Space-Time Projection on Space

[RCopernicus]

I know that Space-Time is curved near the source of stress, but I'm not quite as clear what that means for the projection onto normal space and I'm trying to get my head around it. Is a kilometer on Mercury the same as a kilometer on Neptune? Is there a relatively simple formula (that is, an approximation of GR) that will give me the difference based on one solar mass and the distance from that mass?

 

[jedishrfu]

Here's a reference for length contraction with respect to SR and speed of rotation that may help:

http://en.wikipedia.org/wiki/Length_contraction

 

[A.T.]

Is a kilometer on Mercury the same as a kilometer on Neptune?

How would you compare them?

 

[RCopernicus]

How would you compare them?

This is not an SR question. I would take two rulers, make sure they were the exact same length. Then I would place one on Neptune, then I would place the other on Mercury. Then I would take my spaceship to a spot about a parsec above the sun, look down on the planetary plane and measure the rulers photometrically.

 

[A.T.]

Then I would take my spaceship to a spot about a parsec above the sun, look down on the planetary plane and measure the rulers.

How do you measure them from about a parsec? By their optical angular size? How do you account for the light bending of the light rays due to gravity?

 

[Nugatory]

Is a kilometer on Mercury the same as a kilometer on Neptune? Is there a relatively simple formula (that is, an approximation of GR) that will give me the difference based on one solar mass and the distance from that mass?

I'm going to answer those two questions in reverse order. Yes, there is a relatively simple formula. The Schwarzschild metric is an exact solution to the equations of general relativity for the gravitational field of a spherical mass like the sun. So we can use it to answer your first question.

As for that first question:
A kilometer is defined to be the distance that light travels in 1000/299752458 seconds. So in principle I can set up a light source and a clock on Neptune, a light source and a clock on Mercury, and use these to mark out points on the surface of the planets that are one kilometer apart. Now I have a kilometer on Neptune and a kilometer on Mercury, and all that's left is to compare them. (if you want to define a "kilometer on Neptune" and a "kilometer on Mercury some other way, that's OK, but you have to tell us what it is - it will probably make a difference).

On each planet I construct a perfectly rigid infinitely strong steel structure that exactly spans the distance between my two marks. I use powerful rockets to carefully and gently lift the two structures off the surfaces and fly them to some common location in deep space, far from the gravitational effects of the sun. There, I put them side by side and compare their lengths. They will be different the same.

 

[A.T.]

A kilometer is defined to be the distance that light travels in 1000/299752458 seconds.

I assume you define a second based on a local caesium 133 probe.

There, I put them side by side and compare their lengths. They will be different.

So, if instead of rigid rods you construct two light clocks based on your marks, they will run at different rates, when brought together? And they will both run out of sync with the caesium 133 reference?

 

[harrylin]

This is not an SR question. I would take two rulers, make sure they were the exact same length. Then I would place one on Neptune, then I would place the other on Mercury. Then I would take my spaceship to a spot about a parsec above the sun, look down on the planetary plane and measure the rulers photometrically.

I would take two rulers, make sure they were the exact same length. Then I would place one on Neptune, then I would place the other on Mercury. Then I would take my spaceship to a spot about a parsec above the sun, look down on the planetary plane and measure the rulers photometrically.

That could work with one modification: avoid the images to pass near the Sun, as that would distort the image (search for "gravitational lensing").

Einstein discussed the question of the length of rulers laid out on a planet in one of his first papers on GR.
And if I correctly understand it, then the answer to your question is a bit disappointing: in this case you should see no difference. See p.197 of the English version here: http://www.Alberteinstein.info/gallery/gtext3.html

 

[pervect]

I know that Space-Time is curved near the source of stress, but I'm not quite as clear what that means for the projection onto normal space and I'm trying to get my head around it. Is a kilometer on Mercury the same as a kilometer on Neptune? Is there a relatively simple formula (that is, an approximation of GR) that will give me the difference based on one solar mass and the distance from that mass?

The proper meter, and the proper second, which are the ones defined by the SI standard , are the same everywhere. Coordinate meters, and coordinate seconds, are NOT the same everywhere. It is important to be able to distinguish proper time and distances and coordinate times and distances for this to make sense.

Note that the SI definition of a meter is "the distance light travels in 1/299,792,458 of a second". The second in question is the the length of time measured on an ideal co-located clock, i.e. a proper second. And the distance light travels in this fraction of a proper second is a proper meter. So given a proper second, the SI defintion tells you what the proper meter is - wherever you may be.

Proper seconds contrast with, for example, the time system that GPS uses, for instance, which is a coordinate time system. TAI time is another coordinate time. Coordinate time and proper time run at the same rate only at sea level (more precisely, on the geoid). At higher or lower altitudes, coordinate time is not the same as proper time.

Compare this to the following: anywhere you go on the curved surface of the Earth a meter is a meter is a meter. However, a degree of longitude (which is coordinate based) does not represent the same distance everywhere.

 

[A.T.]

A kilometer is defined to be the distance that light travels in 1000/299752458 seconds. ... I put them side by side and compare their lengths. They will be different.

The proper meter, and the proper second, which are the ones defined by the SI standard , are the same everywhere...
Note that the SI definition of a meter is "the distance light travels in 1/299,792,458 of a second".

Am I misunderstanding what you both say, or is there a direct contradiction between you two?

 

[harrylin]

Am I misunderstanding what you both say, or is there a direct contradiction between you two?

I also have that impression - and of course perfect is right about proper length. However, RCopernicus clarified that his question was not about the proper length.

PS. RCopernicus, in my earlier post I forgot to add the web archive address to the paper by Einstein which has become a dead link - and I can't correct that post anymore... So here's a working link: https://web.archive.org/web/20080306023634/http://www.Alberteinstein.info/gallery/gtext3.html
On p.196, 197 you can find the equations you asked for. However, I do not know how even in theory any gravitational length contraction can be observed without ambiguity.

 

[Nugatory]

Am I misunderstanding what you both say, or is there a direct contradiction between you two?

There is (or was until I corrected mine) and of course pervect is right.

 

[RCopernicus]

I'm beginning to understand the difference between coordinate time and proper time, but I could still use a little help. Let me try the question this way:

I understand that seconds will become longer and a meter shorter, as measured by an outside observer, as you approach the speed of light. I also understand that traveling deep into a gravity well is similar to traveling near the speed of light, so I'm tempted to believe that an observer high above the sun would see similar effects as an object is dropped into a gravity well. So if I take two perfectly rigid rods, put them next to each other (far from any gravitational effects) to make sure they're both exactly 1 km long, and then placed one on Neptune and another on Mercury, then move away from the sun and take pictures of these rods, will they be the same length? If not, what formula could I use to calculate the difference?

 

[A.T.]

I would take two rulers, make sure they were the exact same length. Then I would place one on Neptune, then I would place the other on Mercury. Then I would take my spaceship to a spot about a parsec above the sun, look down on the planetary plane and measure the rulers photometrically.

That could work with one modification: avoid the images to pass near the Sun, as that would distort the image (search for "gravitational lensing").

The whole point of the question seems to be to ask about the effects of spatial distortion of gravity. But the remote optical comparison method proposed by RCopernicus is itself affected by this distortion. So I am not so sure if that is really a meaningful comparison.

 

[RCopernicus]

I'm assuming that we know the mass of the sun, have a good understanding of GR and have a powerful computer that can correct for any gravitational lensing effects.

 

[harrylin]

The whole point of the question seems to be to ask about the effects of spatial distortion of gravity. But the remote optical comparison method proposed by RCopernicus is itself affected by this distortion. So I am not so sure if that is really a meaningful comparison.

I supposed that the optical distortion "straight up" as he asked about will be negligible as long as the light path does not come near a star; however this effect should be zero. In contrast, the effect "from the side" (which he did not ask about) may be significantly affected by the optical lensing. That's why I wrote that I don't know of a way to observe any such effect without ambiguity (it's a combination of effects).

 

[A.T.]

I'm assuming that we know the mass of the sun, have a good understanding of GR and have a powerful computer that can correct for any gravitational lensing effects.

I thought you wanted a relatively simple formula, not something that requires powerful computers. Sure you can calculate the visual sizes, but the results will depend on factors like the perspective, distance etc.

What is wrong with the comparison method proposed by Nugatory? It doesn't require any corrections.

 

[harrylin]

I'm assuming that we know the mass of the sun, have a good understanding of GR and have a powerful computer that can correct for any gravitational lensing effects.

Assuming that a computer does the gravitational lensing corrections for you, I have answered your question. :)

 

[RCopernicus]

I thought you wanted a relatively simple formula, not something that requires powerful computers. What is wrong with the comparison method proposed by Nugatory? It doesn't require any corrections.

There is nothing wrong with his answer but there was apparently something wrong with the way I posed the question. As I understand it, the effects of time dilation and space contraction will cancel out. So when we construct our rigid structure on Mercury using the time it takes light to go a certain distance, of course that structure will be the same as one built on Neptune because clocks run faster on Neptune.

I am trying to visualize a coordinate system of "space" superimposed on the solar system (or galaxy). If I draw concentric circles radiating out from the sun and draw them each one "space" apart, will the gaps near the sun be shorter than the gaps far away from the sun (i.e. a meter depends on location in a gravity well) or are they all going to be the same distance apart (i.e. a meter is a meter is a meter)? I think prevect has given me the right answer but I'm trying to put it in terms my tiny ape brain can understand.

 

[harrylin]

There is nothing wrong with his answer but there was apparently something wrong with the way I posed the question. [..]
I am trying to visualize a coordinate system of "space" superimposed on the solar system (or galaxy). If I draw concentric circles radiating out from the sun and draw them each one "space" apart, will the gaps near the sun be shorter than the gaps far away from the sun (i.e. a meter depends on location in a gravity well) or are they all going to be the same distance apart (i.e. a meter is a meter is a meter)? I think prevect has given me the right answer but I'm trying to put it in terms my tiny ape brain can understand.

pervect did not answer that question. However I gave you in post #8 (with outdated link) the answer that Einstein gave to your question as you now rephrased it, and in post #11 the corrected link to his paper and additional info. Just in case you were not able for some reason to download the paper: "With the tangential position, therefore, the gravitational field of the point mass has no influence on the length of a rod".
Once more, the transformation for the direction that you asked is 1 but it's different from 1 for other directions.

 

[A.T.]

Assuming that a computer does the gravitational lensing corrections for you, I have answered your question.

If you correct for the gravitational lensing effect, don't you just and up with the proper lengths that pervect and Nugatory described?

 

[A.T.]

If I draw concentric circles radiating out from the sun and draw them each one "space" apart, will the gaps near the sun be shorter than the gaps far away from the sun

If you draw them the same distance apart, the gaps will be all the same, obviously.

 

[harrylin]

If you correct for the gravitational lensing effect, don't you just and up with the proper lengths that pervect and Nugatory described?

pervect simply stressed that one should not mix up proper lengths and coordinate lengths. By chance, the coordinate length in the direction that RCopernicus happened to ask about, which is tangential to the field lines, is equal in value to the proper length. In contrast, Einstein found that in radial direction the measuring rod is shortened ("The unit measuring rod thus appears a little shortened in relation to the system of coordinates by the presence of the gravitational field, if the rod is laid along a radius"). And for sure, that is without accounting for the effect of weight on the rod.

 

[HallsofIvy]

I'm going to answer those two questions in reverse order. Yes, there is a relatively simple formula. The Schwarzschild metric is an exact solution to the equations of general relativity for the gravitational field of a spherical mass like the sun. So we can use it to answer your first question.

As for that first question:
A kilometer is defined to be the distance that light travels in 1000/299752458 seconds. So in principle I can set up a light source and a clock on Neptune, a light source and a clock on Mercury, and use these to mark out points on the surface of the planets that are one kilometer apart. Now I have a kilometer on Neptune and a kilometer on Mercury, and all that's left is to compare them. (if you want to define a "kilometer on Neptune" and a "kilometer on Mercury some other way, that's OK, but you have to tell us what it is - it will probably make a difference).

On each planet I construct a perfectly rigid infinitely strong steel structure that exactly spans the distance between my two marks. I use powerful rockets to carefully and gently lift the two structures off the surfaces and fly them to some common location in deep space, far from the gravitational effects of the sun. There, I put them side by side and compare their lengths. They will be different the same.

The problem is that you can't do that. There is, even "in principle, no "perfectly rigid" structure.

 

[A.T.]

The problem is that you can't do that. There is, even "in principle, no "perfectly rigid" structure.

He is not using the rigidity to transmit information faster than light, just to preserve a shape.

The rods could just as well be elastic. The key is that you cut them to length under the same stresses under which you compare them. For example, build them in free fall, under zero internal stress, just as they will have in deep space when compared. Alternatively you use the difference in stresses and known material properties to compute the length a perfectly rigid would have. Talking about "perfectly rigid" is just a shorthand for such procedures.