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The Schwarzschild Solution and Timelike Geodesics

The Schwarzschild Solution and Timelike Geodesics

Utah State University DigitalCommons@USU

Physics Capstone Project Student Research

1-28-2016

The Schwarzschild Solution and Timelike

Matthew Ross Utah State University

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Recommended Citation Ross, Matthew, "The Schwarzschild Solution and Timelike Geodesics" (2016). Physics Capstone Project. Paper 31. https://digitalcommons.usu.edu/phys_capstoneproject/31

This Report is brought to you for free and open access by the Physics Student Research at DigitalCommons@USU. It has been accepted for inclusion in Physics Capstone Project by an authorized administrator of DigitalCommons@USU. For more information, please contact [email protected]. The Schwarzschild Solution and Timelike Geodesics

Matthew B. Ross and James T. Wheeler, Utah State University January 28, 2016

Abstract is the standard theory of the gravitational interaction. It allows us to cal- culate the motions and interactions of particles in a non--time. This presentation will present the derivation of the field by finding a solution of the Einstein for a non-rotating, static vacuum. A general form of the metric for a static, spherically symmetric will be used to calculate the and sub- sequently the Ricci tensor and Ricci scalar which will then be used to find a vaccum solution to the Einstein Equation. Once the solutions of the Einstein equation are found, we can study the equation. This lets us find the for massive particles moving around and into a . Overall, this presentation provide an examination of the basic calculations that are done in General Relativity and shows how matter moves in a curved space-time.

1 Einstein’s General has produced a massive change in how we view the universe and how we calculate the motion of objects in the universe. Einstein was revolutionary in that he was able to develop an accurate theory of time and space that was later confirmed by observation. This theory involved using Riemannian to describe as the motion of particles in a curved spacetime. This is quite different from the classical theory of outlined by Issac Newton and others. Perhaps the biggest paradigm shift caused by Einstein’s General Theory of Relativity is that particles move in the straightest possible lines (geodesics) within this curved spacetime rather than particles being affected by gravity while moving in a Euclidian spacetime. We begin by finding the field for a static, non-rotating Schwartzchild black hole. A metric tensor field is symmetric bilinear form on the that satisfies the conditions of an inner product at each point [1]. Once this metric has been found, of motion – the geodesic equation – around this black hole can be calculated. There are three types of geodesic for which can be calculated: timelike, spacelike, and null. Two of these, the timelike and null, correspond to known types of particle. In this paper we solve the geodesic equation for a radially infalling particle, showing that the particle reaches the horizon in a finite them. Then we compute the solution for circular orbits around a Schwartzchild black hole, showing that there is a minimum radius related to the mass of the black hole.

The Schwarzschild solution Using a static, spherically symmetric metric with all components being functions of the radius allows us to put the element in the form ds2 = f 2(r)dt2 +h2(r)dr2 + 2 2 2 2 2 r d✓ + r sin (✓)d . The coeffecient , denoted gµ⌫ is as follows:

f 2(r)0 0 0 0 h2(r)0 0 g = (1) µ⌫ 0 00r2 0 1 B 000r2 sin2(✓) C B C @ A In order to solve for the functions f 2(r) and h2(r), the for this metric will need to 1 be found. The Einstein tensor is Gµ⌫ = Rµ⌫ + 2 gµ⌫ R, where Rµ⌫ is the tensor, gµ⌫ is the metric tensor field, and R is the Ricci scalar. The Ricci curvature tensor and the Ricci scalar are contractions of the Riemannian curvature tensor. The Riemannian curvature tensor gives a measure the curvature of a Riemannian by encoding the transformation of a vector V~ under transport along a closed loop in the manifold [2]. The method used to find the Riemannian curvature tensor is to first find the Christoffel connec- tion of the metric which is found from a given metric tensor field and its derivatives [3]. Although the Christoffel is not present in the Einstein equation, it is an essential step on the way to calculating the Riemannian curvature of a given metric tensor field. Specifically, the Christoffel connection is given by [3]: 1 ↵ = g↵(g + g g ) (2) µ⌫ 2 µ,⌫ ⌫,µ µ⌫,

Substituting in the metric gµ⌫ above, we find that the Christoffel symbols are:

2 f t =t = 0 tr rt f ff r = 0 tt h2 h r = 0 rr h r r = ✓✓ h2 r sin2(✓) r = (3) h2 1 ✓ =✓ = ✓r r✓ r ✓ = sin(✓) cos(✓) cos(✓) = = ✓ ✓ sin(✓) 1 = = r r r

where these equations are in terms of the functions f,h in the metric gµ⌫ .Aprimedenotesa df derivative with respect to r, f 0 . Using these expressions for the Christoffel symbols, the Riemann ⌘ dr curvature tensor is given by:

R↵ =↵ ↵ +↵ ⇢ ↵ ⇢ (4) µ⌫ ⌫,µ µ,⌫ µ⇢ ⌫ ⌫⇢ µ Substituting to find the components of the Riemann curvature tensor,

f 00h f 0h0 Rt = Rt = rtr rrt fh f r Rt = Rt = 0 ✓t✓ ✓✓t fh2 f r sin2(✓) Rt = Rt = 0 t t fh2 (f 00h f 0h0)f Rr = Rr = trt ttr h3 rh Rr = Rr = 0 ✓r✓ ✓✓r h3 2 r r r sin (✓)h0 R r = R r = 3 h (5) ff R✓ = R✓ = 0 t✓t tt✓ rh2 h R✓ = R✓ = 0 r✓r rr✓ rh sin2(✓)(h2 1) R✓ = R✓ = ✓ ✓ h2 f r sin2(✓) R = R = 0 tt tt fh2 h R = R = 0 rr rr rh h2 1 R = R = ✓✓ ✓✓ h2

3 By contracting the Riemann curvature tensor on its first and third indices, the Ricci curvature tensor can be found. The resulting is the first object to appear in Einstein’s field equation. Using the components found for the Riemann curvature tensor, the Ricci curvature tensor can be calculated as the contraction:

↵ Rµ⌫ = R µ↵⌫ (6) It is worth noting that the contraction on the first and third indices is the only contraction that we care about for these purposes. All other contractions either vanish, or simplify down to R [5, 6]. ± µ⌫ Carrying out the sums, we find

f R = (hf 00 rh0f 0 +2hf 0) (7) tt rh3 1 R = ( hf 00 + rh0f 0 +2fh0) (8) rr rfh

1 3 R = fh + rh0f rf0h fh (9) ✓✓ fh3 2 sin ✓ 3 R = fh + rh0f rf0h fh (10) fh3 Now we are ready to write and solve the Einstein equation [7], 1 G R g R =0 (11) µ⌫ ⌘ µ⌫ 2 µ⌫ µ⌫ where R is the Ricci scalar defined by R = g Rµ⌫ .BytakingthetraceoftheEinsteintensor,wecan show that R =0. Thus, solving the Einstein Equation can be done by setting Rµ⌫ =0. The equations to be solved are therefore,

f 0= (hf 00 rh0f 0 +2hf 0) rh3 1 0= ( hf 00 + rh0f 0 +2fh0) rfh

1 3 0= fh + rh0f rf0h fh fh3 2 sin ✓ 3 0= fh + rh0f rf0h fh fh3 Since Rµ⌫ =0, Rtt + Rrr can be added together to get:

Rtt + Rrr =2(fh0 + hf 0) d =2 (fh) dr which implies fh = K where K is a constant. Cancelling the prefactor on the final two equations, we see that R✓✓ =0and R =0give the same equation. Because the Bianchi identity relates the four equations, there is only one more equation to solve.

4 K Eliminating h = f from R✓✓ =0, K3 Kf r Kf r 0= 0 0 K f 2 f f K2 2f r = K 0 1 f f ✓ ◆ 2ffr˙ = K2 f 2 2f df 1 = (K2 f 2)dr r 2f 1 df = dr (K2 f 2) r

(12)

Integrating yields:

2f 1 df = dr (13) ˆ (K2 f 2) ˆ r r ln f 2 K2 =ln (14) C Solving: C f 2 = K2 + () r where C is the integration constant. The function h2(r) can be found by squaring the equation f(r)h(r)=K and substituting our solution for f 2(r).

K2 f 2(r)h2(r)=K2 h2(r)= (15) ) C K2 + r Now the components of the original can be written as

C K2 ds2 = K2 + dt2 + dr2 + r2 d✓2 +sin2 ✓d2 (16) r C ! K2 + r 2 W 2 C It is beneficial to simplify the time part of the equation to K 1+ dt where W = 2 . Then r K rescaling the time, t¯= K2t dt¯= K2dt puts the line element in the simpler form, ) W dr2 d⌧ 2 = ds2 = 1+ dt2 + + r2d⌦2 (17) r 1+ W ⇣ ⌘ r To find the constant W , we compare to Newton’s law of universal gravitation. At a large enough radius from the spherical mass, we can model the motion of a particle as if the particle was under the influence of Newtonian gravity, neglecting smaller effects evident in stronger gravity. We follow a particle with mass much less than the mass of the spherical body from rest at a large . The geodesic equation is du↵ = ↵ uuµ (18) d⌧ µ

5 and the initial 4- of the particle is: dt u↵ = c , 0, 0, 0 (19) 0 d⌧ By substituting this initial value into the geodesic⇣ equation⌘ for the metric, only the radial component remains. Looking at this radial component of the geodesic equation with the initial condition we get:

r du 2 = 1 ut (20) d⌧ 00 0 Having already calculated the value of ff r = 0 tt h2 W W = 1+ 2r2 r ✓ ◆ t we need to only find the value of u0. The norm of the 4-velocity is

1=g uµu⌫ (21) µ⌫ W dt 2 1 dr 2 d✓ 2 d 2 = 1+ + + r2 + r2sin2✓ (22) r d⌧ 1+ W d⌧ d⌧ d⌧ ✓ ◆✓ ◆ r ✓ ◆ ✓ ◆ ✓ ◆ dr d✓ d dt 1 For an initially stationary particle, d⌧ = d⌧ = d⌧ =0,itistrivialtoseethatd⌧ = W . p1+ r Therefore, we can insert this value into the geodesic equation and find:

dur W W c2 c2W = 1+ (23) d⌧ r 2r2 1+ W ⇡ 2r2 ⇣ ⌘ r At a large enough radius, dt 1 which means that dt d⌧ and d⌧ ⇡ ⇡ dur c2W dt ⇡ 2r2 From Newton’s universal law of gravitation, the due to gravity between two masses is radial with magnitude Mm F r = G (24) grav r2 Substituting this into the second law, m cancels to give dur GM = dt r2 Given that this is a system of a starting at a large radius from a spherically symmetric mass, a comparison of the radial acceleration of the point particle found from the geodesic equation with the radial acceleration found from Newton’s universal law of gravitation can be made. Equating the accelerations,

GM c2W = (25) r2 2r2 where M is the mass of the central spherical body. Solving for the constant W: 2GM W = (26) c2

6 We see that W is the for a black hole. 2GM 2M Now, writing rc2 as r ,wehavethecompletelineelementofaSchwartzchildblackhole, 2M dr2 d⌧ 2 = ds2 = 1 dt2 + + r2d⌦2 r 1 2M ⇣ ⌘ r geodesic calculations can be made.

7 Particles do not each other due to some gravitational force, but rather masses the spacetime around them particles move in the straightest possible lines within this curved spacetime. These straight lines are called geodesics. The geodesic equation produces four differentiable equations that describe the motion of a particle in a space with a given metric tensor field and some initial conditions [8]. We calculate two motions will be calculated by having a particle of negligible mass compared to the black hole: radial infall from rest, and a circular orbit. These will be calculated with GM units in which G = c =1so that M = c2 . The first motion will be calculated with the particle starting from rest. Essentially, the particle will be dropped from a radius r r (where r is the Schwartzchild radius of the black hole) and fall 0 s s radially inward. The initial conditions of this particle will be:

↵ ⇡ x0 = 0,r0, , 0 2 (27) ↵ t u0 = u0, 0, 0, 0 t µ ⌫ t The value of u can be found by looking at the norm gµ⌫ u u = 1.Solvingforu : 0 0 0 0

t 1 u0 = (28) 1 2M r0 Using the Christoffel symbols from above, theq geodesic equations are:

d2xt 2M = utur d⌧ 2 r2 1 2M r 2 r d x M 2M 2 M 2 2M 2 2M 2 = 1 ut + ur 1 u✓ 1 r sin2(✓) u d⌧ 2 r2 r r 1 2M r r r 2 ✓ ⇣ ⌘ ⇣ ⌘ ⇣ ⌘ d x 2 2 = uru✓ +sin(✓) cos(✓) u d⌧ 2 r d2x 2 2 cos(✓) = uru u✓u d⌧ 2 r sin(✓) (29)

d2xt dut r dr By writing d⌧ 2 = d⌧ and u = d⌧ ,integratingthefirstofthegeodesicequationsusingtheinitial conditions becomes:

dut 2M dr = ut 2 2M d⌧ r 1 r d⌧ t u dut r 2M = dr t 2 2M ˆut u ˆr r 1 0 0 r 2M (30) ut 1 r0 ln t =ln 2M u0 1 ! ⇣ ⌘ r 1 2M ut = ut r0 0 1 2M r ! d2x✓ d2x ✓ Integrating d⌧ 2 and d⌧ 2 with the initial conditions shows that u = u =0and therefore d2x✓ d2x d⌧ 2 = d⌧ 2 = constant which is understandable because intuitively the particle is expected to fall radially only and is not expected to have any angular changes in this reference frame. The radial component of the geodesic equation can be found by looking at the normal of the 4-velocity and inserting the values for ut, u✓,andu.

8 r 2 2M 2 (u ) 2 1= 1 ut + + r2 u✓ + r2 sin2(✓) u r 1 2M r 2M r 2 1 r (u ) = 0 + 1 2M 1 2M r r (ur)2 1 2M = r0 1 (31) 1 2M 1 2M r r 2M 2M ur = 1 1 r0 r r⇣ ⌘ ⇣ ⌘ 2M 2M = r r r 0 This yields all components of the 4-velocity,

1 2M ut = ut r0 0 1 2M r ! 2M 2M ur = (32) r r r 0 u✓ =0 u =0

The radial equation can be integrated to find the for the particle to start at r0 and fall to the horizon at r =2M.

dr 2M 2M = d⌧ r r r 0 r dr ⌧ = ˆ 2M 2M r0 (33) r r0 q r0 ⇡ 1 2M 2M = r sin + 1 0 2M 2 r r r r 0 ! r 0 ! Assuming that the initial radius is much larger than 2M, r 2M,thepropertimethatitwould 0 take for the particle to fall from r0 to 2M would be approximately

r0 ⇡ 1 2M 2M c⌧ = r sin + 1 0 2M 2 r r r r 0 ! r 0 ! r r ⇡ ⌧ = 0 0 +1 c 2M 2 r ! (34) 3 2 ⇡ r0c ⇡ 2cr2GM ⇡2r3 = 0 r8GM Using this, a value for r0 can be substituted and the proper time elapsed for the particle to fall into the horizon of the black hole can be found. For example, take r0 =8M. Then we have

9 8M ⇡ 1 2M 2M ⌧ =8M sin + 1 r2M 2 r8M ! r 8M !

⇡ 1 1 3 ⌧ =8M 2 sin + 2 2 4! (35) ⇣ ⇣ ⌘⌘ r 2⇡ p3 =8M + 3 2 ! 9.83M ⇡ GM Notice that in this calculation, M has not been expanded and therefore M = c2 .

10 For the second basic geodesic calculation, we will be looking into a circular orbit. Unlike the previous calculation, the radius will be held constant and the particle will only have a time component and angular components. Additionally, being that the orbit is circular, the particle will only be moving ⇡ in the -direction (The geodesic equation shows that a particle starting at ✓ = 2 will stay there). For this, the initial conditions will be: ⇡ x↵ = 0,r , , 0 0 2 (36) ↵ ⇣ t ⌘ u0 = u0, 0, 0,u0 Knowing that the particle position is constant⇣ in the r and⌘ ✓ direction, the equations of motion can be found from the t and components of the geodesic equation as well as the norm of the 4-velocity. dut 2M = utur d⌧ r2 1 2M r du 2 = uru (37) d⌧ r 2M 2 2 1= 1 ut + r2 sin2(✓) u r ⇣ ⌘ Solving for the time component yeilds:

dut 2M = utur =0 d⌧ r2 1 2M r dut (38) =0 d⌧ t t u = u0 Solving for the phi component yeilds:

du 2 dr = u d⌧ r d⌧ u du r 2 = dr ˆ u ˆ r u0 r0 u r ln = 2ln u r0 (39) ⇣ 0 ⌘ ⇣ ⌘ u r 2 = 0 u r 0 ⇣ ⌘ 2 2 u r = u0 r0 = L L u = r2 Since r is constant, u is constant and the constant L is the per unit mass. These values can be plugged into the norm of the 4-velocity to find the value for ut.

2 2M t 2 L 1= 1 u0 + r0 r0 2 ⇣ ⌘ ⇣ ⌘ L 2M t 2 1+ 2 = 1 u0 r0 r0 (40) ⇣ ⌘ ⇣ ⌘ L2 1+ r2 ut = 0 0 v 2M u 1 r2 u 0 t 11 The value for u can be found by substituting these values into the r component of the geodesic equation. r du M 2M t 2 M r 2 2M ✓ 2 2M 2 2 = 2 1 u0 + 2M u0 1 u0 1 sin (✓)r0 u0 d⌧ r0 r0 r0 1 r0 r0 ⇣ ⌘ r0 ⇣ ⌘ ⇣ ⌘ L2 M 2M 1+ r2 2M L 2 0= 1 0 1 r r2 r 2M r 0 r 0 0 1 r2 ! 0 0 ⇣ ⌘ 0 ⇣ ⌘ ⇣ ⌘ M M L2 L2 L2 2 + 2 2 = 3 2M 4 r0 r0 r0 r0 r0 M L2 M L2 L2 2 = 3 2 2 2M 4 r0 r0 r0 r0 r0 L2 M = r4 r2 (r 3M) 0 0 0 L M = r2 r2 (r 3M) 0 s 0 0 2 Mr0 u0 = sr0 3M (41) 4 Mr0 From equation 39, it can be seen that L = 2 and this value for L can be substituted into r0 (r0 3M) t the equation for u0 to find a value without theq L-term.

L2 1+ r2 u0 = 0 0 v 2M u1 u r0 t 4 1 Mr0 1+ r2 r2(r 3M) = 0 0 0 v 2M u 1 u r0 t M r0 + r0 r 3M = 0 s r 2M 0 2 r 3Mr0 Mr (42) 0 0 2M r + r 3M = 0 0 s r 2M 0 r2 2Mr 0 0 r 3M = 0 sr 2M 0 r r 2M = 0 0 r 2M r 3M r 0 0 r u0 = 0 0 r 3M r 0 With all of the components for the 4-velocity found, the final 4-velocity is:

1 u↵ = pr, 0, 0, pMr) (43) r 3M r This diverges at r =3M showing that there are no timelike orbits that exist at or within r =3M. These circular orbits at 3M are not stable. If the particle were to be bumped or jostled either way in the radial direction, the particle would either fall into the hole or fly off. Other orbits are stable,

12 however, depending on the initial conditions. By doing a calculation such as the one done above, but not leaving the radius fixed, one can show that small perturbations of circular or elliptical orbits lead to nearby orbits. In the elliptical orbits, the perihelion and aphelion precesses. Such a calculation shows, famously, that the excess perihelion precession of is a consequence of General Relativity [9]. This has been an exercise in the basic calculations that can be made in Einstein’s General Theory of Relativity. This is just the beginning of what can be done. There are many more complex calculations that can be made. General Relativity has given rise to an entirely new idea of how particles in our universe interact with one another. In the future, I would like to calculate motions around more complex objects such as non-static, rotating, and charged black holes as well as other, more theoretical, masses. It would be interesting to calculate the geodesics of a particle moving around a or a non-orientable object such as a Klein bottle.

[1] M. Fels, An Introduction To Differential Geometry through Computation (Logan, Utah, 2014), p. 89. [2] S. Carroll, Spacetime and Geometry (Addison Wesley, CA, 2004), p. 121. [3] C.W. Misner, K.S. Thorne, and J.A. Wheeler, Gravitation (W.H. Freeman and co. , US, 1973), p. 210. [4] G. Arken, Mathematical Methods for Physicists,3rded.(AcademicPress,Orlando,FL,1985), p160-165. [5] S. Carroll, Spacetime and Geometry (Addison Wesley, CA, 2004), p. 129. [6] B. Schutz, A First Course in General Relativity, 2nd ed. (Cambridge, Cambridge, UK, 2009) p 164. [7] S. Carroll, Spacetime and Geometry (Addison Wesley, CA, 2004), p. 158-159. [8] B. Schutz, A First Course in General Relativity, 2nd ed. (Cambridge, Cambridge, UK, 2009) p 117-139. [9] Kraniotis, G. V., and S. B. Whitehouse, Compact calculation of the perihelion precession of Mercury in general relativity, the and Jacobi’s inversion problem.Classicaland 20.22 (2003): p 4817.

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