Let's see if I can describe this in words only:

  As we all know, the speed of light in vacuum is the same for all inertial
frames of reference, and because of that, movement in inertial frames of
reference is subject to Lorentz transformations.

  For this reason if there are two spaceships travelling to opposite
directions at speed s (from an external observer's point of view), if
ship A measures the speed of ship B as it perceives it, it's not 2*s,
but something less (and always less than c). It happens because ship B
appears contracted in the direction of the movement.

  Let's assume that ship A measures the speed of ship B by starting a clock
when the bows of both ships coincide, and stops the clock when the bow
of ship A coincides with the stern of ship B. For the sake of simplicity,
let's assume that, while both ships have the same length, ship A measures
the apparent length of ship B to be exactly half of its own length (due to
Lorentz contraction), and that the time measurement was exactly one second.
This gives some speed s' which is less than 2*s, as mentioned earlier.

  Now, let's assume that ship A wants to estimate what ship B would measure
(by using the same method) for the speed of A in this situation.

  Since it took 1 second for the bow of ship A to go from the bow of ship B
to its stern, and since A measured B to be exactly half the length of A,
it would mean by reciprocity that it would take 2 seconds for the bow of
ship B to go from bow to stern of ship A.

  This would seem contradictory (as the ships have the same length and
are travelling at the same speed). However, this would be from the time
reference of ship A, not the time reference of ship B. If ship A wants to
estimate what ship B is measuring, it has to use the time reference of ship
B, not its own.

  If there was a big clock visible on the hull of ship B, which ship A
could observe, it would see that it would seem to go at half speed
compared to the clock in ship A. Hence by using the clock of ship B,
it would take one second for the bow of ship B to go from the bow to the
stern of ship A, hence reconciling the speed measurement.

  However, by using the clock of ship B, it takes only a half second for
the bow of ship A to go from the bow to the stern of the ship B. Since
this is from the time reference of ship B, it would seem that ship B is
measuring a half second for this, rather than two seconds (which is what
ship A measured for the bow of ship B to go from the bow to the stern of
ship A, when using its own clock).

  How is this apparent contradiction between two seconds vs. a half second
reconciled?

-- 
                                                          - Warp

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After a rapid reading of your post (sorry, not much time), it seems that 
it might be some variation of known 'paradox' of special relativity:

See for example:
http://www.phys.ncku.edu.tw/mirrors/physicsfaq/Relativity/SR/barn_pole.html
http://www.einsteins-theory-of-relativity-4engineers.com/ladder-paradox.html

Thibaut




>   Let's see if I can describe this in words only:
> 
>   As we all know, the speed of light in vacuum is the same for all inertial
> frames of reference, and because of that, movement in inertial frames of
> reference is subject to Lorentz transformations.
> 
>   For this reason if there are two spaceships travelling to opposite
> directions at speed s (from an external observer's point of view), if
> ship A measures the speed of ship B as it perceives it, it's not 2*s,
> but something less (and always less than c). It happens because ship B
> appears contracted in the direction of the movement.
> 
>   Let's assume that ship A measures the speed of ship B by starting a clock
> when the bows of both ships coincide, and stops the clock when the bow
> of ship A coincides with the stern of ship B. For the sake of simplicity,
> let's assume that, while both ships have the same length, ship A measures
> the apparent length of ship B to be exactly half of its own length (due to
> Lorentz contraction), and that the time measurement was exactly one second.
> This gives some speed s' which is less than 2*s, as mentioned earlier.
> 
>   Now, let's assume that ship A wants to estimate what ship B would measure
> (by using the same method) for the speed of A in this situation.
> 
>   Since it took 1 second for the bow of ship A to go from the bow of ship B
> to its stern, and since A measured B to be exactly half the length of A,
> it would mean by reciprocity that it would take 2 seconds for the bow of
> ship B to go from bow to stern of ship A.
> 
>   This would seem contradictory (as the ships have the same length and
> are travelling at the same speed). However, this would be from the time
> reference of ship A, not the time reference of ship B. If ship A wants to
> estimate what ship B is measuring, it has to use the time reference of ship
> B, not its own.
> 
>   If there was a big clock visible on the hull of ship B, which ship A
> could observe, it would see that it would seem to go at half speed
> compared to the clock in ship A. Hence by using the clock of ship B,
> it would take one second for the bow of ship B to go from the bow to the
> stern of ship A, hence reconciling the speed measurement.
> 
>   However, by using the clock of ship B, it takes only a half second for
> the bow of ship A to go from the bow to the stern of the ship B. Since
> this is from the time reference of ship B, it would seem that ship B is
> measuring a half second for this, rather than two seconds (which is what
> ship A measured for the bow of ship B to go from the bow to the stern of
> ship A, when using its own clock).
> 
>   How is this apparent contradiction between two seconds vs. a half second
> reconciled?
>

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Thibaut Jonckheere <tua### [at] mapsonyahoofr> wrote:
> After a rapid reading of your post (sorry, not much time), it seems that 
> it might be some variation of known 'paradox' of special relativity:

> See for example:
> http://www.phys.ncku.edu.tw/mirrors/physicsfaq/Relativity/SR/barn_pole.html
> http://www.einsteins-theory-of-relativity-4engineers.com/ladder-paradox.html

  I know the barn pole paradox, but I don't know how to apply the
explanation to the case I described.

-- 
                                                          - Warp

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> Thibaut Jonckheere <tua### [at] mapsonyahoofr> wrote:
>> After a rapid reading of your post (sorry, not much time), it seems that 
>> it might be some variation of known 'paradox' of special relativity:
> 
>> See for example:
>> http://www.phys.ncku.edu.tw/mirrors/physicsfaq/Relativity/SR/barn_pole.html
>> http://www.einsteins-theory-of-relativity-4engineers.com/ladder-paradox.html
> 
>   I know the barn pole paradox, but I don't know how to apply the
> explanation to the case I described.
> 

Usually, these paradox comes from the (mis-)use of 'simultaneous 
events'. In your case, I think simultaneity is hidden there:

>  However, by using the clock of ship B, it takes only a half second for
> the bow of ship A to go from the bow to the stern of the ship B.

Even if A looks at clocks which are on ship B, measuring the time of a 
given event (e.g. Bow of ship A at bow of ship B) means that the event 
"the clock needle is pointing at this number' is simultaneous with the 
given event. And so, for B, these events would not be simultaneous (so, 
  for B, the clock needle would not point towards this number when bow 
of ship A is at bow of ship B).

These paradox can be tricky to solve, so I may be wrong with this. But I 
don't doubt that it can be tracked down to the use of simultaneity, thus 
the link with the barn pole paradox and others...

Thibaut

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Warp wrote:
>   As we all know, the speed of light in vacuum is the same for all inertial
> frames of reference, and because of that, movement in inertial frames of
> reference is subject to Lorentz transformations.

<Sheldon> I just thought it would make a good T-shirt. </Sheldon>

-- 
Darren New, San Diego CA, USA (PST)
    C# - a language whose greatest drawback
    is that its best implementation comes
    from a company that doesn't hate Microsoft.

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