# Yet another Relativity Buster Question

Discussion in 'Physics & Math' started by Singularity, Feb 20, 2006.

1. ### CANGASRegistered Senior Member

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Is it asking too much to expect that the prologue of a thread question should match the question itself? If we are willing to accept fuzzy science, why bother at all?

In the prologue we do not see "R". At least, I do not yet see it.

3. ### SingularityBannedBanned

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Thanks that was an incredible answer. For the first time I knew that not just space and time are relative but measurements are also relative.

So by your explainations, man at A1 will see the Rockets meter stick stretched to 7 meters of his measure tape.

5. ### SingularityBannedBanned

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I am sorry the R was a mistake I meant Rocket by R, U can assume R1 as the point where rocket touches the man at A1.

7. ### PeteIt's not rocket surgeryRegistered Senior Member

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No. According to Einstein's special relativity, length contraction occurs both ways.

Relative to the Rocket, the man's measure tape has 7 marks within the length of the stick (a tape "metre" is shorter than a stick "metre").
Relative to the man, the rocket's stick is only one seventh as long as the distance between marks on the measuring tape (a tape "metre" is longer than a stick "metre").

But how can this be?

The great discovery that Einstein made is that not only is position and velocity relative, but that time and simultaneousness is also relative.

For example:
Relative to the Rocket, the near end of the stick passes mark 1000 on the measuring tape at the same time (simultaneously) as the far end of the stick passes mark 1007.

But relative to the man, these events are not simultaneous - the far end of the stick doesn't pass mark 1007 until some time after the near end passes mark 1000.

Last edited: Feb 21, 2006
8. ### przyksquishyValued Senior Member

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Nice and intuitive, isn't it? The amazing thing is it all turns out to be consistent...

Though some might argue against that...

9. ### SingularityBannedBanned

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I hope U people use an electric connections to confirm which will be the longer stick.

U people are insane nuts.

10. ### RosnetPhilomorpherRegistered Senior Member

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Distance between A and B in rest frame:

l<Sub>AB</Sub> = 10 light-minutes.

Distance in R-frame:

l'<Sub>AB</Sub>/Gamma = l<Sub>AB</Sub>*Sqrt[1-(.99)<Sup>2</Sup>]

Let the time taken be t' minutes.
Within t' minutes, B1 moves a distance of .99t' lightm-minutes towards the laser beam. The laser beam (with speed 1 light-minute/minute) moves a distance of t' towards B1.

l'<Sub>AB</Sub> -.99t' = t'
t' = l'<Sub>AB</Sub>/1.99
t' = l<Sub>AB</Sub>*Sqrt[1-(.99)<Sup>2</Sup>]/1.99
t' = 10*Sqrt[1-(.99)<Sup>2</Sup>]/1.99

t' = 0.7089 minutes (approx.)
t' = 42.53 seconds

11. ### RosnetPhilomorpherRegistered Senior Member

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Yes the distance between A1 and B1 is about 85 light-seconds. But the time at which the beam hits B1 in the rocket's frame is not merely 85 seconds. Because in this frame, B1 is not at rest. If you can't follow my calculations, work it out from the beginning yourself. You can seldom read through physics.

12. ### SingularityBannedBanned

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1,287
Here

we

go

again

There are MileStones between A1 and B1, these show to R that Laser from A1 to B1 reaches in 85 or something seconds, yet residence of R are not shocked.

U all sound like Zombies from Resident Evil.

13. ### SingularityBannedBanned

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U r right, now I dont have to read thoes books anymore, thanks.

14. ### PeteIt's not rocket surgeryRegistered Senior Member

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Rosnet is correct - I answered too hastily in my initial reply.

15. ### RosnetPhilomorpherRegistered Senior Member

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As I said, if the answer is confusing, work it out yourself. Or you can read it about a hundred times till you follow it.

Now, the residents of R do <B>not</B> see the laser beam reaching B1 in 85 seconds. OK? The distance between A1 and B <I>is</I> 85 light-seconds. Correct. But, for the residents of R, A1 and B1 are moving backwards at a speed of 99% that of light. Right? If you are moving forward at a certain speed, then you see everything around you move backwards with the same speed. If you don't believe that, here's a simple experiment:

Apparatus required:
You.

Principle:
Relativity theory.

Procedure:
(1) Run short distance at an approximately uniform speed.
(2) Observe everything around you while running.

Result:
While you are running forward, you see everyhing else move backwards.

Convinced?

So you have to account for this when looking at things form the rocket's reference frame.

16. ### SingularityBannedBanned

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The Rocket never sees the laser fired from behind. For the rocket the laser always travels in front of it. U r confused.

17. ### ZephyrHumans are ONERegistered Senior Member

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At their peril

18. ### Neddy BateValued Senior Member

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It took me quite awhile to realize that the SR theory is internally consistant inasmuch as all clocks in a relatively moving frame can tick slower than all of the synchronised clocks in the local rest frame (a reciprocal condition due to equivalent validity of all inertial frames). The reason this non-intuitive 'reciprocity' works is due to length contraction, and also because the relatively moving clocks are systematically out-of-synch with each other. By systematically, I am referring to the way that the clocks toward the 'leading' end of a (relatively moving) reference frame are behind in time, while the clocks toward the 'trailing' end are ahead in time. (For example, in R1's frame, A1's clock is behind and B1's clock is ahead).

However, these unsynchronized clocks are all synchronized in their own rest frame, and presumably they will remain so after the two different frames are joined together by a period of acceleration. Since time cannot go backwards, we must assume that all of the clocks which are 'behind' will have to catch up to the ones which are 'ahead'. So the question becomes, "Which clock is most ahead in time?" Clearly, the answer would be that it is the clock that is furthest toward the trailing end of the reference frame. The problem is, then, that the reference frame can be considered to be of infinite length. The clock that is 'most ahead in time' is undefined! So how can the clocks ever reach a synchronized concensus after a period of acceleration?

I may not have "busted relativity", but I do think I have demonstrated yet another (at least apparent) paradox. I am sure someone will have an explanation, though.

Likewise, Singularity may not have "busted relativity" either, but at least they have provided some rather entertaining comments:

LOL. Maybe it's just the cute avatar...

19. ### PeteIt's not rocket surgeryRegistered Senior Member

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There's the kicker... I struggled with this for a while back in that thread with the train and embankment clock tickets, but couldn't find a way to accelerate the train clocks to the embankment frame without desynchronizing the clocks.

20. ### DaleSpamTANSTAAFLRegistered Senior Member

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That's a very interesting problem. Given two clocks synchronized in their proper frame that accelerate to a new proper frame, is there any way to accelerate so that they remain synchronized (or at least end up synchronized). I hadn't even considered it.

I will work on it a bit. It sounds like fun.

-Dale

21. ### geistkieselValued Senior Member

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The lack of critical thinking.

Singularity, I've toned it down a tad here.
Pete's, Neddy Bates' and DaleSpam's posts reflect a problem that no one SRTist on this forum has addressed properly, or even improperly. They are gleefully stealing this thread for their own selfish and flawed motives. Innocence? These are adults, educated, sophisticated and supposedly have a sense of justice, reality and objectivity, Yeah, sure!.

The "twin paradox" that poses the question how can the space ship twin and the earth borne twin see the other as existing on equivalent frames of reference? Both cannot age slower than the other, can they? However, the equivalence of inertial frames postulates say they can, don't they, the postulates I mean, do not they say that either/both twins sees the other as aging less than herself?.
This is another form of MacM's 'reciprocity problem' he discussed at length.

The paradox has been resolved, Feynman, Bohm et al, simply said that the space ship is the frame that accelerated hence is the frame that is "really moving", hence the space ship twin ages less than the earth borne twin. Or so the resolution goes.

What the resolution does, and the acceptance has been universal, is to negate the concept of "equivalent coordinate reference frames". Hence the embankment, never accelerating is always at rest with respect to trains moving relative to the embankment.

This leaves special relativity theory in an embarrassing position - there are no equivalent coordinate frames of reference.

Einstein's "equivalence" description has all moving coordinate reference frames measuring the relative velocity of frame and photon the same as distinguished from absolute velocity which only has meaning with respect to the embankment, or the vacua, or simply, with respect to zero velocity.

All this crap about "accelerated sycnchronization of clocks" is mere special relativity bs that is physically meaningless.

Neddy Bates discussion of reciprocity, that MacM dealt with in some depth is especially annoying as he, like the rest, have refused to look at the very basic errors Einstein made in arbitrarily selecting out light as an exception to the laws of motion that require that the laws be applied the same in all inertial frames of reference.

Ducks flying .9999c in the embankment frames are measured flying .9999c. In a frame moving v wrt the embankment the duck is measured as moving .9999c - v relative velocity. Light in the embankment frame is measured as c, while in the inertial frame moving at v wrt the embankment the relative velocity of rame and photon is c - v with v set "at rest" i.e. v = 0. SRT negates the very concept of motion.

Don't be distracted by arguments of "Gallilean transformations", "Lorentz transforms", as these are what follow from the assumption that the relative velocity of frame and photon are always measured as c in inertial frames moving wrt Ve = 0.

Circularity is seen in its most corrupt form when "Gallilean" and "Lorentz Tranbsform" arguments are imposed to substantiate the assumption that c is always the measured relative velocity of frame and photon.

Geistkiesel​

Betcha a doughnut Singularity, that neither of these fellows have looked at Einstein's development of relativity critically. They went to graduate school, wanted to learn SRT, they learned it and now, "they are one."

Geistkiesel​

Last edited: Feb 24, 2006
22. ### MacMRegistered Senior Member

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Do you have a link to such Feynman, Bohm et al, statements? This is precisely what I have claimed and that it is only the observer (clock) which has undergone physical F = ma acceleration (not the pseudo-acceleration of SRT claimed by James R) that undergoes any relavistic affect.

I of course was told to go read a book and learn relativity. It would be nice indeed if such statements by such recognized names could be posted.

Thanks.

23. ### Neddy BateValued Senior Member

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If you were to read my previous post more carefully, you would see that I was only discussing what the SR theory claims about reciprocity; not any of my own theories. Furthermore, my previous post illustrated yet another apparently large, gaping paradox within the constructs of the theory.

Allow me to summarize:
The reason SRT is internally consistent is, in part, because the relatively moving clocks are systematically out-of-synch with each other. I am referring to the way that the clocks toward the 'leading' end of a (relatively moving) reference frame are behind in time, while the clocks toward the 'trailing' end are ahead in time.

These unsynchronized clocks are, however, synchronized in their own rest frame, and they must remain so for any one frame that does not experience acceleration.

Assuming the obvious, that time cannot go backwards, we must assume that all of the clocks which are 'behind' will have to catch up to the ones which are 'ahead'. However, if we assume that a reference frame can be infinitely long, The clock that is 'most ahead in time' is undefined! So how can the clocks ever reach a synchronized concensus? In this case, they cannot, and this presents a problem for SRT.

As you would say, my dearest Geistkiesel, QED.

Last edited: Feb 24, 2006