Path: gmdzi!unido!mcsun!uunet!wuarchive!wuphys!ihr

From: i...@wuphys.wustl.edu (Ian H. Redmount)

Newsgroups: sci.physics

Subject: Re: Black holes and plumbs.

Summary: Lowering a string into a black hole--an old problem (long).

Message-ID: <1990Aug16.195705.24758@wuphys.wustl.edu>

Date: 16 Aug 90 19:57:05 GMT

References: <13807@megatest.UUCP>

Reply-To: i...@wuphys.UUCP (Ian H. Redmount)

Organization: Physics Dept, Washington U. in St Louis

Lines: 83

Posted: Thu Aug 16 20:57:05 1990

In article <1...@megatest.UUCP> bbo...@megatest.UUCP (Bruce Bowen) writes:

>

>Suppose we slowly...reel out a string,

>on the end of which is connected a plumb bob. Also assume that the mass of

>the string is negligible compared to the mass of the plumb bob. The reel

>upon which the string is suspended...is stationary

>with respect to our black hole.

>

>In the relativistic case, would the tension, measured locally along the

>string, be constant along the length of the string? I am inclined to

>believe not....In the limit as the plumb bob

>gets arbitrarily close to the event horizon, the tension in the string

>near the plumb bob goes to infinity, but the tension at the other end

>of the string at the reel goes to a finite value, which is a function

>of the mass of the black hole, mass of the plumb bob, and distance of

>the reel from the event horizon.

>

Apparently the good Mr. Bowen has not read Appendix B of my 1984 Caltech

Ph. D. thesis, though I cannot imagine why not :).

The dynamics---in this case, statics---of the string follows directly from

the law of conservation of stress-energy for the string. The string's

stress energy is, in the ideal approximation, given by two quantities:

its proper linear mass-energy density U, and its proper tension T.

(As is customary, I assume units such that the speed of light c is 1.)

Stress-energy conservation implies the equation of force balance

dT/dl=(U-T)g ,

where l is proper distance up the string and g is the local proper

"gravitational acceleration" as measured by static observers. Except

for the T on the right side, this is exactly the equation you'd expect

on Newtonian grounds: The difference in tension across each infinitesimal

segment of string is equal to the segment's weight. The -T contribution

to the weight density is a relativistic effect.

Relativistically, no (classical) string can have zero mass density; in

fact, the familiar classical energy conditions (e.g., positive mass

density in all reference frames) require U>T. Since the transverse-wave

speed on the string is c*sqrt(T/U), this also follows from causality---no

signals faster than light. The best we can do is string with T=U, and

that is mighty string indeed. Ordinary steel piano wire has a breaking

strength T about 10^(-12)U!

For string with T<U, the above force-balance equation implies that the

tension in the string increases as you go up, from its value at the

plumb bob T(0)=Mg(0), M being the bob mass and 0 denoting that end of

the string. The tension is highest at the top of the string, unlike

Mr. Bowen's conjecture. It could not be otherwise, since the tension

at the top must support the weight of the string as well as the bob.

The top tension is a function of the bob mass and its position in the

black-hole spacetime. Since g goes to infinity at the horizon, there

will be some position above the horizon at which the top tension exceeds

the breaking tension of the string---at most, U---the bob cannot be

suspended any closer that this.

For string capable of sustaining T=U, the force-balance equation implies

that the tension is constant along the string. Such a string is effectively

weightless. Again, it cannot support the bob any lower than the position

at which Mg(0)=U, but for arbitrarily small M this can be arbitrarily

close to the horizon.

These results are mentioned in the article ``Membrane viewpoint on black

holes: Gravitational perturbations of the horizon," by W.-M. Suen,

R. H. Price, and I. H. Redmount, Physical Review D 37, pp. 2761-2789

(see p. 2772), 15 May 1988, and in the book ``Black Holes: The Membrane

Paradigm," K. S. Thorne, R. H. Price, and D. A. Macdonald, eds., Yale

University Press, 1986, p. 240. An early work on the problem is the

article by G. W. Gibbons in Nature Phys. Sci. 240, p. 77 (1972), but

Dr. Gibbons inadvertently omitted a term from one equation and got the

wrong answer.

As an aside, note that string with T=U is remarkable stuff. (Both cosmic

and super string have this property.) If you strung a meter of it between

posts and plucked the center a few millimeters, the waves produced would

travel at the speed of light---clearly a relativistic situation---but no

part of the string would move at more than a few times 10^(-3)c. String

a guitar with this stuff and it (the guitar) would break. But if it didn't,

it would play at frequencies of hundreds of megahertz!

Ian H. Redmount,

Sole Proprietor of any of the above which is not fact

From: i...@wuphys.wustl.edu (Ian H. Redmount)

Newsgroups: sci.physics

Subject: Re: Black holes and plumbs.

Summary: Lowering a string into a black hole--an old problem (long).

Message-ID: <1990Aug16.195705.24758@wuphys.wustl.edu>

Date: 16 Aug 90 19:57:05 GMT

References: <13807@megatest.UUCP>

Reply-To: i...@wuphys.UUCP (Ian H. Redmount)

Organization: Physics Dept, Washington U. in St Louis

Lines: 83

Posted: Thu Aug 16 20:57:05 1990

In article <1...@megatest.UUCP> bbo...@megatest.UUCP (Bruce Bowen) writes:

>

>Suppose we slowly...reel out a string,

>on the end of which is connected a plumb bob. Also assume that the mass of

>the string is negligible compared to the mass of the plumb bob. The reel

>upon which the string is suspended...is stationary

>with respect to our black hole.

>

>In the relativistic case, would the tension, measured locally along the

>string, be constant along the length of the string? I am inclined to

>believe not....In the limit as the plumb bob

>gets arbitrarily close to the event horizon, the tension in the string

>near the plumb bob goes to infinity, but the tension at the other end

>of the string at the reel goes to a finite value, which is a function

>of the mass of the black hole, mass of the plumb bob, and distance of

>the reel from the event horizon.

>

Apparently the good Mr. Bowen has not read Appendix B of my 1984 Caltech

Ph. D. thesis, though I cannot imagine why not :).

The dynamics---in this case, statics---of the string follows directly from

the law of conservation of stress-energy for the string. The string's

stress energy is, in the ideal approximation, given by two quantities:

its proper linear mass-energy density U, and its proper tension T.

(As is customary, I assume units such that the speed of light c is 1.)

Stress-energy conservation implies the equation of force balance

dT/dl=(U-T)g ,

where l is proper distance up the string and g is the local proper

"gravitational acceleration" as measured by static observers. Except

for the T on the right side, this is exactly the equation you'd expect

on Newtonian grounds: The difference in tension across each infinitesimal

segment of string is equal to the segment's weight. The -T contribution

to the weight density is a relativistic effect.

Relativistically, no (classical) string can have zero mass density; in

fact, the familiar classical energy conditions (e.g., positive mass

density in all reference frames) require U>T. Since the transverse-wave

speed on the string is c*sqrt(T/U), this also follows from causality---no

signals faster than light. The best we can do is string with T=U, and

that is mighty string indeed. Ordinary steel piano wire has a breaking

strength T about 10^(-12)U!

For string with T<U, the above force-balance equation implies that the

tension in the string increases as you go up, from its value at the

plumb bob T(0)=Mg(0), M being the bob mass and 0 denoting that end of

the string. The tension is highest at the top of the string, unlike

Mr. Bowen's conjecture. It could not be otherwise, since the tension

at the top must support the weight of the string as well as the bob.

The top tension is a function of the bob mass and its position in the

black-hole spacetime. Since g goes to infinity at the horizon, there

will be some position above the horizon at which the top tension exceeds

the breaking tension of the string---at most, U---the bob cannot be

suspended any closer that this.

For string capable of sustaining T=U, the force-balance equation implies

that the tension is constant along the string. Such a string is effectively

weightless. Again, it cannot support the bob any lower than the position

at which Mg(0)=U, but for arbitrarily small M this can be arbitrarily

close to the horizon.

These results are mentioned in the article ``Membrane viewpoint on black

holes: Gravitational perturbations of the horizon," by W.-M. Suen,

R. H. Price, and I. H. Redmount, Physical Review D 37, pp. 2761-2789

(see p. 2772), 15 May 1988, and in the book ``Black Holes: The Membrane

Paradigm," K. S. Thorne, R. H. Price, and D. A. Macdonald, eds., Yale

University Press, 1986, p. 240. An early work on the problem is the

article by G. W. Gibbons in Nature Phys. Sci. 240, p. 77 (1972), but

Dr. Gibbons inadvertently omitted a term from one equation and got the

wrong answer.

As an aside, note that string with T=U is remarkable stuff. (Both cosmic

and super string have this property.) If you strung a meter of it between

posts and plucked the center a few millimeters, the waves produced would

travel at the speed of light---clearly a relativistic situation---but no

part of the string would move at more than a few times 10^(-3)c. String

a guitar with this stuff and it (the guitar) would break. But if it didn't,

it would play at frequencies of hundreds of megahertz!

Ian H. Redmount,

Sole Proprietor of any of the above which is not fact