1. Jan 11, 2007

### moe darklight

ok, this is realy bugging me that i cant figure it out and it just agitates me all day long. I've heard explanations of these, but those explanations only cause the following questions in my head:

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(u can skip this part, just kinda telling where i come from since this is my first post in this forum)
first off, I just finished highschool and am going into film school so my understanding of physics is not very profound. I've always been very good at logic and science (biology, philosophy, chemistry, etc...) ... not in school - i don't consider school education lol - but spend a lot of my free time studying these subjects under the philosophy that life is too short and the universe too beautiful to not try and understand it.

but math/algebra is another story, i honestly think i have some sort of brain damage when it comes to numbers (couldnt do multiplication in my head if you payed me), so please try not to use equations when explaining but images or analogies instead.

so I may be (and probably am) way off... tyring to understand einstein's book without looking at a formula was challenge enough.

aanyway here's my problems

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the solution that is always given is solved by explaining the different relative times at twhich the barn doors would open.. but isn't that kind of a cheat from the main problem?

what would happen if:
-both doors were closed, and the man with the pole was standing inside the garage holding a pole that is, say, 3/4 the size of the garage.
-this man accelerates (while he is still IN the room) to a speed fast enough for the pole to (in his view) be longer than the room he is in (for a split moment, as to not run through the wall and out of the room, but remain in it)

here's my problem: from the point of view of the man, the pole would crash right through the room, but form the point of view of the room, the pole has shrunk and fits perfectly inside the room... so which is it? the walls can't both collapse and not collapse...

2) with the twin paradox (I'm gonna use the way it's explained here http://www.phys.vt.edu/~jhs/faq/twins.html ; there are different versions)
it explains that the difference happens because one of the twins changes its trajectory while the other one remains in inertia.... but isn't inertia relative too???

-if one twin changes it's trajectory, or accelerates, or decelerates, twists, turns, whatever... would the same not be true of the other twin in that twin's point of view?
if twin 1 is going in one direction and then turns back in the view of twin 2. isn't twin 2 also going in one direction and returning form the view of twin 1?

also

-it states that when twin 2 arrives to the station, his clock is different form the one at the station because he has traveled there at a slower relative speed... but wouldn't the "outside world" also be going at a slower relative speed and, also, the distance be relatively smaller for him (that is: he is traveling 1L at proper distance, but not relative distance from his point of view: wouldn't the distance between the planet and the station have shrunk from his view?), having the 2 effects cancel each other out once he arrives?

i hope i was clear... it's 3 AM.

thanks to whoever takes the time.

Last edited: Jan 11, 2007
2. Jan 11, 2007

### my_wan

This is a bit ambiguous making it very hard to answer. However if you look closely at the relativity of rigidity you will notice the end result will always be the same if you assumed that no shortening or lengthening took place. The apparent warping observed by different observers results because different observers cannot agree on when the different points of the objects where at certain places. Nature does not allow you to say what "the same time" means for all observers simultaneously. We can only define how much two observers will disagree on what "the same time" means. A third observer can disagree with the first two and a fourth then disagree with the first three, etc, etc..

No. Inertia is relative, acceleration is not. All observers will agree on who accelerated regardless of the frame of reference. They can disagree on how much acceleration took place though.

The two effects did not cancel because because all observers can agree that the train decelerated, decelerated is physically the same as acceleration. Note that the train rider was the only one that felt G forces from the changing motion of the train.

3. Jan 11, 2007

### Janus

Staff Emeritus
To add upon what has already been said, assuming that the pole is being held in the middle, the two ends will not accelerate forward simultaneously with the person holding it. His forward Push on the rod has to propagate down the pole to the ends, and this can not happen faster than the speed of light. This means that the person will move quite a bit forward before the back or front of the pole even start moving. If you factor this in you can show that either the pole will punch through the front wall or it will stop before it hit the front wall from both the persons and the barn's perspective.

4. Jan 11, 2007

### moe darklight

ok.... some i understand and some I dont... so bare with me for one sec lol. (and thanks for answering by the way)

- so the changes in lengths of things are only perceptual?
- but the changes in time ARE literal (or else the time difference wouldn't happen)

but what I don't get now is, if the changes in length are only perceptual, then why is the traveling twin (according to the description I read of the paradox in that website) actually traveling a lesser distance? (it says for the traveling twin, the distance was 0.6L while for the stationary one it was 1L)

and how can length contraction be only perceptual but not time dilation? wouldn't light be traveling at a slower speed if time dilated but length didn't?

lol this is hurting my brain.

Last edited: Jan 11, 2007
5. Jan 11, 2007

### Janus

Staff Emeritus
No, it is not that the length contraction is only perceptual, It is that the pole cannot be inifinitely rigid. The ends will not respond instantly to a force applied at the middle.

6. Jan 11, 2007

### moe darklight

(thanks first of all lol)

and i got all that right (i thought i had it wrong, but my original understanding turned out to be right) but i figured what threw my whole understanding off (and still does) is this part:

"Once Prime stopped at the distant station, he rejoined Unprime's frame of reference: Prime saw all the clocks in that frame to be synchronized again (but reading 1.25 years, while his own watch said 0.75 years)."

here's my problem... how can the clocks at the station and at the distant station be synchronized when looked at from both locations (seeing as they are 1Ly away?)
if the clock at the original station reads 0 when the distant one reads 0, seeing as it is 1Ly away, it is really reading 1... so when the travelling twin arrives at the distant station and reads 1.25 on the distant station clock, wouldn't the original station be reading 0.25 from that point of view?

or maybe i should just go watch dr. phil over a bag of chips.

Last edited: Jan 11, 2007
7. Jan 11, 2007

### Fredrik

Staff Emeritus
I should probably begin by saying that I'm not 100% sure about some of the things I'm saying here, so I would appreciate comments from the people here who know SR well.

What Janus said about how the push will propagate through the pole no faster than the speed of light is true if we imagine a guy holding a pole at its center, but that's not how I think about problems like this. It is perfectly ok to imagine instead that some sort of machine pushes every part of the pole forward at the same time.

What you need to understand is that "at the same time" is a relative concept, so we have to specify if the different parts of the pole are to begin their acceleration at the same time in the frame that's moving at a high velocity relative to the room, or at the same time in the room's rest frame? Both options are possible in principle, so we should examine both of them.

Let's specify some numbers first...

Proper length of the room: L
The length of the pole when it's at rest in the room's frame: 3/4*L
Speed after the acceleration: 0.8c (This makes the length contraction factor 0.6).
Duration of the acceleration period: 0 (Instantaneous acceleration is impossible, but SR says we can get arbitrarily close).

Suppose that the machine that accelerates the pole is doing it so that the different parts begin to accelerate simultaneously in the room's frame. At one instant in the room's frame, every part of the pole begins to move with speed 0.8c. Note that this implies that the pole will have the same length in the room's frame after the acceleration. But how is that possible if the moving pole is contracted by a factor of 0.6? The answer is that the machine has to physically stretch the pole by an amount that exactly compensates for the Lorentz contraction to make this happen! Since the length of the pole remains unchanged, there is a time period when the entire pole is moving with velocity 0.8c inside the room.

In the frame that's moving with velocity 0.8c relative to the room, the pole is initially moving with velocity -0.8c, and its length is 0.45L. Suddenly the front of the pole (the end with the higher x coordinate) stops moving, while the rear continues to move at -0.8c until the length has expanded to 1.25L. This is the combined effect of the physical stretching and the Lorentz expansion. The length of the room is only 0.6L, so the pole is more than twice the length of the room. The front of the pole breaks through the wall at some time between those two events.

This is not a contradiction. In one frame the different parts of the pole started moving at the same time. In the other frame some time passed between the moment when the front stopped moving and the time when the rear stopped moving.

Now suppose instead that the machine that accelerates the pole is doing it so that the different parts begin to accelerate simultaneously in the frame that's moving at 0.8*c relative to the room. In that frame, the length of the pole remains constant. But how is that possible? It should expand when it's no longer being Lorentz contracted, shouldn't it? Yes it should, and it does! The machine is squeezing the pole (hmm...that sounds dirty) so that its length is reduced to compensate for the Lorentz expansion. Immediately after the pole has stopped moving in this frame its length is 0.45L and the entire pole is inside the room. (The length of the room is 0.6L).

In the room's frame, the rear end of the pole will change it's velocity from 0 to 0.8c at one instant, while the front end remains stationary. When the rear is at the distance 0.27L from the front, the front end will change its velocity from 0 to 0.8c. The length is 0.27L because it's being squeezed as well as Lorentz contracted. This is less than a third of the room's length, so the pole fits comfortably inside the room immediately after every part of the pole has been accelerated to 0.8c.

It appears that when we accelerate the pole the first way, we get the usual pseudo-paradox (if you understand what I said above you also understand that it isn't a paradox at all), but if we accelerate it the second way, the pole fits inside the room in both frames.

Now I have a question that I hope someone else can answer: It seems that no matter how the pole is pushed, there will be some stretching or squeezing involved. I assume that the internal forces between the molecules in the material will try to restore the pole to its original state when its released from the machine that accelerates it. What will happen to the pole? What will its final state be?

Edit: Maybe the answer is quite simple: If the acceleration didn't damage the pole, it will successfully get back to its original state (in particular it's original proper length 0.75L) in its new rest frame. Yes, I think that must be the correct answer.

Last edited: Jan 11, 2007
8. Jan 11, 2007

### JesseM

The definition of "synchronization" in a given frame is designed to factor out light-speed delays, under the assumption that light signals travel at exactly c in all directions in that frame. For example, if my clock reads 12 midnight Jan. 1 2007, and I look through my telescope and see a clock at rest relative to me which is exactly 1 light-year away and which reads 12 midnight Jan. 1 2006, then I define these clocks to be "synchronized" in my frame, since a light signal traveling at c would take 1 year to travel from that clock to me, and 1 year ago my clock also read 12 midnight Jan. 1 2006.

Last edited: Jan 11, 2007
9. Jan 11, 2007

### JesseM

There is something called Born rigid motion where an object accelerates in such a way that the object will have a constant length in the instantaneous inertial rest frame of an observer sitting on the object from one moment to another (of course the observer's instantaneous rest frame will be different from one moment to another). The observer will also experience a constant G-force, although I'm pretty sure that observers sitting at different points along the object will experience different G-forces (and it may be that the constant length each percieves is different as well, I'm not sure).

So what would happen if a rod started out at rest in a closed room, with the rod 3/4 the length of the room, and then quickly underwent Born rigid acceleration to a point where, in the instantaneous inertial rest frame of an observer on the rod, the length of the rod was greater than the length of the room? I'm not sure how Born rigid motion looks from the point of view of a non-accelerating observer, so maybe in the room's frame the length of the rod would actually expand as it accelerated rather than contract. Anyone know the answer?

10. Jan 11, 2007

### moe darklight

omg, i feel so dumb for missing something so obvious lmao.
*slaps forehead with palm of hand (or is the forehead slapping the palm of the hand? dun dun dun)

thanks guys! now it all makes sense (well, at least the basics, i wouldn't even dream of thinking about concidering to try to understand the more complex aspects of relativity)

11. Jan 12, 2007

### Demystifier

12. Aug 23, 2011

### sasebt

Hi people,

I was reading up on SR, and I ran into this thread. I know it is old, and I dont know if anyone will still read it, but, it is a bit unclear to me how the paradox is resolved.

I will try to describe a slightly different setup:

a barn, 10 meters long, a ladder, 10 meters long, and the barn has 2 doors where the ladder passes through. There are 2 observers, one riding on the ladder, and one standing outside the barn. the doors are rigged with 2 long poles of exact length, connecting at the observer outside the barn, so that, he can really quickly push both poles at the same time (since their ends are bolted together where he stands) and close both doors simultaneously.

from the barn observer: the guy on the ladder flies through the barn doors (so no acceleration is involved on the ladder). as soon as the back end of the ladder passes through the entry door, the barn observer pushes the poles and captures the ladder inside the barn. This is because in his ref. frame the ladder is moving, and it is shorter than 10m, where the barn is still and 10m wide. This means that the ladder would eventually crash into the closed exit door, and break up into pieces. All of these pieces would be inside the barn (the ladder is much more fragile than the barn doors).

from the observer on the ladder: he is sitting still on the ladder, and all of a sudden an barn runs into him with lighting speed. as the barn passes around him, as soon as the entry door clears the ladder, it is closed. but, since the barn is moving, it is less than 10m wide, and the ladder is exactly 10m long, the exit door would slam into the ladder and break it in 2. the ladder will break, and some pieces would be inside the barn, and some will be outside the barn.

as for the rigidity of the poles closing the doors, since they are the same length and same material (rigidity), the force from the barn observer would not close the doors instantaneous, but, in any case, those 2 events should be simultaneous from whatever reference frame you observe.

Anyone care to elaborate on this situation? Or, point out where my thinking has gone wrong?

Thanks,
Saso

13. Aug 23, 2011

### Staff: Mentor

The part in bold is where your thinking has gone wrong. The events are only simultaneous in one reference frame. In all other reference frame they are not simultaneous. This is what is meant by the relativity of simultaneity.

http://en.wikipedia.org/wiki/Relativity_of_simultaneity

This is the single most difficult concept in special relativity, and it is the source of almost all "paradoxes" and mistaken analyses in relativity. Whenever you see something that seems to be contradictory look for the relativity of simultaneity mistake.

14. Aug 24, 2011

### sasebt

@DaleSpam:

Dale, I still cannot understand how those 2 events, closing the doors, will be dependent on the frame of reference, since the barn observer will be pushing the poles (which are identical) both at the same time, pushing them where they are supposedly connected together.

Also, where would the debris from the crushed ladder be? will it all be inside the barn? Part inside, part outside?

Thanks

15. Aug 24, 2011

### Staff: Mentor

In the barn frame the impulses would move at some speed v in both directions, where v is equal to the speed of sound in the pole material. In all other frames the speed will be given by the relativistic velocity addition formula. If you work out the math, you will get non simultaneous arrival.

Inside. Your barn-frame analysis is correct. Only the pole-frame analysis was wrong.

16. Aug 24, 2011

### ghwellsjr

Don't you meant the ladder-frame analysis was wrong? We can stipulate that both doors close at exactly the time the ladder enters the barn according to the barn-frame (by whatever mechanism we choose) and when we transform this scenario into the ladder-frame, it will illustrate the relativity of simultaneity and show that the second door closed before the first door and before the ladder was completely inside the barn.

17. Aug 25, 2011

### Fredrik

Staff Emeritus
I recommend that you read about simultaneity in a book about SR. You will find that any two events A and B that are simultaneous in some frame F, are not simultaneous in any frame that has a non-zero velocity component relative to F along the line in space between the locations of A and B.

18. Aug 25, 2011

### harrylin

19. Aug 25, 2011

### Fredrik

Staff Emeritus
I prefer an approach based on spacetime diagrams, like pages 6-9 here. Unfortunately page 7 doesn't show up in the preview (at least not for me), but there are other ways of getting a copy of the book. The most important detail is figure 1.4 on page 9. (If you're having trouble viewing some pages, it might help if you change the .se in the URL to .com or your own country code).

20. Aug 25, 2011

### K^2

Maybe the diagram would help. See attached image.

t and x are coordinates according to static observer (solid black axis). t' and x' are coordinates according to the guy moving with the ladder (solid blue axis). L and R are the locations of left and right doors of the barn, respectively (solid gray lines). A and B are the rear and the front ends of the ladder, respectively (solid green lines).

There are 3 important events to consider. Where A crosses with L, the rear end of the ladder has entered the barn. Where B crosses R, the front end of the ladder exits or hits the right door of the barn. The observer chooses to close the doors somewhere in between. Suppose, that happens at t=0. So the doors are closing where L and R cross the x axis.

Lines of equal time are parallel to the appropriate x axis. So from perspective of static observer everything is clear. A crosses L, doors L and R close, B crosses R.

I added dashed lines that pass through these event points and are parallel to x' axis to help track these events from perspective of the guy moving with the ladder. Black dashed lines correspond to door closing events, and green dashed lines correspond to ends of the ladder entering/exiting barn. From bottom and moving up we see first the black dashed line that passes through intersection of R and x, then green through B and R, followed by green through A and L, and finally, black through L and x.

In other words, door R closes first, then end B hits the door R, then end A enters the barn, and only then door L closes.

The sequence of events in two coordinate frames are completely different, yet the actual points in space-time are the same. All of this weirdness comes to be because the t' and x' axes appear to be "squished" together. This has to do with the way space-time distances are measured, and time being a special coordinate. That's basically the only thing that makes relativity a bit counter-intuitive.