Why we won't notice anything special when crossing the horizon?

[PAllen]

One observer observes a rocket towing two identical asteroids, using different length ropes. He says the force and the acceleration is larger on the asteroid end of the longer rope. (because ropes are lorentz contracting, which increases the acceleration)

From another frame an other observer is seeing the same event as rocket slowing down the speeds of two asteroids. He says the force and the acceleration is smaller at the asteroid end of the longer rope. (because ropes are lorentz-expanding, which decreases the acceleration)


So why can't the rocket say the forces are the same, at the rocket end of the ropes.

You can call a tail a leg, but that doesn't make it a leg. Proper acceleration is an SR invariant - it is not observer or coordinate dependent. Proper acceleration is what accelerometers measure. You want something different? Pick a different universe that isn't governed by SR.

 

[pervect]

When Joe, in an accelereting rocket, holds a string, that has a mass hanging on the other end, Joe will notice that the force exerted on his hand is the same, regardless of if the mass is hanging low or close to Joe's hand. Also with enormous acceleration this is true.

Now Joe can easily calculate the redshift: For every inch of upwards climb a photon redshifts the same amount. For example 10 percents per 1 meter.

(Because Joe's hand does a constant amount of work per inch when lifting the weight, Joe deduces that a photon loses a constant proportion of its energy per inch when climbing)

There is some precedent in at least one textbook for your point of view, but I've mostly seen it used as an informal "motivational" guide (in Wald, for example) rather than as anything rigorous. Frankly, I still find the idea of the "string" a bit murky, it's never been quite clear to me how an ideal string is defined (especially when we know there is no such thing as a rigid body). So it's difficult for me to comment in detail here.

And yes, an accelerometer shows different readings at different positions in the rocket.


An additional note:

Problems in this scenario, like lack of event horizon, stem from the incorrect idea that a climbing photon loses energy.

A photon that is lowered down loses energy. It loses energy x joules per inch. The energy goes to zero at some distance, where there is an even horizon.

If you are using the definition of "energy" that I think you are, energy-at-infinity (which you could renormalize to energy-at-Joe, I suppose), then a photon never gains or loses energy, regardless of whether it rises or falls. This is the notion of energy that would be compatible with the notion of "force at infinity" or "force at joe" which you were using earlier.

(This notion of energy is different from the local observers notion of energy - the local observer measures energy and everything else with their own clocks, rulers, test masses, etc.)

The "energy-at-infinity" or the "energy-at-joe" is a constant of motion, thus it remains constant for a falling photon - or anything else. (At least as long as the photon or other falling object is following a geodesic path.)

 

[PAllen]

There is some precedent in at least one textbook for your point of view, but I've mostly seen it used as an informal "motivational" guide (in Wald, for example) rather than as anything rigorous. Frankly, I still find the idea of the "string" a bit murky, it's never been quite clear to me how an ideal string is defined (especially when we know there is no such thing as a rigid body). So it's difficult for me to comment in detail here.

This is good point about what is meant by a string. I am assuming a string maintains constant proper length computed in the same simultaneity surface as for Born rigidity.

However, as long as there is SOME adequate definition of string behavior, the result of any measurement made using it in some given apparatus can never be observer dependent. You cannot have a specific pair of instruments (scales in a rocket towing weights on strings of different lengths) that have different readings for different observers.

 

[pervect]

I think my current best guess at a "string" is that it is frame dependent in that we assume it has T^00 is zero in some frame. If T^00 isn't nonzero, it contributes to the force via its weight, which we don't want.

But it's under tension, so the pressure terms are nonzero. The non-zero pressure/ tension terms mean that if we transform the stress-energy tensor to some nonzero frame, T^00 won't be zero anymore in these other frames. This is similar to Rindler's comment about the "mass" of bar under pressure or tension.

So I think a strings definition winds up being frame dependent. At least mine does.

I haven't worked it all out, but I'm currently thinking a "string" is something that obeys $\nabla_a T^{ab} = 0$ and has $T^{00}$ zero, or at least small enough to be ignored.

 

[PAllen]

I think my current best guess at a "string" is that it is frame dependent in that we assume it has T^00 is zero in some frame. If T^00 isn't nonzero, it contributes to the force via its weight, which we don't want.

But it's under tension, so the pressure terms are nonzero. The non-zero pressure/ tension terms mean that if we transform the stress-energy tensor to some nonzero frame, T^00 won't be zero anymore in these other frames. This is similar to Rindler's comment about the "mass" of bar under pressure or tension.

So I think a strings definition winds up being frame dependent. At least mine does.

I haven't worked it all out, but I'm currently thinking a "string" is something that obeys $\nabla_a T^{ab} = 0$ and has $T^{00}$ zero, or at least small enough to be ignored.

This is an explanation of how "my string" is different from "your string". This is true for any given string definition (including the simple one I used, where we assume SR not GR, and no gravity effects on or from the string). My comments are about given some definition of a rocket's string, there is no frame dependence on the behavior of the rocket's string [the rocket determines the frame for its definition] or predictions of what the rocket's string will measure.

If you and I are in relative motion (any type), my Born rigid objects are not the same as your Born rigid objects. But you don't disagree with me about how my Born rigid objects behave.

 

[jartsa]

These statements are all frame-dependent, since energy itself is.

Also, the idea of "energy goes to zero...where there is an event horizon" is incorrect even if we leave out the issue of frame dependence. For example, outgoing photons at the event horizon remain at the horizon forever, but they don't have zero energy in anyone's frame.

I did not say everything at event horizon has zero energy. I said something like:

if you pack some photons into a backpack, and start descending towards an event horizon, using a ladder, not dumping any mass-energy, then you feel the weight of the backpack increasing, while your own weight seems to increase even faster, at the event horizon the weight of the backpack is zero percents of your weight.

 

[PeterDonis]

if you pack some photons into a backpack, and start descending towards an event horizon, using a ladder, not dumping any mass-energy

This is not possible, at least not the way I think you are imagining it.

If you descend "using a ladder", this seems like you are imagining a very slow descent, which could be approximated as a series of static states at decreasing radius. This cannot be done without "dumping any mass-energy". The only way to descend without dumping any mass-energy is to freely fall.

then you feel the weight of the backpack increasing

If you descend slowly, as I just described, so that you are basically occupying a series of static states at slowly decreasing radius, then the proper acceleration you feel will increase, and so will the proper acceleration of the backpack; but they will both increase by the same amount. The proper acceleration of *any* object that is static at a given radius must be the same, by the equivalence principle.

while your own weight seems to increase even faster, at the event horizon the weight of the backpack is zero percents of your weight.

Not correct; see above. Are you actually doing math or analysis to obtain these results? Or are you just waving your hands?

(I'm not sure what you mean by "weight" here, btw. If you mean "force", you need to be more specific about whether you mean locally measured force or force at infinity. Your description is wrong in either case, but it's wrong in different ways depending on which kind of force you mean.)

 

[jartsa]

This is not possible, at least not the way I think you are imagining it.

If you descend "using a ladder", this seems like you are imagining a very slow descent, which could be approximated as a series of static states at decreasing radius. This cannot be done without "dumping any mass-energy". The only way to descend without dumping any mass-energy is to freely fall.



If you descend slowly, as I just described, so that you are basically occupying a series of static states at slowly decreasing radius, then the proper acceleration you feel will increase, and so will the proper acceleration of the backpack; but they will both increase by the same amount. The proper acceleration of *any* object that is static at a given radius must be the same, by the equivalence principle.



Not correct; see above. Are you actually doing math or analysis to obtain these results? Or are you just waving your hands?

(I'm not sure what you mean by "weight" here, btw. If you mean "force", you need to be more specific about whether you mean locally measured force or force at infinity. Your description is wrong in either case, but it's wrong in different ways depending on which kind of force you mean.)

I will rewrite the story:

A robot, carrying a bag of photons, descends towards an event horizon, using a ladder, charging its batteries with the energy released when descending.

The sensors on the robot measure an incresing weight of the photon bag and really fast incresing weight of the batteries.

When the robot approaches the event horizon, the local weight (mass times gravitational acceleration) of the batteries divided by the local weight of the other stuff that descented towards the event horizon, approaches infinity.

 

[PeterDonis]

A robot, carrying a bag of photons, descends towards an event horizon, using a ladder, charging its batteries with the energy released when descending.

Ok, I think I see what you're saying; see below for more details on my understanding of the scenario.

The sensors on the robot measure an incresing weight of the photon bag and really fast incresing weight of the batteries.

When the robot approaches the event horizon, the local weight (mass times gravitational acceleration) of the batteries divided by the local weight of the other stuff that descented towards the event horizon, approaches infinity.

Ok, you've defined what you mean by "weight"; now you need to define what you mean by "mass". I'm going to assume you mean rest mass, measured locally, since you use the term "local weight". I'm also going to assume that the battery has negligible locally measured rest mass when uncharged; i.e., all of the locally measured rest mass of the battery comes from the energy stored in it as charge.

So as you describe it, we start at infinity with a robot of locally measured rest mass m and a battery with zero locally measured rest mass; thus the total energy at infinity of the robot + battery is m. We end up at some finite radius r with the robot + battery system having the same total energy at infinity, m; however, as measured at infinity, this is now divided up into a robot of energy at infinity mV (where V is the "redshift factor" at radius r) and a battery of energy at infinity E = m(1 - V). The closer r is to the horizon, the smaller V is, and therefore the larger the ratio E / mV of the battery's energy at infinity to the robot's.

Locally, at radius r, we have a robot of rest mass m (no redshift), and a battery of rest mass E / V (energy at infinity divided by redshift factor) = m(1 - V)/V. So the ratio of battery rest mass to robot rest mass is the same.

Since the proper acceleration of both the robot and the battery is the same (since they are at the same radius r), the "local weight" of the battery will get larger compared to the "local weight" of the robot. (It's worth noting, though, that if r is closer to the horizon than 9/8 of the horizon radius, the only way for the robot and battery to remain static at the same radius is with a rocket engine or something similar, since no static equilibrium (like that of a planet) is possible; and we haven't included the energy expended by the rocket in our analysis.)

Now that I understand what you were getting at, I'm not sure how it relates to any other issues raised in this thread.

 

[pervect]

I will rewrite the story:

A robot, carrying a bag of photons, descends towards an event horizon, using a ladder, charging its batteries with the energy released when descending.

The sensors on the robot measure an incresing weight of the photon bag and really fast incresing weight of the batteries.

When the robot approaches the event horizon, the local weight (mass times gravitational acceleration) of the batteries divided by the local weight of the other stuff that descented towards the event horizon, approaches infinity.

This violates the principle of equivalence, so it's not what happens according to GR.

Let's suppose you have a super strong shell, containing some matter and antimatter. It will have some particular "weight".

Now you combine the matter and antimatter. The insides which used to contain matter, will now contain various exotic particles and a lot of photons. But the perceived weight won't change, as long as the super-strong shell is strong enough to contain the explosion (all the photons and exotic particles) - such as the bag your robot was carrying.

If you have two packs, one which contains the matter and antimatter, and the other which contains the exploded version, they will both have the same weight initially, and they will continue to have the same weight as you descend into the black hole.

This is true in GR, because of the principle of equivalence. While there are theories of gravity that don't follow this principle, most of them have been falsified. As I recall, many of them were falsified just because they DIDN'T follow said principle, which has been tested to a high degree of accuracy. But I'd have to look it up to be 100% positive.

 

[PeterDonis]

This violates the principle of equivalence

I'm not sure it actually does. Basically he is describing a mechanism for transferring energy at infinity from the robot to the batteries, keeping the total energy at infinity of the robot + battery system constant. The latter is what you are insisting on (correctly); however, energy at infinity is not the same as his usage of the term "weight", which he is defining as proper acceleration times locally measured (i.e., *not* "redshifted") rest mass.

Another way of seeing that his process doesn't violate the EP is to consider an alternative process that arrives at the same end state: we let the robot + battery system free fall from infinity to some finite r; at that finite r, we stop the robot + battery with an apparatus that captures all of its kinetic energy, bringing it to rest locally, and stores the captured energy in the battery.

 

[PAllen]

Pervect was talking about the bag of photons, and I agree with his argument. As for the battery, if you have some mechanism where as robot and battery are lowered, work done by them on some 'generator' as they are lowered via cable (for example) [thus converting PE to electrical energy], and this energy transferred to battery, then battery will increase in weight compared to robot (measured locally to the robot).

Alternatively, imagine free fall. everything has a lot of KE relative to some stationary observer. Imagine a magic process to convert all of said KE to energy of the battery, while stopping robot. Again, battery increases in weight compared to robot. However, bag of photons does not.

 

[PeterDonis]

Pervect was talking about the bag of photons, and I agree with his argument.

I agree too, as far as the bag of photons is concerned: basically, the bag of photons will behave the same as the robot in the robot-battery scenario.

Alternatively, imagine free fall. everything has a lot of KE relative to some stationary observer. Imagine a magic process to convert all of said KE to energy of the battery, while stopping robot. Again, battery increases in weight compared to robot. However, bag of photons does not.

Yes, this is what I described at the end of my last post, and I agree the bag of photons will act like the robot.

However, I also think that jartsa would agree that the bag of photons acts like the robot in these scenarios; I think that's the point he was making when he said the energy of the bag of photons goes to zero at the horizon. (He can correct me if I'm wrong, of course.) It's just saying that, by lowering an object (robot or bag of photons) closer and closer to the horizon, one can extract more and more of its energy at infinity and put it somewhere else (the battery just being one example of a somewhere else). This is true, but as I said a couple of posts ago, I don't see what it has to do with the other questions raised in this thread.

 

[jartsa]

I agree too, as far as the bag of photons is concerned: basically, the bag of photons will behave the same as the robot in the robot-battery scenario.
Yes, this is what I described at the end of my last post, and I agree the bag of photons will act like the robot.

However, I also think that jartsa would agree that the bag of photons acts like the robot in these scenarios; I think that's the point he was making when he said the energy of the bag of photons goes to zero at the horizon. (He can correct me if I'm wrong, of course.) It's just saying that, by lowering an object (robot or bag of photons) closer and closer to the horizon, one can extract more and more of its energy at infinity and put it somewhere else (the battery just being one example of a somewhere else). This is true, but as I said a couple of posts ago, I don't see what it has to do with the other questions raised in this thread.

Exactly!

And how is this related to anything?

Well let's see. I assume it would take energy to winch the robot up a short distance, I mean the force required woud not be zero. Rather the force would 1000 Newtons, if the robot weighed 1000 Newtons on the surface of the Earth, and the black hole has surface gravity of 1 g.

http://en.wikipedia.org/wiki/Surface_gravity#Surface_gravity_of_a_black_hole(the above mentioned winching up was perfomed from "infinity", using a long rope)

 

[PeterDonis]

I assume it would take energy to winch the robot up a short distance, I mean the force required woud not be zero. Rather the force would 1000 Newtons, if the robot weighed 1000 Newtons on the surface of the Earth, and the black hole has surface gravity of 1 g.

More precisely, if the proper acceleration required to "hover" at the radius, above the horizon, at which the robot starts being winched, redshifted to infinity, were 1 g. (This also assumes that the robot is winched up slowly enough that its motion can be approximated by a series of static states at gradually increasing radius, and that the distance through which the robot is winched is small enough that there is no detectable change in the redshifted proper acceleration.) No object can be winched up from the horizon itself, and the surface gravity is the redshifted proper acceleration at the horizon. The redshifted proper acceleration at any point above the horizon will be less.

I still don't see what this has to do with the rest of the thread.

 

[jartsa]

More precisely, if the proper acceleration required to "hover" at the radius, above the horizon, at which the robot starts being winched, redshifted to infinity, were 1 g. (This also assumes that the robot is winched up slowly enough that its motion can be approximated by a series of static states at gradually increasing radius, and that the distance through which the robot is winched is small enough that there is no detectable change in the redshifted proper acceleration.) No object can be winched up from the horizon itself, and the surface gravity is the redshifted proper acceleration at the horizon. The redshifted proper acceleration at any point above the horizon will be less.

I still don't see what this has to do with the rest of the thread.

Why did I say the winching distance must be short? Oh yes, I thought the gravity field is quite inform but the gravitating energy increases rapidly, so the force increases.

I forgot that "unifom" gravity field is not uniform.

So when everyting is taken into account the force is quite constant, over quite large distance, like there was some kind of force field that is quite uniform. Right?

Earlier in this thread me and PAllen were arguing about this matter.

 

[PeterDonis]

So when everyting is taken into account the force is quite constant, over quite large distance, like there was some kind of force field that is quite uniform. Right?

The term "uniform gravity field" is somewhat problematic, as you have seen. Strictly speaking, the term does not apply in the presence of an actual gravitating body; it applies to the apparent "gravity field" seen by a family of accelerated observers in flat spacetime that all maintain a constant proper distance from each other. The term "Rindler observers" is often used to describe such a family of observers. However, as you note, the acceleration felt by such a family of observers is actually not uniform; the observers "lower down" feel more acceleration than the ones "higher up".

In the presence of an actual gravitating body, tidal gravity is present, which makes the field of an actual gravitating body vary with distance in a different way than the apparent "field" seen by Rindler observers in flat spacetime. (Also, the variation with distance depends on the mass of the body, whereas there is only one possible variation with distance for the acceleration felt by Rindler observers in flat spacetime.) I believe the term "uniform field" was used to describe the flat spacetime case to emphasize the fact that there is no tidal gravity in flat spacetime; but it can be confusing to realize that even in the absence of tidal gravity, the "gravity field" seen by Rindler observers still varies with position.

Since you have set your scenario in the presence of an actual gravitating body, the criterion for being able to treat the force felt by an object as constant is that tidal gravity is negligible over whatever distance you are considering. The larger the mass of the body, the larger the distance over which the force can be treated as uniform.