3 clock problem

Discussion in 'Physics & Math' started by ryans, May 12, 2003.

  1. MacM Registered Senior Member

    Messages:
    10,104
    What? Continued

    ryans,

    The picture and the statement below the picture repeats that case.

    Also read the last sentenance.

    He goes on to discuss various geometries, starting with your sphere with a circle on it. so you attitude is BS. You haven't shown me s___. I know that arguement but you continue to avoid the f____g question.


    Question: "Assuming some form of relavistic dimensional change either at the circumference or at the radius and/or both, what is your explanation for claiming Pi changes when measured by a ruler? That is to say any dimensional affect on the rotating object, regardless of geometry imposed, is going to also affect the ruler and no change can be measured and hence no change in Pi.?"

    Just repeating once more so you might catch the point made at the outset. "I do not challenge relativity as to dimensional contraction, I am challenging the erroneous claim that Pi changes due to a change in measurement of a rotating system.
     
    Last edited: May 14, 2003
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  3. ryans Come to see me about a dog hey Registered Senior Member

    Messages:
    995
    Lets say that your arguement is correct, and remember this Mac, I am answering this from within YOUR framework, and it in no way represents my position.

    You are saying that the circumference nor the radius changes for the observer on the merry-go-round right. Lets introduce the concept of frequency here, where the frequency of the merry-go-round is how many times it completes a revolution in one second. If the observer doesn't measure a different value for the circumference of the m-g-r, then Pi stays the same right(remember I am answering this from within your framework.

    The relation of velocity to frequency is.

    v=rw where r is the radius of the m-g-r, w is the angular frequency (f times 2Pi) and v is the tangential velocity. thus you can see that if the radius does not shrink, then as w gets bigger(the m-g-r spins faster) the observer will see himself as travelling faster than light!

    Bit of a Paradox hey!

    There is more to this, but I think this will be enough to totally confuse you.
     
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  5. MacM Registered Senior Member

    Messages:
    10,104
    No Confusion at all.

    ryans,

    Why would that be confusion at all. I am well aware that spinning around with a laser pointed into space does not cause the light beam to become v=c , etc.

    Lets try answering the question shall we. Wouldn't that be more productive.?

    Question: "Assuming some form of relavistic dimensional change either at the circumference or at the radius and/or both, what is your explanation for claiming Pi changes when measured by a ruler? That is to say any dimensional affect on the rotating object, regardless of geometry imposed, is going to also affect the ruler and no change can be measured and hence no change in Pi.?"
     
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  7. ryans Come to see me about a dog hey Registered Senior Member

    Messages:
    995
    Whatever Mac.

    1/ Chroot and I gave you the reason, you chose to ignore it.

    2/If the radius does not contract relative to the observer, he will calculate himself as going faster than c. Since this is not possible, the observer on the m-g-r MUST measure Pi to be smaller.

    You cannot see the wood from the trees Mac.
     
  8. MacM Registered Senior Member

    Messages:
    10,104
    Question

    ryans,

    And neither you nor ch answered the question. You try to spit out a lot of unrelated garbage to confuse the issue. Nothing you say about me alters that fact damnit.

    If you can't answer the question just be man enough to say so.

    Your example above is childishly stupid and has no bearing on the question.

    Answer the question:

    Question: "Assuming some form of relavistic dimensional change either at the circumference or at the radius and/or both, what is your explanation for claiming Pi changes when measured by a ruler? That is to say any dimensional affect on the rotating object, regardless of geometry imposed, is going to also affect the ruler and no change can be measured and hence no change in Pi.?"


    You make the assumption above that somehow I have denied contraction. My position on contraction is neither the question or at issue. I haven't even told you my view.

    The question gives you the options of having perpheral contraction, radial contraction and/or both. What more do you want just answer the question.

    PS: In the event you or any other reader has difficulty reading the attachments, you can click on it and you will get a symbol. Click on the symbol and it enlarges and becomes more readable.

    If you still can't read it you could print it out and it would be readable. But to save you and the others some trouble I will quote page 65, Fig 3.1, here:

    *************************************************
    Fig 3.1 Slim's ruler is contracted, since it lies along the direction of the rides motion. But Jim's ruler lies along a radial strut, perpendicular to the direction of the rides motion and therefore its length is not changing.
    *************************************************
    Just wanted to clear that up since it became an issue and you remarked about it but never came back to acknowledge that I had quoted things correctly.

    Now since the only thing I have done here is pose the above question and have not given you my view of relavistic contraction it seems to me more appropriate to address the submission I made than to attack me for what others have said.

    Doesn't that seem to be logical to you. Why do you assume attacking me rather than addressing the words of others is how to answer the question. Especially since I start this by saying I didn't like the book because he was screwed up.

    And from that you accuse me of being wrong. Does that then not make Mr Brian Green right and therefore you wrong? Or am I screwed up on english as well?

    Quit screwing around and answer the question.

    You opened this can of worms with the following statement:

    **************************************************
    ryans
    3 clock problem
    MacM is convinced Special relativity is flawed.

    I plan to f*** him up in terms of his perceptions of the 3 clock problem.
    ****************************************************

    So let go fishing. Then we can resolve 3 Clocks. What do you say.?
     
    Last edited: May 14, 2003
  9. ryans Come to see me about a dog hey Registered Senior Member

    Messages:
    995
    For you the 3 clocks problem will never be resolved because you will never accept our responses. Further the question cannot be resolved on a philosophical level, you need to have calculated proofs.

    Again, the m-g-r problem has been resolved. Don't accept the answer if you don't like, that's not my problem.

    Stop wasting my time.
     
  10. MacM Registered Senior Member

    Messages:
    10,104
    Fine

    ryans,

    Fine. I agree it is a waste of time but I disagree that I don't accept answers. I only rejected answers that have not been on point. You NEVER answered the question. Not once. You made numerous attempts to side step the issue and give information that was not related. I am content that others can see that and it is clear you either can't or won't answer the question for obvious reasons. the queston requires that you break ranks with Relativity.

    As far as 3 Clocks, you haven't even tried to give one answer. Even after you initiated this string for the purpose as you stated above.

    Your effort to achieve that goal has been one collosal failure.

    For anyone following this miserable thread that is interested in the facts about 3 Clocks the computed data for the case and an outline description of the test can be found under topic "Relativity, by "apolo", message 5/13/03 @ 5:15PM.

    As they say in the biz "That a rap" Bye
     
  11. MacM Registered Senior Member

    Messages:
    10,104
    Same

    ryan,




    Your illustration seems to me to be exactly the same as what is common. That is to watch the world go by while spinning around. Obviously if one takes the distance to the moon (apprx 236,000 miles) then at 7.5375 Rpm it would appear to traverse the sky from your frame of reference at v = c.

    So according to you just how close (shortening of the radius, distance to the moon) must change to preclude such a result?

    It is a nonsense concept, to create a radius which keeps the moon at a real physical motion, the radius would have to equal "0".

    So your view seems to say when one spins around that we collasp the universe instantally. Maybe you have discovered the secret of Particle Entanglement. Just give it spin and it is erverywhere in the Universe at the same time. No wonder the response is instantaneous.

    What do you think? Sssshhh. Can't you address the question?
     
  12. ryans Come to see me about a dog hey Registered Senior Member

    Messages:
    995
    What? Can't understand you Mac, your mumbling again.
     
  13. MacM Registered Senior Member

    Messages:
    10,104
    Simple

    ryans,

    I was just saying that your example explanation of why Pi "Must" change seems simular to the arguement that if I spin around art 7.5 Rpm, then the moon is exceeding the speed of light. Which we know is a false conslusion because the moon has its own locality.

    Let me make one final attempt to resolve this issue.

    Don't let me put words in your mouth but it is my impression that where you are is this:

    "I can show you that Pi MUST change, therefore it does".

    Having said that and not answering the question of how that can occur since any relavistic affect in any geometry imposed would also affect the rotating frame not just the ruler; which mitigates any possiblitiy that Pi can be measured differently hence change.

    That you can't answer that question.

    If that is the case, I accept that as an answer. Not necessarliy that it is correct but that you have given a fair answer.

    Is what I have just said a fair summary or not?
     
  14. river-wind Valued Senior Member

    Messages:
    2,671
    ok, I honestly did not follow this thread 100%, but my impression about measuring the radius of a revolving disk makes me think these three things, I don't know if they are the case, but they are logical.

    1)if you look down on the disk, hold the ruler still above it, and align one end with the center of the disk, you should see the mass on the outside of the disk become more dense, volumous, and it's time should slow down. You will not see the disk shrink in size. I see no reason for the radius of the disk to shrink, because the disk is not moving, the atoms which make up the disk are. and these atoms are not spinning, they are accelarating in a straight line (from their POV). For every instant they move in a straight line, atomic attraction pulls then, resulting in a curved path. *the disk is not what is moving.* It's source material is.
    for every atom that is sequentially closer to the center of the disk, you should see less speed, less density, less volume, and faster time (ie, these differences are how the outside of the disk can move faster then the inside, but remain connected)

    2)if the ruler is attached to the outside edge of the disk, then the atoms which make up the ruler will change in dentisty, volume, and time rate as the disk spins. The effects will be slightly more for the ruler than it will be for the outside edge atoms of the disk, because the atoms of the ruler are slightly father out from the center of the disk. no measurement difference will be noticeable, because by attaching the ruler to the disk, it can now be considered part of the disk. again *the disk is not moving*. it's costituant matter is. So the ruler is not rotating, it is travelling in a forced curved foreward path.

    3)if the ruler was attached to the face of the disk, such that one end lines up with the outer edge of the disk, and the other sits in the center of the disk, then you will find that the outer atoms of the ruler become more dense, more volumous, and time will move slower. it again, has become part of the disk, and by spinning the object, you, as an outside observer will see the effect on the conponant matter of the spinning disk; as the componant matter accelarates foreward.


    the circumfrence does not shrink when you rotate a disk, because the mass which is effected by the accelaration at the edge of the disk is the atoms themselves, not the bonds between them or their subatomic particles. To shrink the circumfrence of the disk, you would have to shrink the distance between the molecular/atomic/nuclear bonds in the componant material, and this does not happen during acceleration.



    Is this accurate?
     
  15. Crisp Gone 4ever Registered Senior Member

    Messages:
    1,339
  16. MacM Registered Senior Member

    Messages:
    10,104
    Rotating Disk

    Crisp,

    Thanks for the link. I'm posting this portion for ryans benefit.


    ****************************************************
    Let's consider a huge disk above which are drawn two concentric circles, one very small and the other one as huge as the disk:

    Our observer is on the disk, which rotates at a very high speed. Another man, in a galilean frame, measures the circumferences (P) of the two circles, and their diameters (d) with a ruler. This man then does the following calculus: P/d. He finds: P/d = p. For him, euclidean geometry is true. (True means here that it describes reality.)

    The observer on the disk measures the circumferences and the diameters WITH THE SAME RULER. For the measurement of the diameters, the ruler is not contracted (from the point of view of the man in the galilean frame) in its length. (See special relativity for this.) Thus, the observer on the disks will find the same results as the man in the galilean frame.

    Now, the observer measures the circumference of the small circle. For the man outside the disk, the rule is not contracted in its length because it goes very slowly (it's close to the center of the disk). Therefore, the observer finds again the same result.

    However, things are going to be different for the fourth measurement. When the observer on the disk measures the circumference of the big circle, he goes very fast compared to the man outside the disk. Hence, from the point of view of this man, the rule is contracted in its length and the observer won't find the same result. For him: P/d does not equal p. Euclidean geometry doesn't describe reality in this case.
    ****************************************************

    Another error I suppose. Since it says the diameter doesn't change and the circumference does.

    And the same error in that it fails to give an explanation as to how the affects of Relativity shortened the ruler without shortening the circumference.

    Since, in my example, the circumference is made of the same material. Indeed the circumference was built using a series of short measuring rods or rulers. So if the circumference is made of rulers, the circumference and the measuring rod both shrink equally and hence there is no measurable change in Pi.

    My complaint is, and has been, that this is a false illustration of Relativity and says nothing about my view of relavistic contraction.

    As yet nobody has addressed the issue of seeing any (measureable) affect at all.

    Do you (ryans) actually not see that any affect by any geometry you impose means the ruler will always measure the same?

    Upto this juncture I haven't given you my opinion. Now I choose to do so, since the fallacy of a changing Pi has been beaten to death.

    I anticipate that the circumference actually contracts but so does the radius. The circumference due to velocity and the radius due to acceleration (gravity). Pi remains the same and the measurement remains the same. to a ruler moving with a rotating body.
     
    Last edited: May 15, 2003
  17. James R Just this guy, you know? Staff Member

    Messages:
    39,426
    MacM:

    <i>Another error I suppose. Since it says the diameter doesn't change and the circumference does.</i>

    It is not an error. Length contraction only happens in the direction of motion. The diameter is not moving outwards or inwards, so rulers placed on the diameter are not contracted. The circumference is, however, moving at right angles to the diameter, so the (stationary) rulers placed to measure it are contracted (from the point of view of somebody on the disk).

    <i>And the same error in that it fails to give an explanation as to how the affects of Relativity shortened the ruler without shortening the circumference.</i>

    An observer in the same frame as a moving object sees no shortening.

    <i>My complaint is, and has been, that this is a false illustration of Relativity and says nothing about my view of relavistic contraction.</i>

    Just because you don't understand it doesn't mean it is false.
    --------

    On a side issue, the value of pi never changes. pi is defined to be 3.14159...
    What changes in this example is the formula for the circumference of the disk. For an observer on the disk, the circumference is not given by C=pi D. Instead C < pi D.
     
  18. MacM Registered Senior Member

    Messages:
    10,104
    Pun

    James r.,


    ANS: Thank you James you have emphasized my point. That statement was a pun aimed at ryans since he had argued with me and even got a bit nasty claiming that the circumference didn't change the it was the radius.




    ANS: Thank you again James. That has been my arguement. "No measurable change by the rotating observer and his ruler. Hence no change in measurable Pi from the moving frame of reference."



    ANS: Finally thank you James R., We can agree. And ryans I hope you understand what James R., has said.

    "No measurable change in Pi". The numerical Pi doesn't change. The circumference, not the radius is affected."


    James R.,

    This arguement ensued simply because I stated I didn't like "the elegant universe", by Brian Greene because in there he state the numberical value of Pi changed becauset the measurement we just went though changed.

    I claimed that was BS and thereafter I have lhad to take a series of slanders from ryans and chroot about how I just was ignorant. As it turns out, if I am ignorant, they have just called you ignorant since we arein full agreement here.

    Case closed in my book. Thanks again.



    __________________
     
  19. ryans Come to see me about a dog hey Registered Senior Member

    Messages:
    995
    Mac, I can garauntee you that for the observer on the disk, the value of the circumference does not change. There is no length contraction of the circumference for the observer on the disk, he is moving with the disk. However he will see the radius change due to radial acceleration. I suspect you tend to avoid acceleration because you have no **** clue how to deal with it.

    And do you always contradict yourself;

    Your words

    "I have no problem with length contraction"

    "If the circumference of the m-g-r changes, so does the rulers length"

    No length contraction of the circumference for the observer on the m-g-r buddy.
     
    Last edited by a moderator: May 15, 2003
  20. MacM Registered Senior Member

    Messages:
    10,104
    ryans, ryans, ryans

    ryans,





    ANS: Wrong on at least two counts. The first one edited by James R., I won't even bother to address.

    The other is a multitude of problems.

    1 - Where is the contridiction? I said the circumference contracted. You said it didn't.

    2 - I said any contraction would be immeasurable to the rotating observer and his ruler. You claimed Pi was changed and went off into explanation about circles on spheres and different geometries and how it was only my ignorance that was the problem.

    3 - Now you are going to guarantee me that the circumference changes but is imeasurable to the moving observer . THAT WAS "MY" ARGUEMENT ALL ALONG.

    4 - As far as the radial change due to "gravity" I don't disagree with that, I have believed that it was necessary all along but it has never, before your introduction of a radial change been presented in Pi changing arguements for Relativity.

    And now you F****** that up (edited pre-emptively) by claiming
    it becomes measureable. You still have the circumstance that the moving observer will see no affect of contraction in the system since his ruler is affected exactly as is the dimensions of the rotating system. Just as I have been saying to you and chroot for days.

    Instead of addressing the issue you have been to busy patting yourself on the back and trying to slam "An Old Ignorant Man" to realise how stupid your statements have been.
     
  21. chroot Crackpot killer Registered Senior Member

    Messages:
    2,350
    So wait a minute -- your entire argument was simply that pi is defined to be 3.1415926535... and therefore does not change?

    - Warren
     
  22. MacM Registered Senior Member

    Messages:
    10,104
    Not Entirely

    chroot,

    Not entirely. That could be one arguement but not necessarily a good one. One could claim (as I did point out) that other geometries, etc, violate the definition of Pi which is the ratio between the circumference of a circle and its diameter. However, I quickly pointed out that the issue was that no measureable change in Pi occurs on the moving frame.

    Any relavistic affect at the circumference that shortens a ruler must also shorten the circumference hence no measureabel change hence no change in Pi by any definition. James R., and everneo seem to agree on this point.

    Added to that arguement (ryan first said the cirfumference didn't change but that the radius changed due to acceleration affects. I then posed the question this way

    "Asssuming any relavistic affect due to motion upon a rotating system, be it the circumference, or the radius and/or both, explain justification for claiming Pi changes since any affect, any geometry imposed by any theory would also necessarily affect the rotating disk in the same manner and magnitude as the ruler, hence no measurable change, hence no change in Pi."

    I also stated in the beginning that I accepted velocity contraction but that the issue was the claims by Brian Greene and the many others (links posted) and by members here, that Pi had a measureable change because the ruler would shrink.

    That position simply ignored the fact that any such affect must also change the rotating object and no such change of measurement could be found.
     
  23. ryans Come to see me about a dog hey Registered Senior Member

    Messages:
    995
    Moderator edit: Post deleted due to unnecessary personal insults.
     
    Last edited by a moderator: May 16, 2003

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