Length Perception & Spacetime Warp: Is it Warped?

[Uchida]

If I make two rods with 1 meter length here on the surface of earth, and send one of them near a black hole that is at rest relative to earth, placing it there with its length alligned in the radial direction of the black hole, would I see the rod close to the black hole with a length shorter than 1 meter, (would the black hole rod be perveived as shorter than the rod here on earth)?

Then, if I go close to the rod near the black hole, would I see it with its original lenght, while the rod on Earth would appear longer than 1 meter?

My question is: does spacetime warp due to gravity field also warp the perception of size?

 

[Nugatory]

does spacetime warp due to gravity field also warp the perception of size?

How exactly are you determining the length of the rod that you’re not close to? Be precise, there are some assumptions hiding in that word “perceive”.

 

[Ibix]

There are significant conceptual difficulties with this apparently simple question.

First of all, if the rod is held above the black hole by its top end then it will be under tension and will be stretched. If supported from below it will be compressed. Even if not supported in any way there will be tidal stresses affecting its length. So it would be an apples-to-oranges comparison since, even in Newtonian physics, you would not expect a stressed rod to be the same length as an unstressed one. Perhaps it's possible to accelerate every point on the rod upwards so that it has no internal stresses, at least in principle.

Another issue is how you intend to measure the length of the distant rod. For example, if you just hold a metre rule next to the rod you will find that your meter rule is the same length as your rod (give or take different material reactions to stress). If you view the rod from the side and compare the apparent angular size of the rod as viewed from a distance to a naive ray-optics calculation you will find that the rod is longer than expected, but this is largely due to the curved path of light - a gravitational lensing effect. And if you view the rod end-on and measure its length using radar you will also find that it is longer than a naive calculation, but again this is due to changed light flight times due to Shapiro delay. I haven't done the maths but it would not surprise me if all three of these approaches gave different answers.

I don't think there's a unique answer to this, in short.

 

[PeterDonis]

Perhaps it's possible to accelerate every point on the rod upwards so that it has no internal stresses, at least in principle.

Not if the rod is supposed to remain stationary. It's impossible in relativity to have a rod that is stationary and also has constant proper acceleration all along its length.

 

[Uchida]

How exactly are you determining the length of the rod that you’re not close to? Be precise, there are some assumptions hiding in that word “perceive”.

There are significant conceptual difficulties with this apparently simple question.

First of all, if the rod is held above the black hole by its top end then it will be under tension and will be stretched. If supported from below it will be compressed. Even if not supported in any way there will be tidal stresses affecting its length. So it would be an apples-to-oranges comparison since, even in Newtonian physics, you would not expect a stressed rod to be the same length as an unstressed one. Perhaps it's possible to accelerate every point on the rod upwards so that it has no internal stresses, at least in principle.

Another issue is how you intend to measure the length of the distant rod. For example, if you just hold a metre rule next to the rod you will find that your meter rule is the same length as your rod (give or take different material reactions to stress). If you view the rod from the side and compare the apparent angular size of the rod as viewed from a distance to a naive ray-optics calculation you will find that the rod is longer than expected, but this is largely due to the curved path of light - a gravitational lensing effect. And if you view the rod end-on and measure its length using radar you will also find that it is longer than a naive calculation, but again this is due to changed light flight times due to Shapiro delay. I haven't done the maths but it would not surprise me if all three of these approaches gave different answers.

I don't think there's a unique answer to this, in short.

Not if the rod is supposed to remain stationary. It's impossible in relativity to have a rod that is stationary and also has constant proper acceleration all along its length.

Thank you sirs for your answers.

Aparently I formulated the question using a too mundane analogy, when all I want is a fundamental answer.
So let me clarifly some assumptions before we can discuss the problem:

• Let assume that the bar is made with a very high youngs modulus and tensile strength material, so that the stretching/rupture due to tidal forces near the black hole be negligible when compared to the total length of 1 meter.
• For calculus purposes, one should consider that the rod is "stationary", because the measurement will me made "instantly" at a given moment. (but we know that the rod will be circularly orbiting the black hole.)
• The distance between the observer and the rod that is far away is known at any given moment, so that at first glance, the length will be measured using apparent angular size from a side view of the rod, not considering gavitational lensing (naive aproach).
• The mass of the black hole and the distance between the rod that its near it and the black hole center is also known, so that in a second time, the length will still be measured using apparent angula size from a side view of the rod, but considering the influence of the gravitational lensing.
• The capacity to measure meter-scale lengths over cosmical distances should not be considered a problem, given that this question must be seen as a fundamental problem instead of a engineering one.

 

[Vanadium 50]

Let assume that the bar is made with a very high youngs modulus and tensile strength material, so that the stretching/rupture due to tidal forces near the black hole be negligible when compared to the total length of 1 meter.

You can't assume this. Relativity imposes restrictions on how strong materials can be.

I think you should try and phrase your questions entirely on local physics - how do you compare a rod far away with a rod next to you?

 

[Uchida]

You can't assume this. Relativity imposes restrictions on how strong materials can be.

I think you should try and phrase your questions entirely on local physics - how do you compare a rod far away with a rod next to you?

Ok, the rod is an analogy for "linear spatial distance".
Since english is not my native language, I could not find a better way to express my problem.

Let me try to put this question on other words:

Does spacetime distortions near a black hole causes apparent length distortions if measured from a region with flat spacetime?(see image attached as reference)

 

[PeterDonis]

Does spacetime distortions near a black hole causes apparent length distortions if measured from a region with flat spacetime?

This is not the sort of question you should be asking in vague ordinary language. You need to ask it in math. You need to be able to say, mathematically, exactly what you mean by "spacetime distortions", exactly what you mean by "apparent length distortions", and exactly what you mean by "measured from a region with flat spacetime".

If you are able to say, mathematically, what all those things mean, I think you will find that your question answers itself. But since there are multiple possible mathematical meanings for all of those terms, there's no way we, who can't read your mind, can tell you which mathematical meaning is the one you intend.