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Apparently a bunch of people totally misunderstood my previous question and choose to ignore the clarifying comments. Let me change the conditions to remove all the confusion.

enter image description here

A group of particles forming a circle of uniform density is moving in a circular path (following that circle shape), at a moderate speed $v_1$ (as observed from a frame of reference bound to the center of the circular path, O1). Let's say this is a cloud of ions filling a cyclotron that had accelerated them to 0.2c and then just maintains their circular trajectory with magnetic field without accelerating them any further. (depicted is momentary speed vector of one of these particles)n accelerator.

An observer (O2) moving at $v_2$ near speed of light (say, 0.9c relative to the previous observer (O1) / the cyclotron) observes the positions and movement of these particles. How will they appear from that observer's point of view - what is their apparent their trajectory and how does their density vary from that observer's point of view?

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SF.
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  • Are you asking what observer 2 will see i.e. including the travel time of the light reaching observer 2's eyes, or are you asking what the shape of the circle will be in observer 2's frame. Also what is the significance of $v_1$? Can we assume $v_1$ is negligable i.e. the particles are effective stationary in observer 1's frame? – John Rennie Jan 25 '16 at 09:03
  • @JohnRennie: Preferably both, regarding your first case. As for the second, v1 is not entirely negligible. I set it to a relatively small value to stop the suspicions of "infinitely rigid rod with ends moving faster than speed of light" (that had murdered my previous question), but as I wrote, let's assume it's 0.2c relative to O1's frame: The relativistic behaviors relative to O1 are negligible, but they are definitely pronounced relative to O2. They are definitely not stationary in any of these frames though. (also, it's a momentary linear velocity in rotary motion.) – SF. Jan 25 '16 at 09:39
  • @SF: how does it matter that the particles in the ring move in a loop? Unless you consider general relativistic effects (which would require specifying if the particles are moving in a gravitational or other field) the problem is equivalent to just looking at how a moving observer sees a moving element of the ring. Or am I missing something? – Wolpertinger Jan 25 '16 at 10:59
  • @Numrok: You're not missing anything. Observing one single particle moving in a circular path would be sufficient. Introducing a ring instead of a single particle is just to avoid the mess with simultaneousness as in any moment of movement of the observer the single particle. And I used a cloud of particles instead of a ring because I definitely wanted to avoid having any solid object here, as it would inevitably invoke the curse of 'infinitely rigid rod'. Assume charged particles in a magnetic field, no significant gravitational effects. – SF. Jan 25 '16 at 11:16
  • Any object seen from a relativistic observer appears rotated, i.e. a round object will appear... round but it will be seen under a different angle than by a resting observer. The circular motion also causes a slight redshift on one side and a slight blueshift on the other, but that's independent of the movement of the observer. – CuriousOne Jan 25 '16 at 11:17
  • @CuriousOne: The caveat here is that particles on one side of the circle move at different speed relative to O2 than on the other side. It's not an immobile round object, but a rotating one. – SF. Jan 25 '16 at 11:19
  • How does the rotation change anything about the shape of the object? I can put a resting particle right next to the rotating ones. Should the moving observer see a separation between the two? Why? The red and blue shifts are trivial doppler effects that already exist for the resting observer. – CuriousOne Jan 25 '16 at 11:20
  • @CuriousOne: There's also the density (spacing) and speed of individual particles. In particular, the "returning" particles would exceed speed of light relative to O2 if something doesn't happen with them in O2's frame of reference, that would prevent them from doing so. Imagine "classic" speed addition: particles on one side move in the same direction as O2, so 0.9-0.2=0.7c. The ones on the opposite side: 0.9+0.2=1.1c. – SF. Jan 25 '16 at 11:29
  • Why would the particles exceed the speed of light in the moving observer system? You know the relativistic formula for the addition of velocities... use it. – CuriousOne Jan 25 '16 at 11:31
  • @CuriousOne: I know the formula. And I know they will be "arriving to the near side of the ring" much slower than they are "departing to the far side" as result. And I'm completely baffled as to what happens in between and the whole thing doesnt get congested... – SF. Jan 25 '16 at 11:34
  • That you are baffled is a statement about yourself, it's not a valid physics question. – CuriousOne Jan 25 '16 at 11:37
  • Let us continue this discussion in chat. – SF. Jan 25 '16 at 11:38
  • Isn't this the: https://en.wikipedia.org/wiki/Ehrenfest_paradox -- Isn't it explained in "The Dynamics of Relativistic Length Contraction and the Ehrenfest Paradox" by Moses Fayngold: https://arxiv.org/abs/0712.3891 or "Proper co-ordinates of non-inertial observers and rotation" by H. Nikolic: https://arxiv.org/abs/gr-qc/0307011 etc. ? – Rob Feb 23 '18 at 09:57
  • @Rob: No, it isn't. The paradox talks about a rigid disc. Which I fought tooth and nail to emphasize the question is NOT about. – SF. Feb 23 '18 at 16:21
  • @SF - I mistook your reference to your previous question, about a "bar" and 'strong tire', and this questions reference to evenly distributed particles to imply that the perimeter (disk or ring) was described in those papers. I did not desire to rehash a point that was closed, thus asking if it was applicable. Thanks for the reclarification. – Rob Feb 23 '18 at 16:34

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I tried to put it into a figure the results from reasoning in the framework of Special Theory of Relativity, this is what the observer would see:

enter image description here

Where you should notice several effects:

  • length contraction along observer's motion causes the ring to look like an elipse
  • equal velocities on equator where particles are crossing from top to bottom or vice-versa their velocities are perpendicular to observer's hence not affected (0.2c on either side)
  • fast transit through "bottlenecks" the apparent concentration of particles on top and bottom, which I mean by bottlenecks, are just an effect of length contraction, but particles are not really stuck there. Actually they are faster than in the rest of the ring (see velocity colorbar). Also the peak velocity on top is higher (.93c) than on bottom (.85c) because on the former they travel towards observer, while on the latter they travel away from him. But the observer sees all of them travelling towards him in both cases with the mentioned peak velocities.
  • spacing asymmetry top vs bottom the different speeds top vs bottom will affect the spacing between particles, the top looking more crowded than the bottom. However the time contraction will counter this effect and and at all times there should be the same amount of them on either side.
  • slanting also due to speed differences top vs bottom there will be a kind of slanting/forward-fall effect on the shape of the circle.

Note The figure is a representation of the effects and might exaggerate some of them, since is not made by calculating the actual values. The speeds-colorbar only shows the value of horizontal component and is also exaggerated, since minimum values should be .9c instead of 0.

rmhleo
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  • Thank you very much! May I ask what software were you using? – SF. Jan 25 '16 at 19:22
  • I used a python and matplotib which I wrote, but as I explained it does not calculate from the equations, but rather creates a plot that shows the effects I mention in a visible way. The functions I used probably differ from the actual solutions that would be obtained from the relativistic equations. – rmhleo Jan 25 '16 at 19:26
  • Thanks. I really wish there was some relativistic Algodoo. It would make visualising what goes on much easier. – SF. Jan 25 '16 at 20:22
  • Ok I will add some equations soon. – rmhleo Jan 25 '16 at 20:28
  • You don't have to. I can look up any equations just fine, it's just that going from an equation to a picture with a setting more complex than a couple points takes unreasonable amount of time and work. – SF. Jan 25 '16 at 21:50