Thought experiment

I have a question for the physics-minded, one that has bugged me for some time. I've argued about this with my son's godfather (an astrophysicist) for some time, now, and neither of us has ever come up with a valid answer. I posited this in another thread, but it's annoying enough that I think it might warrant its own thread.

Here's the thought experiment:

three ships are placed at rest, relative to each other, in three predetermined locations. Each ship has a clock on the outside. Two of ships (Ship A and Ship B) begin to accelerate. Ship C does not accelerate.

Ship A accelerates to .99c. Ship B accelerates to .5c. The clocks have been preset so that, when the accelleration shuts off, they will all read exactly the same time: 00:00. These ships will continue to travel without acceleration for three hours. At the end of that three hour span, they will all pass within easy sight distance of each other at the exact same time, and each ship will take a photo of the other. As soon as the photo is taken, it is transmitted to a central location.

Once those photos are received, they will be compared. The question is, what time is it, in the photos?

For ship A, ship b is traveling at .5c and ship c is traveling at .99c. For Ship B, both ship A and C are traveling at .5c. For Ship C, Ship A is traveling at .99c and Ship B is traveling at .5c. The problem here is that all three ships are traveling at all three speeds, relative to each other. Time dialtion, therefore, is different for each ship depending on the frame of reference. Once the photos are compared, how can all three be at all three states simultaneously?

This 3 ship problem explores an inherent paradox in relativistic theory. If everything is relative to the frame of reference, every particle is in an infinite number of time states relative to everything else.

Do any minds which are more physically savvy than mind care to chime in and make me feel like a moron?

reply to post by herrw

Do any minds which are more physically savvy than mind care to chime in and make me feel like a moron?

You made me feel like a moron while I was reading it, so congrats if someone makes you feel like a moron from an answer....

Originally posted by Chrisfishenstein
reply to post by herrw

 

Do any minds which are more physically savvy than mind care to chime in and make me feel like a moron?

You made me feel like a moron while I was reading it, so congrats if someone makes you feel like a moron from an answer....

i agree lol

That was not the intention, but thanks for the response. It really does bug me, right along with quantum shimmer, and an accurate description of the mechanics of charge. So I'd like to shut down the processes that run on these questions constantly, so that I can burn the neurons on something more productive... like Firefly, or learning Klingon, or something equally important.

Well, if the universe will only last 13 trillion years
then ship C arrives in 13 trillion years,
ship B in half that,
and the time in the picture from ship A
(the one that didn't leave)
would be three hours later that day.

My god man,
why would you go .99c for more than a few nanoseconds!??


Mike Grouchy

reply to post by mikegrouchy


The actual relative speeds are unimportant. Ship A could accelerate to .7c, if it suited the experiment. The end result of the photo is what is most important. What time is it, when the ships snap the picture?

But you do offer an interesting problem. Three hours will be in relation to what? Let's say that the control ship, Ship C, is the timekeeping ship, and the other ships have been preprogrammed to snap their pictures and transmit them at just the right moment.

This 3 ship problem explores an inherent paradox in relativistic theory. If everything is relative to the frame of reference, every particle is in an infinite number of time states relative to everything else.

Do any minds which are more physically savvy than mind care to chime in and make me feel like a moron?

some thing to consider,
if you were on earth, looking at a bright light source and had a gravitational lense along the line of sight of the same light source, so that you can see both grav lens and direct line of sight to object,

if the light source were to flicker would you see the change in light first through the grav lens or would the flicker be noticed at the light source and at the grav lens at the same time?

1/grav lens light flicker first followed by light source flicker
2/both grav lens and light source flicker simultaneously
or is there a third option?

because if the grav lens flickered first how would that reconcile with a C limit from your perspective?
superluminal travel? or a violation of relativity?

i ask the question to help answer the three ship problem

xploder

The time of all three ships will have to be the same for this to occur. They all have to be there at the same time or how can the pictures be taken. If one returns there ahead of time than it will sit there till the next ship arrives and all the clocks would be syncronized. You cannot cheat time in that case.

I think the puzzle is faulty.

c does not move. there is no distance given from the beginning position to the others relative to it's location, while you can compute distance traveled by the other two, you cannot determine distance to C.

Since you do not know the position to C, the assertion that all 3 are close together after 3 hours seems completely manufactured.

If they had started at 12 and they were all assigned a 3 hour trip. The time would necessarily be 3 pm.

REMEMBER: that the set up says "At the end of that three hour span, they will all pass within easy sight distance of each other at the exact same time,"

But do not forget that C didn't move. It seems like it was sitting at the end point for either A or B.

At the computed distance end point, A is nearly twice as far away as B, so it does not seem possible/probable that all three would be within sight distance.

The problem with the puzzle is NOT that it is unsolveable. It is poorly put together.

Thus it is with ALL seeming paradox:

1. The Cretin says: All cretins are liars. THIS is merely how language gets us into trouble and quandaries with misuse whether accidental or intentionally.

2. The Barber shaves those and ONLY those men who do not shave themselves. WHO shaves the Barber?


So we can put seeming puzzles together all day long that seem unsolvable BUT the set up creates impossible/untenable conditions from the start. there IS NO PARADOX

I think this has happened here. So in effect, the puzzle might be pure nonsense.

There is no goofy "magic" here.

Is this three hours travel time relative for each ship as well? 3 hours traveling at .99C will be a very long time compared to the stationary ship.

The setup was faulty, in that I never gave a reference point for the time. In this case, the time of travel would be relative to one location, because time is passing differently for each ship. So let's delete the time aspect of this problem entirely. The ships accelerate to their set speed, and coast until they all pass the same location at the same instant. At that point, they all take a picture.

The underlying paradox remains. All three should have different readouts on their clocks, relative to the perspective of the photograph. Relative speeds remain constant, once the acceleration ends. So the paradox remains.

reply to post by herrw

At the end of that three hour span, they will all pass within easy sight distance of each other at the exact same time,"

That doesn't look right to me. Are you asking to grant you that? If so, then the time is still 0:00. (Based on the spectators time, each picture would arrive at different times because they would be take at a different time from his perspective - I do not think anyone can share the same time as another - it is only that the time is so close that they seem the same - but in your example they would be so far apart, that they would seem nowhere near the same.)

I think.


edit to add: maybe it helps if you think of it like the person who went the fastest, actually slowed his clock the most, and in effect, he would be forced to travel much further than the others, because from the spectator's perceptive, his cargo is ass and he is hailing it away from the others. lol

"because time is passing differently for each ship"

reply to post by herrw




There is no real evidence for that. It is what the theory says. (There's no GOOD way to test it.)
The theory SOUNDS good (the verbal paradigm ala G.E.Moore)but like many unproven things it must stay in the realm of metaphysics because of the intangibles. When you can't really prove anything, you have to go with what seems to be logically True. (But it is still only a talking point) When you can't test a theory, it remains unproven and only a conjecture or supposition until not only you can get some mileage with some empirical testing, but you also account for the falsification. If you can NEVER test it, it becomes a failed theory by it's own untestability.*
The honest physicists will tell you that, the histrionic/egocentric ones will not.
There are no paradoxes there are only misperceptions. (sorry you didn't get that)
So no the seeming paradox does not "hold".

So don't believe that and if it pleases you keep spinning around discussing the Nothing, believing you are right.
I might just happen to actually know something about that, but that shouldn't matter to you.

*Lakatos, Popper, Kuhn, Thaggard

You needed to supply POSITION, not time. (3 hours just IS 3 hours.. ANY relative start time would work) You would also need to use a grid co ordinate system to completely work it out.

reply to post by herrw


I'm hoping the 'c' stands for speed of light, otherwise I'm totally lost. If it is the speed of light, how can the ships going .5c take a picture of the one traveling 99c? I doubt it would even be a blur.

reply to post by akalepos


Time dilation and relativity have very much been proved, GPS is just one of many examples that would not work without it.

If all the ships where traveling a relative three hours then all the clocks would say 3:00:00. The faster you travel the more time slows down with the closer you get to the speed of light the more time comes to a stop. If all the ships travel an absolute 3 hours with the stationary ship as the reference:

Ship A - 0:01:49
Ship B - 1:30:00
Ship C - 3:00:00

reply to post by kwakakev


Thank you, kwakakev.

Now, here's the question: Once the acceleration phase is over, they are all at a rest state with no other energy working on them (assuming a perfect vacuum, which doesn't exist, but this is a thought experiment). Would the time be exactly the same for every ship's photo of the other ships? The speed of ship C, relative to ship A, is .99c as well. Would that not mean that ship C's time dilation, relative to ship A, would be different?

I understand the concept of time dilation, and do not dispute it.

reply to post by herrw

Would the time be exactly the same for every ship's photo of the other ships?

If all the ships started their clocks at the same time, no. The clock times given in the previous answer is what I would expect to see with your though experiment. It is possible for all the ships to pass each other and all show the same time of 3:00:00, but the other ships would have to start their clocks relative to ship C:

Ship A -300:00:00
Ship B -6:00:00

This also means that ships A and B would also have to travel a lot further.

Would that not mean that ship C's time dilation, relative to ship A, would be different?

Your use of the word 'not' in the question is confusing. Ships A, B and C time dilation are all different. It is an objects velocity or meters/second that determines the amount of time dilation present.

Originally posted by herrw
Ship A accelerates to .99c. Ship B accelerates to .5c. The clocks have been preset so that, when the accelleration shuts off, they will all read exactly the same time: 00:00. These ships will continue to travel without acceleration for three hours. At the end of that three hour span, they will all pass within easy sight distance of each other at the exact same time, and each ship will take a photo of the other. As soon as the photo is taken, it is transmitted to a central location.

Ok, so I'm feeling nostalgic for college and dug out my undergrad physics book and will give a go at the problem, besides, I like relativity.

First of all, you didn't really define your problem completely. But, assuming that all speeds and times are measured relative to spaceship C, this is quite answerable. (You need that codicil because none of the ships involved would agree on "when the same time" is. I don't mean that they would argue over what the time would be labeled, but, rather, that A would say "B's clock was at X when we all passed each other" and no one else would agree. The keyword to look up here is "simultaneity" -- something else that is relative. As my old textbook put it: "If two observers are in relative motion, they will not, in general, agree as to whether two events are simultaneous. If one observer finds them to be simultaneous, the other generally will not". Likewise, traveling for "a three hour span" is not enough - you need to know who's holding the watch.) In fact, because you have everyone going at constant speeds when the clocks are started, you only need special relativity, not general (which is a lot harder math all around).

So, on to math: t(0) is called "proper time". It's the shortest amount of time experiencable for any given measured event (ie, the time observed by the fastest moving observer in the system). The equation that governs the relationship between delta-t and delta-t(0) (that is, change in time and change in proper time) for any two observers is delta-t = delta-t(0) / (1 - (v/c)^2)^.5 AKA gamma or the Lorentz time dilation factor times proper time. So the first step is to work out the gamma of each ship. gamma-a = 1/(1-(.99c/c)^2)^.5 = about 7.088, gamma-b = 1.155 and gamma-c = 1 because v=0 (remember: everything is relative to Ship C because it is at rest relative to the observer).

So we know by how much time is altered by their speeds: spaceship A gets about 1/7th the amount of time as spaceship C. So, since we know that delta-t (the amount of time spaceship C saw pass between ships A and B shutting off their engines and when they arrived) was three hours, we can calculate how much time passed on ships A and B. 3hrs*60 minutes/7.088 = 25.4 minutes and 3hrs*60 minutes/1.155 = 155.8 minutes

So, Clock C would read "03:00", Clock B would read "02:35", and Clock A would read "00:25".

I suspect much of your confusion was due to holding pre-Einsteinian ideas about the "3 hours" and "same time" part. There IS NO SUCH THING AS "NOW". It's not something that applies to the universe at large. It's a feature of your local situation and not a universal truth. As my philosophy of physics teacher put it: "Reality is NON-TRANSITIVE!" (Meaning that I can say A caused B and you can say B caused A and we can both be right.) I hope this cleared up your confusion.

Originally posted by akalepos
There is no real evidence for that. It is what the theory says. (There's no GOOD way to test it.)

This statement is factually, provably incorrect. We have mounds upon MOUNDS of evidence. The one that's easiest to grasp is the fact that if you take two atomic clocks and fly one at high speeds for awhile then bring them back together, the one that traveled will have lost time relative to the stationary one. The one that's most convincing when you know the actual science is muon decay rates -- we know how long they last and when we accelerate them in a particle accelerator, they last a lot longer from our perspective. Longer in a way exACTLY predicted by theory. Heck, the fact that your GPS works means that general relativity is correct -- which means that special relativity is correct. No, you are simply wrong.

reply to post by Stunspot


I have not formally studied physics, just a bit of back yard stuff so your gamma relationship instead of the liner one I calculated with sounds right. It does come across that there is a two dimensional aspect to time using the square root of a square, which does make sense when looking at time dilation. Is this how you see it or able to explain the equation you used?