Relativistic bike

Austin0

If you track the mark on the wheel from point A where it is in contact with the track , through one full revolution to point B where it is again in contact, wouldn't it be inevitable that the distance between marks on the track would be equal to the distance traveled by the axle independant of any intermediate motion or contraction.
That both frames would agree on these points , in the bike frame the track distance between A and B being contracted, the distance would be greater according to the track's spatial measurement.
SO the difference would be the gamma factor, yes?

If there were a line of track observers proximate to the wheel point as it traveled between A and B they would be colocated both with an observer at the mark on the wheel as well as bike frame observers at every point.
WOuldn't they agree that there was one complete revolution?
WOuldn't they agree that during the parts of the path generally transverse to the motion of the bike that the point would be uncontracted relative to the bike axle in its frame but contracted equivalently to the bike itself in the track frame??

JesseM

Sure, but why are you worried about time intervals? I'm just concerned with the distance the rolling polygon moves along the track, and it will be equal to the sum of the length of each segment when it was instantaneously flat on the track (and instantaneously at rest relative to the track). That's because the polygon is rolling without sliding, so if you have two successive segments S1 and S2 with a sharp corner C1 between them, then whatever the position of the corner C1 at the instant that S1 is flat, C1 will maintain the same position relative to the track as S1 rises up and S2 begins to move down, and continue to maintain that same position on the track until S2 is lying flat on the track, after which the polygon will pivot on the next corner C2 which lies between S2 and the next segment S3. You can imagine each corner leaves a different-colored mark at the point on track it's contact with while the wheel is pivoting on it, in that case you can see the distance between marks C1 and C2 will be equal to the length of S1 at the instant it lay flat on the track, the distance between C2 and C3 will be equal to the length of S2 at the instant it lay flat on the track, and so forth. So, the distance between successive marks a single corner makes after a complete revolution is just the sum of the length of each segment when that segment lay flat on the track.
The point here is to consider an example which in the limit as the number of segments goes to infinity would reduce to the rolling wheel which is perfectly circular in its rest frame, so we don't have to worry about making the motion of each point match that of a "realistic" rolling polygon made out of some semi-rigid material. Instead of imagining the polygon as a continuous solid object you can imagine each point on each segment is moving independently of the other points with its own little rocket to control its acceleration, and they are all moving in concert in just the right way to ensure that in the rest frame of the center of the polygonal wheel, the polygon maintains exactly the same shape at all times (with all segments of equal length and straight in this frame, and the angle at each corner being equal).

psmitty

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)?

kamenjar

Omg how much mumbo-jumbo 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 re-drawn 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.

JesseM

Yes, we assume that the wheel is circular in the rest frame of its center, since each point on the rim has the same speed in this frame.

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 non-rotating 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 too--therefore, the shape of the wheel in other frames must be the same as the shape of the non-rotating 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.

JesseM

No, you're misunderstanding length contraction in relativity--the laws of physics must work the same way in all frames according to the first postulate of SR, so if it's true in your frame that Earth's ruler is shrunk by a factor of 2 when Earth has a velocity of 0.866c in your frame, then it must also be true that in the Earth's frame your ruler is shrunk by a factor of 2 when you have a velocity of 0.866c in the Earth's frame. So, you don't appear elongated in the frame of an observer on Earth, rather you appear shrunk, just like things at rest in Earth's frame appear shrunk in your frame.

kamenjar

Yeah, I was figuring I had something wrong there. That's why I said "I think that...".

psmitty

I agree, that seems like the only logical explanation, but what appears strange is that wheel also retains its circular shape and its radius, while all points along the rim are contracted "separately" (in lack of a better description - as I cannot describe them contracting, and retaining the same circular form at the same time).

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?

JDoolin

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 three-dimensional helix in space-time. The general form for this helix is a two-variable 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 "re-parameterized" 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'=c-0.1 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 three-dimensional space.

The actual space-time 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.

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Attached Files

my helix 3.nb (7.9 KB, 7 views)

yuiop

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.

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yuiop

Hi JDoolin. I like your animation. It is a pity not many people have viewed it because it very nicely shows the "bizarre" but accurate curving of the wheel spokes in realtime. Nice work!

JDoolin

Thanks, yuiop. I also remembered where I've seen another gif animation of the wheel,here: http://casa.colorado.edu/~ajsh/sr/contraction.html, on Anderw Hamilton's website.

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 line-of-sight. 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 break-time 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 past-light-cone 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.)

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yuiop

For what it is worth, this animation http://www.physicsforums.com/showpos...2&postcount=13 that I posted in another thread, relating to how a relativistic binary star system looks like, is also what I imagine what two particles on the rim of wheel rolling away from the observer looks like, when the wheel is "edge on" (road on the left of animation). What do you think?