I When Does a Train Passenger Experience Complete Darkness in a Tunnel?

[ESponge2000]

TL;DR

We all know what this is: Very high speed train close to c moved through a tunnel, for a tunnel observer, they calculate entrance and exit doors shut and open closely in sync and train length is smaller than tunnel length .

For train passenger the tunnel is smaller than the length of train but the entrance and exit doors don’t shut and open simultaneously, question time

Assume it’s extremely bright and sunny outside the tunnel, but very very dark inside the tunnel

If you are a passenger riding in the “center” of the train , and the shutting and opening of each door (the tunnel entrance and tunnel exit door) happens very fast and occurs at a point where one standing in the center of tunnel would see a moment of pitched black as both doors are briefly shut with train contained inside the tunnel…. They should calculate the full closure of both ends occurred when center of train passed the center of the tunnel …. But they still even can “see” both doors shut with the train still not out the tunnel yet.

So we get a moment of pitched blackness in a tunnel with a train moving inside it.

Now…..Explain what moment (if any) there is a presence from a person riding in the train through the tunnel where train is already past the tunnel exit before the entrance to the tunnel door shuts …. When do THEY “see” that pitched black moment which did take place?

Or otherwise how can both reference frames have different answers to whether or not there was a moment of darkness when they crossed the tunnel?

 

[Ibix]

Have you tried using the Lorentz transforms to see what happens?

 

[ESponge2000]

For a person standing in the middle of the tunnel the activity of the front and end of tunnel would both be calculated to and with a delay, visually seen to occur at similar clock time on there watch . If they can see both doors shut it has to coincide with when any residual light beams commenced travel from each end to the center of tunnel so it should appear dark

But a traveler would not get this dark moment ?

 

[Ibix]

But a traveler would not get this dark moment ?

Are you guessing, or calculating?

 

[ESponge2000]

Have you tried using the Lorentz transforms to see what happens?

Lorentz transforms gives me the X’ and the T’ but it doesn’t help answer this question . Not sure how to incorporate what the person “sees” .
But even if we slow motion down carefully …. There would have to be for center of tunnel , a moment where there’s a gap in sunlight beam front to back right where there’s a gap in sunlight beam back to front and occurs closely at when the shut doors are visible to the center

 

[Ibix]

Not sure how to incorporate what the person “sees” .

You calculate the coordinates of the event where the light from whatever the observer is looking at arrives at the observer. In this case work in whichever frae the door closure is simultaneous at time $0$. Your observer is at some position $x_0$ when the doors shut at positions $0$ and $l$, and they see the doors shut when $x_0+vt=ct$ and $x_0+vt=l-ct$. Solve for $t$ in each case, calculate the observer's $x$ and transform to whatever frame you want.

 

[ESponge2000]

Are you guessing, or calculating?

How can the traveler get a dark moment if the train is too big to fit in the tunnel on the time line of the traveler? So at all times for him there’s at least one side open?

 

[ESponge2000]

You calculate the coordinates of the event where the light from whatever the observer is looking at arrives at the observer. In this case work in whichever frae the door closure is simultaneous at time $0$. Your observer is at some position $x_0$ when the doors shut at positions $0$ and $l$, and they see the doors shut when $x_0+vt=ct$ and $x_0+vt=l-ct$. Solve for $t$ in each case, calculate the observer's $x$ and transform to whatever frame you want.

I’ll try this. Let’s assume the train is 100 light nanoseconds long in its resting frame …(30 meters) Let’s assume the tunnel at rest is identical length. Assume travel at 80% of c

For tunnel frame : a unit 60 train in a 100 length tunnel occupying all but the first and last 20 units of the tunnel when doors instantly slam shut

Traveler train will have 10 cars and perceive first 2 cars already out of the tunnel and back 2 cars haven’t yet entered inside it (wait hold on )…. This can’t occur when any door shuts ! I’m confused here

1 unit length = 1 light nanosecond = 30 centimeters

1 unit time = 1 nanosecond

X0 = 50

Let’s run the transform. Maybe the center train traveler in front of car#5 aligns with an earlier timed end tunnel closure just before the front is out the tunnel

 

[ESponge2000]

Per tunnel frame, It takes front of train 10/0.8 = 12.5 nanoseconds to clear the tunnel and some part of the train is in the tunnel for (10+6)/0.8 = 20 nanoseconds (time from front of train enters tunnel till Back train exits it)

At t = 9.95 to 10.05 nanoseconds doors shut almost instantly and then reopen so at 10 they are shut , while x=5

Tunnel observer sees the doors shut when t = 10 + 5 = 15, at which time 1/3 of train is already out the tunnel and at position 12 but the light from the reopen hasn’t yet reached the middle of tunnel observer , ok that makes sense… in the back it’s at position 6 and seen to have already passed the observer at the moment of apparent dark

The train appears even shorter to the mid tunnel observer …

I’m having trouble getting the Lorenz for the time of visibility to the middle … if I use x=0 t=0 at pitch black I get nothing from this

 

[ESponge2000]

I think I am getting somewhere . What I am gathering is there are always light in this tunnel .
( )
( )
Let this ———— Mean a light beam
And this X mean a dark spot

Tunnel right when both doors shut then reopen
( X———————)
(———————- X )

Tunnel a little time after
(———X -—————)
(——————— X——)
Here still light covers all of tunnel either from one end of tunnel if not the other

When middle of tunnel sees black
When there’s emptiness of light from both ends of the tunnel together
(————-X—————)
(————-X —————)

Near the end
(————————-X)
(X————————)

Ok this means the tunnel is always not empty of light but only the middle gets a glimpse when there’s a gap in both beams from each end

This means that for the train traveler …. Since timing of events appear to happen at different time intervals , The traveler will “see” the absence of the ————-> Light beam at a different clock time than when seeing the absence of the <—————- Light beam! Instead, the distribution of light relative to the apportioned time under tunnel passage will come out similarly over the time but with a higher % of time in the traveler’s reference frame of dimmer sunlight in place of that moment of 0 sunlight

To have no light beams at all in all of the tunnel , it would need to be shut for a long enough time that either the train would need to be moving slower or the the tunnel much lengthier , or kaboom!, in both the first 2 cases would lead no longer to the traveler’s reference frame being one with a train larger than the tunnel , And the kaboom one the front gets hit and quickly after the traveler airbag comes out would be not surprising either

 

[ESponge2000]

I messed something up. I need to run the transform . Merely in a resting frame where both doors never closed in simultaneity doesn’t rule out the absence of light in a part of the tunnel

Take for example the standard tunnel rest frame, being just past entrance of tunnel … some time after the exit tunnel door already closed and re-opened in this frame , Suppose The entrance door now closes … But being situated right next to the entrance door …. The dark break in light beam from the entrance coincides with finally seeing the absence of light caused by the earlier shut and open at the other end … causing darkness

So while I don’t think the traveling train gets to see all dark , it wouldn’t be ruled out just by the fact that at no simultaneity exists for both doors shut together

I hope it doesn’t come off as rude to ibix I didn’t finalize a transform , I am stuck on it and need help… (or maybe the back of the train does get the dark? )

 

[Nugatory]

How can the traveler get a dark moment if the train is too big to fit in the tunnel on the time line of the traveler? So at all times for him there’s at least one side open?

Do remember that bright or dark at any point inside the tunnel depends on light that travelled from the ends of the tunnel while the gates were open - and that light was traveling at speed $c$. So it’s possible for a spot on the train not to be getting light from an open gate, if the gate hasn’t been open long enough.

This situation will be way easier to visualize if you draw a spacetime diagram.

 

[ESponge2000]

Do remember that bright or dark at any point inside the tunnel depends on light that travelled from the ends of the tunnel while the gates were open - and that light was traveling at speed $c$. So it’s possible for a spot on the train not to be getting light from an open gate, if the gate hasn’t been open long enough.

This situation will be way easier to visualize if you draw a spacetime diagram.

Right but if we can gather
1) there is a dark “point” in the center of the tunnel …. At a certain time , And for any moving frame when they are on a point the simultaneity on that point is invariant it’s dark it’s dark

2) some part of the traveling train is covering that dark point , while in the tunnel …. When the distance is 0 they are on the same point then that point would incur the dark point I now think

Time 15 …. Tunnel clock

Time 0: train enters tunnel

Time 10 : doors are shut

Time 12.5: front of train exits

Time 15 : dark in middle of the tunnel moment

For the train any trip through tunnel is 7.5 nanoseconds

At time 0 the front of train is simultaneous with time 8 at other end of tunnel . , at time 3.3333 for train clock the front is simultaneous with the back end tunnel closing at which time it sees the light of the end of tunnel coinciding with the light that originated at time 0

the front of train “sees” the back tunnel door shut at time 6.666667 for the train while in the back it will take a whollllle 23.333333 more nanoseconds additional till the train front seeeeees the entrance tunnel door shut and open

Back of train waits 4.5 nanoseconds to enter and then “sees” the Back tunnel door shut at time 8.666667 with that light beaming towards the reverse direction of the train in motion . Very close to this time mid tunnel goes dark !
But back of train at this point is (8.66667-4.5) / (12-4.5) = 55% into the tunnel and MISSES the dark spot!!!!! At dark spot it’s actually 60% through the tunnel but would need to be 50% through the tunnel to observe it.

I didn’t have to go this far to figure that out.

Tunnel perspective : it’s momentarily black in the middle of the tunnel at Time 15, and front of train in this frame left the tunnel at time 12.5 and the back enters at time 7.5 and exits at find 20.
At time 15 it is
(15-7.5) / (20-7.5) through the tunnel …that’s 60%,

See at the time it’s dark in the middle of the tunnel, all parts of the train passed the midpoint of the tunnel . The darkness in the middle of tunnel doesn’t take place when both doors are shut , it takes place after the “absence of light” from each end , travels at c to conspire in the middle of the tunnel , and while that happens , the train is already mostly beyond the tunnel . Since all parts are past the tunnel in all frames there’s agreement at that point over each part the train that there’s sunlight

Lastly, it is at time 12 for the back of train … roughly when it just about exits the tunnel that it “seeeees the front entrance door finally shut” ,

This means the back of train “sees” a gap of 3.33333 nanoseconds between the tunnel doors closing

Front of train “sees” a gap of a 16.66666667 nanoseconds between the tunnel doors closing

End of tunnel “sees” a gap of 10 seconds between the doors closing

Middle of tunnel “sees” simultaneously both doors shut

Entrance of tunnel “sees” the gap of 10 seconds but with the opposite sequence as the end of tunnel !

Conclusion: To see the darkness in the tunnel means the same thing as to “see” the doors on each end of the tunnel close at the same clock time (I don’t mean calculating them to be simultaneous but “seeing” them visually close together “ And since even the back of the moving train “sees” the end tunnel shut 3.33333 nanoseconds before the back one shuts , It is too far ahead to see the point of darkness in the tunnel that would be due to the shutting of the doors.


But wait …. The view for the middle of the tunnel that coincides with the time the train is not in the middle of the tunnel , Contains a view of the whole train inside the dark tunnel , a little train all inside it , a little train that has not seen the dark spot . Ok my brain is getting tired. Hopefully I didn’t mess this up.

Recapping the parameters
A train traveling 80% of c through a tunnel appears relative to the stationary tunnel to be length 6 light-nanoseconds and tunnel of length 10 light-nanoseconds ,
Where on tunnel’s frame of reference, train entered the tunnel at time 0 and a door on each end of the tunnel “entrance” and “exit” closed briefly and reopened at 10 nano-seconds after the train entered the tunnel.

 

[PeterDonis]

This situation will be way easier to visualize if you draw a spacetime diagram.

@ESponge2000 this is extremely good advice. Trying to analyze relativity scenarios without drawing a spacetime diagram is like trying to do difficult arithmetic in Roman numerals. Yes, technically it is possible to do it, but the extra effort involved is pointless.

 

[ESponge2000]

The math for calculating times and distances is very chart-friendly once I can draw it out because pretty much everything in constant velocity differences between objects involves sloped lines. The quick method with or without charting is 1) find a point on the line and proper slope based on Lorentz transform 2) trace the slope to other points

Also, it’s way easier to deal with the train’s reference frame once we pose it as “this is a length 10 train” that is not in motion, a shrunk tunnel just happens to be moving backwards at 80% of c . Whatever point is x distance ahead of the train from the perspective of itself is “a point that is not fixed to a distance from the tunnel except for at that moment of measurement.

Or simply said, per the train, the length of the train occupies a train’s length while the train is not needing to move across a train length to travel one train length so long as we “mistakenly” use the other reference frame to mark our landmarks, which to the train’s perspective are NOT landmarks, some other train to the side or in front or behind that’s at rest with the train is a landmark

 

[Vanadium 50]

It is certainly your choice not to draw a diagram, just as it was your choice to use your own equations and not the Lorentz transformation in the "math doesn't work out" thread. But you seem to be ignoring some excellent advice you have received here, and it seems not to be working for you. Maybe its time for another strategy.

Draw the diagram.

 

[ESponge2000]

But how on this forum can I draw one ? I need a better YouTube on drawing it. Things that still confuse me :

When to use the horizontal coordinstes parallel to the vertical t axis

What to do when the x’ and t’ origin isn’t on the x and t origin do you stop using the slanted axis for points referencing the other frame do you stop the diagonal parallel at the prime axis or all the way keep it going to the vertical t? still trip this up. Even trying to start with spacetime diagram the twin paradox
Round trip 4 light years each way at 80% of C beginning at time 0

Maybe I need to start there since it’s the case I don’t have to think so most as I know it cold ….

t’ line is also the velocity line (0,0) to (4,5) 4 light years in 5 years , For the traveler it needs to cover 3 dots , X’ is shadow of t’ so it goes (0,0) to (4,3.2) to commence time for ship we draw lines parallel to x’? And where does the 2.4 get denoted?

 

[Ibix]

But how on this forum can I draw one ? I need a better YouTube on drawing it. Things that still confuse me :

If you're using the editor in WYSIWYG mode, paste it in. Otherwise use the "Attach files" button below the reply box.

 

[Nugatory]

What to do when the x’ and t’ origin isn’t on the x and t origin

You can always choose to place the origin of both coordinate systems at the same point, and you pretty much always should - any other choice will complicate the math horribly without contributing any new understanding.

Also note that the Lorentz transformation as they are usually written:$$x’=\gamma(x-vt)$$ $$t’=\gamma(t-\frac{vx}{c^2})$$assume that you are using this convention. The easy way to see this is that these formulas transform $(x=0,t=0)$ to ($x’=0,t’=0)$; the same point is the origin of both coordinate systems.

 

[ESponge2000]

The Lorentz multiple in the graphic does look like an independently applied operation. I mean the Graph doesn’t make it obvious where the units fall on x’ and t’ lines it just the v determines that

I know obviously how to calculate Lorentz , I’m saying the diagram doesn’t make that process any simpler than the formula

Also when I master all this in diagram I have questions about how to graph 2 objects that are moving linearly but not directly apart or towards each other , such that they may 1) be traveling at an angle from each other and/or 2) despite the distance closing between them they won’t come into contact

Both of these are unrelated …. You can travel towards each other at an angle and cross paths , You can also travel the same linear but a distance to the side of each other , You can also travel say northeast horizontal and other object northwest horizontal and also they are at different constant heights , but let’s start basic

Or both apply , I only know how to get graph for the twin paradox type of travel

 

[Ibix]

I’m saying the diagram doesn’t make that process any simpler than the formula

You don't need a map to navigate around the world, but it makes it easier - especially when you try to communicate about it. Minkowski diagrams are maps of spacetime. They let you draw the other frame's axes and coordinate grids. And because the Lorentz transforms preserve straight lines on these maps they make it obvious that paradoxes are impossible.

 

[Ibix]

I'm not totally clear what scenario you are considering. I've drawn a few spacetime diagrams for one that's something like it, which might help you. In this scenario, a train of rest length 10 light nanoseconds ("lns") enters a tunnel of rest length 10 lns, travelling with a speed of 0.8c. There are doors at each end of the tunnel; the one at the entrance is initially open and shuts as the train rear enters the tunnel, while the one at the exit is initially closed and opens just in time to let the front of the train out.

Here is the spacetime diagram in the tunnel frame. It shows the ends of the tunnel as red lines (vertical because the tunnel is stationary in this frame) and the train ends as thick blue lines (slanted to the left because the train is moving in this frame). It also shows the middle of the train as a fine blue line, and I've shaded the region occupied by the train.

We're interested in when the tunnel is dark. Thinking only of the left hand end for a moment, this door is initially open and closes just as the rear of the train passes through. The tunnel darkens from that end from that moment. We can add shading to the diagram to show the regions of spacetime that see the rear door open and closed (I removed the shading of the train for clarity):

The boundary has a speed of $c$, of course, and observers at events in the shaded region will directly see (literally see) the door at the left hand end of the tunnel is shut and they are in darkness from that end. The rear of the train is always in shade, but the train is going fast enough that most of it has left the tunnel before the shade reaches the end of the tunnel. In fact, an observer in the middle of the train is out of the tunnel before seeing that door shut (this would not be the case at a lower speed or if the tunnel were much longer).

Now let's think about the door at the right hand end of the tunnel. The door here is initially shut and must open just before the front of the train gets there. Again, we can show the region that sees the front door shut:

Again, the boundary is moving at $c$, but in the other direction. Observers at events in this shaded region see (directly) the door at this end to be shut, and you can see the moment when the observer at the middle of the train crosses the boundary into light from this end.

And now we can show both doors' shade regions on the same graph:

Only at events in the darker shaded region (the overlap of the two regions illustrated in the previous two graphs) do we see (directly see, with our eyes) both doors to be shut. Only a quite small fraction of the train actually sees that at this speed.

Finally, here's the same scenario from the train rest frame - naturally the blue lines are vertical and the red ones slope to the left:

You can see all of the same regions, lines and intersections, and the same line/boundary crossings happen for the timelike worldlines in the same order. So you can see there's no inconsistencies.

 

[ESponge2000]

Right but if we can gather
1) there is a dark “point” in the center of the tunnel …. At a certain time , And for any moving frame when they are on a point the simultaneity on that point is invariant it’s dark it’s dark

2) some part of the traveling train is covering that dark point , while in the tunnel …. When the distance is 0 they are on the same point then that point would incur the dark point I now think

Time 15 …. Tunnel clock

Time 0: train enters tunnel

Time 10 : doors are shut

Time 12.5: front of train exits

Time 15 : dark in middle of the tunnel moment

For the train any trip through tunnel is 7.5 nanoseconds

At time 0 the front of train is simultaneous with time 8 at other end of tunnel . , at time 3.3333 for train clock the front is simultaneous with the back end tunnel closing at which time it sees the light of the end of tunnel coinciding with the light that originated at time 0

the front of train “sees” the back tunnel door shut at time 6.666667 for the train while in the back it will take a whollllle 23.333333 more nanoseconds additional till the train front seeeeees the entrance tunnel door shut and open

Back of train waits 4.5 nanoseconds to enter and then “sees” the Back tunnel door shut at time 8.666667 with that light beaming towards the reverse direction of the train in motion . Very close to this time mid tunnel goes dark !
But back of train at this point is (8.66667-4.5) / (12-4.5) = 55% into the tunnel and MISSES the dark spot!!!!! At dark spot it’s actually 60% through the tunnel but would need to be 50% through the tunnel to observe it.

I didn’t have to go this far to figure that out.

Tunnel perspective : it’s momentarily black in the middle of the tunnel at Time 15, and front of train in this frame left the tunnel at time 12.5 and the back enters at time 7.5 and exits at find 20.
At time 15 it is
(15-7.5) / (20-7.5) through the tunnel …that’s 60%,

See at the time it’s dark in the middle of the tunnel, all parts of the train passed the midpoint of the tunnel . The darkness in the middle of tunnel doesn’t take place when both doors are shut , it takes place after the “absence of light” from each end , travels at c to conspire in the middle of the tunnel , and while that happens , the train is already mostly beyond the tunnel . Since all parts are past the tunnel in all frames there’s agreement at that point over each part the train that there’s sunlight

Lastly, it is at time 12 for the back of train … roughly when it just about exits the tunnel that it “seeeees the front entrance door finally shut” ,

This means the back of train “sees” a gap of 3.33333 nanoseconds between the tunnel doors closing

Front of train “sees” a gap of a 16.66666667 nanoseconds between the tunnel doors closing

End of tunnel “sees” a gap of 10 seconds between the doors closing

Middle of tunnel “sees” simultaneously both doors shut

Entrance of tunnel “sees” the gap of 10 seconds but with the opposite sequence as the end of tunnel !

Conclusion: To see the darkness in the tunnel means the same thing as to “see” the doors on each end of the tunnel close at the same clock time (I don’t mean calculating them to be simultaneous but “seeing” them visually close together “ And since even the back of the moving train “sees” the end tunnel shut 3.33333 nanoseconds before the back one shuts , It is too far ahead to see the point of darkness in the tunnel that would be due to the shutting of the doors.


But wait …. The view for the middle of the tunnel that coincides with the time the train is not in the middle of the tunnel , Contains a view of the whole train inside the dark tunnel , a little train all inside it , a little train that has not seen the dark spot . Ok my brain is getting tired. Hopefully I didn’t mess this up.

Recapping the parameters
A train traveling 80% of c through a tunnel appears relative to the stationary tunnel to be length 6 light-nanoseconds and tunnel of length 10 light-nanoseconds ,
Where on tunnel’s frame of reference, train entered the tunnel at time 0 and a door on each end of the tunnel “entrance” and “exit” closed briefly and reopened at 10 nano-seconds after the train entered the tunnel.

I posted this many months ago and reread my own post. I need to revise an incorrect statement.
Should be: ,

“But wait …. The view for the middle of the tunnel that coincides with the time the middle of the train is aligned in the middle of the tunnel , Contains a view of a train that’s Actually not evenly centered inside the tunnel since the view of the front and back are of the photons from an earlier snapshot in time, but yet for the middle of the train the photons are merely realtime photons. So one noteworthy point in time from perspective of middle of the tunnel stationary observer will be where both ends of the tunnel appear to be shut simultaneously, a merely symmetric alignment of a train inside a tunnel, but where the portion of the train crossing the middle of the tunnel is NOT the center of the train.

This has to mean the “apparent” length of the train, as opposed to the “logically determined length of the moving train and also from the tunnel’s perspective” would be different from each other. Independent of special
Relativity, we have that the view of the train is time backwards the farther the distance from the naked eye of a part of a train displaced from the center of itself.
The view from the middle of a moving train would have to be even before accounting for special relativistic effects, that the distance from the middle to the front appears shorter than the distance from the middle to the back, when outside the train and viewing each side at the time the middle passes by.
And this is weirder than I thought because at the time the brief absence of light from each end of tunnel penetrates the center of the tunnel view, this doesn’t account for the sources of light reflecting off the train itself and back to the center. This piece has nothing to do with special relativity but i realize this question needs to be revised
“At time 15 in this example, this is the middle of tunnel time upon which both doors “appear” to be closed and upon which no light is being emitted from either end of the tunnel. But this STILL doesn’t mean the view is of complete darkness because the light from other parts of the train and other parts of the tunnel can reflect back onto the middle…. much like how we get moonlight at nighttime when the sun isn’t shining. Yep this eases confusion when I said to myself how can the middle see dark if the train is in view and the train doesn’t get dark? Because light reflected from the train I forgot to account for!

But in summary correct me if I’m wrong

For a moving train at a high velocity, the special relativity piece is that if we constructed a train to be a certain length and then somehow accelerated it to 80% of c , based on our observations and then performed basic math to adjust for the time it takes light to travel to and from …. We would logically conclude a train is 60% of its original length.
But if we just took the raw of what we “see” then if we were in the true middle of the train we constructed, it would still appear to be a train 60% of its original length, but that train would be shifted asymmetrically to where more of the train appears to be in the rear than in the front. Am I right?
If so then also would there almost be a “washout” in the impact between some of the special relativity length contraction and the appearance of expanded length in the back due to the uneven lengths in time for the image of the back of train photons to travel to the eyes to see the image ?

 

[Ibix]

The view from the middle of a moving train would have to be even before accounting for special relativistic effects, that the distance from the middle to the front appears shorter than the distance from the middle to the back, when outside the train and viewing each side at the time the middle passes by.

That depends very much on what your model of "non-relativistic light" is. If you like an ether theory then the speed of light is isotropic in the ether rest frame. If you like an emission theory then the speed of light pulses is isotropic in their emitters' test frames. These produce different results for who sees (as in, receives light pulses) what simultaneously unless all emitters happen to be at rest in the supposed ether frame.

Basically, talking about light without involving relativity will always give results that depend on which wrong theory of light's behaviour you choose. So you need to specify which wrong theory you're using. Or, better, don't bother trying to force something inherently relativistic into a non-relativistic model.

 

[ESponge2000]

This is confusing to me because I’m not trying to contradict special relativity , I am just trying to understand what a stationary observer will “see” visually when a long train passes at 80% of c, . We know the part that the stationary observer will
Infer that the traveler’s clock will
Be 1/0.6 times slower relative to the stationary observer and that the length of the train will be 60% of the original length . These are the inferences of the stationary observer

What I am now asking is to include the impact of what the train length appears like visually, to the stationary observer .

For instance , We infer that the clock on the Sun is merely ticking at the same pace as the clock on earth and that clocks can be close to synchronized such that the Sun will see an earth from
8 minutes ago and the earth will see a Sun from 8 minutes ago, we will then calculate that the sun and earth are in sync, maybe not perfectly but the relative velocity difference is small enough that it’s not a major component of astronomical models at the level
Of hours and minutes and seconds.

So here … we calculate that the 80% of c train is contracted to 60% of its length, But what do we see not calculate but see is the length scalar from middle to back of the moving train when the known to be center of it passes us by?

excluding only red-shift and blue-shift throwing away that part

 

[jbriggs444]

This is confusing to me because I’m not trying to contradict special relativity , I am just trying to understand what a stationary observer will “see” visually when a long train passes at 80% of c, .

So you are asking about Terrell rotation?

 

[PeterDonis]

This is confusing to me because I’m not trying to contradict special relativity

Then you shouldn't say this:

even before accounting for special relativistic effects

There is no such thing. Relativity is the correct theory, not something else plus relativistic effects. The relativistic effects are not added on to something else. They are the correct physics, period.

 

[Ibix]

This is confusing to me because I’m not trying to contradict special relativity

I think, then, that what you wrote was confusing. In particular, even the answer to "what do people see directly" depends on what model of light you are using, and the bit about "independent of special relativity" is incorrect.

I now think you are just trying to work out where the front and back of a train appear to be according to direct observation by an observer who sees it passing at some speed $v$.

Let the train have length $L/\gamma$ in this frame, and let the observer be at $x=0$ and the middle of the train pass that location at $t=0$. The front of the train is therefore at $x=vt+L/2\gamma$ and the back at $x=vt-L/2\gamma$. The past light cone of the observer at time $t=T$ satisfies $x=\pm c(T-t)$. You can eliminate $t$ between each of the first pair of equations and each of the last pair, which will yield four solutions for $x$. Compute the corresponding $t$ values and select the pair with $t\leq T$. Those are the $x$ positions of the train that you see at time $T$, and you can compute the apparent length.

Note that this is strictly only accurate for a 1d train viewed at zero distance from it. A more complete analysis is the Terrell rotation linked by @jbriggs444.

 

[ESponge2000]

Ok running this math I get that the tunnel observer in a tunnel of unit 10 standing at 5 from the entrance , sees a unit 10 train (length contracted to 6 units) when t=10 ,
Train rear end at X=2
Train front end at X =8

At t = 0 the mid is at x=-3 And the front at x=0

But the observer “sees” a front end as it would have reached at t = 8.3333 when T = 10 Which is where X .= 6.666667 (1.66667 away from the observer’s marked position )


The back, however, the is X = -20 (25 away from the observer’s marked position)?

 

[Halc]

Ok running this math I get that the tunnel observer in a tunnel of unit 10 standing at 5 from the entrance , sees a unit 10 train (length contracted to 6 units) when t=10 ,
Train rear end at X=2
Train front end at X =8

At t = 0 the mid is at x=-3 And the front at x=0

But the observer “sees” a front end as it would have reached at t = 8.3333 when T = 10 Which is where X .= 6.666667 (1.66667 away from the observer’s marked position )

Sounds good. It puts the front at x=8 1.33 time units later.

The back, however, the is X = -20 (25 away from the observer’s marked position)?

That seems wrong. In 25 (seconds?) the back will move 25*0.8 = 20 putting it at x=0, not x=2 at t=10

Notice that the train back (any part approaching) appears to move faster than c to the mid-tunnel observer.

Now at t=10, is that mid-tunnel observer in the light or in the dark? I presume both entrances close and reopen simultaneous relative to the tunnel frame, closed for the duration of the entire train being contained,.