Transforms within a shared frame.

In the quest to understand the transforms a little better as suggested by relativity I wanted to shift the focus on to length or distance instead of time.

So the following scenario was constructed as an aid in this understanding.
With the help of this little diagram I drew up:

<img src=http://www.paygency.com/sharedframetrans.jpg>

We have a space port for ships in a stationary position relative to the sun. [A]
We also have a space port 10 lys away also stationary to it’s star.[B]

Over the last 50 years the position has been determined as stable relative to us by routine light signals bouncing back and forth over this 10 ly distance.

On New years eve, the start of the year 2020, a signal was sent to tell our outpost that a ship [SS]will be leaving our space port here and that at the time of receipt of this signal the ship will start it’s journey at the velocity of 0.8c.

Ten years later the outpost receives the signal [2021]and knows that as they received the signal a ship [SS]was starting it’s journey towards them.

At [A] exactly 12.00am the year 2021 [SS] flys by at the rate of 0.8c

Now we have two outposts [A and B] both sharing the same frame of reference in time and space.

The ship [SS] at this velocity will reach [B] in 12.5 years according to [A and B] arriving and flying past [B]at v=0.8c in July the year 2033 which date is shared exactly by both out posts.

1] What distance does [SS] travel according to the ships observer frame?

2] What is actually happening to the space in between our outposts A and B
according to [A and B's] perspective ?
and
3] according to [SS] perspective.?

4]How long does space port [B] have to wait before the ship arrives and flys past?

1] What distance does [SS] travel according to the ships observer frame?

For a speed of 0.8c, the gamma factor is 1.667.
In the spaceship frame, the distance travelled is 10 ly/1.667 = 6 ly.

2] What is actually happening to the space in between our outposts A and B
according to [A and B's] perspective ?

Nothing happened to the space. The distance between A and B in this frame is 10 ly, as specified.

and
3] according to [SS] perspective.?

I'm not sure what happened to the space. However, the distance between A and B was measured to be 6 ly.

4]How long does space port [B] have to wait before the ship arrives and flys past?

12.5 years.

And to answer your next question, the time taken for the trip according to a clock on the spaceship is 7.5 years.

hmmmm....thanks for that shall think on my next question

so according to the ship it takes 7.5 years to go from [A] to [B] which in normal space is 10 ly.

it is interesting to note at this point that a photon will take 10 years.

Say we put inside this ship a special box that allows light to accumulate 10 lys of travel inside it [inside the ship].....I assume that the light inside the box would travel 10 ly equivelent to normal space.?

So that when the ship arrives after it's 7.5 year of travel the light inside the box would have traveled 10 years.....

doesn't that show that if light is invariant that we have a way of absolute reference? Or does that show the transforms as unworkable?
edit: oops......got it wrong.....

so according to the ship it takes 7.5 years to go from [A] to [B] which in normal space is 10 ly.

it is interesting to note at this point that a photon will take 10 years.
In the ship reference frame it takes 6 ears for the photon to travel fromA to B.

Furthermore, I don't know what you mean by "normal space"

Say we put inside this ship a special box that allows light to accumulate 10 lys of travel inside it [inside the ship]

Relative to who?

This isn't possible if the 10 years is measured inside the ship, since the ship only travels for 7.5 years in that frame.

Of course, if you're timing the light in the box from outside, then a 10 year travel time is fine.

so just to confirm you are sayng that in the ships frame light takes 6 years to travel how far?

so james : How far would a beam of light travel in our ship by the time it passes [B]?

the answer to me should be 12.5 ly.............

so just to confirm you are sayng that in the ships frame light takes 6 years to travel how far?

6 light years.

How far would a beam of light travel in our ship by the time it passes [B]?

As measured by who?

jamesr, frame SS =

since the ship only travels for 7.5 years in that frame.

here in lies my argument

The ship only travels 7.5 years and light inside the ship according to the ship has travelled only 7.5 years, when it passes [B]

However for light to get from [A] to [B] it takes 10 light years.

How do you account for the difference of 2.5 years of light travel?

B's clocks run slower than A's.

ha...you have the twin paradox in mind.....A and B are a shared frame in this scenario......

SS's clock runs slower I guess is what you meant....
But I though the reason for teh transforms is to maintain invariance of lihts velocity.

The ship travels for 7.5 light years but achieves a fixed distance of 10 light years.

the distance is contracted to 7.5 years for the ship but not the A and B frame of 10 ly.

However frame A and B is determined by light to be 10 light years.

determined by light....is the important point.......

Invariance would state that the light on board the ship travelling at 0.8c would have to travel 12.5 Ly when it passes [B] by A and B's distances.

Do you see what I am getting at?

QQ:

ha...you have the twin paradox in mind.....A and B are a shared frame in this scenario......

SS's clock runs slower I guess is what you meant....

Yes. Sorry.

But I though the reason for teh transforms is to maintain invariance of lihts velocity.

All frames measure the speed of light to be the same.

The ship travels for 7.5 light years but achieves a fixed distance of 10 light years.

You're mixing frames.

In the Earth frame, the ship travels 10 light years. In the spaceship frame, the spaceship doesn't move, and the Earth travels 7.5 light years.

the distance is contracted to 7.5 years for the ship but not the A and B frame of 10 ly.

That's right.

However frame A and B is determined by light to be 10 light years.

In frame A/B.

Invariance would state that the light on board the ship travelling at 0.8c would have to travel 12.5 Ly when it passes [B] by A and B's distances.

Light on the ship travels at c, not 0.8c.

Do you see what I am getting at?

Not totally, no.

Ok.....lets go to the abstract for a minnie

We still have a rest frame with a distance of 10 ly

but this time our ship is traveling at 'c'

Now normally from A an B perspective the ship woudl take 10 years to get to B.

But from the ships perspective how long would it take?


And our beam of light on board our light ship has travelled how far by the time the ship passes B?

Now if I am not mistaken our beam of light hasn't travelled any distance on board our SS frame if velocity is 'c'.


well if we work back from this absolute in increments of say 1% of 'c' we will find that light is variant and not invariant using the transforms.


The whole issue revolves around the fact that Light takes 10 years to get to B and it is the invariance of light that determines how far it travels upon the ship..........
if the ship is travelling at 0.8 c then the light must because of invariance travel 12.5 light years.distance...regardless of frame....by the time it gets to B.....

But here we have a ship that has a ship time of 7.5 years by the time is get's to B if ship time is 7.5 years then light has only travelled 7.5 years aboard the ship frame.....when it passes the 10 light year distance marker....this indicates that our ships light beam is traveling considerably faster than 'c'

because we are dealing with a distance determined by light in the first place

QQ:

Ok.....lets go to the abstract for a minnie

We still have a rest frame with a distance of 10 ly

but this time our ship is traveling at 'c'

You can't really extrapolate to a reference frame travelling at c, so I will take the speed to be something very very close to, but a tiny bit less than c.

Now normally from A an B perspective the ship woudl take 10 years to get to B.

But from the ships perspective how long would it take?

Essentially zero time.

And our beam of light on board our light ship has travelled how far by the time the ship passes B?

As measured by A/B: 10 light years.
As measured by SS: essentially zero distance.

Now if I am not mistaken our beam of light hasn't travelled any distance on board our SS frame if velocity is 'c'.

Right.

well if we work back from this absolute in increments of say 1% of 'c' we will find that light is variant and not invariant using the transforms.

No. The speed of light is always c.

The whole issue revolves around the fact that Light takes 10 years to get to B and it is the invariance of light that determines how far it travels upon the ship..........

You're mixing the ship frame and the Earth frame. They are different.

if the ship is travelling at 0.8 c then the light must because of invariance travel 12.5 light years.distance...regardless of frame....by the time it gets to B.....

Wrong.

According to A/B, the light travels 10 light years.
According to SS travelling at 0.8c, the light travels 7.5 light years.

But here we have a ship that has a ship time of 7.5 years by the time is get's to B if ship time is 7.5 years then light has only travelled 7.5 years aboard the ship frame.....when it passes the 10 light year distance marker....this indicates that our ships light beam is traveling considerably faster than 'c'

You're mixing frames.

Guys,
I have posted a full set of formulas of SRT and explained the logic of application of them - see lecture "The practical usage of the Lorentz transformations" in the separate thread "One lecture on Twins Paradox" (10-31-04) specially to gave to anybody the tools to calculate exact answers on such problem as the present one. Why you still are keeping the attention of this Forum with the same kind of problems again and again? Why you are trying to force other memebers to do your elementary job of a simple set of algebraic calculations? What a sense it has?

so according to the ships captain youare stating that he would know the distance he has covered as 7.5 ly and not 10 ly..... so as far as he is concerned distances measured by light are irrelevant.?

the actual light distance between A and B is of no value to the SS frame.

so according to SS it has taken 7.5 years to cover 10 light years or does the light distance between A and B have no relevance.?

If the light distance between A and B are not releavant then what is the purpose of the transforms that end up in a FTL result.?

does the SS frame and frame [A and B] agree that the light distance to be covered is 10 ly?

QQ,
What do you mean by "light distance"?

In the spaceship frame, the distance between A and B is 6 light-years, and a beam of light takes 6 years to cover that distance.

You obviously can't see the fact that light is now variant over the two frames?

Please explain what you mean.

I believe Quantum Quack is beginning to see what I have seen. Possibly to make it a
little easier to see:
We will make frame 'A' the frame in which the distance traveled is 10 light years.
We will make frame 'B' the frame in which the distance traveled is 6 light years.

Frame 'A' = 95,000,000,000,000 kilometers.
What is frame 'B' in distance, 95,000,000,000,000 'short' kilometers (shortened meter)
or, 57,000,000,000,000 kilometers? Remember, the speed of light has to remain constant, and there is a followup question.

In the spaceship frame, the distance between A and B is 6 light-years, and a beam of light takes 6 years to cover that distance.

Sorry Pete, to understand what is happening, you have to state the 6 light-year
distance in METERS.

How pointless.

6 light-years = 57 x 10¹⁵m

Side note - Your terminology is a bit confusing.

I am assuming that when you say "What is frame 'B' in distance" you actually mean "What is the distance between the spaceports in frame B".



I also think you're getting yourself confused over the shortened metre issue. Strictly speaking, the metre is not shortened, but moving metre sticks are.

Here's a description which may help:

In the spaceport frame (frame A), the two spaceports are 10 light-years apart. Since the spaceports are stationary in that frame, this means we could put 10 sticks of length 1 light-year each end to end between the spaceports. Alternatively, we could put 95x10¹⁵ metre sticks end-to-end between the spaceports.

In the spaceship frame, those metre sticks are still there, and there are still 95x10¹⁵ of them. But since the sticks are moving in the spaceship frame, they are all shortened in that frame. Each metre stick that is stationary in frame A is only 60cm long in frame B. Naturally, this makes the total distance 95x10¹⁵ x 60cm, or 57x10¹⁵m

First, you then agree that 'B's clock is not running at a different tick rate than 'A's
clock, it is simply recording a different total elapse time due to FEWER meters traveled?
In either frame, for instance, 9,500,000,000,000 kilometers would be covered in one
light year on their respective clocks? Second, the distance is reduced (fewer meters)
in the direction of travel only for 'B's frame, and the meter remains the same (not contracted) in both 'A' and 'B's frame of reference, in all directions for both?

quote:

"In the spaceship frame, those metre sticks are still there, and there are still 95x1015 of them. But since the sticks are moving in the spaceship frame, they are all shortened in that frame."
================================================== ========

This is what I was asking, Pete. If there are still 95x1015 meters the spaceship travels,
how can the spaceship travel that distance in less than 12.5 years without changing
its velocity a great deal. 95x1015 meters in 7.5 years? The spaceship frame HAS to
measure the speed of light as constant, 299,792,458 of its meters (contracted or not)
in one (dilated or not) of its seconds.

This is what I was asking, Pete. If there are still 95x1015 meters the spaceship travels, how can the spaceship travel that distance in less than 12.5 years
It doesn't. In the Earth frame (the frame in which those 95x10¹⁵ metre sticks equals 95x10¹⁵m), the spaceship spends 12.5 years in transit.


95x1015 meters in 7.5 years?
No, 57x10¹⁵m in 7.5 years.
95x10¹⁵ short metre sticks doesn't make 95x10¹⁵m. It makes 57x10¹⁵m.

First, you then agree that 'B's clock is not running at a different tick rate than 'A's clock, it is simply recording a different total elapse time due to FEWER meters traveled?
B's clock ticks over 7.5 years during transit.
57x10¹⁵m in 7.5 years. That's shorter than 12.5 years, right?
Synchronize ship-frame clocks and Earth-frame clocks before leaving Earth.
When the ship arrives at the other end, the ship-frame clocks will read 7.5 years later, and Earth-frame clocks will read 12.5 years later.
How is that not a different tick rate?

In either frame, for instance, 9,500,000,000,000 kilometers would be covered in one light year on their respective clocks?
Covered by what? a beam of light? Yes, but you'll have to explain the relevance.

the meter remains the same (not contracted) in both 'A' and 'B's frame of reference, in all directions for both?
Metres are never contracted. A metre is a metre is a metre. Moving metre sticks are contracted - a moving metre stick is less than 1 metre long.

quote:

"Covered by what? a beam of light? Yes, but you'll have to explain the relevance."
================================================== ==========

Sorry, Pete. Imeant to state that in either frame, 9,500,000,000,000 x .8c =
7,600,000,000,000 meters would be traveled by the spaceship in one of either frames
years. No time dilation. The shorter elapse time would have to be due to FEWER METERS traveled, not the contracted meters of the 'moving' universe frame.

The shorter elapse time would have to be due to FEWER METERS traveled, not the contracted meters of the 'moving' universe frame.
To me they're the same thing.
Fewer metres travelled means that metre sticks of the moving frame are contracted.

I notice that you're implying that only one frame can claim to be stationary in the Universe. Interesting!

No time dilation. The shorter elapse time...
To me, they're again the same thing.
I'm not sure what you think that relativists mean when they say "time dilation"... but to me it equates directly to a shorter time for the spaceship's transit in the spaceship's frame.

If you agree that the spaceship clocks tick over less time during the trip than Earth clocks do, then you're agreeing to what I mean by time dilation:

The time between two specified events is shortest in the frame in which those events are in the same place, and longer in a moving frame.

In this exercise, I am not addressing the issue of only one stationary frame. That was
in my paradox thread. A shorter elapse time does not equate to 'moving clocks run slow' and the Special Relativity gedankins where one frame seeing the moving clock
as accumalating time at a slower rate than their own. The time differential would have
to occur because the end of the trip was reached more quickly because it contained
fewer meters. The tick rate of the moving clock would not change and the 'shorter
distance' traveled would be left dangling on its own with no 'slow running' clock to
back it up.

Pete,

Oh, oh. We are about to agree. that could be scary. :D

I'm not sure what you think that relativists mean when they say "time dilation"... but to me it equates directly to a shorter time for the spaceship's transit in the spaceship's frame.

If you agree that the spaceship clocks tick over less time during the trip than Earth clocks do, then you're agreeing to what I mean by time dilation:

If you are agreeable to the following language change:

"If you agree that the spaceship clocks accumulate less time during the trip than Earth clocks do, then you're agreeing to what I mean by time dilation:"

Recognizing that "What you claim is time dilation, is in reality nothing more than a shorter trip", then we can agree but it is in appropriate, infact flat wrong to infer and/or claim as otehrs do that "time dilation means the clock tick rate has slowed. It doesn't.

A strobe light set to flash once per second at both observer postions would show that the tick rate remained the same the flash interval will remain synchronized while the clocks accumulate different times.

Time has not been affected. Aging differential did not occur. That is the differance and it is a BIG differance.

In this exercise, I am not addressing the issue of only one stationary frame. That was
in my paradox thread. A shorter elapse time does not equate to 'moving clocks run slow' and the Special Relativity gedankins where one frame seeing the moving clock
as accumalating time at a slower rate than their own. The time differential would have
to occur because the end of the trip was reached more quickly because it contained
fewer meters. The tick rate of the moving clock would not change and the 'shorter
distance' traveled would be left dangling on its own with no 'slow running' clock to
back it up.

Your position is my position exactly. d = vt mandates that the tick rate (interval) remain equal otherwise the final reading on the clocks would not be as claimed.

The lesser clock reading is a function of length contraction and no affect on time has occured. If you consider that Relativity requires that the recipocal situation occurs from the other view, i.e. the other clock accumulates the lesser time that means no aging differential in the Twins Paradox, etc.

hey guys, it really is very simple.....not that hard to see...

The captain of [SS] gets off his ship and reprts to the captain of spaceport . They copmpare notes:

Captain B says that light has taken 10 years to get there from [A]
Captain [SS] says that light in his ship has only travelled 7.5 years to get there from [A]

end of stoiry.


If light is invariant then how can light travel only 7.5 years to get to were light normally takes 10 years to get to.

Now I repeat this statement read it carefully please:

If light is invariant then how can light travel only 7.5 years to get to were light normally takes 10 years to get to :confused:

You're wrong, QQ. Light always travels at the same speed. It never goes slower or faster.

You're wrong, QQ. Light always travels at the same speed. It never goes slower or faster.
well I know that JamesR but how do you explain the difference in the speed of light then?

if you say there is no difference then how is it that it takes only 7.5 years over the same distance.

lets refine the scenario a little:

IN our ship and at our destination are two boxes of mirrors that accululate light travel distances.

Box SS and Box B

the captains of both are looking at both boxes on the table in front of them at Space port B.

Box SS states distance travelled only 7.5 ly
Box B states distance travelled 10 ly

Both boxes are sitting on a desk 10 ly away from Space port A

so either the light distance is 7.5 or it is 10 ...which should we choose?

maybe we should just average them out andd call it quits hey? [chuckle]

if time was absolute and no dilations existed the light in box [ss] would travel 12.5 ly.

so we replace our boxes with a light clock that measures distance travelled regardless of time........now I know from teh many threads I have been involved in that light clocks reflect the contracted and dilated state of the clock according to the transforms......so the paradox remains......

What normally happens in a SR paradox thread is that the focus is on a rather intangible thing called time.....endless confusion.....MacM has been talking about perception illusions and such....now I know that the same paradox can be demonstrated for distance as well, howver distance is a little harder to disregard becasue it is very tangible. Especially if the distance is determined by a ray of light.

In this thread the distances are determined by light and have no regard to time.....Space port B is at a distance of 10 ly from A the distance is determined by light travelling in a vacuum. not by other arbitary means.

Time could be what ever you want it to be as long as the same time is applied to 'c' consistantly.....299792kpersecond.

Light travels from A to B in 10 light years.

In our ship the transfroms will suggest 7.5 years.....clearly the ship must be wrong in it's calculations or light is now deemed variant.

light for light..... distance for distance.......

so either we have two distances simultaneously therefore one is illusionary or I have stuffed up some where.....agin!!

so instead of synchronising clocks we have synchronised our distance

In this exercise, I am not addressing the issue of only one stationary frame. That was
in my paradox thread. A shorter elapse time does not equate to 'moving clocks run slow' and the Special Relativity gedankins where one frame seeing the moving clock
as accumalating time at a slower rate than their own. The time differential would have
to occur because the end of the trip was reached more quickly because it contained
fewer meters. The tick rate of the moving clock would not change and the 'shorter
distance' traveled would be left dangling on its own with no 'slow running' clock to
back it up.
Semantics.

If you accept that the clocks in one frame have shorter elapsed time than the other frame, then we agree.

pete you do understand that if this transform in a shared frame can't be resolved the clock issue is nul and void?

QQ:

Relativity says that time and distance are relative to the observer. You're starting to make the same mistakes as MacM, mixing up your calculations in multiple reference frames.

In your spaceship problem, in the Earth frame the distance travelled by the spaceship is 10 light years. It takes light 10 years to travel that distance at a speed of 1 light year per year. In the spaceship frame, the trip takes 7.5 years for the spaceship, and in that time it covers a distance of 6 light years at a speed of 0.8c. In the spaceship frame, light can cover the same distance in 6 years, at a speed of 1 light year per year.

In both frames, the speed of light is 1 light year per year. The speed of light is constant for all observers. It never changes.

QQ:

Relativity says that time and distance are relative to the observer. You're starting to make the same mistakes as MacM, mixing up your calculations in multiple reference frames.

In your spaceship problem, in the Earth frame the distance travelled by the spaceship is 10 light years. It takes light 10 years to travel that distance at a speed of 1 light year per year. In the spaceship frame, the trip takes 7.5 years for the spaceship, and in that time it covers a distance of 6 light years at a speed of 0.8c. In the spaceship frame, light can cover the same distance in 6 years, at a speed of 1 light year per year.

In both frames, the speed of light is 1 light year per year. The speed of light is constant for all observers. It never changes.


So is this the same as saying that the light on board the ship has travelled 7.5 light years?

I take it then that a lightyear is frame dependent...yes?

the point i ma tryuingto convey in all this is no matter what the frame light is supposed to be traveling the same distance, if length is contracted does this mean that light can still travel 299792kms of contracted kilometers?

Say we have 299792kms contracted by 20% due to velocity does this still equal 1 seconds travel.
I was under the impression that the transforms were designed to compensate for contractions an diilations so that a light meter stays a light meter regardless of ocntractions and dilations.

In other words a contracted 299792kms is travelled in >1 second to give us a universal constant or light meter? thus compensating for contraction and dilations.

QQ:

What stays the same is the speed of light. In any reference frame, the distance travelled by a light signal divided by the time taken will be equal to c = 299792458 m/s.

The distance travelled and the time taken can both vary, but will always do so in such a way that the speed of light is constant.

and the fact reamins that on the desk the two light distance clocks show 7.5 and 10 light years respectively.......why is that considered as ok?
[we are talking about light distance not ship distance]

The frame I am concentrating on is the light frame not the ship or the space ports.

The frame of the light [ss] and the frame of the light [ab] seem to be different?

maybe just to clarify the question I should ask:

JamesR:
How far has light aboard the ship travelled by the time it reaches [B]?

Now the answer I am expecting is 7.5 light years beacuse that's how long [time]the ship took to get to [B]

What stays the same is the speed of light. In any reference frame, the distance travelled by a light signal divided by the time taken will be equal to c = 299792458 m/s.

The distance travelled and the time taken can both vary, but will always do so in such a way that the speed of light is constant.

exactly my point......

off topic post deleted

the thing is we have two light frames.....and this is why I see a problem?

One light frame is 7.5 and the other is 10

Where's the problem? Distances don't have to be the same in each frame. Neither do times.

Semantics.

If you accept that the clocks in one frame have shorter elapsed time than the other frame, then we agree.

I think we have agreed that the contracted (shorter) meters belong in the
moving universe frame of reference. In the spaceship frame of reference,
how did its distance become less (fewer meters)? Are BOTH frames of
reference contracted?

so the invariant speed of liht has no grounding at all.....of course it is invariant they say becasue a meter can be any length necessary to make it so.....

JamesR

The length of 10 light years is a length that light travels.
YOu are now saying that 7.5 light years equals 10 light years simply because distance and time can change.....well tell me James Just what does invariant light speed mean....?

it also suggests tha the transforms produce meaningless results

According to [AB] How far has the light traveled inside the ship [SS]?

JamesR what I see you you suggesting is that a light year can be contracted...and if that is the case how does invariants becomes a reality?
Now I am talking about LIGHT DISTANCES, of distances traveled by light.

IN the ship teh distance travelled by light is 7.5 LIGHT years.
In the frame [AB] the lighttravels 10 LIGHT years.

you see no problem with that ?

QQ:

so the invariant speed of liht has no grounding at all...

It is a postulate of special relativity. That's its status in the theory. Experimentally, it has been verified often, so that is the real "grounding".

of course it is invariant they say becasue a meter can be any length necessary to make it so.....

No, that's not true. Any problem in relativity has one and only one correct solution.

The length of 10 light years is a length that light travels.
YOu are now saying that 7.5 light years equals 10 light years simply because distance and time can change.....well tell me James Just what does invariant light speed mean....?

I have never said 7.5 light years equals 10 light years. I have said that people in different frames measure 7.5 ly or 10 ly, depending on which frame.

Invariant light speed means the speed of light is the same in all reference frames.

it also suggests tha the transforms produce meaningless results

I suggest that you don't understand the transforms.

According to [AB] How far has the light traveled inside the ship [SS]?

10 light years, if it has been travelling continuously as the SS went from A to B.

JamesR what I see you you suggesting is that a light year can be contracted...

Well, not exactly. I am saying that the distance between two given objects can be contracted.

and if that is the case how does invariants becomes a reality?

Time dilates as well, in such a way that the speed of light is constant.

Now I am talking about LIGHT DISTANCES, of distances traveled by light.

IN the ship teh distance travelled by light is 7.5 LIGHT years.
In the frame [AB] the lighttravels 10 LIGHT years.

you see no problem with that ?

No. They are different frames.

by JamesR:


Well, not exactly. I am saying that the distance between two given objects can be contracted.


Time dilates as well, in such a way that the speed of light is constant.
================================================== =============

And that is the whole point of this thread, JamesR. Exactly what is this "in such a
way"? Contract the meter and dilate the second and neither distance nor travel time
changes, unless the ratio violates the constancy of the speed of light. Saying light
travels 1/2 meter in 1/599,584,916 seconds is no different than saying light travels
1 meter in 1/299,792,458 seconds. It still takes the same amount of time to travel
10 light years distance.

Just a shortnote this "paradox" if you like is now being worked through in a thread titled "SR - distance paradox"

This thread was originally posted to clarify the length contraction issue within a common frame and I think it has served it's purpose.

The issue of possible conflict has been spelt out in the other thread.

2inquisitive:

And that is the whole point of this thread, JamesR. Exactly what is this "in such a way"?

In the way dictated by the Lorentz transformations. It's not arbitrary. Any relativity problem has one and only one correct solution.