Clarification about length contraction question

[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?

 

[Ich]

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?

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.

 

[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?

 

[Dale]

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

 

[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!

 

[Austin0]

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?

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.

 

[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...

 

[Austin0]

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...

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.

 

[pervect]

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

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.

 

[Dale]

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.

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.

 

[Dale]

My favorite is the Fermi Normal frame.

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.

 

[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 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

 

[Austin0]

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 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

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?

 

[yuiop]

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 0.78 ly at 0.5
I.e. an increase. Is this not so?

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).

 

[Austin0]

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).

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.

 

[bahamagreen]

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.

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.

 

[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.

 

[Ich]

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

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.

 

[Dale]

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.

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.

 

[yuiop]

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.

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.

 

[Austin0]

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.

If you calculate the remaining distance or time in the rest frame of the destination how do you convert those figures into meaningful measurements for the traveler except through transformation to ICIRFs?
I thought what you were doing in your last calculation was calculating coordinate translation resuting from coordinate acceleration in the rest frame into transformed measurements for the traveler. Are you now thinking that those results are not what would be calculated onboard ?
As the Lorentz math is the fundamental evaluation of spacetime I find it hard to picture how calculations in an accelerating system could differ from the calculations of that spacetime in the related IMIRFs as regards distance and time outside the system [with the inherent internal problems of clock synch etc.} How is this possible??

 

[Dale]

As the Lorentz math is the fundamental evaluation of spacetime I find it hard to picture how calculations in an accelerating system could differ from the calculations of that spacetime in the related IMIRFs as regards distance and time outside the system [with the inherent internal problems of clock synch etc.} How is this possible??

Have you read the Dolby and Gull paper I linked to? Pay particular attention to figures 5 and 9.

 

[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?

Have you read the Dolby and Gull paper I linked to? Pay particular attention to figures 5 and 9.

Hi I am trying to get a handle on the paper which is addressed to the mathematically advanced so is hard for me .
Since you understand it I have a question.
In this simple scenario would the radar distances be different from the above??
If it is given that the system was inertial before deceleration and after ,would the radar distances be different from the above??
Thanks

 

[Dale]

Yes, the radar distances would be different. Although you can see it in finite acceleration cases, it is easier to see in the instantaneous case. In that case, the MCIRF distance suddenly "jumps" whereas the radar distance is continuous even in that case.

 

[Austin0]

Yes, the radar distances would be different. Although you can see it in finite acceleration cases, it is easier to see in the instantaneous case. In that case, the MCIRF distance suddenly "jumps" whereas the radar distance is continuous even in that case.

Hi Yes the jump definitely seems to conflict with the absence of any actual displacement
in between. But still isn't the MCIRF distance the true spacetime distance?
If the system continues inertially to the destination, wouldn't the elapsed proper time agree with the time calculated from that distance and velocity?
If the radar distance has a different value does that mean that it would work as a basis for predicting arrival time? If that is the case then it would seem the MCIRF distance is invalid and would open up a lot of questions wouldn't it??RADAR TIME AND RADAR DISTANCE
Consider an observer traveling on path [itex]\gamma[/itex]
with proper time [itex]\tau[/itex]. Define:
[itex]\tau[/itex]⁺(x) ≡ (earliest possible) proper time at which a light ray
(technically, a null geodesic) leaving point x could intercept .
[itex]\tau[/itex]^-(x) ≡ (latest possible) proper time at which a light ray
(null geodesic) could leave [itex]\gamma[/itex], and still reach point x.

[itex]\tau[/itex](x) ≡ 1/2 ([itex]\tau[/itex]⁺(x) + [itex]\tau[/itex]^-(x)) = ‘radar time’.

p(x) ≡ 1/2 ([itex]\tau[/itex]⁺(x) - [itex]\tau[/itex]^-(x)) = ‘radar distance’.


Could you explain this definition of radar distance. The radar time makes sense but since time and distance are equivalent measured as a factor of c (ls) what is the meaning of the separate definition and the subtraction involved??
This doesn't seem to make sense as a factor of c.
Thanks

 

[Dale]

Hi Yes the jump definitely seems to conflict with the absence of any actual displacement
in between. But still isn't the MCIRF distance the true spacetime distance?

There is no such thing as the true spacetime distance. Distances depend on the frame.

If the system continues inertially to the destination, wouldn't the elapsed proper time agree with the time calculated from that distance and velocity?
If the radar distance has a different value does that mean that it would work as a basis for predicting arrival time? If that is the case then it would seem the MCIRF distance is invalid and would open up a lot of questions wouldn't it??

You can use all frames to make this calculation, and they will all agree. An inertial object follows a path called a geodesic, and geodesics can be calculated for any frame, inertial or not. Once you have the geodesics calculated then the proper time is the spacetime interval along the geodesic, which is frame invariant.

Could you explain this definition of radar distance. The radar time makes sense but since time and distance are equivalent measured as a factor of c (ls) what is the meaning of the separate definition and the subtraction involved??
This doesn't seem to make sense as a factor of c.

They are using units where c=1. The subtraction is correct. That is how radar, laser range finders, and even ultrasound all work. Try to just work out the following problem intuitively and see if you agree with the formula.

Two synchronized inertial clocks are at rest wrt each other. The first fires a radar pulse at t=100ns and receives the echo at t=120ns. How far apart are the clocks and what time did the second clock read when the pulse reached it? (use c=1ft/ns)

 

[Austin0]

Yes, the radar distances would be different. Although you can see it in finite acceleration cases, it is easier to see in the instantaneous case. In that case, the MCIRF distance suddenly "jumps" whereas the radar distance is continuous even in that case.

The more I think about it the more interesting the whole idea of radar time and the OP's question becomes.
Do you know of a more simplified explication of the principle and application of radar time?
I didn't find any with google
Thanks

 

[Austin0]

Hi Yes the jump definitely seems to conflict with the absence of any actual displacement
in between. But still isn't the MCIRF distance the true spacetime distance?

There is no such thing as the true spacetime distance. Distances depend on the frame.

Yes I misused a word again. I meant spacetime interval.as I attempted to make clear from the distinction in the same sentance between distance and spacetime distance and in the following.(below)

But isn't a spacetime interval just another way of saying a spacetime distance on a worldline? The spacetime difference between two points? That being said I will remember in the future.

If the system continues inertially to the destination, wouldn't the elapsed proper time agree with the time calculated from that distance and velocity?
If the radar distance has a different value does that mean that it would work as a basis for predicting arrival time? If that is the case then it would seem the MCIRF distance is invalid and would open up a lot of questions wouldn't it??

You can use all frames to make this calculation, and they will all agree. An inertial object follows a path called a geodesic, and geodesics can be calculated for any frame, inertial or not. Once you have the geodesics calculated then the proper time is the spacetime interval along the geodesic, which is frame invariant.

This is what was meant by true spacetime distance. A frame invariant measure.
So this seems to mean that the radar distance would not be the basis of an invariant interval??

Boy do I feel like an idiot . I looked at the Definitions and for some reason got the idea that the proper times they were referring to were intervals for travel to x and for return ,not proper clock readings. Once fixed on that interpretation I couldn't make sense of it or see the correct interpretation
Thanks
ps I thought non-inertial frames did not travel on geodesics . ?

 

[yuiop]

If you calculate the remaining distance or time in the rest frame of the destination how do you convert those figures into meaningful measurements for the traveler except through transformation to ICIRFs?

I guess meaningful is a subjective term. At least the distance in the destination rest frame is continually getting shorter as the rocket heads towards the destination which is the natural expectation, as is the remaining proper time to estimated time of arrival. If we define "heading towards an object" as relative movement in which the distance may or may not be getting shorter over time, then that expression becomes meaningless.

I thought what you were doing in your last calculation was calculating coordinate translation resuting from coordinate acceleration in the rest frame into transformed measurements for the traveler.

I thought that is what I was doing and still do.

Are you now thinking that those results are not what would be calculated onboard ?

What calculations they choose to carry out onboard is entirely up to them. I would assume most people would traveling long distances would be most interested in the remaining proper time to their destination.

As the Lorentz math is the fundamental evaluation of spacetime I find it hard to picture how calculations in an accelerating system could differ from the calculations of that spacetime in the related IMIRFs as regards distance and time outside the system [with the inherent internal problems of clock synch etc.} How is this possible??

By calculating the remaining distance in the ICIRF, we are continuously comparing distances in different reference frames and since those measurements are relative, we cannot in an unambiguous way, say those distances are getting shorter or longer. It seems to be much more sensible to choose a single inertial reference frame to make all measurements of remaining distance and if we do this the remaining distance never gets longer as we "head towards" or approach the destination. The remaining proper time is another sensible measurement that always gets shorter as we approach the destination and this quantity is observer independent.

 

[Austin0]

I guess meaningful is a subjective term. At least the distance in the destination rest frame is continually getting shorter as the rocket heads towards the destination which is the natural expectation, as is the remaining proper time to estimated time of arrival. If we define "heading towards an object" as relative movement in which the distance may or may not be getting shorter over time, then that expression becomes meaningless..

I thought that is what I was doing and still do.What calculations they choose to carry out onboard is entirely up to them. I would assume most people would traveling long distances would be most interested in the remaining proper time to their destination..

well I am glad to hear I understood your method correctly although I misinterpereted your later comments. I got th impression you were perhaps thinking some other metod was more appropriate. Radar distance or some other method. Or that the MCIRF figures didn't accurately reflect reality.

By calculating the remaining distance in the ICIRF, we are continuously comparing distances in different reference frames and since those measurements are relative, we cannot in an unambiguous way, say those distances are getting shorter or longer. .

It seems to me that simply by expressing the current position as a percentage of the total trip distance you have a sensible chart that shows steady movement toward the destination That this would be true even with radical accelerations of any magnitude.
That the current MCIRF may show an increased distance but the total distance would increase by the same factor so the charted position would still be closer to the destination.
SO the passengers in the relativistic jet liner ,with a reduction in velocity, would see not only the changing time remaining that we are used to with bad winds, but would also see an increased distance to destination
. But the computer graphic of trip position would continue to progress.

It seems to be much more sensible to choose a single inertial reference frame to make all measurements of remaining distance and if we do this the remaining distance never gets longer as we "head towards" or approach the destination. The remaining proper time is another sensible measurement that always gets shorter as we approach the destination and this quantity is observer independent.

Yes a single frame such as destination frame would measure a decreasing distance.
But I don't think that is true of decreasing proper time to destination.
With a decrease in velocity it seems inevitable there would be an increase in proper time relative to the destination (or any point) independent of the reference frame with which you measure distance. Is this not so?
The only way I have figured to have a steadily decreasing time to arrival is through calculating the whole trip. In the conditions of your calcs: (with constant proper acceleration to midpoint and comparable deceleration to destination)
The whole trip could be charted in advance. A table relating any instant of proper time to velocity relative to destination frame. Position in that frame . MCIRF at that instant. etc.
SO ETA and current proper time to arrival would be accurate to the degree of exact acceleration and total calculated proper time for the trip. WIth a continuously decreasing time to arrival (barring unforseen changes in acceleration).
An astrogation computer should be able to use accelerometer measurements to correct for deviations from scheduled accelerations if there are changes during flight. yes?.

Anent your earlier mention of the problem of using radar when approaching the moon:The problem of decreasing dilation between emission and return skewing measurements
Wouldn't it be possible to have the computer integrating over the time between emission and return ,to arrive at an accurate proper time for that distance and then extrapolating from that point with the relevant acceleration to project a current distance and a time distance arrival relative to that acceleration? Or something.


Thinking over your figures with the increase in apparent distance reduction from v=0.82 down. It seems like the subjective experience of the passengers would be exactly the opposite. Subjective trip time relative to progress would be continuously expanding from the point of maximum velocity I.e. time would seem to be slowing down and boredom rising ;-)