shape of a relativistic wheel

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

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  1. Tach Banned Banned

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    I have already proven your claim false.


    So you cannot read the post just above? You have had the tools to do this calculation all by yourself for a couple of days now. Had you done it instead of waving your arms, you would have known that your claims are empty.
     
    Last edited: Nov 4, 2011
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  3. Tach Banned Banned

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    did your armchair teach you that as well? you got to stop taking lessons from it , it is giving you bad information

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  5. RJBeery Natural Philosopher Valued Senior Member

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    Yes, sometimes the things I find under this cushion are quite valuable. I've said it before, you have an unusual blindness if you MUST see the math to see the truth.
     
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  7. Tach Banned Banned

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    well, you should fire your armchair as your tutor, it's been teaching you wrong stuff

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  8. RJBeery Natural Philosopher Valued Senior Member

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    You're killing me with the post edits. I don't know if I should reply by editing my responses or making a new post...

    Anyway, what I say is true.
     
  9. Tach Banned Banned

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    The fact that it is wrong should not preclude you from posting more of the same, it has never stopped you in the past. seriously, you should stop taking cues from your armchair.

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  10. RJBeery Natural Philosopher Valued Senior Member

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    Behold the power of thought: colored objects are colored because they only reflect certain frequencies. Subtle shifts in the the spectrum of white light used to illuminate it will not change the emitted color of the object under consideration. Only a mirror would perfectly reflect the incoming frequencies.
     
  11. przyk squishy Valued Senior Member

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    What claim? That energy increases monotonically with velocity? You haven't proven anything false - just dismissed everything I said out of hand.

    Yes. What about it? You haven't done any calculations. Just asserted results. And from your results it's clear that either you've gotten things horribly messed up or you aren't even considering the same problem everyone else is. The energies of the top and bottom halves should not be time-dependant for instance.

    In any case I'm at work now, so I'll take a closer look at it later.
     
  12. Tach Banned Banned

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    Err , all the steps are in the previous posts, you only need to do a simple integral. You have been provided the formulas and their derivations. You had about two days to complete the LAST step, instead of waving your arms you could have done so.

    Because you say so? The tangential speed is time dependent, therefore, so is the total energy. In the frame of the axle, the energies are equal and they are not time dependent. In any other frame, the energies ARE time dependent. They oscillate in time , as I pointed out to you quite a few posts ago. But you wouldn't listen.

    Please do so. You had all the time in the world and you had the derivation up to the last step. I am sure that you will come back with a last gasp effort to prove yourself right and me wrong. I am going skiing, so I will be out until Monday, so you have at least two days to figure how you could possibly miss. I won't be responding until Monday.
     
    Last edited: Nov 4, 2011
  13. Tach Banned Banned

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    And those frequencies still behave like good citizens and they do get Doppler shifted, no?

    You are confusing "not reflecting all frequencies" with "non having the reflected frequencies Doppler shifted". Why do you keep pushing your foot deeper and deeper in your mouth? I am going skiing, so I won't be answering. In the meanwhile, reflect on the errors in your posts. Maybe it is time to take a break?
     
  14. RJBeery Natural Philosopher Valued Senior Member

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    Jesus Tach if we can't agree on this then you're truly lost. A green wheel will emit green light, period. Move the wheel relative to a light source and you're only shifting the average frequency of the illuminating white light. As long as the wheel continues to be illuminated by the proper (green) frequency of light it will continue to emit that green light, and absorb the rest.

    The apparent color of the wheel is indeed dependent on the the movement of the observer (i.e. camera) relative to the wheel. Conclusion: the objects under consideration in this thread do appear Doppler shifted. Enjoy the snow!
     
  15. Tach Banned Banned

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    Correction, when illuminated with white light it will REFLECT ONLY frequencies in the green spectrum.

    White light is made of ALL frequencies, INCLUDING green.


    Nope, because the frequency (green , in this case) is equally blueshifted when reaching the object and redshifted (by the same amount) when arriving at the camera, as I explained in the file that pete had so much trouble understanding.

    Thank you, bye. Reflect on what I've been showing you.
     
  16. RJBeery Natural Philosopher Valued Senior Member

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    I hesitate to respond until I know you're done fiddling with what you've written...

    Anyway, the incident blueshifted green light would not get reflected! The color that got reflected would be the incident yellow light (for example) that happened to get blueshifted into the proper frequency so as to appear green to the wheel. The wheel would emit green, period. The camera would determine the wheel's color based upon the Doppler shifting caused by the relative motion between the objects.
     
  17. przyk squishy Valued Senior Member

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    No Tach, you don't just post "steps" and tell people what calculations to do. Giving people orders, which you are in absolutely no position to do, does not constitute a proof or justification of any kind. If anything I'm inclined not to do them just because it's incredibly arrogant (you show no openness to the possibility you might be doing the wrong calculations in the first place) and plain rude.

    Funny how you leave out any qualifications in that. The tangential speed of some given element of the wheel, as you follow it around while the wheel rolls along, has a constantly varying speed. But the velocity profile of the wheel is static. If the wheel is rolling at velocity \(v\), the point at the top of the wheel always has a speed \(\frac{2v}{1+\frac{v^{2}}{c^{2}}}\) while the point at the bottom in contact with the ground has zero speed for instance.
     
  18. RJBeery Natural Philosopher Valued Senior Member

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    I find it difficult to believe that TACH believes that anyone was actually discussing different energies for each ROTATING half. Intentional obfuscation.
     
  19. Neddy Bate Valued Senior Member

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    Trying to intellectually engage Tach is an exercise in futility. Those who have dared to try have my utmost sympathy.

    He has made it clear in this thread that he does not understand the difference between visual effects, and Relativity of Simultaneity (ROS) effects. The reason there are more spokes in the top half of the wheel, (according to the reference frame of the road), is because of ROS, not because of visual (ray tracing) effects.

    I'm not even sure if he believes in the Doppler effect at all, based on his inane argument about mirrors.

    He also seems to think that, according to reference frame of the road, there is just as much energy in the top half of the wheel as there is in the bottom.

    It's rather amazing, because he certainly is excellent at maths. How he can do the maths without understanding the concepts behind them is a mystery to me.
     
  20. Masterov Registered Senior Member

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    How much change diameter and length of rim of the static wheel for variable speed of rotation (in the plane of the monitor screen) of its.
     
    Last edited: Nov 5, 2011
  21. Pete It's not rocket surgery Registered Senior Member

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    Good point, Mr Beery.

    Proposition for you, Tach.

    As you have pointed out with all the humility and politeness we've come to expect from you, my criticism in this thread of your doppler shift analysis has been woefully inadequate. Clearly, I have something to learn.

    So, after my exams are done, how about you and I explore this moving mirror doppler shift problem properly in the formal debates forum?
    I'll take the position that light reflected from a moving mirror will generally be doppler shifted.
    You take the position that light reflected from a moving mirror is never doppler shifted.
    Loser (as judged by a mutually agreed adjudicator - please suggest someone) has to be perfectly polite to all posters at all times for 3 months, or be banned for that time.
    Or perhaps you can think of a different penalty involving avatar images and associated text?

    What say you?
     
    Last edited: Nov 5, 2011
  22. James R Just this guy, you know? Staff Member

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    Does Tach really think that light reflected from a moving mirror (say one that is heading directly towards the source) will not be Doppler shifted when it arrives back at the source?

    Really?

    I'll take Tach on right now in a formal debate on that topic. It will only take one post to demolish him and a few lines of maths.
     
  23. Pete It's not rocket surgery Registered Senior Member

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    His proof, such as it is, is here:
    https://docs.google.com/viewer?a=v&...N2E4MC00NDc3LWE1ZjctNDA5YjVjZmM5NWIy&hl=en_US

    After a couple of half-assed dismissals, I had a better look and said...
    Tach's response is in [post=2850347]post 214[/post]. Excerpt:
    He also gave the impressions that's it's impossible for the mirror to be approaching both the source and the camera ([post=2850195]post 194[/post], but I didn't get that clarified.
    He might also try to hedge by talking about raytracers vs reality.
     
    Last edited: Nov 5, 2011
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