Looking closely at the descent into a black hole

[.Scott]

TL;DR

I'm taking along a "gravitational altimeter" and trying to make sense of what I would see.

So, as with the many fictitious characters that have preceded me, I am going to drop into a black hole - and I will bring along Alice, that "outside observer" who hangs around black holes waiting for "Bob"s to drop in.

I am going to take along a gravitational altimeter. And I'm going to use a gravitational altimeter design that I believe is easier to model in the BH environment. That altimeter will just be side-by-side pendulums of equal length. At the top they will be a fixed distance apart (symbol W). At the bottom, they will be very, very slightly closer - by a distance dH. And the length of the pendulums will be of equal length L. So, my altitude above the center of whatever mass I am hovering over will be r=L dH / H.
For convenience, I will calibrate my altimeter so the readout (R) is expressed in Schwarzschild radii. So, R = r/S = L dH /(SH).
I do not believe this altimeter will be reporting my areal radius as seen by an outside observer. I will call that "A" (as in Alice's Areal report of my Altitude).

I will descend in luxury - using a Dyson sphere with these specialized accoutrements:
1) It will be made of that sturdy unobtainium alloy that exhibits a speed of sound that matches the speed of light.
2) It will have an adjustable diameter - so, at least initially, I will be able to descend towards the even horizon steps and wait at each step.
3) My box seat will be along the interior surface of this sphere - providing me with a full view of the black hole, the interior of the Dyson, and any other box seats.

When the Dyson sphere is very large compared to the event horizon, then (except for its mass) the effects of the black hole are negligible. And from my box seat, I see this:

I can see the pendulums from all the other box seats and I can see them all pointing directly at the black hole.
But as I descend further and approach A=1.5, the photon sphere, things no longer fall so simply into place.
To see what happens to those pendulums, imagine how the interior surface of the Dyson sphere would appear if it was allowed to descend that far.
At A=1.5, light would be circling the black hole and traveling across the interior surface of the Dyson sphere. So all of the box seats would appear to me to be resting on a wide plane. All of those pendulums would be pointing in the same direction, at a right angle to that plane, as if the black hole was at infinity. And similarly, my gravitational altimeter would be pinned at infinity. So, at A=1.5, R is pinned.

Also at A=1.5, it does not appear to me that there are any compression forces on the Dyson sphere at all. Each part of the sphere has its own unobstructed trajectory towards the black hole that does not conflict with any other part. And, if that's the case, the diameter of the Dyson sphere will not have the same control over the altitude above the black hole that is did when A>1.5.

The problem I have is with that altimeter. It is never going to read between R=0 to about R=3. I believe it will bottom out at a reading of roughly R=3 when A is about 2.

And if I go beyond A, my altimeter goes negative (R<0) - as if the mass of the black hole was in back of me.

And about that altimeter design. The conventional gravitational altimeters measure the difference between gravitational accelerations between a higher and lower part of the instrument. Would they give the same readings as my pendulum-style meters? This is a question for someone who understands GR better than me - but I can't see how they could be different. If that's the case, if you survive A=2 in your drop into the BH, you should have just survived the worse of the speghetification forces.

 

[PeterDonis]

That altimeter will just be side-by-side pendulums of equal length. At the top they will be a fixed distance apart (symbol W). At the bottom, they will be very, very slightly closer - by a distance dH. And the length of the pendulums will be of equal length L. So, my altitude above the center of whatever mass I am hovering over will be r=L dH / H.

What the "altimeter" is actually measuring is the deviation of radial paths from being parallel. That's really a measure of radial tidal gravity; it's only indirectly a measure of altitude since radial tidal gravity is a known function of altitude. But I'm not sure your function $r = L dH / H$ captures that.

At A=1.5, light in a purely tangential direction would be circling the black hole and traveling across the interior surface of the Dyson sphere.

See my addition in bold above. Light is not constrained to travel only tangentially on the photon sphere. So I don't think your description of what you would see there is correct.

at A=1.5, it does not appear to me that there are any compression forces on the Dyson sphere at all.

That's not correct. There is still nonzero proper acceleration on every piece of the sphere, pushing outward.

 

[Ibix]

At A=1.5, light would be circling the black hole and traveling across the interior surface of the Dyson sphere. So all of the box seats would appear to me to be resting on a wide plane. All of those pendulums would be pointing in the same direction, at a right angle to that plane, as if the black hole was at infinity.

I don't think so. You are correct that you can (in principle) see infinitely many arrow tails, but if you look at an arrow tip your light path is inside the photon sphere and will reach the event horizon in short order. So you will see the arrows shrinking rapidly with distance, and gravitational lensing will, I suspect, affect their apparent verticality.

And similarly, my gravitational accelerometer would be pinned at infinity.

It's not clear to me how you are measuring the distance between your pendulums. I presume you're imagining it as the proper distance between pivots/weights along a surface of constant $r$. That will be less at the weight than the pivot at all finite altitudes, including around the photon sphere. The only way it reads zero is if it's not sensitive enough to detect the deflection, in which case it will measure zero at all higher altitudes too.

 

[PeterDonis]

if you look at an arrow tip your light path is inside the photon sphere and will reach the event horizon in short order

The light path from an arrow tip to an observer right on the photon sphere will be moving radially outward if the arrow is near enough to the observer (and with some tangential component unless it's the tip of the arrow right where they're standing). Such light paths don't necessarily fall into the horizon. For an arrow far enough around the photon sphere, of course, the light path to the observer will be radially inward and will fall into the horizon.

 

[Ibix]

The light path from an arrow tip to an observer right on the photon sphere will be moving radially outward if the arrow is near enough to the observer (and with some tangential component unless it's the tip of the arrow right where they're standing). Such light paths don't necessarily fall into the horizon. For an arrow far enough around the photon sphere, of course, the light path to the observer will be radially inward and will fall into the horizon.

Yes - sorry - I got some time reversal symmetry issues in that sentence.

Assuming the observer is on the photon sphere with arrow tails, rays from arrow tips (inside the photon sphere) will reach the observer and continue to infinity, but back-tracing them will lead to the event horizon. This means that the further away an arrow is the shorter it is, more than the usual perspective effect familiar in everyday life.

 

[PeterDonis]

Assuming the observer is on the photon sphere with arrow tails, rays from arrow tips (inside the photon sphere) will reach the observer

This is only true, as I said, for arrows close enough to the observer.

For arrows somewhat further around the photon sphere, the "naive" light path from the arrow tip to the observer starts out going radially inward (think of a chord to a circle), and those light rays will fall into the horizon instead of reaching the observer (because a radially inward light path inside the photon sphere will never change to radially outward). However, there will be a light path that goes radially outward, reaches an altitude somwhat above the photon sphere, and then falls back down to the observer. So the observer will see those arrows in a different direction from arrows closer together, and it will be a counterintuitive direction.

For arrows far enough around the photon sphere, I think there will be no light paths to the observer because they will be blocked by the hole. Observers further out from the hole will see light rays from the photon sphere on the opposite side, gravitationally lensed by the hole, but an observer on the photon sphere itself is too close to the hole to see lensed light. At least, I think that's the case.

 

[.Scott]

What the "altimeter" is actually measuring is the deviation of radial tangential paths from being parallel. That's really a measure of radial tangential tidal gravity; it's only indirectly a measure of altitude since radial tangential tidal gravity is a known function of altitude. But I'm not sure your function $r = L dH / H$ captures that.

Then "Tangential tidal gravity" it is. The equation is derived from drawing a triangle that goes across the tops of the two local pendulums and then down their length to the intersection point - which would be presumed to be the center of the gravitational mass. $r/L = dH/H$ just by similar triangles. I realize that this is the local "feels like" altitude, but it should also be an indication of the horizontal "roominess".

See my addition in bold above. Light is not constrained to travel only tangentially on the photon sphere. So I don't think your description of what you would see there is correct.

I did not state that well. What I meant was that the only light that travels from any pendulum to me will follow a path that hugs the interior wall of the Dyson sphere. So, all such images will appear to approach me from a single plane. It would be the view I would have on a flat surface of other objects on that same flat surface. I believe the distortions I would see in these distant pendulums would be conformal. If so, they would all appear to me to be pointing orthogonality from that flat surface toward what they saw as the center of mass. I understand that if I look above or below that plane, I will see the interior wall of the Dyson sphere or the black hole, resp.

That's not correct. There is still nonzero proper acceleration on every piece of the sphere, pushing outward.

Certainly from Alice's view, the Dyson sphere is moving into more and more cramped conditions - so she would certainly agree it is being compressed and that compression is what is holding back (or slowing down) the fall.

But I do not see any local mechanism for it.
As soon as I cross below A=1.5, I see those tidal effects causing my neighbors to fall away from me - not towards me. And nothing that I can do with the Dyson sphere's radius (except perhaps reducing its radius) would slow me down.

 

[.Scott]

I don't think so. You are correct that you can (in principle) see infinitely many arrow tails, but if you look at an arrow tip your light path is inside the photon sphere and will reach the event horizon in short order. So you will see the arrows shrinking rapidly with distance, and gravitational lensing will, I suspect, affect their apparent verticality.

I had two purposes in describing what I see of other pendulum sets: The first is to provide an overall sense of the experiment. The second was to explain why my gravitational altimeter would be reading (or inferring from tidal effects) a much larger altitude than Alice. It's easier to see where the effect comes from when considering a path along the photon sphere that stretches for a large distance. Once that effect is explain for distances from box to box (which might be thousands of KM), I'm hoping it would be easier to see it a single pendulum pair (which I would imagine to be tens of cM). Since my focus is on local conditions, effects that build up as the photons circle multiple times are not critical.

It's not clear to me how you are measuring the distance between your pendulums. I presume you're imagining it as the proper distance between pivots/weights along a surface of constant $r$. That will be less at the weight than the pivot at all finite altitudes, including around the photon sphere. The only way it reads zero is if it's not sensitive enough to detect the deflection, in which case it will measure zero at all higher altitudes too.

This part is very critical to my overall questions. The two pendulums are hanging from two points firmly attached to the chassis of the meter. That is the fixed horizontal distance "H". From each of those points, I am hanging a pendulum of fixed length L. Then I measure the distance between the bottom ends of the pendulums. That distance will be approximately the same as the distance on the top, H. But, under normal conditions like hovering over a planet, there will be a difference dH (delta H). I am defining dH to be the shortening (vs lengthening) of the distance H because, because under normal conditions they would be pointing directly at a common point in the distance (the center of the gravitational mass of the planet) and therefore deflected slightly towards each other.
But in the vicinity of a black hole, there are local relativistic effects. Light will be traveling along a path that curves inward towards the black hole - and this will affect the measurement of dH as well as the tangential tidal gravity.
I am not concerning myself with the engineering involved. Certainly there would be issues with the sensitivity of the instrument, but I am presuming they are perfect or perfect enough.

 

[PeterDonis]

the intersection point - which would be presumed to be the center of the gravitational mass

There is no such thing for a black hole. $r = 0$ for a black hole is not a place in space; it's a moment of time. The geometric picture you are describing is simply not valid for a black hole.

the only light that travels from any pendulum to me will follow a path that hugs the interior wall of the Dyson sphere.

This is not correct, because the pendulum tips are not on the photon sphere; they are a little bit inside it.

I do not see any local mechanism for it.

I have no idea what you're talking about. The proper acceleration of the worldlines describing each little piece of the Dyson sphere are dictated by your problem specification. Something has to be pushing on each piece of the sphere to provide that proper acceleration. That something would be each little local piece of the sphere being pushed on by neighboring pieces, providing a net outward proper acceleration that holds the sphere in place against gravity. If that's not possible, then the sphere can't even be constructed in the first place. And if it is possible, then it doesn't suddenly stop working at the photon sphere.

As soon as I cross below A=1.5, I see those tidal effects causing my neighbors to fall away from me

Um, what? I have no idea where you're getting this from. Two objects separated tangentially in Schwarzschild spacetime have tidal gravity causing them to converge, not diverge, at any value of $r$.

I think you need to do some math instead of waving your hands.

 

[PeterDonis]

Light will be traveling along a path that curves inward towards the black hole - and this will affect the measurement of dH as well as the radial tidal gravity.

I don't understand this. You can measure dH locally: just put a ruler between the bottom ends of the two pendulums. And I don't see any basis for a claim that light paths affect tidal gravity.

 

[PeterDonis]

That's really a measure of radial tidal gravity

Actually, I was mistaken here; it's a measure of tangential tidal gravity, since the two pendulums are separated tangentially, not radially.

 

[PeterDonis]

All of those pendulums would be pointing in the same direction, at a right angle to that plane, as if the black hole was at infinity.

Even if we grant that things would appear this way to an observer looking only at light signals, this...

And similarly, my gravitational accelerometer would be pinned at infinity.

...is not correct. Two pendulums separated by a small amount tangentially, and both hanging radially, will still have a small convergence at the bottom as compared with the top. Or, to put it another way, tangential tidal gravity does not go to zero at the photon sphere. The math is clear on that; in the orthonormal basis of an observer "hovering" at areal radius $r$, the tangential tidal gravity is $M / r^3$, where $M$ is the mass of the hole and the positive sign means convergence.

 

[.Scott]

There is no such thing for a black hole. $r = 0$ for a black hole is not a place in space; it's a moment of time. The geometric picture you are describing is simply not valid for a black hole.

I am simply computing what report you would get from the gravitational altimeter.
My point is only that the gravitational altimeter isn't reporting the tangential tidal effects that would be expected from a regular gravitational mass.
To avoid the missing r<1, areal radius values are used. What I am wondering how "functional" is this apparent areal radius. For example, should speghetification be calculated based on Alice's view or on this altimeter?

This is not correct, because the pendulum tips are not on the photon sphere; they are a little bit inside it.

That's fine. I was trying to work around the exact A=1.5 case anyway. It is a peculiar case. I have a done a schematic for A=2.1.

I have no idea what you're talking about. The proper acceleration of the worldlines describing each little piece of the Dyson sphere are dictated by your problem specification. Something has to be pushing on each piece of the sphere to provide that proper acceleration. That something would be each little local piece of the sphere being pushed on by neighboring pieces, providing a net outward proper acceleration that holds the sphere in place against gravity. If that's not possible, then the sphere can't even be constructed in the first place. And if it is possible, then it doesn't suddenly stop working at the photon sphere.

I had thought that it was already accepted that a Dyson sphere below the photon sphere was doomed because the speed of sound required to support it would exceed the speed of light - although Google is totally failing my attempts to get a reference. I am not challenging those world lines. But I do wonder if the local view of those worldlines appear to converge. Once we cross below A=1.5, the problem gets more involved. So, at this point I will just stick to the question about the local view of those lines well above A=1.5.

Um, what? I have no idea where you're getting this from. Two objects separated tangentially in Schwarzschild spacetime have tidal gravity causing them to converge, not diverge, at any value of $r$.

As I said, I don't like the A<1.5 case, because as soon as you cross A=1.5, I am uncertain how to interpret the numbers. For one thing, neighboring pendulum pairs become much harder to view.

I think you need to do some math instead of waving your hands.

The math is easy at A=1.5, that's why I described it. One critical assumption I am making is that my distorted view of distant pendulum pairs is a conformal distortion. If it is, then in my local view (at A=1.5) all pendulums hang parallel to each other - or at least that in the limiting case where the pendulum separation is 0, the dH is zero even when it is expressed as a ratio of the pendulum distance. The reason that is critical, is that I am presuming that the two pendulums in my local pair see each other using the same rules are they use when being compared to the more distant ones. So if my side view of distant pairs does not appear to be straight on, then I have no reason to expect that my local pairs will appear to be parallel to each other (at A=1.5).

 

[.Scott]

I don't understand this. You can measure dH locally: just put a ruler between the bottom ends of the two pendulums. And I don't see any basis for a claim that light paths affect tidal gravity.

But I want to take into account local relativistic effects. So I used the method that I described in my last post. That is my basis.

 

[PeterDonis]

I am simply computing what report you would get from the gravitational altimeter.

As far as I can see, you're not computing anything. You haven't shown any math. You're just waving your hands.

the gravitational altimeter isn't reporting the tangential tidal effects that would be expected from a regular gravitational mass.

At this point I don't understand what you think your gravitational altimeter is "reporting".

I had thought that it was already accepted that a Dyson sphere below the photon sphere was doomed because the speed of sound required to support it would exceed the speed of light

That's not correct. The value of $r$ at which this happens is $9/8$ of the Schwarzschild radius of the hole, not $3/2$ of it, which is where the photon sphere is. The result that shows this is called Buchdahl's Theorem.

The math is easy at A=1.5, that's why I described it.

Given that you have made at least two incorrect claims about what happens at A = 1.5, I am not confident that you are using the correct math. And since you haven't shown the actual math you are using, there is no way for me or anyone else to check.

One critical assumption I am making is that my distorted view of distant pendulum pairs is a conformal distortion

I don't know what you mean by this, or what math you think it implies.

I am presuming that the two pendulums in my local pair see each other using the same rules are they use when being compared to the more distant ones.

I have no idea what you mean by this either. It seems to me that you are confusing the effects of light propagation with actual local distances. They're not the same thing.

I want to take into account local relativistic effects.

What do you mean by this?

 

[.Scott]

Even if we grant that things would appear this way to an observer looking only at light signals, this...
...is not correct. Two pendulums separated by a small amount tangentially, and both hanging radially, will still have a small convergence at the bottom as compared with the top. Or, to put it another way, tangential tidal gravity does not go to zero at the photon sphere. The math is clear on that; in the orthonormal basis of an observer "hovering" at areal radius $r$, the tangential tidal gravity is $M / r^3$, where $M$ is the mass of the hole and the positive sign means convergence.

Sorry, I meant gravitational altimeter, not gravitational accelerometer. (I will fix that) But you seem to have interpreted correctly.

If what you are saying is right, then I should be able to see that - no matter how far apart the pendulums are. It means that there is some break down in my logic.

 

[PeterDonis]

If what you are saying is right, then I should be able to see that

If by "see" you mean purely by analyzing the light rays coming from the pendulums and looking at things like angles subtended in your visual field, I have no idea why you would think this. There is no guarantee whatever that an observer at a distance will be able to see, in this way, the same things that are being measured by local rulers.

 

[.Scott]

Here's a schematic of the A=2.1 case.

The red is a rough estimate of what someone on the inner side of the Dyson sphere ("Bob") would draw based on what he sees outside his window. It is schematic, but the key feature is that because of the curved space time geometry, that inside surface would appear flatter that expected. If that is true, and those pendulums still appear to hang perpendicular to that inner Dyson surface (a conformal distortion as viewed by "Bob"), then the altimeter will compute a higher altitude for Bob.

 

[.Scott]

If by "see" you mean purely by analyzing the light rays coming from the pendulums and looking at things like angles subtended in your visual field, I have no idea why you would think this. There is no guarantee whatever that an observer at a distance will be able to see, in this way, the same things that are being measured by local rulers.

That's exactly what I mean. And if the local guy is seeing different angles, he will be seeing different tidal effects. But does he see meaningfully different angles? That A=2.1 diagram I just posted shows my reasoning. But there may very well be holes in it.
As far as why I would think that, how else is he going to measure the local effect?

 

[Ibix]

However, there will be a light path that goes radially outward, reaches an altitude somwhat above the photon sphere, and then falls back down to the observer.

This cannot happen for a Schwarzschild black hole, because for a null geodesic $\frac{d^2r}{d\lambda^2}=\frac{L^2}{r^4}\left(r-\frac 32r_S\right)$, which means that there are no maxima of $r$ on null paths at or above $3r_S/2$. Light can only cross the photon sphere once, either inbound or outbound.

There are light paths just below the photon sphere that can do a lot of orbits (loosely speaking, because the required velocity for a circular orbit is only very slightly above $c$), so you can see arbitrarily many of the pivots of the pendulums if you sit on the photon sphere. But light from the pendulum bob starts out below the photon sphere, so in order to reach you it needs to be on a trajectory that either grazes the photon sphere at your location or passes through it outwards. The optical effects will be weird.

 

[Ibix]

The second was to explain why my gravitational altimeter would be reading (or inferring from tidal effects) a much larger altitude than Alice.

It shouldn't be. The arc length between two points at equal altitude is $r\int_{\phi_0}^{\phi_1}\ d\phi$, as you can read off the Schwarzschild metric. Your pendulum bobs have the same $\phi$ coordinates as their respective pivots, so the proper distance between the bobs is smaller because $r$ is smaller. Thus the ruler distance between the bobs is always less than the ruler distance between the pivots.

Working out exactly what you would get for something like radar distance between the bobs versus the pivots is fairly messy. You could approximate it with Rindler coordinates in flat spacetime, however, by considering your two pendulums to be placed at fixed $y$ coordinates and undergoing Rindler motion in the $+x$ direction. For simplicity consider the echo to happen at $t=0$ and compute the Rindler time between pulse emission and echo reception as a function of the proper acceleration of the Rindler observer. I strongly suspect this decreases monotonically as the proper acceleration increases - i.e., as you get closer to the horizon the radar distance between nearby bobs is always lower than that between their pivots.

 

[A.T.]

... based on what he sees outside his window ...

You seem to be conflating visual effects (how the pendulums look from a distance in curved space time) with what your altimeters actually measure locally all around the sphere (which could be in principle read out locally without any use of light signals). I think you should concentrate on the latter, and leave the visual distortions for later, once the local measurement is clear.

 

[.Scott]

You seem to be conflating visual effects (how the pendulums look from a distance in curved space time) with what your altimeters actually measure locally all around the sphere (which could be in principle read out locally without any use of light signals). I think you should concentrate on the latter, and leave the visual distortions for later, once the local measurement is clear.

I am conflating. If I try to do that with the two pendulums in my local pair, then I need to show that when I visit either of those two pendulums, I see it pointing directly at the BH. But when I look back across to it from the other pendulum, it appears to be off. In fact, it will be off by a little because the light is following a longer path through the non-Euclidean time-space ... but I believe there is another effect that really makes the difference. If instead of measuring the dH, I made the pendulums reflective and tried to bounce photons back and forth between them, then at A=1.5, it would show appear to be parallel - because the arced, non-Euclidean path of the photons would strike both pendulum surfaces at a right angle.
It's that same logic that makes me think that those other pendulums will appear to be hanging perpendicular to the apparent inner surface of the Dyson sphere.

 

[.Scott]

It shouldn't be. The arc length between two points at equal altitude is $r\int_{\phi_0}^{\phi_1}\ d\phi$, as you can read off the Schwarzschild metric. Your pendulum bobs have the same $\phi$ coordinates as their respective pivots, so the proper distance between the bobs is smaller because $r$ is smaller. Thus the ruler distance between the bobs is always less than the ruler distance between the pivots.

In my last post (responding to @A.T. ), I suggested an alternative method for measuring the pendulum angles. Using that second method, it's easy to make quick sense of the disappearing dH at A=1.5. If we have two different ways of making the same measurements and get two different answers, then there is more to explain. I am very uncertain of which explanation to use. After all, the interior surface of the Dyson sphere is acting like the ruler for the pivots, and if I look out the window at it, it is telling me that its full circumference of that inner surface is insufficient to wrap around the BH.

Working out exactly what you would get for something like radar distance between the bobs versus the pivots is fairly messy. You could approximate it with Rindler coordinates in flat spacetime, however, by considering your two pendulums to be placed at fixed $y$ coordinates and undergoing Rindler motion in the $+x$ direction. For simplicity consider the echo to happen at $t=0$ and compute the Rindler time between pulse emission and echo reception as a function of the proper acceleration of the Rindler observer. I strongly suspect this decreases monotonically as the proper acceleration increases - i.e., as you get closer to the horizon the radar distance between nearby bobs is always lower than that between their pivots.

Thanks! Sounds like a homework assignment. I will tackle that this week.

 

[Ibix]

In my last post (responding to @A.T. ), I suggested an alternative method for measuring the pendulum angles. Using that second method, it's easy to make quick sense of the disappearing dH at A=1.5.

But your analysis is wrong. If the Dyson sphere is at $1.5r_S$ then the pendulum bobs are at some smaller $r$. Thus in order to reflect light off a nearby bob, the emitting bob would have to aim upwards slightly and the reflector would need to be angled upwards sightly to reflect back to the source. If you aim horizontally you will miss low.

You seem to be approximating the pivot and bob as both being at $1.5r_S$, which is fine, but means you've approximated the tidal effects you're trying to measure as having zero effect on your measurement. That's why you're seeing zero effect, and you would also see zero effect at higher altitudes with an instrument with precision such that this approximation makes sense.

 

[A.T.]

.... makes me think that those other pendulums will appear ....

OK, but why does it matter how something appears visually? You can have markings next to the pendulums, indicating their actual relative angles.

Just like you can measure correct angles with a protractor, even when looking obliquely at the table so visually everything appears distorted.

 

[PeterDonis]

Here's a schematic of the A=2.1 case.

How are you obtaining this? Based on what math?

 

[PeterDonis]

because of the curved space time geometry, that inside surface would appear flatter that expected.

What math are you basing this on?

 

[PeterDonis]

if the local guy is seeing different angles, he will be seeing different tidal effects.

No, he won't. There is only one spacetime geometry. If the local guy thinks he's seeing different tidal effects, it's because he's not doing the calculations correctly based on his observations, to correct for whatever optical effects the light he's seeing goes through on its way to him.

 

[PeterDonis]

how else is he going to measure the local effect?

I told you how in post #10.