# What is the shape of length-contracted space?

1. Apr 21, 2014

### BitWiz

What is the shape of an "isodistance" volume from the point of view of an observer traveling at relativistic speeds? (EDIT: ... compared to that of an outside observer)

If it is not spherical, how do the distorted (?) distances of all mass objects surrounding the observer affect the observer's path through spacetime? Is there more gravity parallel to the direction of travel? If a rocket points at a massive destination object, will that object tend to align the relativistic rocket in that direction as velocity increases?

Thanks!

Last edited: Apr 21, 2014
2. Apr 21, 2014

### phinds

I'm not sure I really understand your question, based on the wording.

You, right now as you read this, are traveling at a speed of .99999c relative to SOMETHING in the universe but your observable universe, for example is still a perfect sphere. If you define a sphere, say 1,000,000 miles in diameter, centered on you, and there is an observer 1LY away traveling at relativistic speeds not directly towards or away from you, then objects at the edge of that sphere will not be appear to be arrange in a perfect sphere based on you but will be deformed a bit, depending on both his angle of travel and his speed.

A rocket traveling at relativistic speeds would pass by a massive object so fast that the amount of time it spent in the serious part of the gravity well, even if it came fairly close, would be too small to make much of a course change unless it was a black hole and you were going really close so no, a rocket traveling at relativistic speeds and far enough away to not be there yet would not be much affected.

EDIT: and by the way, space does not appear length-contracted since space is not a thing. Stuff IN space appears length-contracted to someone moving at relativistic speeds relative to the stuff.

Last edited: Apr 21, 2014
3. Apr 21, 2014

### BitWiz

Thanks for the reply, phinds. I agree I didn't word this very well. Let's try this:

Two observers are co-located in space at the same point. One of the observers is at rest with respect to the stars, and the other is traveling at relativistic speed. Each observer is at the center of an isodistance sphere of the same arbitrary radius. With respect to either observer, what would the other's "sphere" look like? What shape would it be?

Again, a badly worded question. Let me try this: Let's say that an observer manages to accelerate very close to the speed of light toward the core of the Andromeda Galaxy over a very short distance. Length contraction will bring Andromeda arbitrarily closer. After the acceleration, will the observer feel a stronger gravity pull from Andromeda?

Thanks!!

4. Apr 21, 2014

### pervect

Staff Emeritus
A stationary sphere , observed by a observer in relative motion, would appear as an ellipsoid. The easiest way to visualize it is to visualize a cube around the observer, apply the lorentz transformation to shrink the sides of the cube parallel to to the relative motion (while not affecting the transverse sides), then embedding an ellipsoid in the resulting box.

More on the second question later...

5. Apr 21, 2014

### pervect

Staff Emeritus
OK - for the second question. I'm going to cheat and give the answer for electromagnetism, because the notion of an electromagnetic force is better defined than the notion of a gravitational force, and because the math is easier.

If andromeda was charged, and you measured the electric field (by measuring the relative acceleration between a charged and noncharged particle, for instance), and the moving observer also measured the electric field in a similar manner, you'd both get the same answer if the moving observer was headed directly towards andromeda.

The math looks like this (from wiki)

where the subscripts indicate fields parallel and transverse to the direciton of motion, repsectively.

If you sketched out the electric field from Andromeda, and compared it from relatively stationary and relatively moving points of view, you'd get something like:

6. Apr 21, 2014

### Staff: Mentor

It's worth pointing out that the word "appear" here is not to be taken literally; what you're describing is the shape of a spacelike slice through the sphere's world tube, taken orthogonally to the observer's worldline. This is *not* the same as the way the sphere would literally appear to the observer's eye, because of light travel time delays; to the observer's eye, the sphere will actually appear spherical, but rotated.

For more gory details, see, for example, here:

http://math.ucr.edu/home/baez/physics/Relativity/SR/penrose.html

7. Apr 22, 2014

### pervect

Staff Emeritus
Right, I'm not talking about the visual appearance, but what happens after one corrects for the propagation delay of light.

I'd agree that that is what I'm describing, but while it is accurate, I wouldn't say it's a "beginner friendly" way to describe it :(.

8. Apr 22, 2014

### BitWiz

Thank you, pervect and PeterDonis.

And thanks for trying to make the concepts accessible. ;-)

... But I still feel I have been unable to convey my question ... or I may have the wrong idea entirely based on what phinds said:

I don't understand this. If a rigid rod connects me and a distant object, and (instantaneously if necessary) the distance between me and the object closes relativistically, does the space(time) inhabited by the rod collapse or not collapse with the rod?

Perhaps I need the answer to this before I can reattempt the other question.

Thank you, sirs!

9. Apr 22, 2014

### phinds

Your problem here, I think, is with your concept of a rigid rod. It is something that does not exist in the sense that I think you mean it in. A rod cannot collapse at relativistic speeds, the way your example would require, by simple having the object at the other end push on it and move itself and the end of the rod towards you at relativistic speeds. That would require compressive forces to move in the rod faster than the speed of sound in the rod and that can't happen. On the other hand, if you think of a rod that COULD compress at relativitic speeds (I can't imagine what that would be made of) it will just get shorter as the object moves towards you. The END of the rod will get shorter and the rest of the rod will be unaffected and the object will appear length contracted.

10. Apr 22, 2014

### BitWiz

Thanks, phinds. What I'm trying to separate is the reduction in distance I get from the derived velocity I integrate from acceleration, i.e. "speed," and the reduction in distance due to relativistic effects. Let me modify the example:

A rigid rod extends from a distant object to a point in space one meter in front of me. I accelerate for one meter. At the point that I just touch the rod, has the length of the rod changed from the time that I was at rest with respect to the rod? If so, what has happened to the space(time) that the rod occupied?

Thanks! ;-)

11. Apr 22, 2014

### timmdeeg

In my opinion nothing happens to the space-time. As long as there is relative velocity you will "see" the rod length contracted parallel to the direction of the motion. As soon as you stop in front of the rod you are in it's rest-frame and you will see it's length accordingly (not contracted).

12. Apr 22, 2014

### BitWiz

Hi, Timmdeeg. I don't stop. In fact, I'm still accelerating. I'm just at that instantaneous point in my travel. What would you say then?

13. Apr 22, 2014

### Staff: Mentor

It depends on what you mean by the "length" of the rod. Length is frame-dependent; the rod does not have a single unique "length". As far as any invariants describing the rod, they do not change at all; physically, nothing about the rod changes, only your state of motion changes.

14. Apr 22, 2014

### phinds

No, the length of the rod has not changed. Nothing happened to the rod at all. Your PERCEPTION of the rod will have changed and you will think it is shorter but that's just an effect of your motion and length contraction and has nothing to do with the rod.

You, right now as you read this, are traveling at a speed of .99999c relative to SOMETHING in the universe and relative to that object you are massively length contracted. Do you feel any different?

15. Apr 22, 2014

### timmdeeg

Well, then the rod is still length contracted in your frame. Do you suspect that something physical happens to the rod? As already mentioned that's not the case. Something else must be the background of your reasoning. What could that be?

16. Apr 22, 2014

### BitWiz

Good question. I think I must be missing something very fundamental.

Say I have a series of rigid cylinders with their long axes aligned along a single line in space, and alternating between those cylinders (their ends) is an equal amount of empty space. I now accelerate in a direction parallel to, and adjacent to, the cylinder axes. As my speed increases, the length of the cylinders will *appear* to contract. Correct? And because of the angle, cylinders further away from me will appear to length contract more than those adjacent to me. Correct? If so, then:

1) Is this merely a visual illusion, similar to a Fun House mirror?

2) If the cylinder lengths contract, does the space in between contract, widen, or stay the same?

3) If I double my relative velocity, will I see *more* than twice as many cylinders go past me?

4) If #4 is true, what the heck is going on? ;-)

Thanks, Timmdeeg.

Bit

17. Apr 22, 2014

### Staff: Mentor

In the special sense of "appear" that pervect used in his earlier post, yes. But as I noted in response to that post, this sense of "appear" does not describe what you will actually *see*; it is what you will *calculate* the length of the cylinders to be, after correcting for light travel time delays.

Not in the sense of "appear" used above, no. Length contraction depends only on relative velocity; at any given instant of your proper time (i.e., time by your clock), your velocity relative to all the cylinders is the same, so they are all length contracted the same.

It is true that, since different cylinders are at different distances from you, what you will actually see at a given instant of your proper time will vary from cylinder to cylinder. But in that sense of "appear", the cylinders do not appear length contracted; they appear rotated, as I pointed out in my earlier post.

I would say yes, because I think the key error you are making here is thinking that length contraction is something that actually happens to the object, instead of something that's purely due to you changing your state of motion. The fact that you are accelerating does not change the cylinders at all. Attach whatever sensors, strain gauges, accelerometers, etc., etc. you like to the cylinders: their readings will *not* change at all as you accelerate. Attach a laser and reflector to each cylinder to precisely measure its length in its own rest frame; that won't change either as you accelerate. The only thing that changes as you accelerate is you.

It depends on what you mean by "space in between". But in any case, you should not be thinking in terms of space; you should be thinking in terms of spacetime. Spacetime does not change just because you accelerate. The only thing that changes as you accelerate is you.

18. Apr 22, 2014

### WannabeNewton

It is not about perception in any sense of the word. Length is simply a frame-dependent quantity. Saying "the length of the rod has not changed" has no meaning in SR. If you want to qualify length using "rest length" then the statement is fine but in doing so, you are singling out a specific Lorentz frame in order to do this even though the rest length isn't any more special than the length of the rod as measured in an arbitrary Lorentz frame.

As such, length contraction is not an illusion. It is simply a consequence of making temporal and spatial measurements functions of space-time foliations.

19. Apr 22, 2014

### phinds

Fair enough. I should have said "proper length" and explained it more fully, to be as accurate as you are being.

20. Apr 22, 2014

### Bill_K

I'd say the proper length is the maximum of the lengths as measured in various Lorentz frames.

21. Apr 22, 2014

### pervect

Staff Emeritus
If you have a rigid rod (to be precise, the sort of rigidity allowed by special relativity, Born rigidity), and you measure it's length from a stationary observer, you get a certain length L. If you measure it' length from a moving observer, moving with some relative velocity v, the length of the rigid rod is L' = L / gamma, where gamma = 1/sqrt(1-v^2/c^2), v being the relative velocity. This is called Lorentz contraction.

The idea that space-time collapses isn't a very good way to think about it. Using the less abstract concept of the rod and its length, the point is that length isn't a property of the rod, it's a property of the rod and the observer.

Using the more abstract concept of space-time, the idea isn't that the space-time itself changes (space-time is basically just geometry, and the geometry isn't changing). What is changing when you change velocity is not space-time, but one's view of it. The new view of space-time given by a moving observer is mathematically related to the old view via the Lorentz transform.

22. Apr 23, 2014

### BitWiz

Thanks!

Thanks, pervect, PeterDonis, phinds, and others, for your comments and time. I'm traveling for the next few days, and it will be a good time to think about what you've said. In the meantime, I'll try to keep my speed below relativistic notice.

Bit

23. Apr 23, 2014

### timmdeeg

Bit, perhaps you stick too much on rods, cylinders on something made of a material. However it is sufficient to imagine the endpoints of the rod to be represented by 2 testparticles which are in rest to each other and which are separated by their proper distance, the length of the (now only) thought rod in their frame. If you move relative to these particles, their distance is Lorentz contracted in your frame, whether or not there is a rod in between.

24. Apr 23, 2014

### BitWiz

Thanks, timmdeeg. You've been a big help as well.

I don't mean to be stuck on physical objects. I think I use them in examples to help tell the difference between something that is a property of the geometry as "inhabited" by the observer vs what is observed. The latter can be a light-travel effect, such as the headlight effect, because the stars don't really move. But something about time dilation, coupled with length contraction, doesn't seem to add up. I really CAN go 10 ly in 5 proper years at gamma=2, but since I could never "inhabit" the event where I measure going at 2xc, the distance to the object must truly shorten, its angular size must increase, and it can't be an illusion if my arrival at that destination isn't.

Perhaps length contraction has nothing to do with it. We're just warping the path that light seems to travel. The illusion is in our observing equipment. Thus objects seem to "rotate" but don't. But time dilation? That seems to be a geometry effect.

Bit

25. Apr 27, 2014

### ghwellsjr

Let me change your scenario a little bit to make it easier for me to illustrate with some spacetime diagrams. Let's say you are at rest with a five-foot long rod with the near end 8 feet in front of you and the far end 13 feet in front of you. After a while, you start accelerating toward the rod and by the time you reach it, you are going about 0.6c.

This analysis is going to be based on your taking measurements with a pair of laser range finders, one aimed at the near end of the rod and one aimed at a reflector mounted on the far end of the rod. A laser range finder works by sending out a pulse of light and measuring how long it takes for the light to reflect back to it. It then divides that time by 2 and multiplies by the speed of light to display the distance. By making two measurements, you can determine the length of the rod. For convenience sake, I'm going to define the speed of light to be 1 foot per nanosecond.

While you are still at rest with the rod, it doesn't matter when you make the measurements to the near and far ends of the rod because you'll always get the same answer. But when you are moving, in order to get consistently meaningful results, you have to make both measurements at the "same time". Einstein's way of doing this is to assume that the laser light takes the same amount of time to get to the target as it takes for the reflection to get back to your laser range finder. In practice, the laser beams would have to carry a coded signal so that the time of emission can be correlated with the time of reception in order to identify a "time of the measurement" half way between them. We have to imagine that each laser range finder (or equipment that is hooked up to them) keeps track of all those times and then later, you, or your equipment can create the spacetime diagram that I'm about to show you.

Here's the first diagram showing your path in blue, the near end of the rod in red and the far end of the rod in black:

You're going to start your acceleration at your time zero but long before that, you are going to start to make your measurements and continue doing the same thing throughout the scenario. I have shown a pair of these measurements to illustrate how you determine that the rod is 5 feet long.

Remember, your laser range finders are constantly sending out coded light pulses and getting their reflections some time later and logging them. You can't do the calculations while it is actually happening. After your trip is done, you look at the logs of one laser range finder for a sent/received pair whose average time matches the average time for a sent/received pair on the other laser range finder. The above diagram shows one such matched set. The first laser range finder sent a laser signal at your time of -25 nsecs and received the reflection at -9 nsecs for an average time of -17 nsecs. The other laser range finder sent out a signal at -30 nsecs and received its reflection at -4 nsecs for the same -17 nsec average. Once you have selected a matched set of timings between the two laser range finders, you next calculate the distance that each one measures. You do this by taking the difference between the logged readings for one of the devices, divide by two and multiply by the speed of light. So for the near device, you take 25-9=16 and divide that by 2 to get 8 nsec and since light travels at 1 foot per nsec, this comes out to be 8 feet. Doing the same thing for the far device gives us 13 feet. Thus we have measured the length of the rod to be 13-8=5 feet. Once you understand this principle, you can follow the rest of this process.

I have uploaded some more diagrams and text files to show how the determination is made for both devices. (The dashed line in the text files marks the point where you pass one end of the rod and have to send your signals the other way.) You can click on the thumbnails and files at the bottom of this post if you are interested in seeing the details but here is the final result of how you establish the length of the rod as it approaches and passes you. This, by the way, is known as the radar method and is just one of many legitimate ways of creating a non-inertial rest frame for an accelerating observer:

You will note that the rod starts contracting before the time of your acceleration (0 nsecs) and that it happens differently for both ends of the rod. Make sure you understand that you measure the length horizontally, for example, along one of the horizontal axis lines. You can think of these as lines of simultaneity.

Note also that when the leading edge of the rod reaches you, even though its speed is about 0.6c, the rod has already contracted to less than 80% of its original length. Other ways of establishing a non-inertial frame would make this happen at the point of contact with the rod.

Does this make sense to you. Any questions?

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Last edited: Apr 27, 2014