
#19
Sep2710, 10:05 AM

Sci Advisor
P: 8,470





#20
Oct1010, 04:58 PM

P: 16

Ok, the bottom part of the wheel is actually stationary relative to road frame.
But how does this wheel look for a moving observer (the observer where this wheel is stationary)? If all points along the wheel are moving at some tangential speed, they should all be contracted in the direction of their movement. After all, bottom point of the wheel is contracted along with the road, because it is moving at speed v. So, how do these tiny parts of the wheel contract? Shouldn't the wheel still have the same circular shape and same radius? Furthermore, shouldn't points closer to the axle contract less (their speed is lower)? 



#21
Oct1110, 02:27 AM

P: 101

Omg how much mumbojumbo math just for one question... Maybe I don't understand what is the questions, but it seems simple to me...
In YOUR reference frame the bike wheels turn the same as they do on earth. The bike is still 2m in length and 1m in height, your C is the same and your 1 second is the same, and all the same like it was on earth. If you took a 1m stick with you, it would be 1 meter in all directions. We are moving towards Andromeda at some pretty high speed and it blueshifted. It doesn't mean that we take different yardsticks with us when we measure distances up or down because of that. It is moving with you. It only the things around you that that need to be redrawn and recalculated. The same way, it is only your bike that someone from earth needs to recalculate. Your clock and bike's clock are the same. It would be different at front or back of the bike if you were accelerating, but you are not, so math is the same as the math on earth. Let's say your bike has huge wheels with circumference as big as light year. You you travel .86 earth c, then you are technically moving 2x earth distances per every T if yours. A star that is 20 earth light years away, will be 10 light years away for you. That means that you make 10 wheel turns on your bike. If you were to move very slowly, your wheels would have to make 20 turns (that is for you to travel near earth time and earth frame of reference). I think that to an observer from Earth, it would appear that the circumference of your wheels is 2 light years. You appear elongated in the direction of your travel to them, the same way they appear shrunk to you. 



#22
Oct1110, 01:17 PM

Sci Advisor
P: 8,470

From this, a simple argument shows what the shape of the wheel must be in other frames. Imagine that right next to the rolling wheel we have a nonrotating circular plate of the same radius, so that at every moment in the wheel's rest frame, each point on the rim of the wheel is right next to a point on the rim of the plate. It's a purely local fact that each point on the rim of the wheel is always right next to a point on the rim of the plate, so this must be true in other frames tootherefore, the shape of the wheel in other frames must be the same as the shape of the nonrotating plate in other frames, which is just an ellipse with the axis parallel to the direction of motion shrunk due to length contraction, and the axis orthogonal to the direction of motion having the same length as in the plate's rest frame. 



#23
Oct1110, 01:22 PM

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#24
Oct1110, 01:35 PM

P: 101





#25
Oct1110, 02:55 PM

P: 16

Also, I am still not sure about what are the consequences of length contraction. For example, I am pretty sure that space itself is contracted when observed from a different frame, and that it obviously shouldn't affect the actual object being contracted. In other words, you don't need to worry that an observer in a fast moving train passing by you will somehow contract you, simply by observing you. :) But then there are claims that rigid bodies are "incompatible" with SRT. Why? I don't think I will actually compress a box of steel by observing it from a train, although I may believe that entire space is somehow contracted in sense of simultaneity with my own coordinates. Also, does this mean wheel should break apart? For example, this page mentions "a serious mechanical problem" with such wheel. But there is also an infinite number of frames where any chosen point along the rim has a relative speed of zero, and undergoes no length contraction. So where is the problem then? 



#26
Jan611, 11:09 AM

PF Gold
P: 706

I've been working on this problem for the past couple weeks, for one, trying to reproduce something like the image at http://commons.wikimedia.org/wiki/Fi...ling_Wheel.png
My version of the wheel has six spokes. A spoke, rotating in space becomes a threedimensional helix in spacetime. The general form for this helix is a twovariable parametric equation in r and t: x,y,t=(r cos(omega t + theta), r sin(omega t + theta), t) The surface is formed by allowing r to go from 1 to 1, and letting t go from some arbitrary time in the past to the future. The advantage of expressing it as a parametric equation is that the Lorentz Transformations can be directly applied to the helix to form the new spacetime surface, from the perspective of someone to whom the wheel appears to be rolling by. After you have done this, the parametric equations need to be "reparameterized" in terms of the new t' variable. I actually didn't complete this task, though I might revisit it later. I found a way to shortcut the process for animation. Instead of showing the entire helix, which extends from about t'=10 to t=+10, I show a very small region from t'=c0.1 to t'=c+0.1, and since Mathematica was able to plot this constricted region, by having c increase, I was able to give the appearance of motion in threedimensional space. The actual spacetime plot is static. All that is changing is the value of t', so you can picture a flat surface sliding up a leaning helix, and the animation just shows that flat (constant t') surface. 



#27
Jan711, 01:12 AM

P: 3,966

The attached image is a screen shot of a java simulation of a relativistic wheel I did a long time ago, based on equation derived by Dalespam. The top half of the diagram (frame S) is in the rest frame of the wheel and the background is length contracted and lower half (frame S') is in the rest frame of the background with a length contracted wheel. The simulation traces the paths of two opposite particles on the rim of the wheel. The traces paths nicely demonstrate the long strides taken by the length contracted wheel. There are clocks in the background and in the wheel frame that demonstrate simultaneity too. If anyone actually wants the simulation so they can run it live, they will have to PM me as it too big (2000 kb) to upload here.




#28
Jan711, 01:21 AM

P: 3,966





#29
Jan811, 10:32 AM

PF Gold
P: 706

I am still working on an a related question. This animation shows what the wheel would look like some long distance away, passing by, perpendicular to the lineofsight. It would also be interesting to imagine what the wheel would look like from the perspective of an observer on the rim, or on the axle. In particular, I'm trying to either confirm or refute the results of Misner Thorne Wheeler's Gravitation Section 2.8 "The Centrifuge and the Photon." Unfortunately, my Christmas breaktime is just about over, so now I'll have to work in time to think about this between grading and teaching, instead of just between eating, visiting, and video games. ;) Attachments: To find the view from the perspective of a point on the rim, one must consider the pastlightcone drawn from the event of observation. This gives the locus of events observed at that moment. To find the event locations you must do the Lorentz Transformation on that locus of events. Then, there is more involved if you want to calculate the relative velocity (which seems like it ought to be zero, but I think that our common sense may actually fail us here.) 



#30
Jan811, 10:57 AM

P: 3,966




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