Maintaining the appearance of fixed distance in spite of length contraction?

[md65536]

The thought experiment you propose sounds like "Bell's Spaceship Paradox":

 

http://en.wikipedia....aceship_paradox

 

Is this the situation you're thinking about?

Yes, I think that the link provided gives an answer to the question.

 

In the "Analysis" section, they set up a good simplification (treating Frank and Bert in our examples as point masses).

 

The applicable bit is: "This implies that xA(t) − xB(t) = a0 − b0, which is a constant, independent of time. In other words, the distance L remains the same. This argument applies to all types of synchronous motion."

 

They go on to show that "when switching the description to the comoving frame, the distance between the spaceships appears to increase by the relativistic factor [gamma]. Consequently, the string is stretched."

 

 

In other words: If the front and back of the ship can be accelerated using synchronous motion, then yes, the length of the ship will remain the same, and will appear to remain the same for observers which see the front and back synchronized.

 

It should be easy to show the following: Iff all points on the ship can be accelerated using synchronous motion, observers which see it synchronized would see no length-contraction deformations of the ship. Otherwise, sections in between the synchronized points would stretch in accordance with the wikipedia article.

[J.C.MacSwell]

Yes, I think that the link provided gives an answer to the question.

 

In the "Analysis" section, they set up a good simplification (treating Frank and Bert in our examples as point masses).

 

The applicable bit is: "This implies that xA(t) − xB(t) = a0 − b0, which is a constant, independent of time. In other words, the distance L remains the same. This argument applies to all types of synchronous motion."

 

They go on to show that "when switching the description to the comoving frame, the distance between the spaceships appears to increase by the relativistic factor [gamma]. Consequently, the string is stretched."

 

 

In other words: If the front and back of the ship can be accelerated using synchronous motion, then yes, the length of the ship will remain the same, and will appear to remain the same for observers which see the front and back synchronized.

 

It should be easy to show the following: Iff all points on the ship can be accelerated using synchronous motion, observers which see it synchronized would see no length-contraction deformations of the ship. Otherwise, sections in between the synchronized points would stretch in accordance with the wikipedia article.

 

...and the bolded will be true for Betty and Fiona on Dock at rest in the original frame only. No synchronization of acceleration is possible for Bert or Frank.

[JVNY]

It is possible for a row of objects to accelerate and maintain their proper distance in a single shared accelerating reference frame, although there is a limit of front end acceleration for a given proper row length beyond which rearward parts of the row cannot keep up and so fall behind (and out of the others' frame). A good explanation is here:

 

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

 

A very comprehensive article (behind a pay wall) is Taylor and French, "Limitation on proper length in special relativity," Am J Phys 51 (10), Oct 1983, page 889.

[md65536]

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

Born rigidity seems applicable. The original question might be expressed as "what happens if a Born rigid system stops accelerating and comes to relative rest?"

 

In that case I think the timing could be set up so that one observer sees the length remaining constant, but both observers couldn't. I think this agrees with what J.C.MacSwell wrote. The two observers couldn't remain in sync, as described in the link: "although the proper distances with respect to the instantaneously co-moving reference frames remain constant, the proper times of the different parts of the object do not remain coherent. In other words, if we contrive to hold the spatial relations fixed during an acceleration, a phase shift is introduced between different parts of the object, just as, if the phase is held constant, there is spatial stretching."

 

 

Edit: But then, according to the main observer, the other suddenly goes from having a relative velocity to being at relative rest. I don't know how the change in length contraction and relative simultaneity would appear or if it could be compensated for. There must be a simpler and more interesting way to look at this question.

Edited December 14, 2013 by md65536

[Toffo]

That was my point, though I don't see it as a contradiction.

 

Any nonzero distance (as measured in every current rest frame) between two objects cannot be maintained while they are accelerating (in the direction of the distance between them) while they stay in shared rest frames.

 

A distance (a maximum of which is limited by their separation and rate of acceleration...ie. Frank can't break the laws of physics) can be maintained with respect to the rest frame of one or the other, but not both.

I don't know if this has been said but:

 

When they think that they are staying in a shared rest frame, then they think the distance between them stays the same.

 

When they think that they are not staying in a shared rest frame, then they think the distance between them changes.

 

 

shared rest frame = no relative velocity = no change of distance

Born rigidity seems applicable. The original question might be expressed as "what happens if a Born rigid system stops accelerating and comes to relative rest?"

 

In that case I think the timing could be set up so that one observer sees the length remaining constant, but both observers couldn't. I think this agrees with what J.C.MacSwell wrote. The two observers couldn't remain in sync, as described in the link: "although the proper distances with respect to the instantaneously co-moving reference frames remain constant, the proper times of the different parts of the object do not remain coherent. In other words, if we contrive to hold the spatial relations fixed during an acceleration, a phase shift is introduced between different parts of the object, just as, if the phase is held constant, there is spatial stretching."

 

 

Edit: But then, according to the main observer, the other suddenly goes from having a relative velocity to being at relative rest. I don't know how the change in length contraction and relative simultaneity would appear or if it could be compensated for. There must be a simpler and more interesting way to look at this question.

 

 

Two spacecrafts accelerate so that they think the distance stays the same.

 

Picture:

=> =>

====>

 

 

Outside observer says the distance shrinks.

 

Then an engine break happens to both crafts at the same time in the outside observer's frame.

 

The last craft collides to the rear of the first craf. The collision speed is the contraction speed at the time of the engine break.

Edited December 15, 2013 by Toffo

[md65536]

When they think that they are not staying in a shared rest frame, then they think the distance between them changes.

Unless their timing and acceleration is set up to exactly counteract length contraction from their relative velocity, which is the condition of "Born rigidity."

[Toffo]

Unless their timing and acceleration is set up to exactly counteract length contraction from their relative velocity, which is the condition of "Born rigidity."

 

Yeah. For example: Constant proper acceleration for a leading space ship, a little bit larger constant proper acceleration for the space ship following the first one.

Unless their timing and acceleration is set up to exactly counteract length contraction from their relative velocity, which is the condition of "Born rigidity."

 

Actually there's no room for any unlesses here.

 

Same frame = no relative motion = rigidity

[Toffo]

Let's do some math.

 

First an accelerating rocket at non-relativistic velocity. We are interested about the shrinking velocity of the rocket according to an outside observer.

 

At non-relativistic velocities: coordinate acceleration = proper acceleration.

 

Let's say the acceleration is constant.

 

shrinking velocity at some time from the beginning of the acceleration = time * (rear's proper acceleration - front's proper acceleration)

 

The term (rear's proper acceleration - front's proper acceleration) can be written as: length*acceleration*some constant.

 

"Some constant" might be the gravitational constant, definitely the gravitational constant must be there.

 

 

Hey I think I can do the relativistic case too:

 

shrinking velocity at time t from the beginning of the acceleration = integral from 0 to t ( (rear's proper acceleration - front's proper acceleration) / gamma^2)

 

 

(gamma squared because the acceleration drops as gamma^2, a velocity change is smaller AND it takes a longer time in the outside observers frame)

 

 

 

ADDITION: Gravitational time dilation can be thought as causing the differences between front and rear, and I think there's should be some very simple equation regarding that.

Edited December 15, 2013 by Toffo

[Sensei]

md65536, do you saw Bell's spaceship paradox?

http://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox

[md65536]

md65536, do you saw Bell's spaceship paradox?

Yes. I think Born rigidity is more applicable, as per your link:

'The mathematical treatment of this paradox is similar to the treatment of Born rigid motion. However, rather than ask about the separation of spaceships with the same acceleration in an inertial frame, the problem of Born rigid motion asks, "What acceleration profile is required by the second spaceship so that the distance between the spaceships remains constant in their proper frame?"'

 

I think that I didn't know enough about what I was talking about in post #1 to even ask the right question. For example, what measure of distance are we talking about when saying that it "appears constant"? I think it would make sense to ask something like, is it possible to keep a ship at 1 lightsecond away, such that the time of any return signal from the observer to the ship is always 2 seconds (ie. using radar distance).

 

Suppose we have the observer sending out signals, and it does the instantaneous acceleration, and then continues sending out the signals. Then the question could be rephrased: say we place mirrors at various points, one for each signal sent, so that it is returned to ship with a return-trip time of 2 seconds; is it possible for a single ship to pass through each of those points at the time the signal is reflected, such that a single mirror could be used?

 

 

But if "same distance" refers to the instantaneous distance given by the Lorentz transformation, that's a different question.

 

 

Edit: I think Bell's paradox and Born rigidity refer to lengths as given by the Lorentz transformation. If the question is asking about radar distance, I think the answer might be easier... I think that as long as the observed ship *appears* synchronized with the observer, both the distance and Doppler shift can *appear* constant ...... but the more I think about it, the less certain I am.

Edited December 15, 2013 by md65536

[JVNY]

md65536,

 

Here is a good exercise to start to work through the questions. It uses instantaneous acceleration, even though it cannot generate exact Born rigid motion (and even though many scientists say instantaneous acceleration is not physically possible). It comes up with answers between Born rigid motion and Bell's spaceship paradox. But it has the advantage of simplicity.

 

Start with Bert (back ship) and Frank (front ship) at rest with respect to each other, separated by 100 in their frame, their clocks synchronized in their own frame showing time 0, and in inertial 0.6c motion relative to the ground. Q1: how do they present in the ground frame (what is the ground distance between them, and what does Bert's clock read in the ground frame if Frank's clock reads 0 in the ground frame).

 

Next, at their time 0 and simultaneously in their reference frame, each accelerates instantaneously to 0.8c relative to the ground.

 

Ground frame questions:

 

Q2: in what order do they accelerate in the ground frame? Q3: how much ground time passes between the first of them accelerating and the second of them accelerating? Q3: how much ground distance apart are they when both have accelerated? Q4: what does each clock read in the ground frame at the instant that the second of them accelerates?

 

Ships' frame questions: After their accelerations (which were simultaneous in their own frame), when they are both at rest with respect to each other and in the same frame in inertial movement relative to the ground at 0.8c:

 

Q5: how far apart are they in their own reference frame? Q6: are their clocks still synchronized, and if not whose clock is ahead in their frame, and by how much?

[md65536]

It uses instantaneous acceleration, even though it cannot generate exact Born rigid motion (and even though many scientists say instantaneous acceleration is not physically possible).

I think that is fine because mathematically there is an equivalent... Continuous acceleration can be handled with integration of infinitesimal instantaneous accelerations. Mathematically, instantaneous acceleration must be acceptable at some level, even if it is also infinitesimal. We still have to be careful not to describe physically impossible situations, but abstractly this is okay, if we don't require any absolute physical instantaneity but are simply speaking of infinitesimal or negligible times, and we can set it up so that arbitrarily large durations can be ignored! ... if we're careful.

 

Start with Bert (back ship) and Frank (front ship) at rest with respect to each other, separated by 100 in their frame, their clocks synchronized in their own frame showing time 0, and in inertial 0.6c motion relative to the ground. Q1: how do they present in the ground frame (what is the ground distance between them, and what does Bert's clock read in the ground frame if Frank's clock reads 0 in the ground frame).

 

Next, at their time 0 and simultaneously in their reference frame, each accelerates instantaneously to 0.8c relative to the ground.

 

Ground frame questions:

 

Q2: in what order do they accelerate in the ground frame? Q3: how much ground time passes between the first of them accelerating and the second of them accelerating? Q3: how much ground distance apart are they when both have accelerated? Q4: what does each clock read in the ground frame at the instant that the second of them accelerates?

 

Ships' frame questions: After their accelerations (which were simultaneous in their own frame), when they are both at rest with respect to each other and in the same frame in inertial movement relative to the ground at 0.8c:

 

Q5: how far apart are they in their own reference frame? Q6: are their clocks still synchronized, and if not whose clock is ahead in their frame, and by how much?

d_rest = 100

v = 0.6

gamma = 1.25

 

Q1:

d = d_rest/gamma = 80, the ground frame's length of the ship.

tBert' = 60, coordinate time at Bert, using Lorentz transformation with t=0, x=-80

 

You can also figure it out by figuring out how long it takes the ship to pass the ground observer in the different frames. How would you figure this out?

 

Q2: They both accelerate at respective proper time 0, so Bert accelerates first (60 units ago in Frank's frame).

 

Q3:

60*gamma = 75

 

Q3b: The original ground distance was 80. Bert has closed that distance at a rate of 0.8 - 0.6 = 0.2 in ground frame, for a time of 75, = 15. The new distance at time 0 is 65.

 

Q4:

Frank accelerates at Frank-time 0.

Bert accelerated 60 Frank-units ago, which is a time of 75 in ground frame, which is a Bert time of 45.

 

Is this what you've got, so far?

 

Sorry, I'm getting sloppier as I go.

Edited December 17, 2013 by md65536

[JVNY]

I get the same, thanks, and I apologize for getting the question numbering wrong (and for stipulating that the ships accelerate at 0 rather than 60 in their frame, making it a bit awkward). I will put together a Minkowski diagram shortly and think about your question about the time it takes for the ship to pass, but here is a summary and finishing off of question 5.

 

The ships initially present 80 apart in the ground frame, with their clocks not synchronized there. The rear (Bert) clock is ahead by the product of velocity and proper length, hence 100 * 0.6 = 60. Events that are simultaneous in the Bert/Frank frame (like clock striking 60) are not simultaneous in the ground frame. Events at the rear happen 60 units of Bert/Frank time earlier in the ground frame than events at the front. So Frank's clock reads 0 and Bert's 60 in the ground frame.

 

Now let's assume simultaneous acceleration in the Bert/Frank frame at their time 60 (since Bert is already at 60, and we don't want to have to go back in time . . .). Bert accelerates first in the ground frame (his clock reads 60 at inception). It takes 60 units of Frank time, with Frank traveling at 0.6, before Frank accelerates. That is a head start in the ground frame for Bert of 75 units of ground time, so Bert closes 15 of the separation in the ground frame as you calculated, leading to a 65 distance in the ground frame when Frank accelerates (thus ending Bert's head start and locking in the 65 ground distance).

 

During Bert's head start, he is traveling at 0.8, so his clock ticks slower in the ground frame than Frank's. As you note, Bert's clock ticks only 45 while Frank's ticks 60. The new gamma for Bert is 1.667 (because he is traveling at 0.8), and 75 ground time / 1.667 = 45. Thus, at the time Frank accelerates, Bert's clock reads 105 in the ground frame (60 at the start, plus 45). Frank's reads 60. The difference in clock times in the ground frame is 45 as you note.

 

The proper distance between the ships is the ground distance times gamma, or 65 * 1.667 = 108.333. Both ships are at rest with respect to each other (in the same reference frame). If the clocks are still synchronized in their frame, then Bert's should be ahead of Frank's in the ground frame by the product of proper distance and velocity, or 108.333 * 0.8 = 86.667. But it is only ahead by 45. So Bert's clock has fallen behind in their frame.

 

So, the example shows that:

 

Ships that accelerate at the same rate cannot maintain their proper distance. The proper distance increased by 8.333 in the example. The front ship has to accelerate at a lower rate to maintain proper distance. If the two ships accelerate Born rigidly, then they will maintain their proper distance at all times. Consistent with another post above, same proper distance at all times = at rest with respect to each other at all times = in the same reference frame as each other at all times = agree on simultaneity with each other at all times.

 

Their clocks will run at different rates in their own frame, admittedly. As the example shows, the rear clock actually runs at a slower rate in the ships' frame (thinking about this leads to the equivalence principle, with Bert being lower in the gravity well than Frank). But if they accelerate Born rigidly, with Frank accelerating at a lower rate than Bert, they can maintain their proper distance, stay at rest with respect to each other, remain in a common reference frame, and as a result agree on the simultaneity of events.

[phyti]

Try this.

In fig.1 A and B are separated by a distance d in frame U with synchronized clocks. At t=0 A and B accelerate to a target speed v. A measures the distance to B with light, which returns at Ut =2γγt, and t = d/c. Time dilation transforms (via the arc) Ut to At=2γt, which per SR convention, places the reflection R at R' relative to A. The gap has increased by a factor of γ.

If A and B were connected as a rigid body via em fields, d would have contracted, removing one order of γ, as required in the MMX. In this case, A and B move independently, so length contraction must be done by another method, such as delaying the acceleration for B.

In fig.2 after B delays acceleration, the A measurement places R at R' relative to A, with a gap=d. Since each have the same velocity, the results are reciprocal.

Note that the A clock will be slightly behind the B clock as a result of the delay.

With a = v/c, the delay t1 = d/a(1-1/ γ)

The acceleration was instantaneous for the purpose of simplification.