I. 19.20 ) circular orbit unstable a) Vmax rmcxwrt.hn ✓ rmnx > rmax = , d¥ speed where in → o . have we where : ftp.lm-ep = 1- 12M¥ T " - K ¥-1 " 3M = . frame Ñ is , = d¥T d¥ v= . where d¥ has no radial component circular orbit ) Not the not and we = radius In Ui EI components of are this orthonormal frame Not at is !÷µ 121^-11? a In observer 's so My I → a. As greatest potential In -4-11-121%19 = plot at A has 4- velocity of the . Ut = - Ui EE eI.eu#--nqy--le-I.e-I=Ret---landeI.e-I--Nu---O particle . - . lift [ 0.0.0.x] v2.x =/ ' __ ✗ = f- e¢✗=[ 0,010 I ] , eÉ=[ g. 0.0.0 -11-2%1.5=-1 ] y=( 1- 2¥) - ± eÉ=[ (1-215%40,0) a. eI=r:¥if=r¥ Hiei = - (1-21) .it#.l1-2M-rj-i----l1-2M-rHddt-- And for circular orbits µ¥T=Y→ , v-r.ctzm-rl-t.de#- rltzY-1-EF splug ihgihr- rmnx- 3MV3M b) - (1-5) Same = ± - NIV radius in § -15--1 - . photon For circular Wmax - - - orbit 3M ) as in a) , . F- 3M . f- speed of light . 2 . Assume the observer is light my the , boat near the neutron star : - - let-1 ¥7 observer • on - - - The farthest point backside on farthest from the axis tangent to the no circle is seen which , through by symmetry is light ray the curve . Sy bending za-184 =L A- Ito = 1¥ 1=(11-84)/2 4¥) = 4¥ . mail.5kmk-wkm-i.cn %) - = A. = ⇐ %) - E- if ° = Go -1718 = 72.8° . 3. For a) the light ray light ray distance 's will be scattered The detection angle So some , b) from away deflected depends ray on larger than 527 M , . the distance back turn can is axis from axis its origin to (b) and M . . < - ? I • - - - - → angle of deflection when it is equal to E. light ray will be directly reflected When it is For the be the , . d { ^ b. • close to IT angle deviated ¥ exceed , and this so , , the emitting area beat # A and is but E) = go has lighting a back with limit that it impact parameter - . Its latentf- a lbltu ¥1T Tv 2b / b ti 2¥ 4h da ble b't approximate ble to 1st order for ¥ ' - - - - is the impact parameter that light my deflects by cannot is still received in the a. = extra a . between trouble with - = will angle that the radius d receiver So light ray The , exactly IT . . By WI 19.78 ) a. zfodw [1- will ¥wÑ - = where plug The number - solve for bl a. and We can And 1- will equation fur back into received N=t* A ¥w) = 0 . b' la numerically area A . . is then - c) By analysis btn Na f-* Mi .dz dimensional Then the buser contains mass Unveil of to M ? M f-* Regardless of to " and b' la should all be , multiples of M . In Box 2.1 The . black hole M Mñ . , 1019 photons is per second approximately (number)/see = . to 'tg - = 1×10 m at 104 ! which is also seconds to receive one f-* to " . 0.742×10-28×1015 cm is in order dz . photon a small figure reflected back , , there need which is not around applicable . . 4. 46.1 ) di gap is -_ dtit ( Hfltrz ) ) diet 11 total - = - ( Htttz 1- tttz , independent of are to 1 , Also So , , , o along killing , ✗ (0 o) t - Z andy 0, I , -0 - , so . dy4 do ) killing vectors with these two 10) the metric is also vector with this symmetry is unchanged 11 , o , o, since 1) flt-2-1 . is unchanged . J . G) spatial so coordinates of particle worldline should be straight line a In ok is invariant at rest , . fitz > € " e- origin - "-2440 > ± C- zip > In 't) ez )4ñ< link >z y zP< stench - - b) rÑt-z< Tdbnkl ¥*=Éh*ltio)=±tlt -01 =±a e-¥ e- the fort >o . So , maximum 8L > ti = happens In . at -1=0 Lt 6. d5= dti-llthxxlt-zydx.tl/thyylt-zddytdE - Measuring distance from @ Not to Hit it , dtio IX. Yit) till )µu,o = " ) where We an ds-jcthxxlt-zxdxl-llthyy.tt#dy4d-= parametrize it from o to the from path L* . Giro ) to 1×1%2-1 Use hi given with parameter d. . L-xt-XYLL-xl-YZCL-x-Z.nl 1¥ ¥* ¥* ) Since ✗( = , we can write . ✗ ill) = hid ds becomes Then d5=( It hxxlt-h44.nxjdi.tl/-hyylt-nZN).lnYTdd-(nZTdii Let To first ↳ dd-H-lhxxlt-nml.at/lthyylt-n7blnY4-ln4Y = = order ± approximation ↳ + I fÉda[4th * it nm - ) . Hit / It hyylt-nnln44-ln.TT 84-11=4-1 ) Slits Since we = Lt - E) Y da -48*-1 hxxlt-nZXHP-lsyythyylt-ntnllnptszil.MY] sxy-ohxylt-h74-hyx-ohzz-ohxz-hzx-hyz-hz.ge can write slot 8kt) compactly as )=Éfo↳dd[sijthijlt - vital ] nini 0 , 7. Without gravitational wave 1 I → light Orient the Along the hull " d. × = an - So , disxi •* To find Assume = ( 1. toward X - axis the coordinates , geodesic 1, is are oixi ⇐ along So . d*= + x-axis d=✗ , solutions of - . geodesic egs : TIP d¥dd¥ ↳ -0 for hnek ground spacetime . , ¥×=¥× -4 ¥m=d¥× 0,01 . ftp.sluiuM -0 -8T¥ -28T¥ -8T¥ 8×+64=8×4×7 hap = µ , need we changes 0 fitz, - fttz ) O to Christoffel symbol ( fit has frequency . w ) o ( gap - perturbation in - . giiTIp=±(°§¥•t%¥→ %¥-i ) Taking ray l -8T¥ d¥ ¥5 = sink null = my TIP %Éd¥ dxz along light →✗ geodesic ñsxi d¥ a d5=o displacement , first order of is diagonal ) hap ftp.t-hii-kptniisl-jp-zfohf! -134¥ -31¥ This -01 Mii -1 strip 1=101*+34,4 -01¥ ) scgii - . for ii-xiy.tt - . ① sT¥=±e¥÷t%÷°-%¥iI=o For ② i-x.y.it 8T¥ . -0 - STIX =±fh¥i+%¥- %¥ i-tx.hn/-o.ST-Ix--o For i-X.hxx-flt-z-8-fx-I-tlt.it ) For ③ - 8T¥ =E(%¥ -101¥ 4¥71 For i=× 8T¥ =±(¥¥-)=±3h¥=±tlez) - , For i=y 8T¥ . For i=z , Thus , 8T¥ 22 SX = - - I Coto -21¥) = = - I 2ft-24 az = Éflttzy 228g 0 = 2×2 azsz = Etty - an z) %¥And - - = we ① = c- - - and , , Z) ) , then I awwscwx) have two 1- ' 1-1-2-1 ¥¥= endpoints set , - awwslwttz) ) = = Ea sinlwx) -1C 8×101--8×14=0 so Edw - E. a-wlwscwxltx.lt ( 1- wslwll )) -11 - 82=-18 Sy 0 ( wscwxltx.CI - awaswx for our null geodesic . S×= 3-za-w.wscwxl -10×+0 IFL 4- wslwll ) 8x= similarly - sin ( WH a = If 't - on - I (0+0-0)=0 have we It fit = ) ( 1- wslwll )) -11 )