Question
Asked 7th Oct, 2014

# What is the length of and force on a tether attached to an object near an event horizon in GR?

A distant observer lowers a mass into a gravitational field using a tether.  The tether may be considered massless and unstretchable (other than by relativity effects).  If we assume the mass gets "stuck" as it approaches the event horizon and cannot be retrieved or even slowed, this implies the force on the tether approaches infinite.  If we assume that due to curvature the proper distance to the event horizon approaches infinity, then the length of tether spooled out approaches infinity.  These two taken together would allow a near infinite amount of energy to be extracted.  The amount of energy I believe is supposed to be limited to the rest energy of the lowered mass.  I am looking for the official GR predictions for the tether which conserve energy.  It should not be a hard question.  I just don't happen to know the answer.  I am not looking for any alternate theories, only for the results using GR.  If possible in with reasonable effort, show me how to do the calculation using the Schwarzschild metric.  Thanks.

## Most recent answer

Here is a new and extremely careful development of static gravity which ends with Klaus' tether demonstration.  It follows as closely as possible the classic steps of the field equation and Schwarzschild's solution as can be done without a tensor formulation.  One might call it a Kretschmann-inspired version of GR.  It is only static and symmetric, however.  Comments appreciated.

## Popular Answers (1)

8th Oct, 2014
K. Kassner
Otto-von-Guericke-Universität Magdeburg
Here is something I cooked up during breakfast this morning. I have not counterchecked with Rindler or some other source, so the calculation may not be flawless. But it is much more fun this way...
What I want to calculate is the force on the tether or rope as perceived by the observer lowering the mass down to the black hole. My assumption is that the rope is "relativistically" inextensible, as Robert suggested. By that I mean that its proper length remains locally unchanged but that otherwise all relativistic effects affecting lengths apply.
If the force on the rope is F and the observer has to yield a piece of length dl of the rope, the work done by the mass pulling at the rope will be dW = F dl. Here dl is a proper length increment. This work may be stored by the observer and since the process can be done quasistatically, he has a lot of time to set up contraptions that do the storing properly (lifting a series of weights, extending a spring, etc.).
The energy of the mass m being lowered in the field decreases, due to time dilation:
E = (g00)1/2 m c2=(1-2M/r)1/2m c2
The decrease in energy is used up to do the work, so we have
F dl = - dE                                                                                   (1)
Moreover,
dl = - (1-2M/r)-1/2 dr
dE = dr dE(r)/dr = dr m c2 (1-2M/r)-1/2 M/r2
Inserting dl and dE into (1), we find that the inverse sqare root cancels and obtain
F = mM c2/r2
If we go back to SI units, i.e. replace M by GM/c2, we get
F = G m M/r2
This is just the force the mass m would experience due to Newton's law. On the one hand, this is a very satisfactory result, as we must get precisely this in the limit of small gravitational field. On the other hand, it is a bit puzzling that the force does not seem to diverge as we decrease r to approach the Schwarzschild radius.
Before considering this problem, let us deal with another one. All of the above is correct for an observer located at infinite distance from the black hole. But infinity is infinitely far away even in Newtonian physics. We would need an infinitte tether. Can we modify the analysis to deal with an observer at R1?
Well, that is not too difficult. Such an observer will experience a time dilation factor of (1-2M/R1)1/2, so he will measure all energies larger by a factor (1-2M/R1)-1/2. The proper length dl is invariant, and F is what we want to calculate. So the result is simply
F = (1-2M/R1)-1/2mM c2/r2,
which is a small modification, if R1 is large.
Aha! This may help solve our other problem as well! What will be the force in the rope measured by an observer at r? Why, this is obtained from the above formula by replacing R1 with r
F(r) = (1-2M/r)-1/2mM c2/r2.
(He would measure this if cutting the rope at r and keeping in place the end of the part going down -- for which he would just need the force F(r).)
So the point is that the tension in the rope will not be constant as it would in a massless rope in Newtonian physics! This is because forces are not relativistically invariant and in this particular case this is due to time dilation. F(r) diverges as r approaches 2M.
However, an observer lowering the mass below the horizon will not feel an increased force at his end. The rope will simply go slack, and if he hauls it up, he will find the mass and part of the rope gone.
If he had an unbreakable rope, he might pull the mass back from beyond the horizon. But unbreakability is as impossible as going faster than light. And if you start from the premise of a theoretical impossibility then you may of course get all kinds of nonsensical results out of the theory.
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## All Answers (20)

Charles' answer, ignoring his snide criticisms, is interesting and provides part of what I was suspecting (one of the infinities has to go away, and here the infinite length becomes finite, presumably because the tether is not free falling).  If GR people think force is not meaningful, they they have simply failed to address a basic question in physics.  Basically, the force at the top of the tether can never approach infinity.  If it did, energy greater than the object's rest mass could be extracted.  This requires a finite distance to the event horizon, AND a force reduction law similar to what I derived for time dilation in flat space in my 2011 paper on inertia, but I have some questions about how to go about doing this in GR, and that's what I'm looking for.  Without the force, the question is not answered.  If the LENGTH IS KNOWN, and the ENERGY IS KNOWN, then the force can be derived, because at the top of the tether, force x distance will be energy.
7th Oct, 2014
K. Kassner
Otto-von-Guericke-Universität Magdeburg
"the infinite length becomes finite, presumably because the tether is not free falling"
No, it remains finite, because in spite of infinite length contraction at the horizon, the integral over all proper length elements down there remains finite (if you start at a finite r). You integrate 1/sqrt(1-2M/r). So the total proper distance never gets infinite up to and including the horizon.
Also, your argument about the force becoming infinite does not hold, because the energy content of a slowly lowered mass goes to zero on approaching the horizon, so the traction by the mass does not necessarily increase. This is because by slowly lowering the mass you take away its kinetic energy (and in free fall, all of the energy would be kinetic at the horizon). The way to get the maximum energy out of the process would be to lower the mass to the horizon and release it there. Then you gain an enery of m c2 (and the mass of the black hole does not increase). In terms of power, it is not a very efficient process, as it becomes very slow towards the end. In terms of energy, it is of course optimal. (In reality, you should release the mass slightly above the horizon, if you do not want to break the rope.)
By the way, this is described in Rindler's book, which you have.
Please someone provide an alternative to the following calculation, or show how it works with energy conservation.  Not interested in philosophical debates on this question, just calculation procedure.
R1 = radius of top end of tether in Schwarzschild coordinates
R0 = gravitational radius R=2GM/c2
Proper distance from R1 to R0 = tether length = ∫R1R0ds
= ∫R1R01/(1-R0/r)1/2dr
=(R0/2)ln((1-R0/r)1/2+1) - (R0/2)ln((1-R0/r)1/2+1) + r(1-R0/r)1/2
where the final formula is evaluated from R1 to R0.  The problem is it appears  this value is infinite (at least the right hand term is infinite, and the ln terms should be much less than the right hand term.
7th Oct, 2014
K. Kassner
Otto-von-Guericke-Universität Magdeburg
R0R1 1/(1-R0/r)1/2dr
=R0 ( (1-R0/R1)1/2 / (R0/R1) + 1/2 ln ((1+(1-R0/R1)1/2 )/(1-(1-R0/R1)1/2 )) )
You can easily check this by taking the derivative w.r.t. R1, by the way.
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I'm not surprised.  I did not compute it personally.  But I noticed that it was inconsistent.  Please do not refer to MTW.  I sold this expensive useless paperweight eons ago.  I found a post on Physics Stack Exchange which works out the integral to a different answer (looks like KK's above)  and they even have a graph.
However, reading the fine print, I see that the distance still approaches infinity.  It does not seem to conserve energy, unless the mass of the BH can be extracted this way???  (to avoid simple moving the BH, set up a tether platform on opposite sides of it)
7th Oct, 2014
K. Kassner
Otto-von-Guericke-Universität Magdeburg
My result does not approach infinity and you can easily verify it. Derivatives are straightforward.
K, sounds interesting, I'll plot it this evening.
And the verdict is ... two of the three integrals do not work (Rennie from link above, and integral-calculator.com).  They blow up when used in Excel due to negative arguments to the LN function, even when R0 is avoided.  Kassner's works, and I confirmed the numbers by performing the integration numerically.  This is a corrected post.  In the first post, I had left out a factor of R0 in Kassner's formula.  (just got lost in counting the parentheses)
I started at 10000 m to build a log scale table approaching the event horizon of a solar mass at 2954 m, Schwarzschild coordinates.  I was intending to plot R1-R0 vs. the value of the integral above.  So the numbers should be in the thousands of meters initially.
The other two formulas cannot be used to go all they way to the event horizon because they blow up there and the R1 to R0 evaluation can't be completed.  Just evaluating them for the R1 values in my table starting at 10000, Rennie's formula produces credible numbers down to 6561 meters.  At that point R1-R0=3439 and evaluating Rennie over that interval produces 5940, which could be reasonable, at least it is greater than 3439.  But then it blows up.  I cannot see any reason why it should.  The last value it produced was over 15000.
The formula from integral-calculator.com is not usable at all because of the term ln((1-R0/r)1/2-1) which is always negative outside the event horizon.  Excel just won't evaluate it.
The spreadsheet is posted below.  (Updated to calculate using a direct integral, and to plot distance from R1 so as to avoid the blowup at R0 ... plot also attached ... looks like in converges very close to 11997.5 meters proper distance to EH compared to dr of 7046.1 meters - as of 10/8 updated with correct implementation of Klaus' formula)
8th Oct, 2014
K. Kassner
Otto-von-Guericke-Universität Magdeburg
Robert,
My formula is correct. I have checked it by differentiation. Why don't you do the same thing, just for double-checking? Numerical integration is only the last resort....
Differentiation w.r.t. R1 must give the integrand at R1, and setting R1=R0 must give zero. If the formula passes these two tests, it is correct. So the only way for it to be wrong is that I have made an error in taking the derivative (the other check is trivial). This should preferably not be checked by myself again but by someone else.
8th Oct, 2014
K. Kassner
Otto-von-Guericke-Universität Magdeburg
Here is something I cooked up during breakfast this morning. I have not counterchecked with Rindler or some other source, so the calculation may not be flawless. But it is much more fun this way...
What I want to calculate is the force on the tether or rope as perceived by the observer lowering the mass down to the black hole. My assumption is that the rope is "relativistically" inextensible, as Robert suggested. By that I mean that its proper length remains locally unchanged but that otherwise all relativistic effects affecting lengths apply.
If the force on the rope is F and the observer has to yield a piece of length dl of the rope, the work done by the mass pulling at the rope will be dW = F dl. Here dl is a proper length increment. This work may be stored by the observer and since the process can be done quasistatically, he has a lot of time to set up contraptions that do the storing properly (lifting a series of weights, extending a spring, etc.).
The energy of the mass m being lowered in the field decreases, due to time dilation:
E = (g00)1/2 m c2=(1-2M/r)1/2m c2
The decrease in energy is used up to do the work, so we have
F dl = - dE                                                                                   (1)
Moreover,
dl = - (1-2M/r)-1/2 dr
dE = dr dE(r)/dr = dr m c2 (1-2M/r)-1/2 M/r2
Inserting dl and dE into (1), we find that the inverse sqare root cancels and obtain
F = mM c2/r2
If we go back to SI units, i.e. replace M by GM/c2, we get
F = G m M/r2
This is just the force the mass m would experience due to Newton's law. On the one hand, this is a very satisfactory result, as we must get precisely this in the limit of small gravitational field. On the other hand, it is a bit puzzling that the force does not seem to diverge as we decrease r to approach the Schwarzschild radius.
Before considering this problem, let us deal with another one. All of the above is correct for an observer located at infinite distance from the black hole. But infinity is infinitely far away even in Newtonian physics. We would need an infinitte tether. Can we modify the analysis to deal with an observer at R1?
Well, that is not too difficult. Such an observer will experience a time dilation factor of (1-2M/R1)1/2, so he will measure all energies larger by a factor (1-2M/R1)-1/2. The proper length dl is invariant, and F is what we want to calculate. So the result is simply
F = (1-2M/R1)-1/2mM c2/r2,
which is a small modification, if R1 is large.
Aha! This may help solve our other problem as well! What will be the force in the rope measured by an observer at r? Why, this is obtained from the above formula by replacing R1 with r
F(r) = (1-2M/r)-1/2mM c2/r2.
(He would measure this if cutting the rope at r and keeping in place the end of the part going down -- for which he would just need the force F(r).)
So the point is that the tension in the rope will not be constant as it would in a massless rope in Newtonian physics! This is because forces are not relativistically invariant and in this particular case this is due to time dilation. F(r) diverges as r approaches 2M.
However, an observer lowering the mass below the horizon will not feel an increased force at his end. The rope will simply go slack, and if he hauls it up, he will find the mass and part of the rope gone.
If he had an unbreakable rope, he might pull the mass back from beyond the horizon. But unbreakability is as impossible as going faster than light. And if you start from the premise of a theoretical impossibility then you may of course get all kinds of nonsensical results out of the theory.
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Klaus, I'm real interested in your weight calculation and I'll look at it in detail this evening.
On the other formula, I found that I was so busy counting parentheses I missed the factor of R0 in the front.  It is quite correct indeed!  Thanks!  (BTW, the online derivate calculators cannot handle it any better than the integral calculators.  I tried a couple of them.)
Klaus, thanks for the simple, one dimensional (two counting time) analysis of the tether consistent with GR.  I see how you treat dl and dr and it makes sense.  dr is used to get the energy delta.  But that energy is distributed over the proper length dl.  That's what I was thinking.
In my 2011 analysis using Γ as the time dilation factor between any two reference frames not relatively moving, in the case you are analyzing here the Schwarzschild factor (1-GM/rc2)-1/2, I concluded that the force at the top of the tether was reduced by this factor as compared to what was "felt" by the time dilated observer lower down. (in formula (12) I am using F' as the top of the tether) I did this by modeling a force as momentum impulses, assuming that momentum is an invariant across the two reference frames, and reasoning with time and the impulse arrival rate.  I was worried that maybe the spatial curvature would act like a mechanical pulley ratio and mess things up, and am relieved to see that it does not.  This seems to be due to the fact that the energy difference is coming from dr.
By postulate, I had gravitational acceleration proportional to 1/r2 relative to the top observer (this was deducible from equivalence, actually, that the bottom observer would be the one to experience relativistically enhanced acceleration).  Now I am sitting here trying to figure out how I made the energy balance without the dl relationship!
OK, I see (there was a long pause there, it didn't come fast).  In fact it is related to the thesis of that paper to accomplish curvature effects with inertia.  In effect, I had a "relative" inertia increase by Γ over dr which is less than dl by Γ.  The Γ's cancel and the energy result is the same.  The lack of spatial curvature fails of course, because experimentally ∫dl is the amount of tether spooled out, not ∫dr, a consequence I realized immediately when I read of Shapiro delay.
Using your approach it is clear what the role of the particular value of the Schwarzschild metric is.  It results in dE/dr that preserves the Newtonian square law of gravity for the distant observer as he measures it at various remote points using the tether.  I had guessed that and was looking for evidence, and here is the smoking gun that says it is so.  I suspect that there were choices having to do with preserving the existing "laws" of physics in the development of the field equation that led to that outcome, but I will continue looking into this.
For those that continue to wonder at the motivation for my poking around, this provides an opportunity to state it clearly.  While for the electrostatic force I might grant that the fundamental law was the inverse square relation, I suspect that for gravity, which is much weaker than other quantum forces, it is an indirect effect and that the 1/r Newtonian potential "law" is the fundamental law.  If so, the field equation does not preserve this law.  It is not possible to preserve both.  One must choose.
BTW I have corrected several previous posts a 2nd time.  Klaus has given two answers to this question (the first about the distance integral) which are worthy of the record books in their clarity and ease of understanding, as Charles also noted.  Thanks to Charles for digging through MTW and determining that they are consistent with that source.
OK, well anyway, thanks for checking.  Perhaps I sold my copy hastily, but I was really disgusted a few years ago and clearing out the bookshelf and selling stuff and it was worth several times what I had paid for it.  :D
@ALL I am posting a chart of time to event horizon in distant observer frames in the other thread (click below).  Unlike the distance to the horizon, it does not appear finite.  But extremely short.  It is a more difficult integral than the distance I believe so I didn't try it analytically.
Klaus' tether derivation continues to lead to new simple explanations for various things.  See links.

## Similar questions and discussions

What is the non-zero Ricci tensor equation for a metric field slightly non-Schwarzschild?
Question
• Robert Shuler
Static gravity, i.e. the Schwarzschild metric, depends mostly on the vacuum field equation: Ricci tensor = 0.  Using notation from https://en.wikipedia.org/wiki/Deriving_the_Schwarzschild_solution we get one thing from symmetry and the definition of metrics, a form in which there are two relevant coefficients, A(r) and B(r) (the time and space coefficients, roughly speaking).
From the Ricci tensor = 0 constraint, which has been given only vague physical interpretations, we get two things:
1. A(r)B(r)=K where K is a constant.  From the limit at great distances where A->B->1, we see that in normal coordinates A(r)=1/B(r) (or -1/B(r) if you are including the sign in the coefficient).  This gives the relation between time dilation and radial length contraction (or spatial expansion).
2. Substituting for B(r) we get a differential equation rA'=A(1-A) where ' means derivative and A=A(r).  This has a solution of the form (1-K/r)-1 where here K is a different constant, and the Newtonian limit gives the usual Schwarzschild time dilation factor 1/(1-2GM/rc2)1/2 and the inverse for radius.
As a friend showed a few months back, this leads to gravity which for a distant observer with a tether amounts to an inverse square force law, oddly enough.  But potential, obviously, is not Newtonian for any observer.
Suppose we want one of the two variations of potential to be the time dilation factor.  These are both approximately equal to the Schwarzschild factor for large r.  They are 1/(1-GM/rc2) and (1+GM/rc2).  Take the second one.  It implies, according to another question I posed https://www.researchgate.net/post/What_differential_equation_has_a_solution_of_the_form_Fx1_1_Kx-2 , that the differential equation in step 2 above must be something like rA'=2A(1-A).
That factor of 2 when backed through to the Ricci tensor means that it cannot be quite zero for small radii.  At that point I get lost trying to draw more definite conclusions.  For example, if we have a different equation:
Ricci tensor = X
Then what is X (probably a complicated function) such that we still have A=1/B, but we get rA'=2A(1-A).  It is a matter for someone who knows all the little summations and conventions by heart and is good at fudging.  Any takers?

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