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What follows is a long self-made example to deal with my conceptual issues of visualizing curved spacetime.

Imagine an observer floating somewhere in space. He feels no strain on his body, indicating he is free-falling and is thus moving along a geodesic.

He is a magician and has the power to conjure up glowing sticks within his hands.

Their construction is as follows (the following are valid in his vicinity of course):

1) Each stick is very thin and of 1m in length when he conjures it in his hands.

2) Each stick is rigid and mass-less

3) It glows along it's cylindrical surface uniformly. The glow varies with intensity such that it goes from complete dark to peak brightness to dark in 1 heartbeat of the magician in his local rest frame. It is thus like a clock synchronized with his heartbeat.

The magician starts conjuring up billions of these sticks in his vicinity and starts assembling them into a cubic lattice in his vicinity. As he starts assembling the lattice, it grows in size and moves further out far beyond his vicinity. He effectively creates a crude cartesian co-ordinate system (without any markings).

In curvature free space-time, his eyes will project the glowing grid lines as straight lines intersecting at infinity and the whole system will glow in the same way.

Now if space-time had some arbitrary curvature (not too high), how would the glowing grid 'lines' be projected onto his eyes?

Alternatively he could create a spherical system by making concentric glowing shells of increasing radii (where the latitudes and longitudes glow) made with uniform curvature in his vicinity. He could fix the spatial origin as wherever he wants. In flat space-time, the concentric shells would be projected onto his eyes as concentric circles. What would he see in curved space-time?

Is this how experimenters construct their 'imaginary' co-ordinates systems for work in astro-physics?

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See Gravitational Lens - "In general relativity, light follows the curvature of spacetime" –  RedGrittyBrick Aug 14 '13 at 14:31
    
The eye collects light from various glowing sticks. This light has traveled along null geodesics but the eye projects all of this into 2 dimensional image. I want to know (roughly) what image would form of this specific grid in the case of low curvature. Most of the literature of gravitational lensing deals solely with the images of point objects when a big object is in the way, not the images luminescent grids. –  dj_mummy Aug 14 '13 at 15:53

3 Answers 3

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First of all, even in special relativity there is a problem with rigidity (see for instance Bell's spaceship paradox and Ehrenfest paradox). So in general relativity one could only aim for reasonably rigid objects, bearing in mind that even while free-falling they could deform or be destroyed under the action of tidal forces.

Let's mention various effects your magician could observe.

In addition to general impossibility of constructing cartesian grid in curved space-time mentioned by Julian it is worth mentioning that the grid would not be stationary even if the space-time is -- the sticks would be drifting apart (or pulled colser) in accordance with laws of geodesic deviation. So after some time the lattice would either has (constantly increasing) gaps or overlaps.

Another effect is red or blueshift of light. If the glowstick would be in the gravity well, the the magician (situated outside of the well) would observe increase of wavelength of light (or lowering of color temperature). On the other hand observer situated inside the well will observe blueshifted light from outside.

If the distance between the oscillating light source and observer is comparable to $c\tau$ ($\tau$ is the period of heartbeat) then the visible oscillation cycle would acquire additional phase. For light sources synchronized along the line starting near the observer it would appear that a wave of varying intensity is traveling along this line.

And gravitational lensing (mentioned by RedGrittyBrick) would apply if the light from the glowing sticks would travel through regions of curved space (such as near massive object). The effect could produce caustics and curve straight lines.

And for the examples of what someone could see in concrete practical cases of noticeably curved space look at the galleries of images and animations at "Space-Time Travel: Relativity Visualised". Note, that some of the renderings ignore blue/redshift of colors. traversable wormhole

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I really like the website you recommended. But I wonder why nearly all the examples are for spherically symmetric distortions? Of course I can imagine that the grid lattice would look all distorted and stuff. Which brings to my issues in the other question: what do the co-ordinate systems even mean physically if we get visual distortions like this? If visual signals fail, then maybe other sensory inputs like touch could work? –  dj_mummy Aug 16 '13 at 19:48
    
Space-times with spherical symmetry are among the simplest solutions of Einstein equations. Visual signals do not fail they get distorted in a way prescribed by general relativity. Observer still can extract information from such a signal both about the light source and about the space-time this light signal has been traveling through. –  user23660 Aug 17 '13 at 5:41
    
Touch, that is interaction through direct contact, is possible only for the immediate vicinity whereas the corrections from general relativity are only mostly noticeable only at large distances from earth orbit up to cosmological scales. –  user23660 Aug 17 '13 at 5:55
    
But what if I had long elastic arms and sent them towards a stick in a highly distorted area? What would my hands measure a small stick as: 1m or something else? I would imagine it would be one metre? I could move my long hands inch by inch along (ignoring time for a moment) the sticks and measure the position of various objects. –  dj_mummy Aug 17 '13 at 6:21
    
So basically what you are saying is that the inspite of all distortions and the lack of rigidity, it is a perfectly acceptable co-ordinate system in theory? –  dj_mummy Aug 17 '13 at 6:27

First, assuming he can make the sticks glow in synchronous with his heartbeat, the information from his heart to the sticks must travel instantaneously. In any case, he will not see them glowing in unison because the light from the farthest sticks will take longer to reach his eyes than the closer ones (assuming they are far enough so the eyes can detect the difference). Still in his frame of reference they will all light at the same time (he can calculate that but is not the way they will perceive them). Regarding the grid, if space time is not flat, then the grid elements will not longer look as squares. In general you just cannot build a square in curved space time, by square I mean a geometric form that has four sides and each one intersects at an angle of 90 degrees. Try drawing a square large enough at the surface of a ball an you will realize why (but you can do built a triangle will all sides perpendicular to each other). From inside, you will be able to perceive this. Euclidean geometry will no longer work for you.

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The sticks are synchronized with his heartbeat in his immediate vicinity. So assuming a very small curvature (concave) the lines will appear curved and the grid will appear 'squashed'? So how do experimenters set up a co-ordinate system? Isn't it like this except that the sticks are also marked and the number of pulses is also communicated by each stick? If not how in the world do physicists make measurements of position? –  dj_mummy Aug 15 '13 at 7:41
    
no, the lines will look straight, it is just that the grid will not look like made of squares but of some other trapezoidal shape (if the curvature is large enough for this to be perceived. The answer to the other issue will take longer and there are different ways to do it. You should ask that as a separate question. –  user16007 Aug 15 '13 at 23:54

My apologies that I can't follow your question enough to answer everything, in particular I can't see how you project your images. If you extended them in what look like straight lines to you, you'll get radial lines expanding from your position, but it seemed like you wanted a Cartesian grid.

However I do want to answer the part about strain and its relationship to geodesic motion. You did mention that the observer feels no strain and you then concluded that this means the observer moves on a geodesic. This is not correct.

An extended observer moving freely will feel strain from a non spatially uniform curvature, but this strain goes away as the extended observer gets smaller and smaller (the curvature then becomes more and more uniform over the region where the observer is), so it is not related to whether the observer is in free fall. But even a small nonuniformity will produce a small strain on an extended body. Imagine a blob of water in space, it will adjust it's volume to minimize it's surface area, so if space is curved nonuniformly it will adapt to that and not be a perfect sphere. But the extended body is still jointly in free fall.

But you are correct that a common way to deviate from geodesic motion is to produce a strain that induces an acceleration. The earth strains your foot, compressing it so that you foot will push on your lower leg to strain it so that it will push on your knee to strain it, etc. All of which result in you not moving on a geodesic when the earth strains your foot. So strains indeed can cause a deviation from geodesic motion, but so do pressure gradients (that's what keeps the atmosphere from moving geodesically), so a strain is just one way among many to try to move non geodesically.

Another issue is that no real systems ever moves on a true geodesic. Geodesic motion is a limit when the object has no mass, no energy, no spin, no charge, etc. Real systems warp spacetime themselves instead of moving on a fixed background determined by everything else. Sometimes those effects are small, other times they are not.

So any extended systems is strained by nonuniform curvature, and real systems cause their own curvature.

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Thanks for answering. I have long since resolved this question myself. As you have alluded to, my question was founded on incorrect assumptions. There is no way to view an 'extended' grid system 'simultaneously' in GTR. The image formed at the event of the eye (point-sized eye) is just an arbitrary local measurement and lends no legitimacy to the notion of an 'extended' frame. –  dj_mummy Sep 4 at 6:29

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