Supplemental Material

Repulsion Dominating in the Gravitational Field of a Black Hole

Return to Modern Relativity

© 2000

 

 

 

Abstract

We derive the geodesic motion of a neutral particle falling along the polar axis of a Kerr-Newman black hole with a static metric. It will be seen that spin and charge each give rise to repulsion. It will also be seen that if the black hole has only mass and electric charge, but no spin there will exist a gravitational repulsion due to the charge that will throw the particle back out past both horizons in a finite proper time.

  1. The equation of motion

    The equation for geodesic polar motion of a neutral particle in a Kerr-Newman black hole space-time is

    (1a)

    And the time travel equation relating proper time to remote frame time is

    (1b)

    Where is the specific angular momentum. For a Kerr-Newman black hole,

    (2)

    J is the angular momentum, M the mass and is the specific charge of the black hole and .

    Eqn. 1 becomes

    (3)

    And for a Reissner-Nordstrom black hole J = 0 Eqn. 3 becomes

    (4)

    Since it is currently thought that the charge, mass and spin are the only distinguishing characteristics of a static state black hole Eqn. 2 can be taken as the definition of a . However, even if a is not restricted to this but is allowed to be any function of r with the metric remaining static for the duration of the observation of the particle's motion Eqn 1a and 1b would still obtain. However, for a expressed as a Laurent series truncated for bm such that m > 2 results in a singularity in Eqn 1a at the origin. Also, for m > 2, the sign of the last term in the series will determine whether the neutral particle in free fall will encounter a final gravitational attraction or repulsion as it approaches r = 0. If a Taylor series is added to a and the series is truncated at any an n > 0, equation 1b vanishes at infinity. Since we typically put our remote observers there, such a generalization should demand a be normalized to 1 at infinity. Therefore, any addition of a Taylor series to the definition of a must not be truncated.

     
  2. Method

In the following, we will derive Eqn. 1 by applying Boyer-Lindquist coordinates to the Kerr-Newman metric.

We relate a to Boyer-Lindquist coordinates as follows:

(5)

(6)

With the following definitions,

(7)

8)

(9)

The invariant interval in this metric becomes

(10)

The covariant metric tensor is thus given by

(11)

where we have adopted the usual ordering of spherical-polar coordinates.

Simple inversion yields the contra-variant metric tensor:

(12)

Since we consider here only polar motion, dq = 0, q = 0 or p and v ® 0, as q ® 0. Eqn. 10 thus reduces to

(13)

ds = dct , so this can be rewritten as

(14)

From here all that is needed in order to find the equation of motion is to find dt/dt . The geodesic equation is used to find this. We use the geodesic equation in the form

(15)

x0 = ct, so the one of these that has the information we want is

(16)

Written out in terms of our coordinates and recalling that dq = 0 for polar motion, this is

(17)

The equation for the affine connection is

(18)

The only ones of these used in Eqn. 17 are G 0m n .

(19)

Referring to the contra-variant metric tensor the only nonzero g0s are g00 and g03 so this becomes

(20)

The particle is a test mass and so the curvature of the space-time due to the falling particle is taken to be small enough so that it doesn't significantly change the metric as it falls, so the metric remains static and gm n ,0 = 0. So Eqn. 20 reduces to

(21)

From here we will look at those of the G 0m n contained in Eqn 17. Starting with

G 033.

(22)

Referring to the covariant metric tensor we notice that none of gm n depend on f so G 033 = 0.

Now we look at G 030.

(23)

Again time and f independence of gm n results in G 030 = 0.

Now we'll look at G 031.

(24)

f independence reduces this to

(25)

Recalling that as q ® 0, p then v ® 0 we notice that g03 and g33 are zero at that those q for any r. Therefore g03,1 = g33,1 = 0 and therefore G 031 = 0.

Now we look at G 000.

(26)

Again time and f independence results in G 000 = 0.

Now we look at G 011.

(27)

Referring to the covariant metric tensor g01 = g10 = g31 = g13 = 0. Also f recalling f independence we find that G 011 = 0.

Now Eqn 7 reduces to

(28)

Now we find G 001.

(29)

4

Again referring to the covariant metric tensor we see that g01 = g31 = g10 = 0. Also for q = 0 or p g30 = 0 at any r and so g30,1 = 0 so Eqn 29 reduces to

(30)

Inserting this into Eqn 28 results in

(29)

Simplifying:

(30)

(31)

(32)

For the case of polar motion we also notice that so Eqn 32 becomes

(33)

Simplifying

(34)

Now we notice that for q = 0, p so Eqn 34 becomes

(35)

Separation of variables results in

(36)

Integrating and simplifying results in

(37)

The g is a constant from integration. Recalling again that g00 = D /r 2 for q = 0, p this becomes

(37)

For q = 0,p this can also be written

5

(1b)

Inserting Eqn 37 into Eqn 14 results in

(38)

For q = 0,p Eqn 14 can also be written

(39)

(These equations are for polar motion, but this is comparable in form to the case of equatorial motion expressed by equation 12.3.24 on page 320 of General Relativity, by Robert M. Wald)

Differentiating Eqn 39 with respect to t and use of the chain rule results in

 

(40)

Simplifying this becomes

(1a)

Placing in and r0 = 2GM/c2 for the Kerr-Newman Black hole and doing the differentiation with respect to r results in

(3)

For a Reissner-Nordstrom black hole J = 0 so a = 0 so this becomes

(4)

 

III. Discussion

A Kerr-Newmann black hole has two event horizons where D = 0. The inner horizon is given by

 

(41)

and the outer horizon is given by

(42)

It is thought that in the case of , the black hole has too much spin and charge for such a hole to form under gravitational collapse10, but lets consider a state of extreme charge where and . Both of the event horizons coincide at and equation 4 becomes

(43)

The inverse distance cubed repulsion dominates over the inverse distance squared attraction for small r. It begins to dominate where .

This occurs at . This distance is also where the event horizons occur. Since the repulsion doesn't become dominant until the particle reaches the location of the event horizons, the particle must fall past both outer and inner horizons. According to the equation of motion the particle is then thrown back out.

IV. Final remarks

We have seen that the presence of charge in a black hole leads to a repulsion in the gravitational field that dominates over the attraction due to mass at small distance from the center. And we have derived the geodesic motion for a neutral particle falling along the polar axis. At a small enough distance the spin's effects on gravitation dominate over both attraction due to mass and repulsion due to charge. And we have seen that in the absence of spin, the particle's geodesic motion leads back out of the hole in a finite proper time.

  Appendix A  

 

© 2000 Samuel Cox