# Velocity vs distance

1. Sep 14, 2006

### jnorman

am i correct in understanding that as one's velocity increases, the distance between it and anything else approaches zero? for example, at the speed of light, there is no distance between you and any place else in the universe (since no time elapses?) - is this right?

if this is the case, one would not really ever need to travel in excess of C to move quickly (in one's own time frame) between galaxies. i recall a past query discussing a spaceship which travels at 0.99C away from earth to a star 10LY distant. an observer on earth would determine that it takes the ship about 11 years to get to the star. however, the clocks on board indicate that only about 1 year passes - indicating to the people on board that the ship has travelled 10LY in only 1 year. (how far is it "really" between earth and the star?)

sorry - i know i must be missing something fundamental here...

2. Sep 14, 2006

### pervect

Staff Emeritus

You've accurately described the state of affairs. For the numerical details, see http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken].

Note that if you are limited to an acceleration of 1g, you can actually cover more distance in n years of your time (aka proper time) than you could with Newton.

Of course, as viewed by Earth time, the trip will take longer, and you'll come back to a much older planet, depending on the details of your trip.

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3. Sep 14, 2006

### WhyIsItSo

pervect,

I read that article you referenced. It contains a section which says:

"Einstein postulated that any experiment done in a real gravitational field, provided that experiment has a fairly small spatial extent and doesn't take very long, will give a result indistinguishable from the same experiment done in an accelerating rocket. So the idea that the rocket's ceiling ages faster than its floor (and that includes the ageing of any bugs sitting on these) transfers to gravity: the ceiling of the room in which you now sit is ageing faster than its floor; and your head is ageing faster than your feet."

This doesn't seem right to me. For one thing, Einstein specifically said "real gravitational field", whereas the rocket's "gravity" is the result of acceleration. Because of the latter, the inverse square rule does not apply, so I find their assertion about different parts of the ship aging at different rates to be in error.

Comment?

4. Sep 14, 2006

### Staff: Mentor

"Einstein postulated that any experiment done in a real gravitational field, provided that experiment has a fairly small spatial extent and doesn't take very long, will give a result indistinguishable from the same experiment done in an accelerating rocket."​

5. Sep 14, 2006

### WhyIsItSo

I must be missing the point. The article is discussing a trip of great distance and duration. It does not seem to agree with the conditions Einstein is stipulating.

6. Sep 14, 2006

### bernhard.rothenstein

have, please a critical look at arxiv
physics/0607030 [abs, pdf] :
Title: Period measurement by accelerating observers

7. Sep 14, 2006

### JesseM

You can just interpret that section to be talking about the results of a brief experiment in comparing the instantaneous rate of aging of the ceiling vs. the floor:
Of course, if the ratio between the rate of aging of the ceiling and the floor is the same at every moment along the entire trip, then it's probably a safe bet to say this implies you'd get the same ratio for the overall aging of the ceiling vs. the floor over the trip's entire duration, so here you can use the equivalence principle to conclude something about an experiment that isn't just over a brief time interval.

The equivalence principle is one of the most fundamental ideas of general relativity, by the way--see http://scholar.uwinnipeg.ca/courses/38/4500.6-001/Cosmology/general_relativity.htm [Broken] for some basic information on it.

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8. Sep 14, 2006

### Staff: Mentor

As JesseM explained, the stipulated conditions refer to the experiment done in a real gravitational field, not to the rocket.

9. Sep 14, 2006

### JesseM

Well, the conditions can also apply to an experiment done on a rocket (the previous paragraph actually starts out by talking about such an experiment), provided they're done in a small region of spacetime (small region of space and short interval of time). The equivalence principle says the results of this experiment on the accelerating rocket should be identical to the results of the same experiment done on a rocket at rest in a gravitational field of the right strength.

10. Sep 14, 2006

### Staff: Mentor

Maybe I'm just asleep today, but I would think that stipulations like "fairly small spatial extent and doesn't take very long" would apply to experiments in real gravitational fields to avoid the non-uniformity that would be apparent. I see no reason why they'd apply to experiments done in a uniformly accelerating rocket in free space.

11. Sep 14, 2006

### WhyIsItSo

I have read the "equivalence" article, yet I still do not find relevance to the original article's contention that time is passing differently in different parts of the rocket. It seems to me that "equivalence" is misapplied to this assertion. The rocket is its own frame of reference, and at any instant it is entirely described by one inertial frame. Where is the suggestion of time difference within this frame coming from?

And more basic than this, the equivalence Einstein was talking about appears to me to be concerned with the path of objects over (a short) time.

I see nothing in his description to permit this broad application into the realm of some time distortion within a given frame of reference.

12. Sep 14, 2006

### JesseM

Ah, I see what you're getting at. I guess what I should say is that the equivalence principle is always about comparing two situations, one in flat spacetime and an equivalent one in a gravitational field; the necessity of limiting yourself to a small region of spacetime in the flat spacetime case is just a byproduct of the fact that you want to find an equivalent situation in a gravitational field, and in the gravitational field you need a small region of spacetime so you don't have to worry about tidal forces.

13. Sep 14, 2006

### MeJennifer

Yes, but only in the direction of travel.

Again, only in the direction of travel. However a photon spins so it is a bit more interesting.

Correct.

That is certainly possible. In relativity time and space are dynamic, in other words there is no absolute notion of time and space it all depends on the paths traveled in space-time.

No, I think you are right on the money!

14. Sep 14, 2006

### JesseM

I think it comes from the fact that if you want to accelerate an extended object such that its length in the instantaneous inertial reference frame of the object's center remains constant from one moment to another, then the instanteous velocity of the object's front and the instantaneous velocity of the object's back cannot be the same in the frame of its center, which means clocks at the front and back must be ticking at different rates in this frame. I don't see that they really explain this in the article though. You could also argue in reverse, if you already know about gravitational time dilation, that the equivalence principle implies you must see the same sort of time dilation in an accelerating rocket.
It's about the laws of physics as seen by a freefalling observer in a small region of space in a small region of time being exactly equivalent to the laws of physics seen by an inertial observer in flat spacetime. So if you have a small windowless elevator in which you perform any experiment (including one involving a comparison of the rates of ticking clocks, not just one involving the paths of objects you throw around) that lasts for a short time, you should have no way to distinguish whether the elevator is moving inertially in deep space or whether it's plummeting downwards in a gravitational field. And it's not too hard to show that this also implies a second version of the equivalence principle which says that any experiment done in a windowless elevator that's at rest in a gravitational field (sitting on the surface of the earth, say, or suspended at a constant height) must give the same results as an elevator accelerating in deep space--just imagine the larger inertial elevator containing a smaller elevator with its own tiny experimenters that's accelerating towards the ceiling of the larger elevator, and then compare with the case of a large elevator falling in a gravitational field which contains a smaller elevator that's hovering at a fixed height in that field--in both cases the first version of the equivalence principle tells you that the observers in the larger elevator should see the same outcomes of any experiments done in the smaller elevator.

Last edited: Sep 14, 2006
15. Sep 14, 2006

### pervect

Staff Emeritus
There isn't any conflict, just a side issue that hasn't been fully explained in this FAQ entry. I believe this issue is explained in the article on "Bell's spaceship paradox", but I'm not sure how clear the explanation is there, I seem to recall that people have had trouble understanding it.

Here are the gory details. If you are in a rocket, and your head maintains a constant distance from your feet (this is called Born rigid motion), then the following two statements will be true:

For a small rocket, the effect is completely ignorable. The effect will be a time acceleration factor on the order of 1+gh/c^2, where g is the acceleration, h is the height, and c is the speed of light.

Thus for h=1 km in a 9.8 m/s^2 field, the effect will be 1 part in 10^13

Both your head and your feet will be in "hyperbolic motion" as described briefly in the Wikipedia article http://en.wikipedia.org/wiki/Hyperbolic_motion_(relativity [Broken])

(Both hyperbola will have the same origin, i.e. the asymptotes of both hybperbola will intersect at the same point.)

If you imagine a very very large rocket, your head will age faster, but will essentially be unaccelerated. The acceleration of your head will not be zero, just very small.

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16. Sep 14, 2006

### WhyIsItSo

Alright, I just know I'm getting myself into trouble here, but...

I'm not comfortable with the notion that acceleration (a force) is equivalent in all respects to gravity (not a force).

Standing on the ground, my head is some distance from my feet; thus a little further from the Earth's center of mass, thus the gravitational effect is minutely less.

Standing on the "deck" of this rocket ship, I do not see the justification for saying the acceleration on my head is other than exactly the same for my feet.

In this accelerating frame of reference, there is a direct physical connection between the thrust of the rocket ship and all parts of the system, including me.

In gravity, it is action at a distance, and does not involve acceleration but rather space-time bending.

pervect, that wiki reference was indeed to trivial. Do you perhaps have a link to a direct translation of Einstein's writings or addresses that covers this issue?

17. Sep 14, 2006

Staff Emeritus
It is an old and false objection to the Principle of Equivalence that tidal forces (which are what you are talking about here with the different acceleration between your head and your feet) can distinguish artificial accelerations from gravitational ones. The Principle is stated in LOCAL terms (not always clearly expressed in popular accounts). And local means "in the limit as the distance betrween measuring points goes to zero". So an ant doing your experiment would find less difference in acceperation than you do, and a microbe less than an ant, and so on. And since this is a classical theory, within it we can take the limit down to zero and find that without the handy help of finite differences of measurment there is no way to tell artificial from gravitational accelerations.

18. Sep 14, 2006

### WhyIsItSo

*lmao* Ok, that is pushing me away from the contentions of others. It seems from what you say that I, being a much larger system than an ant, can detect the difference between real gravity and artificial gravity induced by acceleration. That suggests to me that gravity and acceleration are not comparable events.

I still cannot see any explanation for why my head accelerates differently to my feet.

Are you supporting what I've said, or criticising what I've said and I'm still simply missing your point?

19. Sep 14, 2006

### JesseM

Are you familiar with the concept of "limits" from calculus? The equivalence principle is a limit principle--in the limit as the region of spacetime in a gravitational field approaches zero size, the difference between experiments in that region and experiments in flat spacetime also approaches zero. For any finite region there will be finite differences between the two, but for any small difference you pick--say, 1 part in 10^30--you can pick a sufficiently small region of curved spacetime such that experiments in that region will differ from identical experiments in flat spacetime by no more than that amount.

20. Sep 14, 2006

### WhyIsItSo

I appreciate your attempt, but sadly it does not help. As for limits, I have a very basic understanding, and I think I follow what you are saying...

...but you describe the mathematical relationship, which unfortunately is not helping me to see that there is a relationship in the first place.

Accleration and gravity are different things, and I feel a red flag about jumping to the direct comparison of their relative effects. I'm not saying everyone is wrong, but I feel it is a proper attitude that I question, and understand, the underlying supposition.

What is the logic upon which this mathematical relationship rests?

21. Sep 14, 2006

### JesseM

Well, it might help to understand how general relativity originated. Einstein wasn't collecting large amounts of empirical data and trying to find a theory that would fit it--much like attempts at theories of quantum gravity today, he knew that there was an inconsistency in previous theories (Newtonian gravity being incompatible with special relativity since it treats gravitational influences as instantaneous), and he wanted a single consistent theory that would reproduce the correct predictions of these previous theories. He also had some theoretical ideas about what a relativistic theory of gravity should look like, and one of the most important of these was the equivalence principle. It would probably be fair to say the only reason for believing this was "elegance"--these sorts of aesthetic criteria are not uncommon in theoretical physics. The point is, he had given himself a bunch of constraints on what this theory should look like, and finally found a mathematical theory which satisfied them...I think most of the theory was determined by such constraints, and that it would be hard or impossible mathematically to find a distinct theory which still satisfied them (not totally sure about this though). Anyway, only after the entire theory had been created could any new predictions differing from Newtonian gravity be made, and general relativity has passed test after test with flying colors. So the idea that the laws of physics reduce to SR in the limit as the region of spacetime gets arbitrarily small is not something you test directly (although you can probably show that the error in SR predictions keeps getting smaller as you reduce the volume and time of an experiment), but it's built into the structure of GR which leads you to many other accurate predictions. I'm not sure if anyone has come up with an alternate theory that reproduces these predictions but where the equivalence principle is false--some version of "modified Newtonian gravity" may qualify--but if they have, I'm pretty sure it'd be a much more contrived-looking theory than GR, with a lot of extra bells and whistles tacked on without any justification from simple principles.

22. Sep 14, 2006

### pervect

Staff Emeritus
Yep.

Which particular reference are we talking about? If we're going to bandy references around, the best discussion of uniformly accelerated motion that I'm aware of is in Misner, Thorne, Wheeler (herafter: MTW) textbook "Gravitation".

The material here is persented at a more advanced level than the topic actually demands, i.e. it's presented using the language of tensors, where this is not strictly necessary.

If you're looking for a quick reference to veryify my informal description of what's going on, check out pg 164-165 of "Gravitation", Box 6.2 "accelerated observers in brief.

You probably won't gain a full understanding until you work the referenced exercise though - but I don't know if you have the math.

The treatment of the problem as Born rigid motion is complex enough, but still simpler than the treatment given in MTW. See for instance

http://www.mathpages.com/home/kmath422/kmath422.htm

You basically need to appreciate that objects in Born rigid motion follow hyperbolic paths. These paths have a constant Lorentz interval from the "pivot point" as described in the above page, which means that x^2 - t^2 = constant, which implies hyperbolic motion.

An important point is that "line of simultaneity" for any observer in hyperbolic motion passes through the origin (i.e. the pivot point). This means that hyperbolic observers not only stay a constant distance away from the pivot point, they stay a constant distance away from each other when they measure distances along the "line of simultaneity", which turns out to be the defintion of Born rigidity.

The "line of simultaneity" is just the set of points that an instantaneous inertial co-moving observer would regard as simultaneous. In SR, simultaneity depends on the observer.

23. Sep 16, 2006

### WhyIsItSo

Pervect,

While much of the math went over my head, I accepted his calculations as correct, and concentrated on the logical flow of his statements.

Unfortunately, what the article means to me is that this desribes a phenomenon an inertial frame would observe, but I don't see that the accelerating object itself actually experiences anything unusual.

If I had two clocks in my rocket, one at the fron, one at the rear, you, an inertial observer, would read a difference, not just in their time, but also in their physical location (I see the "bending" being talked about). In fact, the time and material distortions sound like you'd be challenged to even identify that it was my Rocket you were looking at :)

BUT, I do not see how this article argues that I would observe anything odd at all.

Sorry. That was an awesone attempt you made, but I still don't see any effects for my frame of reference.

24. Sep 16, 2006

### JesseM

Instead of a rocket, it might be easier to just imagine 3 clocks attached to individual rockets and accelerating in formation, with each clock's rocket programmed to accelerate in such a way that, in the instantaneous inertial rest frame of the middle clock at any given moment, the distance to the front and back clock always remains constant under that frame's definition of simultaneity and distance. Since the instantaneous inertial rest frame of the middle clock is constantly changing, and each successive frame has a different definition of simultaneity, you can see intuitively why this might force you to pick paths for the front and back clock such that their velocity as a function of time would not be identical, and thus their rate of ticking would not be identical either.
But without doing the math, do you think there's any reason to think you would't observe the front and back clocks ticking at different rates? Do you have any argument as to why you think they wouldn't?

To figure out how the rate of ticking of the front and back clock will look in the instantaneous rest frame of the center one, you'd probably first have to figure out their paths and velocities as a function of time in a single inertial frame, and then you can use that to figure out each clock's rate of ticking as a function of time in this frame, and then you can do a Lorentz transformation to figure out each clock's rate of ticking in the instantaneous inertial rest frame of the center clock at any given moment. This wouldn't be an easy calculation, so again, without actually doing it yourself, why would you have any particular reason to think the result must be that both clocks are ticking at the same rate in the instantaneous inertial rest frame of the center clock?

25. Sep 16, 2006

### Tomsk

Is this time difference something to do with there being a time delay between a signal travelling from one end of an accelerating body to the other? There would be a small delay as your feet push on your legs, your legs push on your body, etc, and by the time your head has a force acting on it to accelerate it, your feet have travelled a bit, and are moving faster than your head. I could be completely wrong, hopefully someone can correct me...