Consider this scenario that RobA posed. Now, instead of the bandit being positioned at the end of the tunnel, let's put him in the middle inside of it. When he presses a button on a remote, both openings to the tunnel will close, and it will take 15 seconds for the light from the remote to reach each door in his frame, and they will close simultaneously from his perspective. Okay, so when the front of the train reaches the bandit, he presses the remote. In the bandit's frame, the light from the remote rushes to the forward opening at c, while the train is moving at less than c, so the light will reach the door before the front of the train does and the door will close before it. Six seconds after the front of the train has passed, the bandit sees the back of the train at the same place as he is positioned. So in his frame, six seconds later, the back of the train is now in the same place he is, well within the tunnel, but the light from the remote has not yet reached the rear opening of the tunnel, so that door has not yet closed. So according to the bandit's perspective, the forward door closes before the train reaches it and the rear door closes well after the train is fully within the tunnel, so the bandit has effectively trapped the train! How is that possible if the tunnel is only 3 light-seconds long while the train is a light-minute long according to the passenger's point of view?

EDIT- The back of the train will pass the bandit just over six seconds later than the front, not precisely six, because the train is travelling at less than c.

__________________
Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

There is something else I have been puzzling over involving Lorentz contraction and time dilation. After the M-M experiment, Lorentz determined that if the apparatus is contracted in the forward direction only by sqrt[1 - v/c)^2], then the results would be null as found. Einstein determined that there would be a time dilation that matches the contraction.

Now, let's say we see a rocket that is travelling away and contracted to half its original length as compared to when stationary. When the passengers of the rocket move from the back of the rocket to the front, they would appear to move at half the rate they might otherwise, so one could say that time dilation is taking place. However, if they move at half the rate over half the distance, it will take them the same amount of time to reach the front as would be measured when stationary. When moving from side to side, there is no contraction or time dilation in that direction, so it would take the same amount of time for that as well. In other words, all clocks within the rocket would still tick at the same rate as in the stationary frame, with no real time dilation taking place.

One can visualize this easier by imagining square tiles on the floor when stationary. When travelling away from an observer that remains stationary, the tiles appear contracted in the forward direction by half the distance, but the same from side to side. A passenger, walking from the rear to the front of the rocket will appear to move half as fast, so covering half the distance in the same time according to the stationary observer's clock, but since the tiles are also contracted, the passenger will walk across the same number of tiles in the same time as when stationary, reaching the front at the same time according to any clock on the rocket or the stationary observer's clock. So it seems that if a real time dilation actually exists between frames, it apparently would not relate to length contraction in any direct way that I can see, anyway.

__________________
Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

One more thing. Length contraction only takes place in the forward direction. The length contraction in the forward direction of the rocket is the same to a stationary observer regardless of which direction the rocket is actually travelling relative to the observer. However, if light were emitted from the rocket at some angle to the stationary observer, it seems that the Relativistic Doppler shift should depend upon the angle between the line between emitter and receiver and the line of travel. In other words, along the length of the rocket, there is a length contraction of sqrt[1 - (v/c)^2], so apparently light emitted along that line will be time dilated by the same amount according to relativity. In addition to time of flight of light effects, the light will be shifted to sqrt[1 - (v/c)^2] / (1 + (cos 0) v/c) over time or (1 - (cos 0) v/c) / sqrt[1 - (v/c)^2] over length. Either way, it comes out to sqrt[(1 - v/c) / (1 + v/c)] when the rocket is moving away from the observer.

Now let's consider that for light emitted across the side of the rocket. There is no length contraction or time dilation in that direction, so no frequency shift. If we figure for the length contraction at some angle to the line of travel of the rocket relative to the observer's frame, it will be

sqrt[1 - (v/c)^2] in forward direction
1 to side

So (cos 0) sqrt[1 - (v/c)^2] at some angle to line of travel for the forward direction of the rocket

(sin 0) at some angle to the side

Total contraction observed along line of sight is

L^2 = [(cos 0) sqrt(1 - (v/c)^2)]^2 + (sin 0)^2
= (cos 0)^2 - (cos 0)^2 (v/c)^2 + (sin 0)^2
= 1 - (cos 0)^2 (v/c)^2

L = sqrt[1 - (cos 0)^2 (v/c)^2]

So any light emitted along this direction should involve this length contraction and/or time dilation, not simply sqrt[1 - (v/c)^2]. The Relativistic Doppler shift, then, would become

sqrt[1 - (cos 0)^2 (v/c)^2] / (1 + (cos 0) v/c)
= sqrt[(1 - (cos 0) v/c) / (1 + (cos 0) v/c)]

When travelling directly away from the observer, this reduces to just sqrt[(1 - v/c) / (1 + v/c)]. When travelling directly toward the observer, the (cos 0) flips the signs and it becomes sqrt[(1 + v/c) / (1 - v/c)]. Is there any reason this should not be the case? I'm not sure if that would help with the "train in the tunnel" puzzle or not, but it might.

__________________
Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

This is a little tricky. Let me try something. The rocket is length contracted in the forward direction by half of that of the stationary frame. The passengers, when moving in the forward and back directions, are also observed moving half as fast. They move half as fast over half the distance, so the time on their clock in the rear and front of the rocket tick at the same rate as in the stationary frame. There is no length contraction to the sides, so those clocks tick at the same rate also. All clocks on the rocket tick at the same rate as in the stationary frame, so a stationary observer will notice no time dilation when comparing clocks.

However, when moving forward to back, the passenger is observed moving half as fast. The passenger emits light along the line of travel to the stationary observer. All processes in the forward to back direction are moving half as fast, so the process that produces the light emits it at half the frequency in that direction according to the stationary observer. The light is observed at half the emitted frequency in addition to time of flight effects, providing a Relativistic Doppler shift. Does that sound reasonable?

__________________
Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

I don't want to get into answer(s), since I have a slightly different view from most of the mainstreamers on this forum WRT this question. However, some caveats:

1. Try to keep it physically possible. There are no light-year long trains or tunnels.

2. The doors cannot close instantly in the real universe.

3. Read carefully about the mechanics of particle accelerators, particularly colliders. They're the only thing we have that makes real physical (although very small) things go very, very fast.

4. Don't be blown away by the math. Nail down the definitions of reference frames, the angle of observation, simultaneity, and the definition of co-moving reference frames and observers.

Lastly, I bet this OP is good for about 90 posts. Oh, and as Richard et al. will surely mention, what happens in one reference frame happens in all reference frames. The results cannot be changed by the choice of frames.

Regards, John M.

John, are you aware of Lorentz's original version of why the "contraction" takes place?

Is the "too big train/car" entering the "too small tunnel/garage" not a first year's relativity problem, where it is shown that it all hangs together with simulteneity not being the same in different intertial frames?

This is explained in any introductory book on special relativity.

__________________

Any comments in glorious red are to be considered in ModeratorMode.


善數, 不用籌策 (shàn shù, bù yòng chóu cè)
He who is good at counting, uses no counting tools
“A good scientist has freed himself of concepts and keeps his mind open to what is”
道德經, 二十七 (dào dé jīng, 27)

True. But wait until you see what kind of answers come in!

Regards, John M.

p.s. Ah, OP, Tus has a good suggestion above on where to look for an answer.

Didn't Lorentz use the contraction hypothesis as a means to preserve the idea of the ether? This was at a time when the speed of light with respect to the ether could not be measured.

Einstein took it one step further.

Yes, that's true. This link explains it. Guess I missed the point in the OP, though, so here's the thing. According to the link, simultaneity of the doors closing to the bandit means the forward one closes, then the rear one to the passengers on the train. From the perspective of the passengers, the train can fit into the barn if it smashes into one end and becomes compressed. But, to the bandit, the forward door closes before the train reaches it, and the rear one after it passes the bandit, so the train is entirely within the tunnel, doors closed, before any physical compression can even take place.

__________________
Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Very good. A lot of scientists in the late 19th Century thought there was something in space which was stationary and fixed with a “fixed” and “stationary” universe (this was before they knew about expansion and fast star and galaxy motion), that served as a “light speed regulating” medium. Some of them described it as being something like a special “field”.

The Michelson-Morley experiment was designed to detect the motion of the earth through that “ether” at a speed of 18.6 miles per hour (the earth’s revolutionary speed around the sun, and they assumed the sun to be “fixed” in the universe and the “ether”).

So many scientists were surprised by the M-M “null” result, two or three of them (including Lorentz) came up with the idea that one arm of the M-M apparatus “shrank” in the direction of its motion “through the ether”, thus giving M-M a false “null” result.

The “contraction” of matter, Lorentz theorized, was caused by some kind of resistance to motion that the “ether” put up when atoms moved through it. So, in his version of the theory, only objects that moved through the “ether” contracted in their direction of motion. Objects that did not move through the “ether” did not contract. So there was no paradox in his version of the theory.

Later Einstein came along and removed the ether and said that both of two “relatively moving” objects contracted in the direction of their “relative motion”.

Einstein’s version has no “force” put on the moving objects, while Lorentz’s version requires a force to cause the contraction.

Seems there is a paradox in the Einstein version, but no “ether”, and there is an “ether” in the original Lorentz version, but no “paradox”.

See:

“Michelson’s Interference Experiment”, by H. A. Lorentz, an excerpt from his 1895 book:

http://www.lawebdefisica.com/arts/lorentz/

See section 2 of the article:

“Thus one would have to imagine that the motion of a solid body (such as a brass rod or the stone disc employed in the later experiments) through the resting ether exerts upon the dimensions of that body an influence which varies according to the orientation of the body with respect to the direction of motion.”

See section 3:

”Surprising as this hypothesis may appear at first sight, yet we shall have to admit that it is by no means far-fetched, as soon as we assume that molecular forces are also transmitted through the ether, like the electric and magnetic forces of which we are able at the present time to make this assertion definitely. If they are so transmitted, the translation will very probably affect the action between two molecules or atoms in a manner resembling the attraction or repulsion between charged particles. Now, since the form and dimensions of a solid body are ultimately conditioned by the intensity of molecular actions, there cannot fail to be a charge of dimensions as well.”

Full 1895 book here:

http://www.historyofscience.nl/searc...age&startrow=1

The excerpt starts here at section 89 and goes to the end of Section 92.

http://www.historyofscience.nl/searc...df=&var_pages=

What physical compression? If we word this in terms of the barn and pole vaulter, the barn is contracted as far as the vaulter is concerned. When you ask the barn help, they'll tell you the pole is contracted. This is not physical, it's relative.

When you say: "But, to the bandit, the forward door closes before the train reaches it, and the rear one after it passes the bandit, so the train is entirely within the tunnel, doors closed, before any physical compression can even take place."

This really makes me wonder what you're trying to say.

Let's reword this...

1 light-second (ls) long train going through a 1.1 ls long tunnel.

Because the passengers' point of view and that of the observer standing next to the tracks is from two different point-times in Minkowski spacetime.

The passengers see a very short tunnel, but see the front door open well before the back door closes, therefore, there is plenty of room.

The bystander sees a very short train, shorter than the length of the tunnel.

They're both right. However, both perspectives are distorted by sqrt(1-v²/c²). That is, the bystander sees a train of length which is reduced by that factor, while the passengers see events happening in a different timeframe.

The reason this works is because what the bystander sees isn't what actually exists - by the time he sees the leading edge of the train at point x, it's no longer actually there.

__________________
If I set the budget, we'd have Ares and more. Unfortunately, I don't set the budget, and Ares is just too expensive and too far out for us to accomplish our goals within the budget we were given.

If we halt the ISS, all versions of Ares, and transport Orion and Altair aboard DIRECTv3's Jupiter family of Shuttle-Derived Launch Vehicles, we just might make it back to the Moon by 2020.

Well, according to the link, if the doors stay closed then the train will smash into the forward door and cause it to become compressed beyond the Lorentz contraction, so that the passengers and bandit will agree that the train now fits inside the tunnel when it comes to rest. But in the stationary frame, both doors close a couple of seconds before the train smashes into the forward one. So if it is the train smashing into the forward door that causes the compression that allows it to fit within the tunnel, then how can it fit into the tunnel before it smashes into the forward door in order for the compression to actually take place to begin with? That would seem to be a violation of causality. The cause should precede the effect in any frame.

In case anybody thinks I might be mixing frames or simultaneity effects in that last paragraph, which I am not as far as I can tell, let's try it from another point of view also. According to the link, if the doors close and then open again, the train can fit into the tunnel without a physical conpression because the passengers see the forward door close and open just before the front of the train reaches it, and then they will see the rear door close and open after the back of the train has entered. But let's place a passenger at the back of the train simultaneous with the closing of the rear door within their own frame just after the back of the train has entered the tunnel. That same passenger will see the tunnel only 3 light-seconds long while the train is 60 light-seconds long, so to the passenger, the front of the train has already travelled far out of the front door.

Now, to the rear passenger, the closing of the rear door is simultaneous since both are in the same place. The closing of both doors is simultaneous in the stationary frame. But that means that for the rear passenger, as the rear door closes simultaneously with them passing, the forward door will not have closed yet, being at a larger distance. The closing of the forward door will occur later to the rear passenger, but the front of the train has already exited the tunnel.

__________________
Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Grav,

It's all about simultaneity as was mentioned above. The notion of "the train being inside the tunnel" has no invariant meaning. Where an extended object is, that is the spatial location of all points that make up the object, where it is "now" depends on that frame's notion of simultaneity. It's where the world lines of all the points intersect an observer's "spatial hyperslices", his surfaces of constant time in the space-time.

To the rear passenger, the front door closes and re-opens well before the rear door closes. The front of the train has indeed exited the tunnel before that happens in the train frame.

If you're getting the simultaneity shift mixed up (which direction causes a forward shift vs a backward shift), review the Lorentz transform carefully.

Another thing to keep in mind is that clocks separated by distance that are synchronized in one frame are not synchronized in others. Say the rear and foward passengers on the train have synchronized their watches. To them, each clock reads the same thing "now". When the front clock ticks 1, the rear clock ticks 1.

But that is not the case to the stationary observer. Both clocks are ticking slow at the same rate, time dilation, but they are also shifted out of sync. They do not read the same time at the stationary observer's "now". That is very in reconciling whose clock read what when some event occured at their location. From the stationary perspective, no wonder the moving observers didn't think they were inside the tunnel at the same time. Their clocks and rulers were all screwed up and they don't know what "at the same time" is and can't see the tunnel is clearly much longer than the train.

And the moving observers, whose clocks are working perfectly, see the stationary observer's clock and ruler all screwed up and that explains why he thought they could be in an obviously too-short tunnel at the same time.

-Richard

Okay, let's say we have three observers, Alice, Bob, and Charlie, lined up in a row. Alice and Bob are stationary to each other with some large distance between them and Charlie is moving toward Bob with some relative speed. Alice's and Bob's clocks are synchronized and they have calculated when Charlie will reach Bob, and they both flash a light at that predetermined time, just as Charlie passes Bob, simultaneously in the stationary frame. Which flash will Charlie see first?

__________________
Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Hmm, lemme look.

Thanks, John M.

Mmmm, how sweet it is!

Thanks, Grav, good link.

Before? After? Ahead? Behind?

Let's try it this way:
--> relative speed
->some speed
0> no speed


Stationary Alice - T=11:32.43
0
V



(scroll down - there's a "large distance between them")



^
0
Stationary Bob - T=11:32.43
^
|
|
Charlie Gonzales (Speedy's brother)

At T=11:32.47, Charlie passes Bob, and precisely at that moment both Alice and Bob flash a light at Charlie - which light will Charlie see first?

Bob's, and at Bob's 11:32:47

__________________
If I set the budget, we'd have Ares and more. Unfortunately, I don't set the budget, and Ares is just too expensive and too far out for us to accomplish our goals within the budget we were given.

If we halt the ISS, all versions of Ares, and transport Orion and Altair aboard DIRECTv3's Jupiter family of Shuttle-Derived Launch Vehicles, we just might make it back to the Moon by 2020.

I have thought of something extremely simple, that unless someone can show me how my logic is faultering, which I don't think it is, would pose a really serious problem to relativity as we know it.

Relativity says that for observers with a relative speed between them, each will observe the other as contracted. They will not see themselves as contracted because their rulers have contracted as well. This is to be expected if we measure distance with distance. But what if we measure it with time? Let's say that two observers, Alice and Bob, are stationary to each other at each end of a rod and moving at some relative speed to a third observer, Charlie, with the same line of travel as along the rod. According to relativity, Alice and Bob should measure some distance along the rod and Charlie will see that distance contracted.

Bob observes the time on Alice's clock. From the time lag between his own clock and her's, he concludes the distance along the rod in the stationary frame to be d_stat = c (T_B - T_A). At that instant, the light from Alice's clock has travelled to Bob's and is now simultaneous with his own, being in the same place. The light from both clocks, continuing from Alice and originating from Bob, will then travel together to Charlie. Both rays of light are simultaneous all the way, following the same path with there being no reason for one to travel faster or slower than the other, especially since Alice and Bob are travelling together in the same manner, so the light travels the same away from each along the same line. So both rays will also reach Charlie simultaneously. This means that Charlie simultaneously reads the time from both clocks in the same way that Bob did, and sees the same time lag between them, so will measure the same distance between them that Bob did. So all observers along that line in that direction will measure and agree upon the same distance between Alice and Bob, regardless of the distance that the light travels or the relative speed.

So if Charlie measures the rod as contracted, then so will Alice and Bob. If there is a time dilation for Alice and Bob, whereas their clocks tick slower, then Charlie will also measure a contraction due to the lesser time lag, but then, so will Alice and Bob. Furthermore, contraction and time dilation in this sense would not cancel each other out, but would instead be additive. No matter which way one cuts it, with whatever time dilation or contraction that might actually take place, the stationary observers and observers with some relative speed will observe the same time lag between Alice and Bob, and measure the same distance between them, so neither frame will measure a contraction any different for the other due to a relative speed. Therefore, certain observations having nothing to do with relative speed in any way, at least as far as an observed contraction due to a constant speed of light measured by all observers is concerned, it would appear that light must travel relative to an absolute frame, ballistically, or some combination thereof. I will continue to study this further.

__________________
Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Hi guys;
I've been away from BA for about 4 months now, ...and now the first post I click on I had to find out the tragic news.... that the earth's orbital speed around the sun has slowed down to an unbearably slow crawl ! No wonder global warming has begun... (Not!).

Nice to see you 'slow-pokes' solving the universe's problems anyway.

G^2

Well, dang! Sorry.

Wouldn't they notice that their rulers appear to be much wider than they used to be? And when they turn them sideways, don't you think they would notice that the length of the ruler changes, being shorter in the direction of its motion and changing to being longer while perpendicular to the direction of motion?

Interesting - but I claim my right as a layman to totally disagree with the final paragraph (horrors - get thee to ATM!! )

This treatment of the paradox says that the entire train fits into the tunnel, but then when the brakes are applied (so bringing the train into the same reference frame as the bandit), the back uncompresses and springs back.

nope

What had me confused when I was considering this was a passenger in the end carriage. From the bandit's point of view, the passenger enters the tunnel with the train, but after the train has braked the end carriage (with the passenger) ends up OUTSIDE the tunnel.

Let's consider what the passenger sees. He sees the train approaching a really small tunnel - and since it is really small, he never goes into it !

So here was my paradox - the bandit sees the small train entirely inside the tunnel. The passengers never see themselves enter the tunnel.

My solution was this :
If the passengers never see the inside of the tunnel in their frame, then they cannot have entered the tunnel in ANY frame. Therefore, my guess at what the bandit saw must be incorrect.

The only scenario that fits is that the bandit sees the front of the train enter, but the end of the train stays way back outside. In other words, he sees a really short train approaching, but then it telescopes forward - so the front enters at full speed, but the back slows down.

grav,

To Charlie, Alice and Bob's clocks are dilated. The distance has to be mulitplied by gamma according to Charlie.

Say A and B are moving apart such that gamma = 2. Alice sends Bob a light pulse snapshot of her clock every second of her proper time. Bob will receive those pulses 2 seconds apart. So when he sees the image of Alice's clock reading T, 2T has passed according to his clock, and the light travelled a distance of c*2T, not cT.

-Richard

I am not finding for the total distance light travels between observers with relative motion, if that's what you are doing. I cannot tell exactly because you have not set a T=0 or included time of flight effects for what time B's clock will read when A's reaches it, which would involve both the distance between observers and the relative speed, but I am using the time lag between two observers that are stationary to each other to find the distance between them, according to what they would observe and measure, and according to that of another observer with some relative speed, which happens to not involve the distance light travels to another observer or the relative speed in any way to begin with after all. I was trying to determine if an observer with some relative speed would measure a contraction that the stationary observers would not. Turns out observers with any relative speed or distance will measure the same contraction between the stationary observers, if one were to take place, and the stationary observers will measure it as well.

As far as a time dilation of the clocks is concerned, if the clocks in a moving rocket are dilated according to a stationary observer, so that they are ticking half as fast, and the passengers appear to be moving about the rocket half as fast also, then the rocket will not appear to have contracted in any direction at all. This is because the times on the clocks must read the same to the observer as they do to the passengers when they coincide with the clocks in the same place, so placing them in the front and back of the rocket and on each side, the times that are read by the passenger when coinciding with them, from when they begin moving about the rocket to when they stop, must read the same as the observer watching the rocket, so it ends up all being just a matter of the measured time lag between the readings on the clocks when the passenger coincides with them.

If there is a contraction along the line of motion, so that the passengers appear to be moving half as fast along that direction, then the clocks will all be observed to be ticking at the same rate as they normally do to the passengers in the stationary frame, since the passengers are observed moving half as fast over half the distance, whereby over the same clock rate. So there is no time dilation of the clocks observed, just as there was no contraction with the time dilation of the clocks. Neither affects the other, so these are two completely separate happenings. They are additive, or well, actually multiplicative. If contraction and time dilation went hand in hand in equal proportions, so that one would observe a time dilation of one half as well as a contraction along the line of travel of one half, then one will actually observe the passengers moving from side to side at half the rate, but one fourth the rate in the forward and back direction.

__________________
Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

Grav,

I'm sorry, I indeed screwed that up. The distance from which it was emitted would be v*2T, not c*2T.

My point is that time of flight measurements in one frame have to be scaled by gamma. Let T be a delta-t, the time of flight for light in one frame. The travel distance there is c*T.

Now, in a frame in relative motion so that gamma = 2, that time interval is dilated by the factor of 2, and so the time difference is 2*T, so the light travelled a distance of c*2T in that frame, agreeing with the Lorentz contraction by the same gamma factor.

That is, what is a distance of 2x in the moving frame looks 1x in the stationary frame, 2cT vs cT. You can't use images of moving clock differences to calculate light distances without taking into account the time dilation of the clocks you're watching.

-Richard

I'm still not quite sure what you are doing there, but let's take the viewpoint of an observer stationary with the tail end of a rocket before it takes off, which will produce an observed contraction, but we will not include any time dilation for now. A quick burst of acceleration occurs at every point along the rocket, so all parts of it quickly (instantly) begin moving at v away from the stationary observer. She sees the tail end immediately speeding away at v. She sees the front end start to move at t=d/c. Over that time, the tail end will have moved to d'=vt. However, the contraction observed also includes the time it takes the light to travel back from the tail end from some position that it has moved to in the same time it takes the light from the front end to travel to the same position of the tail end. In other words, while the light travels from the front end at c, the tail end moves forward at v. At some point, the light from the front end will coincide with the tail end and both rays will travel together to the observer. That occurs when v for the tail end travelling forward from the original d=0 and c travelling back from d coincide, so vt + ct = d. The time to coincide, then, is t = d / (c+v), so this is the time lag between front and tail ends, and the original time lag across the length of d was t=d/c. The optical contraction, then, is the ratio of these times for light to cross the length, or 1/ (1+v/c). After the front end begins moving at v, this contraction will continue to be observed constantly.

Now look at what happened there. In order to find the contraction, we had to find where the light travelling from the front coincides with the tail end. If we placed clocks at the front and tail ends, then the light from both clocks will coincide at that point and the rays then travel together to the observer, so that the observer measures a contraction. But notice that those rays also coincide for a passenger at the tail end of the rocket in the same way, so the passenger will also measure the same contraction. If we now figure that the rocket's clocks are time dilated, then we can make it so that the passengers notice no contraction at all in their own frame, and the passengers will meaure it at its original length before motion began. But then, since the same times will coincide for the passenger at the tail end of the rocket, and those same times will travel together to the stationary observer, she will measure the original length of the rocket as well, and measure no contraction either. So one cannot measure a contraction or no contraction without the other doing the same thing. Furthermore, the clocks on the rocket would have to tick faster to cancel out the apparent contraction, not slower, since they are additive, so a larger time cancels out a contracted distance, but again, that would also cancel it out for the observer.

__________________
Let's put together the pieces of The Grand Puzzle . (website - now revised)

"Let's define another operator, Sz, which we won't pay any attention to."
"This transformation will automatically make zero equal zero."
"It may be true that zero equals zero -- and that is certainly an equality -- but I don't want to go into the details at this time."

RobA wrote the following:

1. Length of train: 60 ls (light-seconds) long
2. Length of tunnel: 30 ls

You wrote:

1. Length of train: 60 ls (light-seconds) long
2. Length of tunnel: 3 ls (passengers' point of view)

Initial Conditions:

Length of tunnel: Ltun=30 ls (light seconds)
Length of tunnel (relative): L'tun=3 ls
Length of train (rest): Ltra=60 ls

Bandit hits button when front of train passes him.

The solution to both is a length contraction, so L'=L*sqrt(1-vsquared/csquared))

Thus, v=c*sqrt(1-(L'/L)^2)

Query: What is the velocity such that L'tra=Ltun=30 ls?

Answer: v=0.866 c

Query: What is the velocity such that L'tun=3 ls?

Answer: v=0.995 c

Here's what happens:

1. Since the train's 0.995 c velocity is < c, the front door will close 0.075 s before the train reaches it. At that point, the front of the train is 14.925 ls past the bandit. We will call this X. Thus, X=14.925 ls.

2. At that same instant, the back door of the tunnel closes. Where is the back of the train? It's 60 ls long, and the front is 14.925 ls past the bandit, so the back is Y = 60 ls - 14.925 ls = 45.075 seconds prior to the bandit.

3. The door is 15 ls prior to the bandit.

4. Thus, the door shuts on the train, chopping the back 30.075 ls of the train...

CLEAN OFF!

Folks - length contractions apply to where the front and back end of the train appears to be. The train remains 60 ls long, just as the tunnel remains 30 ls long.

The paradox is one of perception, not one of reality.

__________________
If I set the budget, we'd have Ares and more. Unfortunately, I don't set the budget, and Ares is just too expensive and too far out for us to accomplish our goals within the budget we were given.

If we halt the ISS, all versions of Ares, and transport Orion and Altair aboard DIRECTv3's Jupiter family of Shuttle-Derived Launch Vehicles, we just might make it back to the Moon by 2020.

And in reality, the back half of that train, travelling at 0.995 c, whacks the entire mountain...

CLEAN OFF!

__________________
If I set the budget, we'd have Ares and more. Unfortunately, I don't set the budget, and Ares is just too expensive and too far out for us to accomplish our goals within the budget we were given.

If we halt the ISS, all versions of Ares, and transport Orion and Altair aboard DIRECTv3's Jupiter family of Shuttle-Derived Launch Vehicles, we just might make it back to the Moon by 2020.