shape of a relativistic wheel

Discussion in 'Physics & Math' started by DRZion, Oct 31, 2011.

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  1. Tach Banned Banned


    Hmm, I just showed you that:

    \(\frac{KE}{m_r}=c^2 \frac{\gamma(V)-1}{\gamma(V)}\)

    Do I need to show it to you again?

    The two lengths of curve aren't equal. The reason is that the line separating the two halves is not an axis of symmetry . Not only that, it departs from the axis of symmetry more as the speed V increases. So your parametrization [0,L] and [L,2L] isn't working.
    Last edited: Nov 2, 2011
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  3. przyk squishy Valued Senior Member

    No. Nor would the gravitational attraction to the wheel be detectable. So what? Did you ever show that the gravitational effects produced by the rolling wheel you talk about would be detectable? You'll love this: calculations please, since I consider this a rather nontrivial result given the complexity of GR.

    I doubt that, somehow.
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  5. Pete It's not rocket surgery Registered Senior Member

    Where? I first explored the concept in a thread [thread=56170]five years ago[/thread].
    Didn't use those. And you never followed through on the equation of the spokes. Have you checked what shape the equation predicts for the spokes at some t?
    Exactly. The aberration of the wheel means that the spokes are distributed unevenly in the ground frame.

    And they were such huge errors, too.... A rendering mistake in the tex, and an irrelevant upper limit for the magnitude of x1 and x2.

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    The aberration of the distribution of the rim is inherent in the thread from the very first post, so you 'pointing it out' is kind of redundant.

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    As late as post 48, you were still questioning whether the transformed midline would be S-shaped or U-shaped.

    Tach, the only one confused about "upper half" is you.
    RJBeery, DRZion, przyk, and myself have consistently been talking about the portion of the wheel above the level of the axle.
    I'm sorry if you misunderstood.

    So do the calculation. It's pretty straightforward.
    [post=2848763]post 67[/post] shows the same thing in a similar but simpler scenario.

    Um, no. We're talking about the top half of the wheel in the moving frame - that's the bits above the level of the axle.
    Last edited: Nov 2, 2011
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  7. przyk squishy Valued Senior Member

    Again, because your memory is apparently dying, all I said about kinetic energy and relativistic mass is that they're "closely related". And they are: you've just derived the relation above[sup]*[/sup]. I only ever said that the total energy and relativistic mass were proportional. Are you claiming the kinetic energy is relevant in GR but the total (kinetic + rest) energy is not? Given your tendency to back peddle I'd actually like to have an anwer to this from you on record.

    [sup]*[/sup] And since you like picking on any nit you can: yes, I know that the velocity also appears in that relation, and I never said they were uniquely related.

    I never said otherwise, and fail to see how that's relevant.
  8. Pete It's not rocket surgery Registered Senior Member

    The notions you're calling BS seem unrelated to the points that follow.
    What frame dependent gravitational effects could anyone possibly deduce from the vague notions I describe?

    The problem is that no one else was using that interpretation of "halves", which means you're on a different wavelength to everyone else, and disagreeing with things that nobody said.
    And you haven't posted any support of your equal KE claim beyond vague armwaving.
    Perhaps you should ask others in the thread. But I'm not about to embark on another pissing contest about communications skills, Tach. Let's just focus on what is right, and what isn't.

    For example:
    Are we all agreed that in the moving frame, the portion of the wheel above the level of the axle has more kinetic energy than the portion of the wheel below the level of the axle?

    Pot, kettle.
    You made a specific claim (that in the moving frame, the upper half of the wheel's mass has the same KE as the lower half of the wheel's mass). So far, this claim has been supported only by armwaving.
    I'm suspicious of that claim. I don't claim it's false, but I'm not prepared to accept it as true, because of my own armwaving reasons.
    You said you did the exact integral earlier. Where is it?

    Yes, that's GR spookiness. Here Be Dragons. I'm completely incapable of calculating whether an event horizon forms in an arbitrary situation like this, or the value of any other frame-dependent effect.
    How about you?
    Do you have calculations, or just armwaving?

    That would make the velocity of the rim in the axle frame exceed c.

    We know that:
    \(\omega R < c\)
    \(-R < x_1' < 0\)
    \(0 < x_2' < R\)

    So given \(R>0\) and \(\omega > 0\)...
    \(0 < \frac{-\omega^2Rx_1'}{c^2} < 1\)
    \(0 < \frac{\omega^2Rx_2'}{c^2} < 1\)
    Last edited: Nov 2, 2011
  9. James R Just this guy, you know? Staff Member


    YOU claim you've done the calculation.

    So, post it!

    Or admit you can't do it.
  10. Tach Banned Banned

    Actually I pointed out to you that there are no gravitational effects to speak of. Repeatedly. Here it is again:

    -there are no gravitational effects in the frame of the axle

    -if you run quickly with respect to the axle, the gravitational effects, whatever they may be , show up

    -the effects increase with the relative speed between the observer and the axle
  11. Tach Banned Banned

    Here, remember?

    Good for you, that wasn't the point.

    Then, you have to deal with the very disturbing issue that , if you covered half the wheel with a fence of height R, you would be seeing 6 spokes (not 4) above the fence. I pointed out this to you twice.

    The point is that there were mistakes.

    Yes, I am questioning this because it results into an absurd situation of being able to see more than half of the spokes above a fence of height R. Apparently you didn't register this , though I pointed it out twice already.

    But what both you and przyk fail to acknowledge is that the notion of what is "upper half" is frame dependent.

    Yep, the part that you fail to accept that it is frame-dependent.
  12. Tach Banned Banned

    Well , for one, you all claim that the "upper half" of the wheel has more energy (kinetic, total) than the "lower half". So, since total energy does gravitate, it follows that any moving frame could detect this disparity while the axle frame will obviously not detect anything.

    That is true, because none of you is willing to accept the idea that the midline separating the two halves is frame dependent.

    NO. None of you posted a formal calculation supporting this claim.

    You are getting personal again. I will post my calculation AFTER you post yours supporting the opposite claim. So far , neither you, nor przyk posted any calculations. You made no attempt, przyk made a sketch that I have shown to contain an invalid parametrization of the length of the halves as a function of the arc length. AFTER you post a complete proof, I will post my rebuttal.

    Once again, once you or anyone else posts their complete proof, I will immediately post mine.

    You mean that you still maintain that you can get gravitational effects by moving by the wheel? And your answer is just "GR spookiness"? Let me ask you this, if you move very fast wrt the wheel, will it collapse into a black hole?

    Yes, you are right. I get that the spokes are on the same side of the diameter, though my math is different since I have the parametrization of x and you the reverse of yours:

    \(x'=R sin (\omega t' +\phi)\)
    \(y'=R cos (\omega t' +\phi)\)

    leading to the exact equation of the spokes given early in the thread:

    \(y-R=\gamma (x-Vt) ctan[\omega \gamma (t-Vx/c^2)+\phi_i ]\)

    t=0 results into:

    \(y-R=\gamma x ctan (\phi_i-\omega \gamma Vx/c^2 )\)
    \(x'=\gamma x\)


    \(y-R=x' ctan (\phi_i-\omega V x'/c^2)\)

    So, for:

    \(y_1-R=x'_1 ctan (-\omega V x'_1/c^2)<0\)

    \(y_2-R=x'_2 ctan (\pi-\omega V x'_1/c^2)<0\)

    which leaves us with the paradox of having the spokes bunched up on one half of the wheel. To make matters even worse, the spokes bunch up "above" the fence of height R in your parametrization of the circle and bunch "under" the fence of height R in my different parametrization of the circle.
    Last edited: Nov 2, 2011
  13. RJBeery Natural Philosopher Valued Senior Member

    Tach: Przyk and Pete understand my question. What you're trying to do is to define the upper and lower semi-circles by their E values, so asking about whether their E values are equal becomes silly.
    Ironically, even though you appear to be confused, you also understand my question but you're simply dismissing the it based on the very same absurd conclusions that I did.
    Also, when you say
    ...I thought we had established that the upper half had more E than the lower half? Did you read this:
    Do you REALLY need me to plug numbers in for you to be convinced?

    \(E_{upper}\) and \(E_{lower}\) would differ due to their moments being unequal. Their moments are unequal because the distances from the point of rotation (S) to their center of masses (\(CoM_{upper}\) and \(CoM_{lower}\)) are unequal.
    \(E_{upper} = 1/2*((({(3\pi+4)R}/{3\pi})^{2}+{R^{2}}/{2})M/2)*\omega^2\)
    \(E_{lower} = 1/2*((({(3\pi-4)R}/{3\pi})^{2}+{R^{2}}/{2})M/2)*\omega^2\)

    The relativisitic analysis would only exacerbate the differential.
  14. Pincho Paxton Banned Banned

    Where do these shapes come from? The problem is that the wheel flies apart ages before any of these images happen, and therefore there are no physics which allow it to happen. The bonding is imaginary, the bending involves imaginary light. I don't see the point of messing with your mind. The wheel just dissolves into quantum energy. Don't screw your head's up.
  15. origin Heading towards oblivion Valued Senior Member

    Bubble man to the rescue. No need for confusion, just wave your arms and spout random scientific words and all will become clear!

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  16. OnlyMe Valued Senior Member

    The fact that frame-dragging is not detectable (given current technologies), cannot be used to set it aside. If frame-dragging does exist, then it does exist, with or with our "our" ability to detect it.

    As Pincho points out, there is a great deal of this discussion that has no basis in reality. The wheel could not remain intact under the conditions required by this hypothetical. Not based on any materials we currently have at our disposal.

    It would be a good idea that everyone keep in mind that the problem is not one that is real. It is hypothetical and beyond the reach of current technologies. It appears that to some extent GR effects are being applied from a SR frame of reference. A very tricky task in the best of circumstances.

    There is nothing wrong with completely hypothetical discussions. They are good exercises in logic and can at times provide insights that are not always otherwise clear. In discussions like this, hypothetical thought experiments very often breakdown, as variations of perspective ( FoR), diverge without clear definition. From the perspective of an observer it does not appear that everyone involved in this discussion is even trying to discuss the hypothetical from the same perspective. It does not even appear that the different perspectives involved are clearly defined.
  17. Tach Banned Banned

    You personally have not established anything. In the frame of the axle, contrary to your claim, \(E_{upper}=E_{lower}\). In the frame of the ground, przyk tried to sketch a proof that \(E'_{upper}>E'_{lower}\) that I have shown to be invaild due to an incorrect parametrization of the integration path.

    I have no idea what you are writing but it is incorrect for the reason pointed in the previous paragraph.
  18. Tach Banned Banned

    OK, let's try this. Let's assume that the frame dragging produced by the rotating wheel is detectable (it isn't really , but , for the sake of the argument let's assume it is). Then, according to all the people claiming that there is some asymmetry in "the matter between the top and the bottom halves as viewed rom the perspective of a frame moving wrt the axle" it would meen that an observer in the moving frame would be able to detect the effect of asymmetric mass distribution. Moreover, since the asymmetry increases with relative speed, the effect should become more noticeable. Prove that this is the case.
  19. przyk squishy Valued Senior Member

    Then you're plain wrong, because there is both gravitational attraction and frame-dragging. Whatever the magnitude of these effects, they're there.

    Derivation or reference please.

    Derivation or reference please.
  20. OnlyMe Valued Senior Member

    Why are you asking me to prove your case?

    Even if you determine the frame dragging effect of a relativistically rotating wheel to be trivial, to the discussion, dismissing it as non-existent is not consistent with what we know of the relationship between matter and space, or spacetime.

    You seem bent on holding others to some technical extreme of interpretation, apply the same hurdle to your own position. In a peer reviewed paper, though the frame dragging effect would likely be discounted as trivial, it would be mentioned as trivial and set aside rather than, an assertion made that it is non-existent.

    However, since the hypothetical involved here, goes far beyond any pratical real situation, how is it possible to know with any certainty what impact frame dragging might play? The hypothetical involves conditions that cannot exist, except within a hypothetical, science fiction or one's imagination. At least within the context of current knowledge and technologies.
  21. RJBeery Natural Philosopher Valued Senior Member

    Tach: So to be clear, you contend that the KE for both semi-circles is equal for all frames? Put it another way: if the wheel itself were a semi-circle (while at rest), and then rotated, your contention would then be that its KE would remain constant at all times from any given frame? You don't see a problem with this?

    You know, just because everyone says you're wrong doesn't make you a martyr. Sometimes it just means you're wrong.

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  22. Tach Banned Banned

    I said "gravitational effects" as in "differences" between the "upper" and "lower" half. Do you think that there are any differences for an observer located in the center of the wheel?
    If there aren't any, why would such differences appear for a moving observer?

    This is what your approach claims, so I am asking you to prove it.
    Last edited: Nov 2, 2011
  23. Tach Banned Banned

    No, this is not what I claim, you should know that KE is frame dependent. What I claim is that the KE's of the "upper" and "lowwer" "halves" are equal in the rest frame of the axle as well as in all the frames moving wrt the axle. The midline between the "top" and "bottom" half moves in such a way as to accommodate this equality.

    You know, if you don't even understand the subject of the debate after so many posts, you should refrain from making personal comments. This is the least you could do.
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