Relativity problem ...

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18 years 9 months ago #16895 by tvanflandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by EBTX</i>
<br />Can anyone see a logical flaw in the following relativity argument?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">All relativity paradoxes ultimately come down to the lack of remote simultaneity. If A & B have a relative motion, they will not agree about what time it is "now" at C even when A & B are at the same place at a given instant. That is because, in relativity, there is no such thing as "now" except within one's own frame. SR has no universal time or universal instant of now.

So here's where your example violates SR:

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">beards (which grow at the rate of 1 inch per year)<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Judging by your numbers, you must have meant "1 inch per <i>month</i>".

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">"B" leaves A1 clean shaven and travels to A2. He arrives about one year later and is still clean shaven while the guy on A2 has a 12 inch beard. Relativity is verified.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">B arrives one year later by A1-A2 time, but only a short time later by his own (equally valid) time. B is not surprised to find A2 with a long beard because, when B left A1 and A1 was claiming that A2 must be shaving his beard "now", B clearly inferred that A2 (who is moving rapidly toward B from a relatively short distance away) already shaved a year ago in B clock time.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">He then shaves again (as does A1 and A2),<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">B and A1 can shave at the same time because they are co-located. But B and A1 cannot agree about when A2 did this second shave because time at A2 is different for the two of them. Moreover, time at A2 changed rapidly in B's frame when B turned around.

The rest of your argument is filled with more instances of the same problem. I captured the essence of this paradoxical dilemma in my "twins" article, wherein the traveler takes along a native clock and a "GPS clock" keeping true Earth time. This makes the paradoxes much easier to resolve. See metaresearch.org/cosmology/gravity/gps-twins.asp

If you patiently work through that article, you will see that A1's beard has a growth spurt in B's frame when B turns around at A1. More amazing yet, if B reverses again to travel in the original direction, A1's long beard will <i>ungrow</i>!

Like most people, you may not like what SR predicts about nature. But SR has been proved internally consistent, so you cannot find a paradox that cannot be resolved. The only way to falsify SR is with a phenomenon that travels faster than light in forward time. That has now happened, and gravitational force is the example that falsified SR (in favor of paradox-free LR) on that very grounds. -|Tom|-

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18 years 8 months ago #13063 by EBTX
Replied by EBTX on topic Reply from
You're absolutely right. I woke up last night with the solution (I've been down this road before, long ago ... just forgot). What I'm really looking for (in dredging up these old, solved conundrums) is a solution to a problem I posed while you were gone ... to which I have no present answer and don't recall ever seeing one.

It involves the effect of relativistic translational velocity on a rotating body. If the axis of rotation is perpendicular to the direction of motion, one side of a rotating sphere is traveling faster than the other side relative to a stationary observer.

Because the equation for centrigugal force is F = mv^2/r ... what happens to the body when the "m" parameter increases on one side of the sphere and decreases on the other relative to a stationary observer?

Superficially, it would appear that the sphere would curve in a manner opposite a pitched curveball ... due to unbalanced centrifugal forces on the two sides. Of course, this violates linear and angular momentum conservation (a "reactionless drive"), so it must not be true. And, it violates special relativity since the sphere would appear to be just a rotating sphere incapable of any anomalous motion to someone traveling along with it. On the other hand, if one could somehow grab hold of the rest of the universe and toss it back, one could move forward without the appearance of throwing something out in the other direction (which is what a reactionless drive is all about).

How does one solve the apparent problem without a technologically unfeasible experiment? Or, how can this be reconciled, logically, with special relativity and momentum conservation without just asserting the truth of both as in "It can't be, therefore it isn't".

The mechanical situation seems right vis a vis special relativity when applied to the individual parts of the sphere but the general outcome appears untenable. Relativistic mass increase has to be factored into particla accelerators that run in circles, so why would it not be evident in the high velocity sphere?

Also, we can suppose that an absolute reference frame exists by Newton's bucket method, i.e. you rotate a bucket of water and it rises on the sides of the bucket. Special relativity invokes Mach's principle, "The universe is rotating in the opposite direction". But if you rotate another bucket in the opposite direction to the first, the universe would then have its Machian rotation cancelled. Yet the water still goes up the sides of both buckets.

Have you encountered this problem before?


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18 years 8 months ago #13064 by tvanflandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by EBTX</i>
<br />If the axis of rotation is perpendicular to the direction of motion, one side of a rotating sphere is traveling faster than the other side relative to a stationary observer.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Yes, this is a well-known SR scenario also. The solution is that the different time dilations and length contractions applicable to the near and far sides of the rotating sphere exactly counter the differing speeds and light-time delays, so that the sphere appears to remain a sphere and in uniform rotation. The phenomenon has even been named: "Thomas precession". Some papers by Ron Hatch go into great detail about his.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Also, we can suppose that an absolute reference frame exists by Newton's bucket method, i.e. you rotate a bucket of water and it rises on the sides of the bucket. Special relativity invokes Mach's principle, "The universe is rotating in the opposite direction".<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">In SR, accelerations are "absolute", not relative. The simplest case is the Sagnac effect, in which Michelson-Morely interferometers do show fringe shifts because they detect the absolute rotation of the laboratory or Earth.

GR attempts go get around this by redefining "acceleration" and "velocity" in 4-space terms. But without going there, the short answer is that rotation is absolute and translation is relative. The fact that rotations can be regarded as consisting of many small translations amounts to another form of the twins paradox, in which there is an asymmetric result over a round trip (i.e., a complete rotation). -|Tom|-

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18 years 8 months ago #14375 by Ryan2006
I'll tell you what the flaw is you say B is A1 and A2 I see your point though rate x time=distance is different in both cases thus th beards are longer or shorter depending on wether particles are negatively charged or positively charged electromagnetic energy I've been told dictates structure not organization rate x time does not equal rate times time in a paradox so when does a paradox begin and when does it end. Time slows down in relation to space the further you travel the younger you are the paradox is to reference frames meeting eachother.

ryan Henningsgaard

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18 years 8 months ago #16900 by EBTX
Replied by EBTX on topic Reply from
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">The phenomenon has even been named: "Thomas precession". Some papers by Ron Hatch go into great detail about his.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

Thanks. I'll see if there is anything on the net about it. Without having a proper name to put into google, it can be difficult to locate that type of arcane info.

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18 years 8 months ago #16875 by Thomas
Replied by Thomas on topic Reply from Thomas Smid
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by tvanflandern</i>
Like most people, you may not like what SR predicts about nature. But SR has been proved internally consistent, so you cannot find a paradox that cannot be resolved.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
I disagree: SR isn't internally consistent and you <i>can</i> find a paradox that could not be resolved with SR (see my pages regarding the Lorentz Transformation and Time Dilation and Twin Paradox ).


www.physicsmyths.org.uk
www.plasmaphysics.org.uk

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