shape of a relativistic wheel

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

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  1. Pete It's not rocket surgery Registered Senior Member

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    Ha. My mistake, I should should have paid more attention.
    Nevermind, I'll have a look and find your mistakes for you.
     
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  3. Pete It's not rocket surgery Registered Senior Member

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    OK, easy mistake to make - in general, the angle of incidence is equal to the angle of reflection only in the rest frame of the mirror.

    Formalism to come, if I have time.
     
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  5. DRZion Theoretical Experimentalist Valued Senior Member

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    No, the Terrell Penrose effect will still be present but it won't as precisely affect the shape of the track as a perfectly circular wheel does. AlphaNumeric as well as the videos show that when the spinning (not rolling) wheel is in the plane of the monitor it will not appear distorted. This will not be the case for the tank tread because of its non-symmetrical shape.

    Tach, if there is no meaning to proofs made of words, then what is the point of talking? If we cannot make sense of anything said without mathematical symbols, then how can we even begin to have a discussion on relativity? If you're correct then we have been spouting gibberish for the past 160 posts!

    If the observer is moving at the same speed as the wheel's tangential velocity, the wheel will still look distorted.

    Yes, but they look differently distorted from different angles! Pete has pointed this out in a later post, quoted here at the end.

    Hmmmmmmmm.. I don't even know enough about it to make a statement. The way I understand, it is now something like an 'electron cloud' ... too bad, because orbiting electrons would provide a possibility to test this experimentally.

    I suppose if there were relativistic tanks I would have known already.

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    I know that accelerators have to work with ions because they use strong electric fields to force these to move. Do they use completely bare nuclei? (and I am too sleepy now to check) Also, isn't it possible that a hydrogen atom with an electron attached will form when particles smash together? This would still conserve charge.

    The animations given earlier are the best I can provide, but these not only account length contraction but other effects such as terrell rotation and ray tracing. This wheel looks 100% normal.

    How can you disprove it? We agree that visually, at least, most of the spokes will be above a line drawn in the rest frame. Mass will intuitively be found in the same place.

    Yep, and this is why I started this thread too.

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    Chinglu brought up an objection which was just too much like my earlier objection - that there will be no way to determine the course of events using relativity. Minds work alike, and they can fail alike as well. I still have to make up my mind about it.


    Even so, if only 4 spokes are present above the edge of the paper, they will still be bunched up!

    This adds even more complexity to the issue. Because now the spokes will be shifting around as the observer moves around. Hence, no force is acting on the spinning wheel, but as the observer moves around the spokes will appear to migrate from one portion of the wheel to another, and quite likely shift the center mass of the wheel.

    These are good questions! I am too tired to think clearly today, I will try tomorrow though.
     
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  7. Tach Banned Banned

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    You will never become a physicist at this rate. If you are closed - minded and you can't follow someone else's bona fide proof, how will you ever learn?
     
  8. Pete It's not rocket surgery Registered Senior Member

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    I should hope not, or this medical degree would be a real waste of time!
     
  9. Pete It's not rocket surgery Registered Senior Member

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    I'm going to have to quit anyway. I have exams next week, and this has just been a big exercise in procrastination.

    Also, it's beginning to feel like I'm kicking a puppy attacking my shoe...

    So I give up, Tach. Just take the goddam shoe!

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  10. Pete It's not rocket surgery Registered Senior Member

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    Last post for me...

    More correctly, the angle of incidence equals the angle of reflection only in reference frames where the mirror's velocity is parallel to the reflecting surface.
     
  11. Tach Banned Banned

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    Good, stick to medicine. Hopefully you are more open minded when it comes to treating people

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  12. Tach Banned Banned

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    Yes, this IS the frame used in the derivation. The ONLY frame.

    If you get a different result from zero shift, you got the wrong result.
     
  13. Pete It's not rocket surgery Registered Senior Member

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    Just one more... (I must be an addict)

    OK, Tach.
    In your diagram, you have the object moving at velocity V.
    Since you said you're working in the object rest frame, I'll call that a typo and pretend that V is attached to the camera and the light source.
    Also, you have the primary ray pointing at the camera instead of the source.
    And you don't need separate scenarios for approaching and receding... it's OK to let \(\phi\) vary from 0 to 2pi.

    Now, \(\phi\) is the angle between V and the primary ray, which I think you've correctly used in your formula for \(f_{mirror}\).

    But you have used \(\phi\) again in your formula for \(f_{s'}\).
    This is not right.
    In the \(f_{s'}\) equation, you should be using \(\phi + 2\theta\), the angle between V and the secondary ray.

    But don't take my word for it... ask your professor. I'm sure they will be happy to help you with the doppler shift from a moving mirror.
     
    Last edited: Nov 4, 2011
  14. James R Just this guy, you know? Staff Member

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    Just like you delivered the math on the proof you said you had that there was not more mass above the fence than below it?

    Ha!

    At this point, it is quite clear you don't even understand the issue you're discussing. For a start, you're confusing the actual coordinate locations of the spokes in different frames with the image that those spokes make on some distant detector or camera. Why you decided to introduce a distant camera at all is a bit of a mystery, since the issue is whether the wheel has more mass above the fence than below it - an issue you pretended you had solved earlier in the thread.

    No surprises there. True to your usual form, the more wrong you are proved, the more you back away. In the end, you resort to blatant denial. In the past, I've even seen you go back and edit previous posts you made to hide your errors. It's as if you can't admit your many mistakes even to yourself. They must be obliterated from your memory for the protection of your extra-large but oh-so-fragile ego.

    Yeah. Sure. 48 hours (or is it 72?) after you said you had a simple proof that the mass above the fence was always equal to the mass below the fence in all frames, we have yet to see anything from you.

    I'm sure your imaginary "file" will be as forthcoming.

    The man who looks down like a self-made god on everybody else apparently has quite a different standard when it comes to the treatment he expects from others.

    This must adversely affect your daily life. I wonder if you'll grow up at some point. I must say, the chances aren't looking good right now, from what I can see.

    How convenient for you. When your own work refutes your arguments, suddenly you just want to forget the whole thing.

    Get out your Doppler-shifted mirror and take a good hard look at yourself in light of your comment, Tach.

    Surely, hypocrisy is one of the most face-palming traits one can have.
     
  15. przyk squishy Valued Senior Member

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    It shouldn't be necessary - the energy density (per unit mass) is a strictly monotonically increasing function in v. If the speed v is higher, so is the energy. Your not going to deny that the speeds are consistently higher in the top half of the wheel, are you? :bugeye:

    No, here it's clear you're trying to redefine what the discussion is about again. The "top half" of the wheel is the part of it above the axle (and the strip of paper mentioned in the OP). Yes, this is "frame dependent". No, that doesn't make it "invalid".
     
  16. Tach Banned Banned

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    Actually, it IS necessary. So, please do the calculation, I have given you (and JamesR, who's still contributing nothing, except stalking) all the tools to do this simple calculation.
     
  17. Tach Banned Banned

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    No, the speed V is attached to the object. "Working in the object rest frame" means using the relativistic Doppler equations expressed in angles measured in the rest frame of the object, once for the ray coming from the source, when the object acts as a target and the second time for the reflected ray , when the object acts as a source. You obviously did not take time to understand the writeup.



    Yes, this is standard raytracing , if you don't understand the notions, just ask.


    Good, you got this part.

    Nope, you did not get this part, see above. The angle between the speed V and the secondary ray and the angle between the speed V and the primary ray are the SAME (\(\phi\)) due to geometric symmetry. Definitely not the \(\phi + 2\theta\) you are claiming. Claiming such a thing shows that you don't even understand the basics of the relativistic Doppler effect.

    The error is yours. Again. If you don't understand things, just ask, I will be more than happy to explain.
     
    Last edited: Nov 4, 2011
  18. Tach Banned Banned

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    So, James, 24 hours after you were given all the tools to calculate a simple thing, you are still not able to perform the calculation, yet you have the time to stalk all the posts for imagined errors. You know, if you used a fraction of the time you are using for stalking and trolling to make the calculation instead you would have found a very interesting fact:

    The total energies for the halves of the wheel are:

    \(E_1=\frac{1}{1-(V/c)^2}(\pi-2(V/c)^2 sin(\omega t))\)
    and
    \(E_2=\frac{1}{1-(V/c)^2}(\pi+2(V/c)^2 sin(\omega t))\)

    So, contrary to all claims by parties that did lift a finger to do any calculations, waved their arms and flapped their wings, it turns out that:

    \(E_1>E_2\) for \((2k+1)\pi<\omega t<2k \pi\)
    \(E_1<E_2\) for \(2k \pi<\omega t<(2k+1) \pi\)

    On average, the energies are the same:

    \(<E_1>=<E_2>\)

    The same goes for the energy density, contrary to all unsubstantiated claims that przyk has been making. Had he done the calculations instead of waving his arms , he would have found out the above.
     
  19. RJBeery Natural Philosopher Valued Senior Member

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    Searching for "Last Edited By Tach" in this thread: 31 results

    What's quite interesting is to note when the edit times are made after there are follow-up comments by other posters.
     
  20. RJBeery Natural Philosopher Valued Senior Member

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    The last time in this very thread you made this "all the tools to do it yourself" comment you were shown to have done the calculation wrong.
     
  21. Tach Banned Banned

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    You really, really need to stop stalking. If you used a fraction of the time you are using stalking my posts for doing some real science, you would be learning new things.



    I gave you all the tools to do the calculations yourself, had you spent 5 minutes, much less than it took you to compose this post, you would have gotten the result.

    Since you couldn't do a simple calculation even in the context of all the materials being done for you, you have the results posted for you, ready made. I am sure that the next question coming from you is going to be : "how did you get that?".



    It isn't my fault that Pete managed to misunderstand every basic notion from the explanation. He imagined errors where there were none.

    Stop stalking. Stop trolling. They are unbecoming traits for a moderator. If you can't stick to science , get a different job.
     
    Last edited: Nov 4, 2011
  22. przyk squishy Valued Senior Member

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    No it's not. It is not always necessary to calculate two quantities to show that one is larger than the other. In this case it simply follows from the fact that energy increases strictly monotonically with speed.

    This calculation that's so simple you never do it?
     
  23. RJBeery Natural Philosopher Valued Senior Member

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    BTW, Tach, this "zero Doppler shift" is only valid for mirrors. An object such as a colored, spoked wheel would indeed appear Doppler shifted. Your comment is therefore irrelevant and in the context of this thread, false.
     
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