[Fruityloop (OP)]

Special relativity....

Ship1 at rest at top:

1 light-year
T--------------------N
N-----------T <-------- .8c
.6 light-year

Ship2 at rest at bottom:

.6 light-year
T----------N -------> .8c
N--------------------T
1 light-year

The spaceships are each 1 light-year long and are passing each other at .8c. So each ship is contracted to .6 light-year when viewed from the other ship.

T1 and T2 both have a lamp with a lightbulb. Both lamps are initially off. When T1 is lined up with N2, T1 turns on his lamp. When T2 is lined up with N1, T2 turns on his lamp. When T1 turns on his lamp is T2's lamp on or off?

[Janus]

I assume you mean the top ship to be ship 1.   Thus according to ship 1, T2 turns on before T1, and according to ship 2, T1 turns on before T2.   This is Relativity of Simultaneity. So you can't say which light turns on first unless you specify the frame you are considering thing from. 

[Fruityloop (OP)]

This is showing 2 ships that are passing each other at the moment that T of ship #1 is passing N of ship #2. Each diagram shows one of the ships at rest and the other ship moving.

Special relativity....

Ship #1 at rest at top:

1 light-year
0sec               0sec
T--------------------N
N-----------T <-------- .8c
0sec   15,147,296sec
.6 light-year

Ship #2 at rest at bottom:

.6 light-year
0sec    -15,147,296sec
T---------------N -------> .8c
N--------------------T
0sec                 0sec
1 light-year

T1 is T of ship #1.  N1 is N of ship #1.  N2 is N of ship #2.  T2 is T of ship #2.

T1, N1, T2, and N2 each have their own clock.

At the moment that T1 is lined up with N2 the clocks at T1 and N1 display 0 seconds in their own rest frame and the clocks at T2 and N2 display 0 seconds in their own rest frame.  T1 is displaying 0 seconds and N1 is displaying -15,147,296 seconds when viewed from ship #2.  N2 is displaying 0 seconds and T2 is displaying 15,147,296 seconds when viewed from ship #1.

Using the math formula for the rear clock ahead effect....
((distance between clocks)*(velocity of reference frames relative to each other))/(c^2) we get ((3.527*10^12 miles)*(149,025 miles/sec))/(3.47*10^10 miles^2/seconds^2) = 15,147,296 seconds.
So the clock in the rear is 15,147,296 seconds ahead of the clock in the front.

Now to the point of this post.  There seems to be a problem here.  It seems that T1 is passing N2 twice.

T1 passes N2 when the clock at N1 displays -15,147,296 seconds.  T1 passes N2 again when the clock at N1 displays 0 seconds.

Also,
T1 passes N2 when the clock at T2 displays 0 seconds.  T1 passes N2 again when the clock at T2 displays 15,147,296 seconds.

What is displayed on the clocks at N1 and T2 when T1 passes N2?

What is the resolution to this problem?

[chris]

I'm confused about what T and N are?

[alancalverd]

Not to mention top and bottom.

[Halc]

I merged this since it's all the same question/scenario.

I'm confused about what T and N are?

Nose and tail of each.  Yes, specification would have been nice.
Top and bottom were very much specified.

Slight edit for clarity:

Ship 1 frame:

1 light-year
T1--------------------N1
N2-----------T2    <----- .8c
.6 light-year


Same picture, Ship 2 frame:

.6 light-year
T1-----------N1     -------> .8c
N2--------------------T2
1 light-year

I have given a clock to T1, N1, N2, and T2 at the moment that T1 is lined up with N2.

'Given a clock'?  What does that mean? You can't be present at all these events, so you can't be there to give gifts.  I will presume that you mean that the clocks are set to zero, but only two clocks are present at that event, so the other two cannot be set without specification of simultaneity convention.

So the clock in rear is 15,147,296 seconds ahead of the clock in front.

But then you say this.  Without a frame specification, how does this occur?  The clock in which rear? (There's one in each ship), and why is it 'rear' now instead of 'tail?  Call the ends F and R, not N and T if you're going to use those words.

You said 'given a clock to T1, N1, N2, and T2 at the moment that T1 is lined up with N2', which suggests that the clocks are synced in this unspecified frame, not one 15m seconds ahead of another. So 'given a clock' apparently does not imply that the clocks are in any way in sync.

What is the resolution to this problem?

Resolution: You need to be more specific about how the clocks are set because all the facts you list cannot all be true. Remember the answer by Janus and take relativity of simultaneity into account. You cannot assert when some remote event occurs without specification of frame.

I have no idea what all the clocks read because your wording is utterly vague about it.

[paul cotter]

I was thinking of attempting a reply but in the absence of clear diagram properly annotated I could not figure out what the question meant.

[neilep]

I was thinking of attempting a reply but in the absence of clear diagram properly annotated I could not figure out what the question meant.

I too was thinking the same but in the absence of any knowledge ,I decided against it.😊

[Bored chemist]

Special relativity....

Ship1 at rest at top:

1 light-year
T--------------------N
N-----------T <-------- .8c
.6 light-year

Ship2 at rest at bottom:

.6 light-year
T----------N -------> .8c
N--------------------T
1 light-year

The spaceships are each 1 light-year long and are passing each other at .8c. So each ship is contracted to .6 light-year when viewed from the other ship.

T1 and T2 both have a lamp with a lightbulb. Both lamps are initially off. When T1 is lined up with N2, T1 turns on his lamp. When T2 is lined up with N1, T2 turns on his lamp. When T1 turns on his lamp is T2's lamp on or off?

Yes.

[Fruityloop (OP)]

Special relativity....

Ship1 at rest at top:

1 light-year
T--------------------N
N-----------T <-------- .8c
.6 light-year

Ship2 at rest at bottom:

.6 light-year
T----------N -------> .8c
N--------------------T
1 light-year

The spaceships are each 1 light-year long and are passing each other at .8c. So each ship is contracted to .6 light-year when viewed from the other ship.

T1 and T2 both have a lamp with a lightbulb. Both lamps are initially off. When T1 is lined up with N2, T1 turns on his lamp. When T2 is lined up with N1, T2 turns on his lamp. When T1 turns on his lamp is T2's lamp on or off?

​T1 lining up with N2 is simultaneous for both ships.  At the moment that T1 is lined up with N2 according to ship1,  ship2 also agrees that T1 is lined up with N2.  Therefore, when T1 turns on his lamp, T2's lamp is simultaneously both on and off which is impossible.  Therefore special relativity is invalid.

[alancalverd]

If it's good enough for Schrodinger's cat, it's good enough for a spaceship.

[Janus]

Special relativity....

Ship1 at rest at top:

1 light-year
T--------------------N
N-----------T <-------- .8c
.6 light-year

Ship2 at rest at bottom:

.6 light-year
T----------N -------> .8c
N--------------------T
1 light-year

The spaceships are each 1 light-year long and are passing each other at .8c. So each ship is contracted to .6 light-year when viewed from the other ship.

T1 and T2 both have a lamp with a lightbulb. Both lamps are initially off. When T1 is lined up with N2, T1 turns on his lamp. When T2 is lined up with N1, T2 turns on his lamp. When T1 turns on his lamp is T2's lamp on or off?

​When T1 turns on his lamp, T2's lamp is simultaneously both on and off which is impossible.  Therefore special relativity is invalid.

No, your grasp of Special Relativity is lacking.  You constantly want to treat simultaneity as being absolute. SR isn't flawed, it just conflicts with your preconceived ideas.