# Clarification about length contraction question

1. May 11, 2012

### bahamagreen

A space traveler is very happy to discover that after a long period of 1 g acceleration the measured distance to his destination has contracted substantially.

At some point he begins to plan for reversing thrust...

It come to him that if he reduces the relative speed between his craft and his destination, the contraction of the distance he has been happy about is going to diminish its effect.

It seems that as he slows the craft down, the space forward will be expanding, and the net distance to his destination might be increasing, in spite of the craft's constant progress toward his destination.

I've spent some time with some relativity calculator sites that show contraction factor scenarios - some of which result in changes of many light years in measured distance. None of them seem to be designed to look at this question...

Does the reduction in contraction effect (net distance expansion) always work out with the relative approach speed deceleration (distance covered) so that the math never allows the space traveler to believe he is not making headway toward his destination at all times?

2. May 11, 2012

### Ich

No.
As always, you can describe this effect in different ways.
For example (boringly) as changing aberration, making the destination look farther away. That describes what you see.
Or (way cooler) as the destination moving backwards in time, as it is behind the event horizon that comes from the immense gravitational field that permeates the universe as you accelerate. That describes coordinate behaviour in an accelerated frame.

Whatever, this is not too strange, as you are accelerating and therefore changing your view of the world. If the destination accelerated and instantly looked as if receding, that woud be cause for concern.

3. May 11, 2012

### bahamagreen

You answered "No", but your elaboration seems to say "Yes", then your conclusion states that "yes" would be a concern...? Could you clarify more?

4. May 11, 2012

### Staff: Mentor

There is no single standard reference frame for non inertial observers. You are going to have to specify exactly how you plan to construct the non inertial reference frame.

My favorite method is the Dolby and Gull approach:
http://arxiv.org/abs/gr-qc/0104077

5. May 11, 2012

### yuiop

Something to consider here is the what you will actually see when the relativistic aberration effect is taken into account. One way we perceive the distance to an object is by the visual subtended angle of the object. For example if we accelerated extremely rapidly to relativistic speeds towards the moon, the moon would shrink visually and give the visual impression of going away from the moon. As we approached the moon and decelerated rapidly preparing to land, the visual size of the Moon increases giving the impression we are accelerating towards it. This visual effect is almost the exact opposite of the calculated increased distance mentioned in the OP.

Also, as we slow down, crude radar measurements of the distance to the Moon will appear to be increasing because of time dilation effects. Altogether very disconcerting for a relativistic pilot!

Last edited: May 11, 2012
6. May 11, 2012

### Austin0

My guess would be that with a 1 g deceleration the calculated distance would continue to decrease but at a diminishing rate. The decrease in the gamma factor and the consequent increase in caculated distance would be less than the actual forward dispacement during any small interval of consideration.
But looking at ICMIF's at the point of maximum velocity and after a radical deceleration it seems like the calculated distance woud increase more than the forward translation during deceleration. Dissapointing to the traveler!
I didn't do any calculations so this is also just a guess.

Last edited: May 11, 2012
7. May 11, 2012

### bahamagreen

OK, good. It is pretty strange.

Dalespam - yes, while the deceleration is on the observer's reference frame will be non-inertial.

With that in mind...

What if the space traveler's navigator suggests that the reverse thrust during the deceleration phase be turn off periodically so that their frame is then inertial and they can take some measurements.

Would subsequent inertial frame measurements (or any portion of the series of them) indicate that the distance to the destination was increasing, in spite of making headway continuously toward the destination during the deceleration phase?

Sort of like the GPS telling you it is more miles to go than the last estimate every time you pause at a rest stop...

8. May 11, 2012

### Austin0

In your great thought experiment it is not necessary to stop acceleration. You can simply do calculations based on an instantaneously co-moving inertial frame (ICMIF)
As others have pointed out hypothetical measurements within such a system are problematic. It is much easier to analyze from the destinatin frame and transform the calculations to two ICMIFs at different points in the deceleration to figure out what they would calculate for distance.
I think the result would depend on the magnitude of acceleration and the remaining distance to the destination.

9. May 11, 2012

### pervect

Staff Emeritus
My favorite is the Fermi Normal frame. The good news is that Newton's laws work correctly in the entire region the frame covers. The bad news is that the frame doesn't (and can't) cover all of space-time.

The Dolby-Gull approach is rather ingenious in that it covers all of space-time, but Newton's laws will not work uniformly with this approach.

MTW covers the Fermi Normal frame in detail, there are a few online references that I don't have handy at the moment that talk about it a little.

10. May 11, 2012

### Staff: Mentor

The traveler's frame is still non-inertial, even if it has inertial segments. So this doesn't avoid the question I posed of specifying exactly how you are calculating the non inertial frame.

11. May 11, 2012

### Staff: Mentor

As far as I know, the Fermi normal coordinates can only be constructed around a geodesic. So I think they cannot be used for a non inertial worldline, although they can be used in curved spacetime.

12. May 11, 2012

### yuiop

For the constant proper deceleration case, the remaining distance calculated in the ICIRF appears to be always decreasing for increasing t, where t is measured in the rest frame of the destination object. I think the qualitative conclusion will be the same when using the proper time of the decelerating rocket.

I concluded this by plotting this equation for the remaining ICIRF distance (D) for various values of a:

D = d*√(1-v2/c2) = c2/a*(√(1+(a*t/c)2)-1)*√(1-(at)2/(c2+(a*t)2))

where v is the instantaneous velocity at time t. For the decelerating case plot negative values of t.

See http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken] for the source of the equations.

It appears that for very high initial velocities and extreme deceleration, the remaining ICIRF distance appears almost constant for a while, until the velocity drops below about 0.8c and then the remaining ICIRF distance appears to drop off fairly quickly as the destination is approached. It appears that at least for constant proper deceleration, that the maths does indicate the rocket is making headway (as measured in the ICIRF) toward the destination at all times

Last edited by a moderator: May 6, 2017
13. May 12, 2012

### Austin0

If the ship is 1 ly from the destination in that frame , with a velocity of 0.8c then decelerates to a velocity of 0.5c in 0.1 ly it would seem that the internal distance calculations would change from 0.6 ly at 0.8 to 7.8 ly at 0.5
I.e. an increase. Is this not so?

Last edited by a moderator: May 6, 2017
14. May 12, 2012

### yuiop

Yes, it would seem to be so. I can only guess that the result of the constant deceleration equations is different because the deceleration rate is designed to bring the rocket exactly to rest at its destination with constant proper acceleration. You deceleration rate is more extreme and if maintained would bring the rocket to rest short of its destination. That may explain the different results (or I got the equation wrong).

15. May 12, 2012

### Austin0

I think you are right. That's why I said "I think the result would depend on the magnitude of acceleration and the remaining distance to the destination".
I assume your calculations are correct which certainly seems to answer the basic question fully.

Last edited: May 12, 2012
16. May 12, 2012

### bahamagreen

Very nice! The mechanics of the "paradox" is the comparing of distance measures to destination from different frames?
If I understand you, the navigator's measurements during an individual inertial segment while coasting are good measurements valid throughout the duration of that individual coasting segment, and while observing from within a particular coastal segment the contraction effect will be observed to be constant.
But, any distance measures from previous coastal segments compared to a present coastal segment measure must be viewed as comparing data collected within different source frames. So the implication of an anomalous comparative increase in distance to destination can't be made.

I'm still trying to catch up with the subsequent posts... like I'm in a different frame.

17. May 12, 2012

### yuiop

This is a new calculation, this time with constant coordinate acceleration (A), velocity (v) and distance (d) all measured in the rest frame of the destination.

The equation for the ICIRF distance (D) for this case is:

D = d*√(1-v2/c2) = A*t2/2*√(1-(At/c)2)

This time it appears that for any constant coordinate deceleration that brings the rocket exactly to rest at the destination, the remaining ICIRF distance increases, while the velocity is greater than about 0.82c.

See the attached example curve which seems to preserve the same shape with a peak distance at about 0.82c for any deceleration A. The red line is the velocity of the decelerating rocket and the blue curve is the remaining distance (with the destination on the right). I wonder if there is a way to work out what this magical critical velocity is?

{edit} Should of worked it out sooner. Just find the derivative of the distance equation and set it equal to zero to find the value of t at the peak, and then multiply by A to get the velocity and it turns out the critical inversion velocity = √2/√3 c ≈ 0.81649 c and is independent of the constant deceleration that brings the rocket to rest at its destination.

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Last edited: May 12, 2012
18. May 12, 2012

### Ich

Sorry for the confusion.
I meant: no, there's nothing that prevents this from happening, there are cases where braking makes the destination "move further away".
The "moving away" part is what I meant when I said "this effect..." later.

I messed up the acceleration part, thanks yuiop for correcting.

The comment on "accelerating frames" referred to the Rindler frame.

"yes" would only be a concern if it happened due to the destination accelerating, because an accelerating body must look the same as an identical momentary comoving (not accelerating) body.

19. May 12, 2012

### Staff: Mentor

No, this isn't at all what I am saying. I am saying that you have to completely specify your method of determining "the reference frame" of an non-inertial observer. There is no standard way of doing so. Furthermore, there is no guarantee that the method will agree with the corresponding inertial frame during the coasting segments.

Once you have specified your method then you can use that method to determine the distance to the destination as a function of time. But you have to specify it first.

EDIT: although your reply wasn't what I was trying to say, it is correct. Instead of using a non inertial frame you can use any inertial frame, including ones where the traveler is at rest during part of the trip. But you cannot compare distance measurements across those frames.

Last edited: May 12, 2012
20. May 13, 2012

### yuiop

I agree with Dalespam that comparing ICIRF distances with changing proper acceleration is not a sensible or meaningful measurement. Better to calculate the remaining distance in the rest frame of the destination or calculate the estimated remaining proper time to arrival (which is probably the measurement the traveller would be most interested in). Both these measurement reduce in a consistent fashion as the rocket decelerates on approach to its destination.