I Measuring Mass of a Bowling Ball Dropped into a Black Hole

[Herbascious J]

In a recent thread we discussed the idea that an object slowly dropped into a black hole, can have its rest mass recovered, as energy, if slowly brought to a halt at the event horizon. Once the object is dropped, it would be unrecoverable, and the BH would gain no new mass. I am under the impression that an observer bound to the object would not detect a loss of mass, so this effect, if measurable, would be relative.

If we imagine that the object is a large bowling ball, held in the hands of an astronaut we can imagine an experiment that she may conduct. While approaching the event horizon, the astronaut is brought to a halt and she then applies a force to the ball, the ball accelerates and moves from one hand to the other. The force would be a set quantity and known to both her and an observer far away. By doing this experiment the astronaut can determine the mass of the ball by observing the acceleration between her hands. The outside observer can see the experiment unfold at a distance and also measure the acceleration, therefore determining a mass.

I am curious; from the point of view of the astronaut I expect she measures the same mass for the ball that was determined while she was far away with the other observer before being lowered into the BH. The distant observer, however, is able to recover the rest mass of the bowling ball as usable energy. Once the ball is dropped into the BH, he measures no change in the mass of the BH. What results would the distant observer measure while watching the astronaut’s experiment from far away? Would he see the astronaut successfully measure a mass for the ball and how would he explain anything that he is a witness to?

 

[phinds]

In a recent thread we discussed the idea that an object slowly dropped into a black hole, can have its rest mass recovered, as energy, if slowly brought to a halt at the event horizon. Once the object is dropped, it would be unrecoverable, and the BH would gain no new mass.

Huh? Please provide a link to that thread.

 

[Herbascious J]

https://www.physicsforums.com/threa...oles-goes-to-infinity-as-they-collide.972615/
Entry #24

 

[phinds]

https://www.physicsforums.com/threa...oles-goes-to-infinity-as-they-collide.972615/
Entry #24

Yeah, thanks. I had already found it by following your posts. I'm still trying to get my head around it.

 

[Herbascious J]

Yeah, thanks. I had already found it by following your posts. I'm still trying to get my head around it.

It's a little off topic from the original post. It's buried a ways down there, but it really made an impression on me and I'm trying to get a better understanding of what it means.

 

[phinds]

It's a little off topic from the original post. It's buried a ways down there, but it really made an impression on me and I'm trying to get a better understanding of what it means.

Exactly. Me too.

 

[jartsa]

I am curious; from the point of view of the astronaut I expect she measures the same mass for the ball that was determined while she was far away with the other observer before being lowered into the BH. The distant observer, however, is able to recover the rest mass of the bowling ball as usable energy. Once the ball is dropped into the BH, he measures no change in the mass of the BH. What results would the distant observer measure while watching the astronaut’s experiment from far away? Would he see the astronaut successfully measure a mass for the ball and how would he explain anything that he is a witness to?

Astronaut: "The mass of the ball seems to be 0.1 times my mass"
Distant observer: "The mass of the ball seems to be 0.1 times the astronaut's mass"
Distant observer: "The mass of the black hole seems to be million times my mass"
Distant observer: "The mass of the black hole seems to be trillion times the astronaut's mass"
Astronaut: "I agree with the above"

 

[Herbascious J]

Distant observer: "The mass of the ball seems to be 0.1 times astronaut's mass"

Is it possible that the force appears to be weaker to the outside observer? Perhaps the acceleration is quicker? I'm referring directly to the experiment that determines the apparent mass of the ball.

 

[jartsa]

Is it possible that the force appears to be weaker to the outside observer? Perhaps the acceleration is quicker? I'm referring directly to the experiment that determines the apparent mass of the ball.

How is it surprising in any way that force appears to be weaker to outside observer?
An accelerating rocket passes us at very high speed. At the nose of the rocket there is a juggler whose forces transverse to the motion are diminished, because transverse forces transform this way: F'=F/gamma.

At the rear there is another juggler whose speed is even faster, because the rear moves faster when the spaceship is length contracting. This person's transverse forces are even more diminished. Also it may be interesting that this person is the "lower" person in the accelerating frame of the rocket.

 

[PeterDonis]

I am under the impression that an observer bound to the object would not detect a loss of mass

That's correct.

The force would be a set quantity and known to both her and an observer far away.

Yes, but the distant observer can't determine the mass of the bowling ball relative to himself by using the force exerted by the astronaut. All he is able to determine from the observations you describe is the mass of the bowling ball relative to the astronaut. But he cannot assume that the mass of the bowling ball relative to the astronaut is the same as the mass of the bowling ball relative to himself.

In order to measure the mass of the bowling ball relative to himself, the distant observer would have to exert a force himself that acts on the bowling ball. For example, he could measure the force he has to exert on the tether he is using to lower the bowling ball, in order to hold the bowling ball stationary. And he will find that, while this force does increase as he lowers the bowling ball, it increases much more slowly than the force measured by a strain gauge on the tether at the point where it attaches to the bowling ball (and observed, for example, by the astronaut). This is an indication that, relative to the distant observer, the mass of the bowling ball has decreased (and he will be able to relate this decrease to the amount of energy he has extracted from the tether during the lowering process).

 

[pervect]

In a recent thread we discussed the idea that an object slowly dropped into a black hole, can have its rest mass recovered, as energy, if slowly brought to a halt at the event horizon. Once the object is dropped, it would be unrecoverable, and the BH would gain no new mass. I am under the impression that an observer bound to the object would not detect a loss of mass, so this effect, if measurable, would be relative.

If we imagine that the object is a large bowling ball, held in the hands of an astronaut we can imagine an experiment that she may conduct. While approaching the event horizon, the astronaut is brought to a halt and she then applies a force to the ball, the ball accelerates and moves from one hand to the other. The force would be a set quantity and known to both her and an observer far away. By doing this experiment the astronaut can determine the mass of the ball by observing the acceleration between her hands. The outside observer can see the experiment unfold at a distance and also measure the acceleration, therefore determining a mass.

I am curious; from the point of view of the astronaut I expect she measures the same mass for the ball that was determined while she was far away with the other observer before being lowered into the BH. The distant observer, however, is able to recover the rest mass of the bowling ball as usable energy. Once the ball is dropped into the BH, he measures no change in the mass of the BH. What results would the distant observer measure while watching the astronaut’s experiment from far away? Would he see the astronaut successfully measure a mass for the ball and how would he explain anything that he is a witness to?

Locally, the bowling ball will have a well-defined stress-energy tensor, $T^{\mu\nu}$. There will be stresses in the bowling ball needed for it to hold it's shape when it - and the person holding it - hover over the event horizon.

The question here is how you want to go from the locally defined stress energy tensor to a notion of the "mass" of the bowling ball. This gets more involved than it appears. You basically seem to want to assume that the bowling ball is a rigid object, and to compute the mass by the ratio of force/acceleration in the limit of a small force.

Rindler performed some calculations that may be a good guide as to how to approach this problem, though unfortunately I don't have the book in question, "Relativity, special, general and cosmological", <<amazon link>>. The approach is somewhat old-fashioned, but it works out exactly what you want to figure out, the ratio of force/acceleration.

From memory, the resulting ratio depends on the direction of the force. If the force is transverse to the pressure, I believe there is no effect on the force/acceleration ratio, that corresponds to my image of your "left hand to right hand", with the person standing holding the bowling ball against the local direction of gravity. Hopefuly we have the same image in mind. But if the force is not transverse (the person judges the weight by your experiment but moves the bowling ball up and down), the answer is different.

Unfortunately, I don't own the book to refresh my recollection, but I believe that's what happens.

There are other logical ways of getting the "mass" that don't involve your experiment, but they got a bit long and detracted from my message, so I snipped any discussion of them.

The problem is much simpler in flat space-time, so finding the "mass" of a bowling ball in the flat space-time of the Rindler metric of an accelerating elevator is a simpler problem.

I will say a few words about the non-flat space-time case, though. In the presence of curvature, parallel transport is path dependent, which means that while the energy density as defined by the stress-energy tensor is always well defined as a geometric object, summing together energy*volume becomes coordinate dependent in the presence of curvature, as the stress-energy tensors are in different tangent spaces.

Let's talk about the connection between the simple flat-space time case and the Schwarzschld case. In one of my old posts, #16 in the thread "Jetpacking above a black hole", I give a coordinate transformation that , sufficiently near a black hole, makes the Schwarzschild metric appear as the Rindler metric <<link>>. So this gives us a way to apply the results from "Einstein's elevator" to an actual black hole.

In the elevator model, we don't have an event horizon anymore, but we do have the "Rindler horizon", where the metric coefficeint of $g_{00}$ vanishes as happens in the Scwharzschild metric. When the approximate conditions of being sufficiently close to the black hole are met, it turns out that the local acceleration of gravity for a static observer is c^2 / d, where d is the "distance away from the event horizon" as measured in the elevator. c, of course, is the speed of light.

This is interesting, because it demonstrates that the "force of gravity" is about 10^13 Earth gravities (9 * 10^13 meters/sec^2) at a distance of 1 km above the horizon, for any black hole of sufficiently large mass that the approximations define. This may give some insight as to the magnitude of the pressures we are talking about, and why we usually don't have to worry about the relationship between pressure and mass. If we assume that pressures are "small enough", we don't have to worry about their effect. One way of looking at the problem you're asking about is to note that you're considering a case where the pressure can no longer be neglected by the way you constructed the problem.

This post is bit long, but the topic of "mass" in general relativity is a rather advanced and tricky topic.

It's not the same as the original problem, but I can't help but point out that there's a close and simple relation between $\rho + 3P$ for an ideal fluid with a stress energy tensor of diag($\rho, P, P, P)$ and the Komar mass of general relativity. However, explaining in more detail would best be left to another thread.

I suppose my quick summary of all of this is that to keep Newton's law working right, it's by far the simplest to assume that pressures are "low" in a relativistic sense. And your very defintion of "mass", as the ratio of Force/acceleration, is motivated by Newton's laws.

If we can assume the pressure contributions to "mass" are negligible, a lot of the need for discussion vanishes. By insisting that the bowling ball be suspended near the event horizon of a black hole, though, one almost inadvertently has to start to deal with the non-negligible pressure case, as the example of 10^3 Earth gravities 1km away from the horizon illustrates.

 

[PeterDonis]

The question here is how you want to go from the locally defined stress energy tensor to a notion of the "mass" of the bowling ball.

That's a question of interest in general, yes (and the rest of your post gives good information on it), but note that it is not necessary to answer this question in order to assess where the difference between the locally measured mass of the bowling ball (the mass measured by the astronaut) and the mass measured by the distant observer comes from. To assess that, it is perfectly fine to treat the bowling ball itself as a test object with some specified rest mass (as measured locally) in the Schwarzschild spacetime of the black hole.

 

[pervect]

That's a question of interest in general, yes (and the rest of your post gives good information on it), but note that it is not necessary to answer this question in order to assess where the difference between the locally measured mass of the bowling ball (the mass measured by the astronaut) and the mass measured by the distant observer comes from. To assess that, it is perfectly fine to treat the bowling ball itself as a test object with some specified rest mass (as measured locally) in the Schwarzschild spacetime of the black hole.

Ah, if you make the bowling ball a point particle, instead of a ball - I suppose you can the norm of it's 4-momentum as the mass.

That is quite a bit simpler.

 

[Heikki Tuuri]

I wrote about this problem in my physics blog on May 23, 2019.

Let the ball mass be 1 kg. Suppose that the redshift is 10% from the astronaut to the far-away observer. If the far-away observer lowers the ball slowly with a tether, he can recover 1/11 of the mass-energy of the ball from the work done by the tether. One may say that the "gravitational mass" of the ball is now only 10/11 kg from the viewpoint of the far-away observer. That is the amount of mass-energy that he gave to the gravitating system.

The astronaut, who is handling the ball, measures locally the gravitational mass and the inertial mass of the ball as 1 kg. This follows from the equivalence principle.

If the far-away observer accelerates the ball through an inelastic tether he has attached to the ball, what happens? We may assume that the gravitational force is canceled locally with a spring attached to the ball. Then the far-away observer can probe the inertial mass of the ball with the tether.

Suppose that the tether does 0.5 joules of work, as measured by the astronaut. The far-away observer only did 10/11 * 0.5 J of work. Why? The astronaut can recover the work as heat and send it back as photons to the far-away observer. Because of the redshift, the conservation of energy requires these numbers.

The astronaut measures that the end speed of the ball is 1 m/s.

The tether is inelastic. The two people agree on the length that it moves (because they both measure 1 meter with identical inelastic rulers), but disagree on the time. Because of time dilation, the far-away observer measures the speed as 10/11 m/s. The far-away observer thinks that he pulled on an inertial mass of 11/10 kg, as he did 10/11 * 0.5 J of work.

Conclusion: the far-away observer measures that the "gravitational mass" of the ball is less than 1 kg, but through pulling the tether, he measures that its inertial mass is larger than 1 kg.

The equivalence principle says that locally, the astronaut will think the ball has the same properties as it had in outer space. But if we do measurements through some method from far away, this is not necessarily true.

 

[metastable]

In a recent thread we discussed the idea that an object slowly dropped into a black hole, can have its rest mass recovered

Isn’t there a presumption here that the person doing the “slow dropping” isn’t initially moving directly away from the black hole with at least escape velocity, or significantly more velocity than escape velocity?

 

[PeterDonis]

an inelastic tether

I don't think you mean "inelastic", since that would mean the tether would deform, and later on you say that it doesn't:

The tether is inelastic. The two people agree on the length that it moves (because they both measure 1 meter with identical inelastic rulers)

I think what you mean is "rigid", in the sense that the tether transmits all forces exerted at either end to the other end, without any dissipation, and that observers at both ends will agree on the length the tether moves.

 

[pervect]

I wrote about this problem in my physics blog on May 23, 2019.

Let the ball mass be 1 kg. Suppose that the redshift is 10% from the astronaut to the far-away observer. If the far-away observer lowers the ball slowly with a tether, he can recover 1/11 of the mass-energy of the ball from the work done by the tether. One may say that the "gravitational mass" of the ball is now only 10/11 kg from the viewpoint of the far-away observer. That is the amount of mass-energy that he gave to the gravitating system.

The astronaut, who is handling the ball, measures locally the gravitational mass and the inertial mass of the ball as 1 kg. This follows from the equivalence principle.

If the far-away observer accelerates the ball through an inelastic tether he has attached to the ball, what happens? We may assume that the gravitational force is canceled locally with a spring attached to the ball. Then the far-away observer can probe the inertial mass of the ball with the tether.

Suppose that the tether does 0.5 joules of work, as measured by the astronaut. The far-away observer only did 10/11 * 0.5 J of work. Why? The astronaut can recover the work as heat and send it back as photons to the far-away observer. Because of the redshift, the conservation of energy requires these numbers.

The astronaut measures that the end speed of the ball is 1 m/s.

The tether is inelastic. The two people agree on the length that it moves (because they both measure 1 meter with identical inelastic rulers), but disagree on the time. Because of time dilation, the far-away observer measures the speed as 10/11 m/s. The far-away observer thinks that he pulled on an inertial mass of 11/10 kg, as he did 10/11 * 0.5 J of work.

Conclusion: the far-away observer measures that the "gravitational mass" of the ball is less than 1 kg, but through pulling the tether, he measures that its inertial mass is larger than 1 kg.

The equivalence principle says that locally, the astronaut will think the ball has the same properties as it had in outer space. But if we do measurements through some method from far away, this is not necessarily true.

Let's talk about the stress-energy tensor $T^{\mu\nu}$ in the tether. Presumably, one is supposed to assume that $T^{00}$ is zero in the tether, otherwise it would be difficult (I think impossible, myself) to justify the claims that the tether only transmits force, rather than "being acted on by gravity". And of course the tehter must has some non-zero tension in the direction of the force. So if we align the $x^1$ coordinate with the tether, then $T^{11} < 0$. I believe this describes the stress-energy tensor of the tether.

The fundamental issue then is that the condition that $T^{00} = 0$ depends on one's coordinate choice. So there is an implicit assumption of some particular "special" set of coordinates when doing thins sort of analysis. I think this assumption needs to be made explicit, otherwise one can get some wrong ideas of when this construct can be applied, and when it cannot be applied.

It's fairly well known, I think, that a tether under tension with $T^{00}=0$ and $T^{11} <0$ and no other terms in the stress energy tensor is a form of exotic matter, it violates the weak energy condition. This means that frames exist in which $T^{00} < 0$. It can also be verified by direct computation, the simplest case is to consider a section of such a tether in a flat space-time, and do the appropriate Lorentz boost to the stress-energy tensor of the tether.

 

[PeterDonis]

Presumably, one is supposed to assume that $T^{00}$ is zero in the tether, otherwise it would be difficult (I think impossible, myself) to justify the claims that the tether only transmits force, rather than "being acted on by gravity".

The stress-energy tensor does not determine whether the object is "acted on" by gravity. The motion of all objects is affected by the spacetime geometry. The stress-energy tensor determines how the object acts as a source of gravity and therefore changes the spacetime geometry. I take it that the scenario is intended to be that the tether does not change the spacetime geometry, so all of its SET components are assumed to be negligible.

However, "negligible" in terms of affecting the spacetime geometry is not the same as "exactly zero". It is perfectly possible for the tether to have an energy density that exceeds the tension in it in its rest frame (which is basically the condition to avoid it being "exotic matter") while still having both of them so small that they produce negligible spacetime curvature.

 

[pervect]

The stress-energy tensor does not determine whether the object is "acted on" by gravity. The motion of all objects is affected by the spacetime geometry. The stress-energy tensor determines how the object acts as a source of gravity and therefore changes the spacetime geometry. I take it that the scenario is intended to be that the tether does not change the spacetime geometry, so all of its SET components are assumed to be negligible.

However, "negligible" in terms of affecting the spacetime geometry is not the same as "exactly zero". It is perfectly possible for the tether to have an energy density that exceeds the tension in it in its rest frame (which is basically the condition to avoid it being "exotic matter") while still having both of them so small that they produce negligible spacetime curvature.

I can see I need to add a bit more context to my remarks. So I'll do that now.

Wald has a construction similar Heikki's in his chapter on energy in his text "General Relativity". Specifically, pg 288.

Thus, the local force which must be exerted on a unit test mass to hold it in place is given by equation (11.2.3). However if we choose to calculate the force which must be applied by a distant observer, at infinity (e.g., by means of a long string) we find that this force differs from the local force by a factor of V (see problelm 4 of chapter 6).

The "factor V" here is just the redshift factor, the norm of the Killing vector. Wals has previously assumed that the space-time in question was stationary, which is why there is a Killing vector, a point that will become important later. But let's go back to the string. What string is Wald referring to? We need to visit problem 4 of chapter 6.

Suppose a particle of mass m is held stationary by a (massless) string...

He then suggests we use "conservation of energy" arguments to show that the force at infinity is lower by the redshift factor V in a stationary spacetime.

Now, as written, this isn't very clear, my remarks were an effort to make things more clear and precise. If we were in flat space-time, I think it's quite reasonable to say that a string is massless if $T^{00}=0$ everywhere. It's not entirely clear what we mean by "massless", I am suggesting that this is the appropriate definition in this specific context.

Of course, there is a fine point here. This proposed defintion of massless string needs an appropriate frame of reference frame to make sense, as the defintion is NOT frame independent. A string that is "massless" by this defintion in one frame in flat space-time won't be massless in a frame moving relative to the first frame. But in the stationary space-time, we have an appropriate notion of "at rest", so we can use this simple defintion of "massless" in the appropriate rest frame.

And the "conservation of energy" arguments are that $\nabla_a T^{ab} = 0$

This is enough to duplicate Wald's result. It also illustrates the limitations of the approach, why the approach only works for stationary space-times.

We can also see that if $T^{00}$ was not zero, we'd have a "heavy string", not a massless string, and the force at infinity would include the weight of the test mass, plus the weight of the string. To not include the weight of the string, the string must be massless. The term "massless" in this context is very vague, but I believe that setting $T^{00}$ everywhere is the appropriate definition.

 

[Herbascious J]

Thank you for the indepth analysis and discussion, it is much appreciated. Is it possible to interpret all of the rest mass energy as sourcing from the BH's mass itself, as opposed to the object being lowered? Does this change the math? I'm assuming you can't do this, and the energy is coming strictly from the lowered object. But I want to verify that the result is not interpretive. I'm curious as it will have an impact on a follow up question I'd like to post.

 

[PeterDonis]

It's not entirely clear what we mean by "massless", I am suggesting that this is the appropriate definition in this specific context.

I think "massless" as Wald was using the term means we are assuming its entire stress-energy tensor is negligible. (Note that if all components are zero in one frame, they are all zero in any frame, so saying that the entire SET is negligible is frame independent.) I think the reason for making this approximation is twofold: first, to avoid affecting the spacetime geometry, as I said in my previous post, and second, what you say here:

we'd have a "heavy string", not a massless string, and the force at infinity would include the weight of the test mass, plus the weight of the string.

In other words, we don't want the string to affect the measurement of force that we're making.

the "conservation of energy" arguments are that $\nabla_a T^{ab} = 0$

This is conservation of the source of gravity, and is valid in any spacetime, whether or not it is stationary. It is not the same as the conserved quantity $p_a \xi^a$ in a stationary spacetime (where $p_a$ is the 4-momentum of a test object and $\xi^a$ is the timelike Killing vector), which is usually called "energy at infinity" and is the conservation of energy that is relevant to the "mass on a string" thought experiment.

 

[pervect]

I think "massless" as Wald was using the term means we are assuming its entire stress-energy tensor is negligible. (Note that if all components are zero in one frame, they are all zero in any frame, so saying that the entire SET is negligible is frame independent.) I think the reason for making this approximation is twofold: first, to avoid affecting the spacetime geometry, as I said in my previous post, and second, what you say here:

Unfortunately, if one assume the stress energy tensor is zero, there's no way for the string to transmit a force. Which is required by the problem statement. We can, however, make the non-tension terms zero, and that's what I believe Wald is suggesting. It's hard to be positive, because it's not discussed in the text, it's only presented as an exercise.

This is conservation of the source of gravity, and is valid in any spacetime, whether or not it is stationary. It is not the same as the conserved quantity $p_a \xi^a$ in a stationary spacetime (where $p_a$ is the 4-momentum of a test object and $\xi^a$ is the timelike Killing vector), which is usually called "energy at infinity" and is the conservation of energy that is relevant to the "mass on a string" thought experiment.

I agree that it's different - but the string is not following a geodesic. It's true that $p_x \xi^a$ is constant for a particle following a geodesic, but that's not the case being analyzed.

I've seen the vanishing of the divergence of the stress-energy tensor called local energy conservation in several places, but the only one I can directly recall is one of the FAQ's about energy in GR. <<link>>

 

[metastable]

In other words, we don't want the string to affect the measurement of force that we're making.

If the string is orbiting the black hole, is it rotating? Can it maintain born rigidity?

 

[PeterDonis]

if one assume the stress energy tensor is zero, there's no way for the string to transmit a force.

We are not assuming that the stress-energy tensor is zero. We are assuming that it is negligible as far as the spacetime geometry is concerned. Of course the string has some nonzero stress-energy; it's just not enough to cause any detectable spacetime curvature.

We can, however, make the non-tension terms zero, and that's what I believe Wald is suggesting.

I disagree. There is no need to deal with the string's stress-energy tensor at all in order to model how it transmits force. The only reason to deal with the SET would be to model the string's effect on the spacetime geometry, and we are assuming that it has no such effect that we can detect. See above.

It's true that $p_a \xi^a$ is constant for a particle following a geodesic, but that's not the case being analyzed.

The case being analyzed, which involves slowly lowering an object in a stationary gravitational field, can be (laboriously) analyzed in terms of a succession of geodesics with gradually decreasing energy at infinity, such that the decrease in energy at infinity of the object is exactly balanced by the energy extracted by the distant observer from the lowering process. So even though the object itself is not moving on a geodesic, energy at infinity is relevant to this case. It is true that the energy at infinity of the object itself is not conserved, since it is not moving on a geodesic. But the total energy at infinity of the system consisting of the object + the energy being stored by the distant observer (in some energy storage device whose details we are leaving unspecified) is conserved.

However, that's actually not the key point here. The key point here is that the covariant divergence of the SET is not relevant to this case, since it has nothing to do with the extraction of energy from the slow lowering process. See below.

I've seen the vanishing of the divergence of the stress-energy tensor called local energy conservation in several places

It's called that in many textbooks and papers. MTW calls it that; Wald probably does too although I don't have my copy handy right now to pick out specific instances. And in all those places, the "energy conservation" they are talking about is conservation of the source of gravity. MTW discusses in detail how the Bianchi identities obeyed by the Einstein tensor, which are what enforce the covariant divergence of the SET being zero via the Einstein Field Equation, ensure "automatic conservation of the source", and why that's a nice thing to have.

But in the thought experiment under discussion, any "conservation of energy" is not conservation of the source, since we are assuming that the string is negligible as a source of gravity (as is the object being slowly lowered). So the only relevant concept of energy has to do with the timelike Killing vector field.

 

[PeterDonis]

If the string is orbiting the black hole

It isn't. It is hanging radially, and the only motion involved is the slow unrolling of the string as the object is lowered. All motion is purely radial.

 

[metastable]

It isn't. It is hanging radially, and the only motion involved is the slow unrolling of the string as the object is lowered. All motion is purely radial.

If this is the description of the scenario, won't the person doing the "lowering" "plummet" "like a rock"?

 

[PeterDonis]

won't the person doing the "lowering" "plummet" "like a rock"?

No. They are assumed to be supported by something; since the central mass is a black hole, the support would presumably be a rocket or something similar that could exert thrust to hold itself stationary. The person doing the lowering is assumed to be very far away from the hole, so the thrust required to remain stationary is minimal.

 

[metastable]

No. They are assumed to be supported by something; since the central mass is a black hole, the support would presumably be a rocket or something similar that could exert thrust to hold itself stationary. The person doing the lowering is assumed to be very far away from the hole, so the thrust required to remain stationary is minimal.

So in the scenario, we are saying the amount of thrust required to "remain stationary" isn't the amount of thrust it would take to keep an "untethered" "lowerer" stationary with respect to the black hole at that coordinate, plus the additional force exerted on the "lowerer" through the string?

 

[PeterDonis]

So in the scenario, we are saying the amount of thrust required to "remain stationary" isn't the amount of thrust it would take to keep an "untethered" "lowerer" stationary with respect to the black hole at that coordinate, plus the additional force exerted on the "lowerer" through the string?

No, it is the sum of those two things. The force exerted at the lowerer's position due to the tether having to support the object being lowered is redshifted, so it does not increase without bound as the object at the lower end of the tether approaches the horizon.

You are correct, though, that unless the black hole is supermassive, the thrust required to hold station even at a very large distance, once the force on the tether is included, will not be "minimal". Since this is a thought experiment, a large amount of thrust being required is not a problem.

 

[metastable]

No, it is the sum of those two things. The force exerted at the lowerer's position due to the tether having to support the object being lowered is redshifted, so it does not increase without bound as the object at the lower end of the tether approaches the horizon.

You are correct, though, that unless the black hole is supermassive, the thrust required to hold station even at a very large distance, once the force on the tether is included, will not be "minimal". Since this is a thought experiment, a large amount of thrust being required is not a problem.

Does this not imply then that the "lowerer" will always be using more power in the experiment than is "gained" back via the tether?

 

[PeterDonis]

Does this not imply then that the "lowerer" will always be using more power in the experiment than is "gained" back via the tether?

So far we have been ignoring any power requirements for exerting thrust. This is typical for these kinds of thought experiments.

Since this thread is about how we would measure the mass of the object being lowered, not about whether any net energy can be extracted from the lowering process once the energy requirements for holding station are included, ignoring the power required to exert thrust for this thread does not seem objectionable. If you want to discuss the net energy extraction question, it would be fine to start a separate thread on that topic.

 

[Heikki Tuuri]

https://en.wikipedia.org/wiki/Ultimate_tensile_strength
I calculated a practical example. Make a 1 x 1 millimeter^2 thick tether of graphene and put it between the Moon and the surface of Earth.

Its mass is 380 metric tons, weight 60,000 Newtons, and it can lift a load of 7 tons from Earth.

The redshift from the surface of Earth is less than 10^-9. That is too small. We cannot measure the increase in the inertial mass of the 7 ton load.