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

[WannabeNewton]

Peter, the link DaleSpam gave explains exactly what you asked about regarding the required exit speed of the cable and the asymptotic approach of the exit speed towards that of light. See the section entitled "Free fall".

 

[PAllen]

Ah, ok, I agree this is certainly possible. The only complication I can see that we haven't covered is the question of how fast the cable would need to be paid out; would the speed of cable exiting the ship have to exceed the speed of light at some point before the object at the lower end of the cable hit the singularity? I haven't done any analysis about that, but I think it's a legitimate question.

The link Dalespam provided (which would cover only the extent to which you could approximate the supermassive BH horizon by a Rindler horizon) suggests:

1) If you allow the string string to play out faster and faster, tension need not grow, and exceeding the speed of light need never occur (just ever closer).

2) If you want to have a constant speed of feeding out string, you can easily get well below the horizon, but at some point your string will break (tension will become infinite)

 

[Dale]

The link Dalespam provided (which would cover only the extent to which you could approximate the supermassive BH horizon by a Rindler horizon) suggests:

That is a good point to remember. The Rindler horizon has no tidal stresses, the only stresses are due to accelerations in the inertial frame. A real black hole would have the acceleration effects as well as tidal effects, particularly for a very long rope.

 

[PeterDonis]

(which would cover only the extent to which you could approximate the supermassive BH horizon by a Rindler horizon)

It's not just that; when you're considering strings that might stretch down to the singularity, you *can't* approximate that with flat spacetime, no matter how massive the hole is. Tidal gravity goes to infinity at the singularity, even if it's negligible at the horizon.

1) If you allow the string string to play out faster and faster, tension need not grow

Agreed.

and exceeding the speed of light need never occur (just ever closer).

Not sure about this part. As I read Egan's analysis, he's saying the speed of payout is unbounded, which means it will at some point exceed the speed of light. I don't think he's saying the rate of payout will approach the speed of light asymptotically.

(And of course, as I noted above, Egan's analysis doesn't account for spacetime curvature between the horizon and the singularity, which will make the required increase in rate of payout even greater, and therefore, I think, cause it to exceed the speed of light even sooner.)

2) If you want to have a constant speed of feeding out string, you can easily get well below the horizon, but at some point your string will break (tension will become infinite)

Yes.

 

[PAllen]

Not sure about this part. As I read Egan's analysis, he's saying the speed of payout is unbounded, which means it will at some point exceed the speed of light. I don't think he's saying the rate of payout will approach the speed of light asymptotically.

(And of course, as I noted above, Egan's analysis doesn't account for spacetime curvature between the horizon and the singularity, which will make the required increase in rate of payout even greater, and therefore, I think, cause it to exceed the speed of light even sooner.)

The plot (just above the heading "String unreeled at a constant rate") seems to show the v(τ) for different initial conditions all have an asymptote of -1 = speed of light playout.

[edit: actually, the pure free fall with slack case is discussed above the heading "dust trail". It is abslutely clear and stated that: "and the velocity she needs to give the rope will asymptotically approach the speed of light"]

I agree that all of this analysis is only reliable for some modest size region near a supermassive BH horizon. Beyond that, new calculation would be required, and I would not hazard a guess about how they turn out (especially for the rope nearing the singularity).

 

[PeterDonis]

The plot (just above the heading "String unreeled at a constant rate") seems to show the v(τ) for different initial conditions all have an asymptote of -1 = speed of light playout.

Ah, ok, I see. The equation for v(τ) given above that plot bears this out; at large $\tau$ the factor $g(\tau)$ dominates everything else, so the ratio goes to 1, and it's always less than 1 because the numerator is always less than $g(\tau)$ while the denominator is always greater.

 

[DrGreg]

As I read Egan's analysis, he's saying the speed of payout is unbounded, which means it will at some point exceed the speed of light. I don't think he's saying the rate of payout will approach the speed of light asymptotically.

Egan refers to "rate of unreeling" rather than "speed", which I take to mean the rate at which the rest length of the unreeled rope increases with respect to the ship's proper time. That's not we'd call "speed" (limited by c) but "celerity" (unbounded).

 

[PeterDonis]

Egan refers to "rate of unreeling" rather than "speed", which I take to mean the rate at which the rest length of the unreeled rope increases with respect to the ship's proper time. That's not we'd call "speed" (limited by c) but "celerity" (unbounded).

Actually, on re-reading, he seems to refer to both things at different points in the article. In the specific plot PAllen was referring to, it looks like he does mean speed, but not unbounded--see my previous post in response to PAllen.

 

[jartsa]

The tug is propagating at the speed of sound in the rope, isn't it?
But why it would never reach the falling observer? It moves faster than the falling observer.

The person sitting in the rocket will see the outside universe becoming very old.

A person falling into a black hole does not see the universe outside becoming very old.

Therefore the tugs, that might be morse coded news about the old universe, do not reach the falling person.

 

[PeterDonis]

The person sitting in the rocket will see the outside universe becoming very old.

If they sit long enough, yes. But remember that light from the outside universe still takes time to get to them.

A person falling into a black hole does not see the universe outside becoming very old.

Only because they hit the singularity in a finite proper time and are destroyed. Up until that point, the same light that reaches the person sitting in the rocket will reach them eventually as well.

Therefore the tugs, that might be morse coded news about the old universe, do not reach the falling person.

They do until he reaches the singularity, when he gets destroyed. The event at which that happens does have a past light cone which only contains a portion of the entire spacetime, so yes, events outside that past light cone will never be seen by the falling person. But that doesn't prevent events within that past light cone from sending signals to the falling person.

 

[jartsa]

Not necessarily. The gravitational redshift depends on the person's height relative to the size of the hole; if the hole is large enough the redshift from feet to head will be very small, even if the acceleration required for the person to hover is very large.

I don't believe that. If a large force is felt, then a large energy loss of a climbing photon must be assumed, and therefore a large redshift.

However, the more important point is that the redshift is present *because* both the head and the feet are accelerating; if they are freely falling it is absent, regardless of altitude. See below.

I would say the redshift is there because things vibrate at different frequencies at different altitudes. And the redshift gradually disappears according to an abserver that starts falling. Let's say all body parts of the falling person start falling simultaneously.

A hovering observer's view is the following:
the information channel from the falling person's foot to his head gradually becomes
1: shorter
2: faster

The head is scooping up the information that was stored in the space between the head and the foot.
That's why, for the head, the gravitational time dilation of the foot seems to gradually disappear, as the head, according to a hovering observer, is gaining speed.

 

[PeterDonis]

If a large force is felt, then a large energy loss of a climbing photon must be assumed

Why do you think this must be the case?

I would say the redshift is there because things vibrate at different frequencies at different altitudes.

This is one way of looking at it, but it has limitations. One of which is that it invites incorrect inferences like the ones you are making. Another is that frequencies of vibration are frame-dependent, and the redshift can be characterized entirely in terms of frame-independent quantities.

And the redsift gradually disappears according to an abserver that starts falling.

No, this is incorrect. A freely falling observer sees zero redshift between his head and his feet, even if he has only been freely falling for an instant when the light is emitted from his feet. I explained why in my post #35.

Let's say all body parts of the falling person started falling simultaneously.

A minor point, but simultaneously according to whom? Simultaneity is frame-dependent. I assume you mean simultaneously according to the hovering observer, but it's good to be explicit.

A hovering observer's view is the following:
the information channel from falling person's foot to his head gradually becomes
1: shorter
2: faster

If by "information channel" you mean "distance", this is true, but as I've already said, it can't have anything to do with the disappearance of the redshift.

The head is scooping up the information that was stored in the space between the head and the foot.

Huh? This makes no sense. Information isn't being "stored" in a fixed location. It's traveling in light beams, moving outward from the feet towards the head. (And anyway, the redshift doesn't gradually disappear, so your explanation doesn't even get the facts right.)

 

[jartsa]

Why do you think this must be the case?

If I build an accelerometer, that works by measuring redshift, my device works in space, away from masses. If in gravity field my device measures different acceleration compared to an accelerometer that works by measuring the compression of a spring, then I have buid an equivalence prinsiple busting device.

No, this is incorrect. A freely falling observer sees zero redshift between his head and his feet, even if he has only been freely falling for an instant when the light is emitted from his feet. I explained why in my post #35.

The head, by accelerating, caused the photons between it and the foot to redshift. So there exist redshifted photons after the head has stopped accelerating. The freshly emitted photons are the ones that are not redshifted.

(This was a case of a freefalling person observerving a hovering person who starts freefalling)

A minor point, but simultaneously according to whom? Simultaneity is frame-dependent. I assume you mean simultaneously according to the hovering observer, but it's good to be explicit.

Yes the hovering observer.

 

[PeterDonis]

If I build an accelerometer, that works by measuring redshift, my device works in space, away from masses.

Huh? What kind of accelerometer are you talking about? There is no gravitational redshift away from masses.

If you mean Doppler shift, that isn't caused by acceleration, it's caused by relative velocity. How do you propose to design an accelerometer that works by measuring it?

The head, by accelerating, caused the photons between it and the foot to redshift.

Which doesn't apply if the head is not accelerating. Which it isn't if the person is free-falling.

So there exist redshifted photons after the head has stopped accelerating.

Huh? How does that work, when you just said the head causes photons to redshift by accelerating?

Anyway, your understanding of how gravitational redshift works is flawed:

The freshly emitted photons are the ones that are not redshifted.

No, the redshift doesn't occur when the photons are emitted. It occurs when the photons are detected. The photons don't "carry" the redshift with them.

 

[jartsa]

Huh? What kind of accelerometer are you talking about? There is no gravitational redshift away from masses.

If you mean Doppler shift, that isn't caused by acceleration, it's caused by relative velocity. How do you propose to design an accelerometer that works by measuring it?

When I said that very close to event horizon the redshift is very large, I said that because I thought the people, the experts, in the forum might not know such thing.

The proper acceleration very near the event horizon is very large, according to a hovering observer, that is hovering near the event horizon.

The equivalence prinsiple says that it is not possible to know whether you are in a closed chamber on a surface of a planet that has large gravitational acceleration, or in a windowless rocket that accelerates a lot.

If there's a large redshift in the rocket, there's a large redshift on the planet, if there's no redshift in the rocket, there's no redshift on the planet.

Here "redshift" means the reddening of light, according to an observer that is placed above the light source.

So does it seem probable now that a person hovering very close to an event horizon feels a large g force, and sees a large redshift, when looking at his feet?

 

[pervect]

As far as redshifts go, if we take a light beam falling from infinity, and it falls to the hovering observer, we do have a blueshift that approaches infinity. This is the same doppler shift that's being called "redshift" by jartsa

But if we look at the infalling observer, there are NO infinite doppler shifts. It requires a detailed calculation, but there is actually a redshift for an infalling observer, the doppler shift due to his velocity more than compensates for the doppler shift due to gravity.

The doppler shift depends on whether or not the ship is infalling - there may be infinite doppler shift (which some people interpret as time dilation) for the hovering spaceship, but this doesn't mean that there is infinite doppler shift for the ship falling into the black hole.

On a related note - "forces" were mentioned. specifically

If a large force is felt, then a large energy loss of a climbing photon must be assumed

The difficulty with this is worth more explanation.

Suppose we have a rocketship in deep space. And we mount accelerometers on the front and back of the ship. And suppose we accelerate the ship in a Born rigid manner.

Then the accelerometers on the front and back of the ship will have different readings. Which we might interpret as "tidal forces". I'm afraid I don't know of any better name to call them, so I'll just enclose them in scare quotes. This is very closely related to the Bell spaceship paradox. In the Bell case, we make the acceleration at the front and back of the ship equal, and we see that the ship pulls apart. In the Born rigid case, we must have the accelerations at the front and the back different to keep the motion Born rigid.

But if we look at the same ship, in flat space-time, there are no "tidal forces". All the accelerometers read zero.

Under normal circumstances the "tidal forces" induced by acceleration are negligible. Falling into a black hole is not one of those cases.

In fact we have the case that the "tidal forces" as defined by the difference in accelerometer readings, approach infinity for the hovering spaceship.

This does not mean that the "tidal forces" experienced by an infalling spaceship are infinite, however. In fact, the tidal forces experienced by the infalling spaceship can be made as low as desired, and the radial components of these forces will be -2GM/r^3. One can find a derivation of this in MTW (and many other GR textbooks). Page numbers in MTW on request.

Because M and r_s, r_s being the Schwarzschild radius, are proportional, the tidal forces at r=r_s will be proportional to 1/r_s^2 and/or 1/M^2. So they can be made as small as desired.

So there isn't any real mystery here, just some incorrect assumptions, the assumption that "tidal forces" are not affected by accelerations.

 

[PeterDonis]

When I said that very close to event horizon the redshift is very large

Redshift for what specific scenario? Redshift is not a property of a single location; it depends on the *difference* in altitude between two locations (more precisely, the difference in altitude between the locations of two observers each "hovering" at a constant altitude). See further comments below.

I said that because I thought the people, the experts, in the forum might not know such thing.

I can't speak for anyone else, but I certainly knew it. It's obvious from looking at the Schwarzschild line element; $g_{tt}$, which is the redshift factor, goes to zero as you approach the horizon. This is a well known property of Schwarzschild spacetime.

The proper acceleration very near the event horizon is very large, according to a hovering observer, that is hovering near the event horizon.

Yes, this is well known too.

The equivalence prinsiple says that it is not possible to know whether you are in a closed chamber on a surface of a planet that has large gravitational acceleration, or in a windowless rocket that accelerates a lot.

Or in a windowless rocket hovering at a constant altitude above a gravitating mass, yes.

If there's a large redshift in the rocket, there's a large redshift on the planet, if there's no redshift in the rocket, there's no redshift on the planet.

Yes, if the conditions are the same in both cases.

Here "redshift" means the reddening of light, according to an observer that is placed above the light source.

Yes.

So does it seem probable now that a person hovering very close to an event horizon feels a large g force, and sees a large redshift, when looking at his feet?

The amount of proper acceleration (or g force) depends on your altitude above the horizon. The amount of redshift depends on the *difference* in altitude between your head and your feet, compared to the mass of the hole.

 

[jartsa]

As far as redshifts go, if we take a light beam falling from infinity, and it falls to the hovering observer, we do have a blueshift that approaches infinity. This is the same doppler shift that's being called "redshift" by jartsa

But if we look at the infalling observer, there are NO infinite doppler shifts. It requires a detailed calculation, but there is actually a redshift for an infalling observer, the doppler shift due to his velocity more than compensates for the doppler shift due to gravity.

The doppler shift depends on whether or not the ship is infalling - there may be infinite doppler shift (which some people interpret as time dilation) for the hovering spaceship, but this doesn't mean that there is infinite doppler shift for the ship falling into the black hole.

On a related note - "forces" were mentioned. specificallyThe difficulty with this is worth more explanation.

Suppose we have a rocketship in deep space. And we mount accelerometers on the front and back of the ship. And suppose we accelerate the ship in a Born rigid manner.

Then the accelerometers on the front and back of the ship will have different readings. Which we might interpret as "tidal forces". I'm afraid I don't know of any better name to call them, so I'll just enclose them in scare quotes. This is very closely related to the Bell spaceship paradox. In the Bell case, we make the acceleration at the front and back of the ship equal, and we see that the ship pulls apart. In the Born rigid case, we must have the accelerations at the front and the back different to keep the motion Born rigid.

But if we look at the same ship, in flat space-time, there are no "tidal forces". All the accelerometers read zero.

Under normal circumstances the "tidal forces" induced by acceleration are negligible. Falling into a black hole is not one of those cases.

In fact we have the case that the "tidal forces" as defined by the difference in accelerometer readings, approach infinity for the hovering spaceship.

This does not mean that the "tidal forces" experienced by an infalling spaceship are infinite, however. In fact, the tidal forces experienced by the infalling spaceship can be made as low as desired, and the radial components of these forces will be -2GM/r^3. One can find a derivation of this in MTW (and many other GR textbooks). Page numbers in MTW on request.

Because M and r_s, r_s being the Schwarzschild radius, are proportional, the tidal forces at r=r_s will be proportional to 1/r_s^2 and/or 1/M^2. So they can be made as small as desired.

So there isn't any real mystery here, just some incorrect assumptions, the assumption that "tidal forces" are not affected by accelerations.

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)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.

 

[PeterDonis]

This does not mean that the "tidal forces" experienced by an infalling spaceship are infinite, however. In fact, the tidal forces experienced by the infalling spaceship can be made as low as desired, and the radial components of these forces will be -2GM/r^3. One can find a derivation of this in MTW (and many other GR textbooks). Page numbers in MTW on request.

Because M and r_s, r_s being the Schwarzschild radius, are proportional, the tidal forces at r=r_s will be proportional to 1/r_s^2 and/or 1/M^2. So they can be made as small as desired.

I notice that you switched here from "tidal forces" in scare quotes to tidal forces proper, with no scare quotes. I don't know if this was intentional, but it's correct. The tidal forces (no scare quotes) that you describe as being experienced by the infalling spaceship are in fact due to spacetime curvature (I would say they *are* spacetime curvature). The "tidal forces" (with scare quotes) that you describe earlier are not, since they can be present in flat spacetime. (I agree it's unfortunate that there is no standard term for "tidal forces" due to differences in proper acceleration from one place to another.)

However, the more important point, to me, is that these "tidal forces" due to differences in proper acceleration are also not the determiners of "gravitational redshift" (which I put in scare quotes because it is also present in flat spacetime in an accelerating rocket). The redshift is there because of proper acceleration, not because of differences in it; the easiest way to see this is to note that it is present in the Bell spaceship paradox as well--the spaceship in front will see light from the spaceship in back redshifted, even though they both have the same proper acceleration.

 

[WannabeNewton]

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.

How did you come to this conclusion? The gravitational field is only uniform in sufficiently small regions.

 

[PeterDonis]

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.

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.

 

[jartsa]

How did you come to this conclusion? The gravitational field is only uniform in sufficiently small regions.

Well I have heard, or read, that inside an accelerating rocket there's an uniform "gravity field".

Then of course I have to guess what that means.

Well obviously as accelerometers show different readings at different "altitudes", the only observer that might possibly observe an uniform gravity field, is an observer that uses a string and a weight to probe the gravity field, while staying in one place.

So uniform "gravity field" means uniform "gravity field" according to that observer.

 

[WannabeNewton]

Again, the gravitational field is only uniform in a sufficiently small region. If the observer is hovering some arbitrary distance above the black hole and is suspending in the Schwarzschild gravitational field a mass attached to an arbitrarily long string then the magnitude of the force he exerts on his end of the string is obviously dependent on the altitude above the black hole that the mass is suspended at (as well as the mass of the black hole) and varies significantly on a global scale.

 

[anorlunda]

Please allow me to phrase the question differently.

I am free falling inside the EH with my hand extended radially inward. At time TA, my hand is at radius R1 and my eye is at radius R2. A photon is emitted from my hand. It follows a geodesic.

My eye is also following a geodesic. My eye reaches radius R1 at a later time TB, but by then the photon is somewhere else with radius <R1.

The question: does a geodesic originating at R2 at time TA ever intercept a geodesic originating at R1 at time TA? (Where R1 is not equal to R2 and both R1 and R2 are inside the EH)

If yes, I can see my hand. If no, I can't.

[Whoops, I see a flaw in my own question. I presume that TA at R1 is simultaneous with TA at R2. Simultaneity is problematic. ]

 

[PAllen]

Please allow me to phrase the question differently.

I am free falling inside the EH with my hand extended radially inward. At time TA, my hand is at radius R1 and my eye is at radius R2. A photon is emitted from my hand. It follows a geodesic.

My eye is also following a geodesic. My eye reaches radius R1 at a later time TB, but by then the photon is somewhere else with radius <R1.

The question: does a geodesic originating at R2 at time TA ever intercept a geodesic originating at R1 at time TA? (Where R1 is not equal to R2 and both R1 and R2 are inside the EH)

If yes, I can see my hand. If no, I can't.

[Whoops, I see a flaw in my own question. I presume that TA at R1 is simultaneous with TA at R2. Simultaneity is problematic. ]

You would see your hand. In terms of r coordinate of SC interior metric, a timelike world line decreases r faster than outward directed light. In terms of the free fall coordinates, of a sufficient small region and time period, NOTHING is detectably different than a free faller in interstellar space.

 

[jartsa]

Again, the gravitational field is only uniform in a sufficiently small region. If the observer is hovering some arbitrary distance above the black hole and is suspending in the Schwarzschild gravitational field a mass attached to an arbitrarily long string then the magnitude of the force he exerts on his end of the string is obviously dependent on the altitude above the black hole (as well as the mass of the black hole) and varies significantly on a global scale.

Ok.

The force is about the same one millimeter above the horizon and two millimeters above the horizon, if one millimeter can be considered a small distance in this gravity field.

 

[PAllen]

Well I have heard, or read, that inside an accelerating rocket there's an uniform "gravity field".

Then of course I have to guess what that means.

Well obviously as accelerometers show different readings at different "altitudes", the only observer that might possibly observe an uniform gravity field, is an observer that uses a string and a weight to probe the gravity field, while staying in one place.

So uniform "gravity field" means uniform "gravity field" according to that observer.

You have to be careful what you mean for a uniformly accelerating rocket. If all parts have the same proper acceleration, then all accelerometers in the rocket will show identical readings. However, such a rocket will soon tear itself apart, per Bell Spaceship paradox. On the other hand, if the rocket is undergoing Born rigid acceleration (or close to it), then the front will have slightly less proper acceleration (and lower accelerometer reading) than the back. [These comments are for flat spacetime, not for the case of rocket hovering near a BH. However, near a supermassive BH, the result would be essentially identical].

 

[jartsa]

You have to be careful what you mean for a uniformly accelerating rocket. If all parts have the same proper acceleration, then all accelerometers in the rocket will show identical readings. However, such a rocket will soon tear itself apart, per Bell Spaceship paradox. On the other hand, if the rocket is undergoing Born rigid acceleration (or close to it), then the front will have slightly less proper acceleration (and lower accelerometer reading) than the back. [These comments are for flat spacetime, not for the case of rocket hovering near a BH. However, near a supermassive BH, the result would be essentially identical].

An astronaut accelerating using a backpack rocket, holding two strings with different lengths, identical lumps of iron attached to the ends of the strings, will report that same force is pulling both strings.

An observer that sees the strings Lorentz contracting, will say that a larger force is pulling the object attached to the longer string.

 

[PAllen]

An astronaut accelerating using a backpack rocket, holding two strings with different lengths, identical lumps of iron attached to the ends of the strings, will report that same force is pulling both strings.

An observer that sees the strings Lorentz contracting, will say that a larger force is pulling the object attached to the longer string.

False. The reading on a given instrument is never observer dependent. It is either / or for both observers, as I explained:

- if the two accelerometers read the same, the string will soon break.
- if the string never breaks, the accelerometers will not read the same.

These are coordinate and observer independent realities. Note, I have used no coordinate quantities, only direct measurements in my description.

 

[jartsa]

False. The reading on a given instrument is never observer dependent. It is either / or for both observers, as I explained:

- if the two accelerometers read the same, the string will soon break.
- if the string never breaks, the accelerometers will not read the same.

These are coordinate and observer independent realities. Note, I have used no coordinate quantities, only direct measurements in my description.

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?