5Lij = -

2G

15 c5mri Q

H5Lij

Substituting this into (31) gives

XL a\ = - 2G15 c5

Yeabgrb m rlQH5Llg\ = - 2G45 c5

Yeabg3 m rb rl QH5Llg] = - 2G45 c5

Yeabg3Ibl QH5Llg]

= -2G

45 c5YeabgAQbl + dbl IkkE QH5Llg] = - 2G

45 c5eabgYQbl QH5Llg] Since

Xeabgdbl QH5Llg\ = X eabgQH5Lbg\ = 0 since XQ ab\ = k Q ab t = k Qab -+ = 0 Therefore

XL a\ = - 2G45 c5

eabgYQbl QH5Llg] = - 2G45 c5

eabgZQ.. bl Q...

lg^ This is the rate of loss of angular momentum. (35)



13/21

III. The PSR 1913+16 Binary Pulsar

Overview

In 1974 Russell Hulse and Joseph Taylor, using the Arecibo radio telescope in Puerto Rico, completed a survey of

pulsars. Among the 40 new pulsars that they discovered was the first to be located in a binary star system. The

two stars orbited around each other once every 7.75 hours, as shown below in the highly eccentric velocity curve

from Hulse and Taylors original Letter to the Astrophysical Journal in 1975 [6].

The pulsar has a very short period of only about 0.059 seconds. The accuracy of this period is very high -- on the

order of10-14 seconds [5]. Since the pulsar is created by a neutron star with a mass of 1.4 M packed into a radius

of only about 10 km, a very intense gravitational field I105 times as strong as the sun at its surface) is created inthe region of interest. This combination of a very accurate clock and strong gravitational field creates a powerful

laboratory to study the validity of Einsteins theory of General Relativity.

The binary system can be analyzed using basic orbital equations from Keplerian mechanics. The two stars will

move around their center of mass in an elliptical orbit as shown in the diagram below. Call the semi minor axis of

the pulsar orbit ap and its eccentricity e. Label the mass of the pulsar mp and the companion star mc. The angle of

inclination of the orbit to our line of site is i and the period of the orbit is Pb.

There are additional Non-Keplerian parameters which are also relevant to this system. These arise out of

General Relativistic effects. First, the orbits period will shrink over time due to gravitational radiation. The rate

of change is P

b. Second, the orbit also precesses, just as Mercurys perihelion precesses about the sun due to the

effects of GR. In this case, the pulsars periastron (the point of closest approach to the system center of mass)

advances at a rate of 4.2 degrees per year! This is Xw 0\ in the table below. Finally, the arrival time of the pulses isaffected by both relativistic Doppler effects and gravitational redshifts. This is captured by the parameter gshown

in the table below.

!

Pulsar

Companion

e ri ast ro n A pa st ro nApastron

Binary Pulsar System

ap



14/21

The PSR 1913+16 system has been studied extensively over the last thirty years. In 2005 the most recent set of

measured orbital parameters were reported as follows [7]:

Table 1

Fitted Parameter Value

Iap sin iMc HsL 2.3417725 (8)e 0.6171338 (4)

T0 (MJD) 52144.90097884 (5)

Pb HdL 0.322997448930 (4)w0 (deg) 292.54487 (8)

Xw 0\ (deg/yr) 4.226595 (5)g (s) 0.0042919 (8)

Pb I10-12 s/s) -2.4184 (9)

mp (M) 1.4414 (2)

mc (M) 1.3867 (2)

Circular Orbit Calculations

As a first approximation we shall assume that the two stars of mass m1 and m2 follow circular orbits about their

common center of mass, as shown below.

m1

m2

a1

a2

If the respective distances from the center of mass are a1 and a2, then

m1 a1 = m2 a2 = m a where a = a1 + a2 and the reduced mass m = m1m2 / (m1 + m2 ) (36)

Each mass orbits in the (x,y) plane according to:

xi = ai cos wt yi = ai sin wt z = 0

Keplers third law gives:

w2 =G m

a3

with m = m1 + m2 (37)

The total energy of the binary system is E = KE + PE. Using vi = wai , (36) and (37) give

E = J m1 a122

+m2 a2

2

2N w2 - G m1 m2

a= -

1

2

G m m

a(38)



15/21

With the orbital period Pb =2

w, the rate of orbital decay is

1

Pb

Pb

t= -

1

w

w

t=

3

2 a

a

t= -

3

2E

E

t(39)

Thus the rate of orbital decay = -3

2E

E

tand we can use (30) to find

E

t. Hence we need to calculate the

quadrupole moments for this orbit.

Quadrupole Calculations

Using the definition ofQij in (23) with a sum over two point masses replacing the integral inIij gives:

Qxx = i=12 H3 xi ximi L - dxxIkk with

Ikk= i=12 Hxi xi + yi yi +zi zi)mi = i=12 ai2Icos2 wt + sin2 wt)mi = a12 m1 + a22 m2 = m aHa1 + a2L = m a2

Then Qxx = 3 Ia12

m1 + a22

m2M cos2

wt - m a

2

= 3 m aHa1 + a2) cos2

wt -m a

2

=m a

2

(3cos

2

wt -1) Similarly

Qyy = m a2 (3sin2 wt -1) and Qzz = - m a

2

Qxy = Qyx = i=12 H3 xi yimi L = 3 Ia12 m1+ a22 m2M coswt sinwt = 32 m a2 sin2wt

The derivatives are as follows:

Q...

xx = - Q...

yy = - 12 m a2 w3 sin 2 wt and Q

...

xy = Q...

yx = - 12 a2 w3 cos 2 wt

Using Ysin2

x] =1

2 , the average squared values over one period are:

ZQ...

xxQ...

xx^ = ZQ...

yyQ...

yy^ = ZQ...

xyQ...

xy^ = 72 m2a4 w6

Circular Orbit Decay Rate

Using (30) we can calculate I = -E

t= -

G

45 c5i, jZQ

...

ij Q...

ij ^

E

t= -

G

45 c5JZQ

...

xxQ...

xx ^ + ZQ...

yy Q...

yy ^ + 2 ZQ...

xy Q...

xy ^N

E

t= -

32 G

5 c5

m2a4 w6 = -32 G4

5 c5

a5

m2 m3 using (37) (40)

Then using (38) and (39) the circular orbit decay rate is:

1

Pb

Pb

t= -

3

2

2 a

G m m

32 G4

5 c5

a5

m2m3 or



16/21

1

Pb

Pb

t= -

96

5

G3 m m2

c5

a4

This is the circular orbit decay rate. (41)

Elliptical Orbits

In 1963 Peters and Mathews [8] showed that for a general orbital eccentricity, e, thatE

(e) =E

(e=0) f(e). The

form off(e) will developed below.

The basic Keplerian orbital equations [9] are:

r =aI1-e2M

1+ e cos qwith angular momentum l= m r2 q

= m G Hm1 m2L a I1 - e2M . This gives

q =

G Hm1+m2L a I1-e2Mr2

(42)

The quadrupole terms are the similar to the circular forms shown above, with a replaced by r and wtreplaced by q.

Qxx = m r2(3 cos2 q-1) Qyy = m r

2(3 sin2 q-1) Qzz = -m r2 Qxy = Qyx = m r

2 3sinqcosq

Then the derivatives are:

Qxx = -

2 m sinHqL He + 3 cosHqLL a H1-e2LG Hm1+m2Le cosHqL+1

Q..

xx =1

2 a Ie2-1M G m Hm1 + m2L H13 e cosHqL + eH4 e + 3 cosH3 qLL + 12 cosH2 qLL and

Q...

xx =1

a4 Ie2-1M4

m sinHqL He cosHqL + 1L2 HeH9 cosH2 qL + 11L + 24 cosHqLL Ia I1 - e2M G Hm1 + m2LM32 or

Q...

xx = b sinHqL He cosHqL + 1L2 (e(9 cos(2q)+11)+24cos(q)) with

b2 =G

3Hm1 m2L2 Hm1+m2La5I1-e2M5

=G

3 m2 m3

a5I1-e2M5

Similarly

Q...

yy = -b sinHqL He cosHqL + 1L2 He H9 cosH2 qL + 13L + 24 cos HqLL

Q...

zz = 2be sinHqL He cosHqL + 1L2

Q...

xy = Q...

yx = -3

2b He cosHqL + 1L2 H5ecosHqL + 3ecosH3 qL + 8 cosH2 qLL



17/21

Using (30) the total power is:

E

t= -

G

45 c5JZQ

...

xxQ...

xx ^ + ZQ...

yy Q...

yy ^ + ZQ...

zz Q...

zz ^ + 2 ZQ...

xy Q...

xy ^N

Squaring the quadrupole derivatives from above, but before taking the average over a period we have:

E

t= -

4G4 m2 m3

15 c5 a5I1-e2M5Y He cosHqL + 1L4 I11 e2 cosH2 qL + 13 e2 + 48 e cosHqL + 24M ] (43)

In order to take the average over one period, t, we convert tto qusing (42):

t =IaI1-e2MM32

GHm1+ m2L

1

H1+e cosqL2 q and

t =2 p a32

GHm1+ m2Lthen

1

t 0tt =

I1-e2M322 p 0

2 p

H1+e cosqL2 q so now (44)

E

t= -

2G4 m2 m3

15 p c5 a5I1-e2M7202 pHe cosHqL + 1L2 I11 e2 cosH2 qL + 13 e2 + 48 e cosHqL + 24M q

= -32G4 m2 m3

5 c5 a5I1-e2M72I1 + 73

24e2 +

37

96e4M (45)

Comparing this with the circular result, (40), we finally have:

E

t(e) =

E

t(0) f(e) where

f(e) =J1 + 73

24e2+

37

96e4N

I1-e2M72

Therefore, using (41), the orbital decay rate for the elliptical orbit with eccentricity e is:

1Pb

Pbt

= - 965

G

3

m m2

c5

a4

f(e) This is the elliptical orbit decay rate.

PSR 1913 +16 Decay Rate Results

Inserting the latest observed data from Table 1 (and using Keplers third law (37) to provide a from Pb) gives a

theoretical decay rate of:



18/21

Ptheory = -2.4021 x 10

-12 s/s or about -7 x 10-5 s/year

The observed decay rate in Table 1 is -2.4184 .00009 x 10-12 s/s. A small correction should be applied to

account for the relative acceleration between the binary pulsar and the solar system. Using the estimate of -.0128

.0050 x 10-12 s/s provided in [7] gives a net P

.

obs = - 2.4056 x 10-12 s/s.

Therefore:

P

.

obs

P.

theory

= 1.0015 .0021 This is a measure of the accuracy of the theory of gravitational radiation.

Circularization of the Elliptical Orbit

From equations (30) and (35) we see that the radiating system will lose both energy and angular momentum (L).

Since an orbits eccentricity is related to its angular momentum and energy via:

e2 = 1 +2E L

2

G2 m3 m2

(46)

we can see that the radiating system will lose eccentricity over time and will eventually convert into a circular

system. This will be shown below.

Equation (46) implies:

e

t

=L

2E+ 2E L L

.

G2

m

3m

2e

(47)

To computee

twe must findL

.

for an elliptical orbit. Following the same process as above for the elliptical

energy calculation, starting with (35) and recognizing that in our case L is along the z direction. Then

XL \ = - 2G45 c5

ezbgZQ.. bl Q...

lg^= -

2G

45 c5J ZQ.. xl Q

...

ly ^ - ZQ..

yl Q...

lx ^N= -

2G

45 c5J ZQ.. xy Q

...

yy ^ +ZQ..

xx Q...

xy ^ - ZQ..

yx Q...

xx ^ - ZQ..

yy Q...

yx ^N= -

2G

45 c5

J ZQ..

xy

JQ...

yy - Q...

xx

N^+

ZQ...

xy

JQ..

xx - Q..

yy

N^NInserting the derivatives from above, but before taking the average over the period, we have:

XL \ = - 4 G72 m2 Hm1+m2L525 c

5a72 I1-e2M72

X He cosq+1L3Ie2 + 3 eHcos2q + 4 cosqL + 8) \



19/21

Taking the average over one period using1

t 0tt from (44) gives:

XL \ = - 32 G72 m2 Hm1+m2L525 c

5a72

J1+ 78

e2N

I1-e2M2(48)

WithL = m G Hm1 m2L a I1 - e2M , E = - G m1 m22 a

and using (45) through (48) gives

e

t= -

304

15mm2 G3

c5a4

e

I1-e2M52I1+ 121

304e2M This is the rate of change of eccentricity. (49)

This equation shows thate

t 0, therefore the elliptical orbit will eventually decay into a circular orbit with e = 0.

Once the circular orbit is reachede

t= 0, and there are no further changes in eccentricity. However, as will be

shown below, by the time the orbit has circularized the separation between the two masses has been reduced to zero as

well.

The Evolution of Orbit Size

From the relation E = -G m1 m2

2 awe can easily find a

fromE

:

a

t=

G m1 m2

2E2

E

t=

2 a2

G m1 m2

E

t then using (45) we have:

at

= - 64 G3

m m2

5 c5

a3I1-e2M72 I1 +

7324

e2 + 3796

e4M (50)

We can now find an interesting relationship between a and e by taking the ratio of (50) and (49):

a

e=

12

19

a

eJ1+ 73

24e2 +

37

96e4N

I1-e2M J1+ 121304

e2N

(51)

The exact solution to (51) is given in Peters [10], where these ideas are developed :

a(e) = Ic0

e1219

MI1-e2M (1 + 121304 e2M8702299 with c0 determined by initial conditions a0 and e0 (52)

A log plot of this function follows on the next page.



20/21

Separation vs. Eccentricity

0.2 0.4 0.6 0.8 1.0

e

0.1

0.5

1.0

5.0

10.0

50.0

100.0

a@eD

If we insert a(e) from (52) into (49) we obtain:

e

t= -

304

15m m2 G3

c5

c04

e-2919 I1-e2M32

J1+ 121304

e2N11812299

(53)

The lifetime (from gravitational radiation decay alone) THa0, e0) can be found by integrating (53):

T

Ha0, e

0) =

15

304c5c0

4

mm2 G3

0e0

e2919J1+ 121

304e2N11812299

I1-e2M32e This is the lifetime of the binary system. (54)

In the case of the PSR 1913+16 system we have a0 = 1.949 109 m and e0 = .6171. With these values we can

determine c0 from (52) as 1.552 109 and then we have the lifetime of our binary system as:

T1913+16 = 5.244 10160

.6171e2919J1+ 121

304e2N11812299

I1-e2M32e seconds

After evaluating the integral numerically we find that:

T1913+16 = 9.489 1015 seconds or 3.009 108 years.



21/21

IV. References

[1] Taylor, J.H., 1994, Binary pulsars and Relativistic Gravity, Reviews of Modern Physics 66,711.

[2] Belusevic, R., 2008,Relativity, Astrophysics and Cosmology, Volume 2, (Wiley).

[3] Cheng, T., 2005,Relativity, Gravitation and Cosmology, (Oxford U. Press).

[4] Landau, L.D. and Lifshitz, E.M., 1975, The Classical Theory of Fields, (Butterworth-Heinemann).

[5] Taylor, J.H. and Weinberg J.M., 1989, Further Experimental Tests of Relativistic Gravity Using the Binary

Pulsar PSR 1913+16, Astrophysical Journal 345,434.

[6] Hulse, R.A. and Taylor, J.H., 1975, Discovery of a Pulsar In a Binary System, Astrophysical Journal 195,

L51.

[7] Taylor, J.H. and Weinberg, J.M., 2005, The Relativistic Binary Pulsar B1913+16: Thirty Years of Observa-

tions, ASP Conference Series 328, 25.

[8] Peters, P.C. and Mathews, J., 1963, Gravitational Radiation from Point Masses in a Keplerian Orbit,

Physical Review, 131, 435.

[9] Goldstein, H., 2002, Classical Mechanics, (Addison Wesley).

[10] Peters, P.C., 1964, Gravitational Radiation and the Motion of Two Point Masses, Physical Review, 136,

p1224B.


grav radiation project

Documents