Relativistic bike
If a bike is running at v=0.866c, and it travels path of length L in time T, when observed from the stationary frame, same path will have length L/2 and be traveled in time T/2 in his, moving frame (because gamma=2 for v=0.866c).
Now, what I don't understand is: If the bike is not contracted in his (moving) frame and its wheels have a circumference C, that means that its wheels will make L/(2C) turns in this frame. So, how can they make the same number of turns in the stationary frame, when wheels are contracted, and path is twice the length? 
Re: Relativistic bike
Quote:
Your intuition that both frames must measure the same number of turns is correct  e.g. you could put a red dot on the bicycle, and put a red dot on the ground every time the red dot passes the ground. Everybody will agree upon where and when to place the dots, and thus upon how many dots there are. But what is the difficulty in accepting this fact? Allow me to guess at the reason you haven't provided: you think that if a spinning roundish object moves along the ground, that in one turn, the object has moved a distance equal to its circumference. Now, this isn't even true in classical mechanics  wheels can deform and slip. Getting a distance roughly equal to the circumference in one turn is a rather special (albeit common) thing! Still refuse to let go of your assumption? Then try producing an argument why this special thing often happens in classical mechanics. If you can do so, there will probably be a step that clearly doesn't work in the special relativistic case. I think it might be a fun exercise to reanalyze the problem with a rolling cube. And I just noticed there are two cases to consider:

Re: Relativistic bike
If you have Mathematica then I can post a notebook where I analyzed exactly this problem.

Re: Relativistic bike
Points on the rim of the wheel as they touch the ground aren't moving in the direction of motion. So why would it be a surprise there isn't any relativistic effect?
I.e. the velocity of the car is +v. In the frame of the care, the wheel is moving backwards with a velocity r*omega, so it's velocity is v. Adding the two together, we get zero. 
Re: Relativistic bike
Quote:
Quote:

Re: Relativistic bike
Quote:
IF you chart the path of any particular point relative to the road wouldn't that path be a periodic curve??? With every point on the curve being forward in space [wrt the road] relative to the preceeding point??? SO am I off somewhere here??? 
Re: Relativistic bike
Quote:
Quote:
If we view the situation on the basis of physics ,with the turning of the wheel being the motive force for the forward motion of the system. The system then, would only move as much as the distance covered by the contact distance of the wheel with the road per revolution. So in the road frame the system velocity would not be a simple linear function of revolutions per unit time. This would seem to mean a greater number of wheel revolutions proportional to system velocity as velocity increased and the distance traveled by the wheel contact decreased. If in the bike frame the relationship of revolutions and velocity remained linear this does seem to present an interesting problem of agreement between frames on number of revolutions as per the OP Or not? more thought 
Re: Relativistic bike
Would an arbitrary contact point on the tire vary in speed in relation to the forward velocity of the bike? And would the circular motion of the tire be affected differently for relativity purposes?

Re: Relativistic bike
Quote:
Spatially, the curved path of a point would be longer than the linear path of the system yet over a traveled distance they both end up at the same place so it would appear the average velocity of the wheel point would equal the constant velocity of the bike. I think maybe the same charted motion path would also be valid as a velocity chart. 
Re: Relativistic bike
Quote:
http://img202.imageshack.us/img202/2...nanimation.gif Quote:
Quote:
What bothers me is that it turns out that wheel's circumference (average or whatever you want to call it) is larger, according to the path it travels in a single turn. I don't see the mathematical explanation for this. Points on the wheel have to be stretched in order for this to happen. 
Re: Relativistic bike
It doesn't really matter what the points do when they aren't touching the ground. The point is that if you look at the point touching the ground, and its closeenough neighbors, they are not moving. So if you put a zillion dots on the wheel, when the dots are on/near the ground, they have the uncontracted distance. Thus counting "dots per second" and multiplying by "distance between dots" gives you the correct velocity, because the distance between dots doesn't change in the important region when the dots are near the ground.

Re: Relativistic bike
Quote:

Re: Relativistic bike
Quote:

Re: Relativistic bike
1 Attachment(s)
Quote:

Re: Relativistic bike
Quote:
Considering the cube wouldn't it be true that all points would be moving at different velocities wrt the track and that the contraction in that frame would be differential according to position and direction at any instant??. As it pivots, the two opposing faces would have greater velocity and be more contracted than the parts closer to the pivot point which travel less distance sweeping the same segment of arc??? Wouldn't the edges then be positively curved?? The relative contraction of the opposing faces also resulting in curvature of the faces next to the pivot point? Looking at the descending edge before contact with the track; with greater relative velocities with increasing distance from the pivot point , it would appear unlikely that it could actually make contact as a straight surface simultaneously at all points. Or would you disagree??? 
Re: Relativistic bike
It does not have to be a circle that moves along:
http://mathworld.wolfram.com/images/gifs/roll3gon.gif http://mathworld.wolfram.com/images/gifs/roll4gon.gif http://mathworld.wolfram.com/Roulette.html 
Re: Relativistic bike
1 Attachment(s)
Quote:

Re: Relativistic bike
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?? 
All times are GMT 5. The time now is 06:45 AM. 
Powered by vBulletin Copyright ©2000  2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums