Looking closely at the descent into a black hole

[PeterDonis]

Light can only cross the photon sphere once, either inbound or outbound.

Ah, yes, I forgot about that feature of the effective potential for null geodesics. Serves me right for not following my own advice to look at the math.

 

[PeterDonis]

@.Scott multiple people now have told you that your hand-waving analysis is wrong and given reasons why. At this point you either need to show us some math to back up your claims, or drop them and re-do your analysis based on better premises, or this thread will be closed as we are just going around in circles.

 

[.Scott]

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.

The point I am making only relies on there being someplace (like $1.5r_S$) where the bobs will be pointing directly down, will reflect the light horizontally, and then those photons move on to the next bob. So, if the Dyson sphere is slightly too low, crank it up a half a meter and repeat the experiment.
As I understand it, photons on the photon sphere follow an unstable circular path that is always perpendicular to the vertical. So if you reflect from a vertical surface, it will reflect from that surface and start a new orbit that is always perpendicular to the vertical.

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.

The only reason I am picking $1.5r_S$ is that the math happens to be easy at that point. Everything there is vertical, the photons are following an easy path to describe, and if the next pendulum over appears to be parallel to the one next to me - or even if it just 50% closer to parallel than it "should" be, the basic problem is demonstrated and my follow-on questions stand.

 

[PeterDonis]

The point I am making only relies on there being someplace (like $1.5r_S$) where the bobs will be pointing directly down

The bobs are pointing radially inward by construction. That is what "directly down" would normally be taken to mean.

But the bobs still converge; their bottom ends are slightly closer together, in terms of tangential proper distance, than their top ends. @Ibix explained to you why. So if by "directly down" you mean that that is no longer true--that the bobs no longer converge because that somehow stops happening at the photon sphere--then that is simply wrong.

As I understand it, photons on the photon sphere follow an unstable circular path that is always perpendicular to the vertical.

But if you are an observer on the photon sphere, and you are looking at the bob of a pendulum that is suspended from the photon sphere, the bob will be slightly below the photon sphere, so light coming from it to your eye will not be traveling on such a path as you describe. @Ibix explained this to you as well.

So your whole analysis appears to be based on two mistaken premises. And you have shown no math to back up your claim that those premises are correct.

 

[.Scott]

In response to: "because of the curved space time geometry, that inside surface would appear flatter that expected".

What math are you basing this on?

That's just the circular trajectory of the photons in the photon sphere. For example, in a Dyson sphere with a small planet, a point on the interior of the sphere that is 90 degrees from my position on the sphere will cross through the interior of the sphere at a 45-degree angle from its source and land in my camera at a 45-degree angle from horizontal. In that same experiment with A=1.5, only photons that launch from the source horizontally will reach me - and those will come at me from the horizontal. The only math here is the geometry of what is happening with a circular orbit.
With A>1.5, the photons will still not follow the Euclidean path, and the launch and intercept angles range from the A=1.5 case (0 degrees) to the A=infinity case (45 degrees). That's the math.

In response to "Here's a schematic of the A=2.1 case."

How are you obtaining this? Based on what math?

That A=2.1 case is only intended as a schematic. It shows a transition step from the A=infinity top the A=1.5. It is intended to illustrate my line of thought, not to be precise. If I looked up the exact trajectory for photons at A=2.1, I could make that diagram more precise, but it would not add to this discussion.

 

[weirdoguy]

I still see no math, only bunch of words.

 

[.Scott]

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.

If it is purely an optical effect, then yes. But how do I know that's what it is? It seems to be a feature of the non-Euclidean space. This is really getting to the meat of my question.

 

[.Scott]

I told you how in post #10.

(how to measure dH)

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.

The reason I do not just "measure the top and bottom" is that I am unsure whether I would be properly considering relativistic effects. For example, if I stretched the ruler across space from one pair to the next, would that ruler follow the curvature that the photons are following? If there is some kind of simplicity to the mechanical measurements that addresses the non-Euclidean issues, then I am all ears.

 

[.Scott]

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.

OK. Then what kind of measurement should he do. Does the dH really go to zero? Or will the altimeter really work correctly? At the moment, you seem to be on the "dH is right - the optical measurement method is wrong". That's clearly possible - but I also think it's just a guess. Are you sure the measurement method is wrong? Why?

 

[PeterDonis]

That's just the circular trajectory of the photons in the photon sphere.

No, it's not. The claim you are making involves more than that.

It seems to be a feature of the non-Euclidean space. This is really getting to the meat of my question.

The "non-Euclidean space" does not have the effects you appear to think it has. See further comments below.

what kind of measurement should he do.

I already told you that in post #10. As I have already reiterated to you once.

Does the dH really go to zero?

No. @Ibix explained to you why quite a few posts ago. As I also reiterated to you.

? At the moment, you seem to be on the "dH is right - the optical measurement method is wrong". That's clearly possible - but I also think it's just a guess.

You are going way off into left field here. What you are calling dH is the "reality"--what local rulers measure. Any "optical measurement method" that does not give the same answer is indeed wrong--because, as I've already said once, whoever is making the measurement is not correctly calculating the optical effects on the light as it travels to him from whatever is being measured. I don't see why this should be an abstruse concept. It's no different from not concluding that a stick that is half immersed in water is actually bent, because it looks bent due to optical effects.

Since you have consistently refused to do any math, I'll do some, to show you what role the "non-Euclidean space" does and does not play in this scenario. We have two pendulums, each suspended from the photon sphere $r = 3M$, and each of proper length $L$. They are separated tangentially, at the top, by a proper distance $D$. We want to calculate their tangential separation at the bottom.

We use Schwarzschild coordinates, so the pendulums are separated tangentially by an angle $\Delta \phi$, and this angle does not change from top to bottom, because by hypothesis the pendulums are hanging purely radially. We put the pendulums on the "equator" of the photon sphere, so that at radial coordinate $r$, the tangential proper distance between the pendulums is $r \Delta \phi$. We know that at the top, $r = 3M$. We want to find $r$ at the bottom; as @Ibix told you quite a few posts ago, this $r$ will be less than $3M$, but we want to calculate how much less.

From the Schwarzschild line element, we have that for a purely radial curve, $ds = dr / \sqrt{1 - 2M / r}$. We plug in $r = 3M$ to get the factor in the denominator on the RHS (strictly speaking this is an approximation, but it's good enough for this problem). We know $ds$ for the pendulums--it's just the proper length $L$. So we have that $r$ at the bottom of each pendulum will be

$$
r = 3M - \frac{L}{\sqrt{3}}
$$

Note that this means that $dr$ is less than $L$--i.e., the difference in areal radius is less than the proper distance. That is the effect of the non-Euclideanness of the space, and it is the only such effect.

Using the fact that $\Delta \phi = D / 3M$, by applying $D = r \Delta \phi$ at the top of the pendulums, we find that, at the bottom of the pendulums, the tangential proper distance between them is

$$
\left( 3M - \frac{L}{\sqrt{3}} \right) \frac{D}{3M} = D \left( 1 - \frac{L}{3 \sqrt{3} M} \right)
$$

So the difference in tangential proper distance from top to bottom is

$$
dH = \frac{L D}{3 \sqrt{3} M}
$$

Note that the above calculation can be generalized to any $r$ above the horizon, and will always show a nonzero difference in tangential proper distance between the top and bottom of the pendulums.

 

[.Scott]

The bobs are pointing radially inward by construction. That is what "directly down" would normally be taken to mean.

But the bobs still converge; their bottom ends are slightly closer together, in terms of tangential proper distance, than their top ends. @Ibix explained to you why. So if by "directly down" you mean that that is no longer true--that the bobs no longer converge because that somehow stops happening at the photon sphere--then that is simply wrong.

No. My concern is that when I look out to the other pendulum pairs, what I see it not a simple 90-degree angle between the pendulum and the bottom of the Dyson sphere - that, for example, the actual view is foreshortened along some axis other than the horizontal or vertical. If that is the case, then that would spoil my entire construction because it would indicate that the same effect would influence my measurement between the two pendulums in my pair. If I saw that in the in the A=1.5 case, I would immediately agree that there is nothing more to be explained - that even if the altimeter is wrong, it could be wrong for inconsequential reasons. Like your suggested benign "optical effect".

But if you are an observer on the photon sphere, and you are looking at the bob of a pendulum that is suspended from the photon sphere, the bob will be slightly below the photon sphere, so light coming from it to your eye will not be traveling on such a path as you describe. @Ibix explained this to you as well.

This is just an issue of the experimental set-up. I can have my box seat below the bottom surface of the Dyson sphere so that when I sit down the photon sphere is at eye level and level with any part of the pendulum that is best to be observed.
But it doesn't have to be that critical. I can crank the whole experiment up a few meters to get a slightly better view of all the other bobs. They will still look like spread out horizontally - almost (but no longer exactly) in a flat horizontal surface. It would still be very apparent that if I want to find the spot on the Dyson sphere that is directly across from my position, I shouldn't look straight - I need to look off to the side. Do I really need to show the math that compares the A=1.5 view to the A=1.50001 view? If it adds to the discussion I will do that, but I don't think it would.

Once the curvature of the bottom of the Dyson sphere starts appearing less curved than it should be, then the problem is posed. How can that apparent distortion, (perhaps effecting our local altimeter) accomodate what we expected to find?

 

[.Scott]

@.Scott multiple people now have told you that your hand-waving analysis is wrong and given reasons why. At this point you either need to show us some math to back up your claims, or drop them and re-do your analysis based on better premises, or this thread will be closed as we are just going around in circles.

I will look at the "homework" that @Ibix gave me (which I think is promising.). And I will also look to get a better calculation for the A=2.1 case - although I do not think that will change this discussion in the least.

As for the "optical effect". There is not more math to that. It's just the light path - which is usually considered a pretty reliable ruler in indicating the relativistic effects.

 

[A.T.]

The reason I do not just "measure the top and bottom" is that I am unsure whether I would be properly considering relativistic effects.

And your way to avoid these potential local relativistic effects, is to instead rely on distant visual observations of completely distorted light signals? Sounds backwards to me. What is your goal with all that?

It's just the light path - which is usually considered a pretty reliable ruler in indicating the relativistic effects.

This is again exactly backwards. Relativistic effects is what is left, after you have corrected for the optical effects which affect the observation, like finite signal speed.

 

[PeterDonis]

the light path - which is usually considered a pretty reliable ruler in indicating the relativistic effects.

Please give a reference for this very surprising claim.

And please read the below from my previous post again--and again and again until it sinks in. I don't think you've grasped the crucial point.

I don't see why this should be an abstruse concept. It's no different from not concluding that a stick that is half immersed in water is actually bent, because it looks bent due to optical effects.

 

[PeterDonis]

Do I really need to show the math that compares the A=1.5 view to the A=1.50001 view?

The math I showed you in post #40 is the kind of math I was looking for from you. I strongly suggest that you read that post very carefully.

 

[.Scott]

And your way to avoid these potential local relativistic effects, is to instead rely on distant visual observations of completely distorted light signals? Sounds backwards to me. What is your goal with all that?

I am not avoiding them. In fact, some revelation about exactly how they are working might be the resolution to this. At least to me, those distant observations tell me something about the non-Euclidean environment that I am it - and alerts me to look for a discrepancy in my altimeter.
As I mentioned before, one thing I am interested in what kind of a distortion I am looking through. If the image is foreshortened in the wrong way, then it is near useless for my purpose.
Basically what my view out the window is telling me is that once I reach the photon sphere, gravity will no longer be pulling me into a collision course with the rest of the Dyson sphere. Its' like Alice can tell me that I'm going to get hammered, but when I look out the window at the hammer, it's not moving towards me and nothing seems to be pulling it towards me. I have no visual evidence of an impending crunch and no obvious mechanism to create such a crunch.

 

[PeterDonis]

At least to me, those distant observations tell me something about the non-Euclidean environment

Whatever they tell you is irrelevant to what you say you are trying to calculate. I showed you in post #40 how the non-Euclideanness affects what you say you are trying to calculate. It has nothing whatever to do with optical effects.

what my view out the window is telling me is that once I reach the photon sphere, gravity will no longer be pulling me into a collision course with the rest of the Dyson sphere.

And this is wrong. It amounts to saying that radial spatial lines no longer converge at the photon sphere. And that is wrong. I gave you an explicit calculation of the convergence in post #40.

At this point I strongly suggest that you stop posting until you have read and understood all of the feedback you have been getting. You have repeated wrong statements that have already been corrected, and we are getting close to the point where this thread will need to be shut down because it is going in circles.

 

[A.T.]

... but when I look out the window at the hammer, it's not moving towards me ...

Where did you get that from? The visually perceived angle between the pendulums is a different matter from whether they appear moving towards the observer.

 

[.Scott]

Where did you get that from? The visually perceived angle between the pendulums is a different matter from whether they appear moving towards the observer.

At a minimum, at A=1.5, I do not see anything in front of me except the black hole. Whether I see anything being pulled toward me is something we should discuss after I lay out some of the math (probably not this week).