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287

11.1

Trajectories in the Schwarzschild spacetime

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

§ 11.7, we see that for such an orbit the impact parameter (b) is small: it is aimed more

 

 

 

 

directly at the hole than are orbits of smaller E˜ and fixed L˜ .

 

 

 

 

 

 

 

 

 

 

Of course, if the geometry under consideration is a star, its radius R will exceed 2M, and

 

 

the potential diagrams, Figs. 11.1–11.3, will be valid only outside R. If a particle reaches

 

 

R, it will hit the star. Depending on R/M, then, only certain kinds of orbits will be possible.

 

 

 

 

 

 

 

 

 

 

 

Perihelion shift

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

A particle (or planet) in a (stable) circular orbit around a star will make one complete

 

 

orbit and come back to the same point (i.e. same value of φ) in a fixed amount of coordi-

 

 

nate times, which is called its period P. This period can be determined as follows. From

 

 

Eq. (11.17) it follows that a stable circular orbit at radius r has angular momentum

 

 

 

 

 

 

 

 

 

 

 

 

 

 

L˜

2 =

 

 

 

 

Mr

 

 

,

 

 

 

 

 

 

 

 

 

(11.20)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

1

−

3M/r

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

and since E˜ 2 = V˜ 2 for a circular orbit, it also has energy

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

2M

2

 

 

 

 

 

 

 

 

 

M

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

E˜ 2

= 1 −

 

 

 

 

 

 

/

1 −

3

.

 

 

 

(11.21)

 

 

 

 

 

 

 

 

 

 

r

 

r

 

 

 

 

 

Now, we have

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

dφ

 

 

 

 

 

 

 

pφ

 

 

 

 

 

 

 

pφ

 

= gφφ L˜

 

 

1

L˜

 

 

 

 

 

 

 

 

:= Uφ =

 

 

 

= gφφ

 

 

 

=

 

(11.22)

 

 

 

 

 

 

dτ

 

m

 

m

 

r2

 

 

and

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

dt

 

 

 

0

=

p0

 

 

 

 

 

00 p0

 

 

 

00

(−E˜ ) =

 

 

 

 

E˜

 

 

 

 

 

 

:= U

 

 

 

= g

 

 

 

 

= g

 

 

 

 

 

 

.

(11.23)

 

 

 

dτ

 

m

 

 

m

 

1

−

2M/r

 

 

We obtain the angular velocity by dividing these:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

dt

 

 

 

dt/dτ

 

 

 

 

 

 

r3

 

1/2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

 

 

 

 

 

 

 

 

 

 

(11.24)

 

 

 

 

 

 

 

 

 

dφ

dφ/dτ

 

M

 

 

 

 

 

 

 

 

 

The period, which is the time taken for φ to change by 2π , is

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

r3

1/2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

P = 2π

 

.

 

 

 

 

 

 

 

 

 

(11.25)

 

 

 

 

 

 

 

 

 

 

 

 

 

M

 

 

 

 

 

 

 

 

 

This is the coordinate time, of course, not the particle’s proper time. (But see Exer. 7, § 11.7: coordinate time is proper time far away.) It happens, coincidentally, that this is identical to the Newtonian expression.

Now, a slightly noncircular orbit will oscillate in and out about a central radius r. In Newtonian gravity the orbit is a perfect ellipse, which means, among other things, that it is closed: after a fixed amount of time it returns to the same point (same r and φ). In GR, this does not happen and a typical orbit is shown in Fig. 11.4. However, when the effects of relativity are small and the orbit is nearly circular, the relativistic orbit must be almost closed: it must look like an ellipse which slowly rotates about the center. One way to describe this is to look at the perihelion of the orbit, the point of closest approach to

288

Schwarzschild geometry and black holes

(a)

(b)

Figure 11.4

(c)

(a) A Newtonian orbit is a closed ellipse. Grid marked in units of M. (b) An orbit in the Schwarzschild metric with pericentric and apcentric distances similar to those in (a). Pericenters (heavy dots) advance by about 97◦ per orbit. (c) A moderately more compact orbit than in (b) has a considerably larger pericenter shift, about 130◦.

the star. (‘Peri’ means closest and ‘helion’ refers to the Sun; for orbits about any old star the name ‘periastron’ is more appropriate. For orbits around Earth – ‘geo’ – we speak of the ‘perigee’. These opposite of ‘peri’ is ‘ap’: the furthest distance. Thus, an orbit also has an aphelion, apastron, or apogee, depending on what it is orbiting around. The general terms, not specific to any particular object, are perapsis and apapsis.) The perihelion will rotate around the star in some manner, and observers can hope to measure this. It has been measured for Mercury to be 43 /century, and we must try to calculate it. Note that all other planets are further from the Sun and therefore under the influence of significantly smaller relativistic corrections to Newtonian gravity. The measurement of Mercury’s precession is a herculean task, first accomplished in the 1800s. Due to various other effects, such as the perturbations of Mercury’s orbit due to the other planets, the observed precession is about 5600 /century. The 43 is only the part not explainable by Newtonian gravity, and Einstein’s demonstration that his theory predicts exactly that amount was the first evidence in favor of the theory.

289

 

11.1 Trajectories in the Schwarzschild spacetime

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

To calculate the precession, let us begin by getting an equation for the particle’s orbit.

 

 

 

We have dr/dτ from Eq. (11.11). We get dφ/dτ from Eq. (11.22) and divide to get

 

 

 

dr

 

2

 

 

E2

−

(1

 

2M/r)

1

 

L2/r2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

=

 

 

˜

 

−

 

 

 

 

+ ˜

 

.

 

 

(11.26)

 

 

dφ

 

 

 

 

 

 

 

 

 

L2

/r4

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

˜

 

 

 

 

 

 

 

 

 

 

 

 

 

It is convenient to define

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

u := 1/r

 

 

 

 

 

 

 

 

 

 

 

(11.27)

 

and obtain

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

du

 

2

 

 

 

E˜ 2

− (1 − 2Mu)

 

1

 

2

 

 

 

 

 

 

 

 

 

 

=

 

 

 

+ u

.

 

 

(11.28)

 

 

dφ

 

L˜ 2

L˜ 2

 

 

 

 

The Newtonian orbit is found by neglecting u3 terms (see Exer. 11, § 11.7)

 

 

 

 

 

 

 

 

 

 

 

du

2

 

 

E˜ 2

1

 

 

 

 

 

 

 

2

 

 

 

 

Newtonian :

 

 

=

 

 

−

 

 

(1

− 2Mu) − u

 

.

(11.29)

 

 

dφ

L˜ 2

L˜ 2

 

= ˜ 2

A circular orbit in Newtonian theory has u M/L (take the square root equal to 1 in Eq. (11.17)), so we define

 

 

 

 

y = u −

 

M

 

 

 

 

 

 

 

 

 

 

 

 

 

,

 

 

 

 

 

 

(11.30)

 

 

 

 

 

L˜ 2

 

 

 

 

 

 

so that y represents the deviation from circularity. We then get

 

dφ

=

 

L˜ 2

1

+

L˜ 4

−

 

 

 

dy

 

2

 

E˜ 2 −

 

 

M2

 

 

 

y2.

(11.31)

It is easy to see that this is satisfied by

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Newtonian : y = %

E2

 

2/L2

−

1

&

1/2

 

 

 

 

 

˜

 

+ M ˜

 

 

 

cos(φ + B),

(11.32)

 

 

L2

 

 

 

 

 

 

 

 

 

 

˜

 

 

 

 

 

 

 

 

 

where B is arbitrary. This is clearly periodic: as φ advances by 2π , y returns to its value and, therefore, so does r. The constant B just determines the initial orientation of the orbit. It is interesting, but unimportant for our purposes, that by solving for r we get

 

1

 

 

M

+ %

E2

+

M2/L2

−

1

&

1/2

 

 

 

 

 

 

Newtonian :

 

 

 

=

 

 

˜

 

˜

 

cos(φ + B),

(11.33)

 

r

L2

 

 

L2

 

 

 

 

 

 

 

 

 

˜

 

 

 

˜

 

 

 

 

 

 

which is the equation of an ellipse.

We now consider the relativistic case and make the same definition of y, but instead of throwing away the u3 term in Eq. (11.28) we assume that the orbit is nearly circular, so that y is small, and we neglect only the terms in y3. Then we get

Nearly circular:

dφ

 

=

˜

+

L˜ 2 ˜

−

 

+

L˜ 6

+

L˜ 2 y − 1 −

L˜ 2

y2.

(11.34)

 

dy

2

 

E2

 

M2/L2

 

1

 

2M4

 

6M3

6M2

 

 

290

Schwarzschild geometry and black holes

This can be made analogous to Eq. (11.31) by completing the square on the right-hand side. The result is the solution

 

 

 

y = y0 + A cos(kφ + B),

 

 

(11.35)

where B is arbitrary and the other constants are

 

 

 

 

 

 

 

k = 1 − 6L˜2

 

 

 

1/2

,

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

M2

 

 

 

 

 

 

 

 

 

 

 

 

 

y0 = 3M3/k2L˜ 2,

L2

˜

 

−

+

L6 − y02

&

.

(11.36)

A = k % ˜ +

 

 

 

1 E2

 

M2/L2

 

1 2M4

 

1/2

 

 

 

 

 

 

 

 

 

 

 

 

 

 

˜

 

 

 

 

 

 

˜

 

 

 

 

The appearance of the constant y0 just means that the orbit oscillates not about y = 0

= ˜ 2 =

(u M/L ) but about y y0: Eq. (11.30) doesn’t use the correct radius for a circular orbit in GR. The amplitude A is also somewhat different, but what is most interesting here is the fact that k is not 1. The orbit returns to the same r when kφ goes through 2π , from Eq. (11.35). Therefore the change in φ from one perihelion to the next is

 

2π

= 2π 1 −

M2

 

−1/2

φ =

 

6

,

k

L˜ 2

which, for nearly Newtonian orbits, is

φ 2π 1 + 3M2 .

˜ 2

L

The perihelion advance, then, from one orbit to the next, is

= 2 ˜ 2

φ 6π M /L radians per orbit.

(11.37)

(11.38)

(11.39)

˜

We can use Eq. (11.20) to obtain L in terms of r, since the corrections for noncircularity will make changes in Eq. (11.39) of the same order as terms we have already neglected. Moreover, if we consider orbits about a nonrelativistic star, we can approximate Eq. (11.20) by

L˜ 2 =

 

Mr

 

 

≈ Mr,

 

 

1

−

3M/r

 

 

so that we get

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

φ ≈

 

M

 

 

 

6π

 

.

 

 

(11.40)

r

 

 

For Mercury’s orbit, r = 5.55 × 107 km and M = 1 M

+

= 1.47 km, so that

 

( φ)Mercury = 4.99 ×

10−

7

 

 

 

(11.41)

 

 

radians per orbit.

Each orbit take 0.24 yr, so the shift is

 

 

 

 

 

 

 

 

 

 

 

( φ)Mercury = .43 /yr = 43 /century.

(11.42)
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