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-   -   Relativistic bike (http://www.physicsforums.com/showthread.php?t=432025)

psmitty Sep25-10 09:09 AM

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?

Hurkyl Sep25-10 09:36 AM

Re: Relativistic bike
 
Quote:

Quote by psmitty (Post 2900173)
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?

Why wouldn't it be able to?

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 round-ish 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 re-analyze the problem with a rolling cube. And I just noticed there are two cases to consider:
  • When pivoting around the edge on the ground, the point of contact is stationary in the moving frame of reference
  • When pivoting around the edge on the ground, the point of contact is stationary in the stationary frame of reference
Rolling motion in one frame is rolling with slipping in the other! I'm not sure how seriously to take the analogy with the wheel, though.

DaleSpam Sep25-10 11:37 AM

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

pervect Sep25-10 04:18 PM

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.

JesseM Sep25-10 05:41 PM

Re: Relativistic bike
 
Quote:

Quote by psmitty (Post 2900173)
So, how can they make the same number of turns in the stationary frame, when wheels are contracted, and path is twice the length?

In the frame where the wheels are contracted in the direction of motion it would no longer be true that the distance the wheels travel is just (number of rotations)*(circumference of the wheel at any given moment in that frame). I gave a conceptual argument for why this is true on this thread where we were imagining a train wheel that left a mark on the track with each full rotation:
Quote:

Quote:

But now I think: the distance of the marks in either frame ought to be the circumference of the wheel in that frame.
That should certainly be true in the tram frame where the wheels are round, though it obviously doesn't still work in the rail frame where the shape of the wheel is distorted into an oval shape with a smaller circumference (because the semi-major axis is the same as the radius of the wheel but the semi-minor axis is Lorentz-contracted), and yet the distance the wheel covers between marks is larger. To think about why it doesn't work in the rail frame, you should consider why it is that we believe the distance a rolling wheel travels in one rotation is equal to its circumference. I think one way of approaching it might be considering the limit of a "rolling" rigid equilateral polygon as the number of sides go to infinity--by "rolling" I mean that it balances on a corner, then falls flat on an edge, the raises up on the next corner (which does not change positions as long as it is touching the ground), then falls flat on the next edge, and so on. Since each corner lands one edge-length apart from the last one, it's not hard to see why after a full rotation the distance the polygon has traveled is equal to the sum of all its edges, i.e. its circumference. But now consider what happens if we no longer require that the polygon remain rigid, so the distance between a given pair of corners is constantly changing as the polygon rotates, and the speed at which each corner rotates is non-uniform too (both of which would be true for nearby points on a rotating wheel in SR). In this case, there's no longer any obvious reason why the distance the polygon moves in one rotation should be equal to its circumference at a single instant.

Austin0 Sep26-10 03:02 AM

Re: Relativistic bike
 
Quote:

Quote by pervect (Post 2900725)
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.

Aren't all points on the wheel moving forward relative to the road??
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???

Austin0 Sep26-10 03:42 AM

Re: Relativistic bike
 
Quote:

Quote by JesseM (Post 2900856)
In the frame where the wheels are contracted in the direction of motion it would no longer be true that the distance the wheels travel is just (number of rotations)*(circumference of the wheel at any given moment in that frame). I gave a conceptual argument for why this is true on this thread where we were imagining a train wheel that left a mark on the track with each full rotation:

Quote:

But now consider what happens if we no longer require that the polygon remain rigid, so the distance between a given pair of corners is constantly changing as the polygon rotates, and the speed at which each corner rotates is non-uniform too (both of which would be true for nearby points on a rotating wheel in SR). In this case, there's no longer any obvious reason why the distance the polygon moves in one rotation should be equal to its circumference at a single instant.
Very apt and convincing analogy. According to this [if I am understanding you correctly] the edge of the polygon in contact and the basis for rotation would be contracted so after a full rotation of n polygonal edges the distance covered would be less than n*(edge L). or with a round wheel less tha 2pi *r.
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

Drakkith Sep26-10 05:15 AM

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?

Austin0 Sep26-10 05:26 AM

Re: Relativistic bike
 
Quote:

Quote by Drakkith (Post 2901529)
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?

Sure seems like a particular point would vary in speed relative to the road.
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.

psmitty Sep26-10 12:22 PM

Re: Relativistic bike
 
Quote:

Quote by Hurkyl
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!

Ok, but this shouldn't be a problem. We can imagine it's a pinion on a rack:
http://img202.imageshack.us/img202/2...nanimation.gif

Quote:

Quote by DaleSpam
If you have Mathematica then I can post a notebook where I analyzed exactly this problem.

By all means, I would be very grateful.

Quote:

Quote by JesseM (Post 2900856)
In this case, there's no longer any obvious reason why the distance the polygon moves in one rotation should be equal to its circumference at a single instant.

Ok, it's clear that parts of the wheel are contracted and dilated as they move around, but they "get back" to the stationary frame at the point where the wheel touches the road (because speed is there zero).

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.

pervect Sep26-10 05:14 PM

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 close-enough 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.

JesseM Sep26-10 06:02 PM

Re: Relativistic bike
 
Quote:

Quote by psmitty (Post 2901930)
Ok, it's clear that parts of the wheel are contracted and dilated as they move around, but they "get back" to the stationary frame at the point where the wheel touches the road (because speed is there zero).

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.

They are stretched, relative to their length in the frame where the center of the wheel is at rest! Again just imagine a polygonal wheel where each straight segment lies flat on the track in succession. Each segment is at rest relative to the track when it's lying flat on it, so in the frame where the center is at rest and the track is moving backwards at high speed, each segment is moving backward at the same speed when it's lying on the track, so it's contracted by the same factor as the track in this frame. It's in this frame that the distance between the marks is the same as the circumference of the wheel. In the frame where the track is at rest, each segment is larger when it's at rest on the track, and thus the distance between marks is larger in this frame.

yuiop Sep26-10 06:19 PM

Re: Relativistic bike
 
Quote:

Quote by Hurkyl (Post 2900199)
I think it might be a fun exercise to re-analyze the problem with a rolling cube. And I just noticed there are two cases to consider:
  • When pivoting around the edge on the ground, the point of contact is stationary in the moving frame of reference
  • When pivoting around the edge on the ground, the point of contact is stationary in the stationary frame of reference
Rolling motion in one frame is rolling with slipping in the other! I'm not sure how seriously to take the analogy with the wheel, though.

If we define the road as the "stationary frame" and the centre of the rolling cube (or wheel) as the "moving frame" with horizontal velocity v relative to the road, then I think your two statements should be:
  • When pivoting around the edge on the ground, the point of contact is moving with velocity v and the centre is stationary, as measured in the moving frame of reference
  • When pivoting around the edge on the ground, the point of contact is stationary and the centre is moving with velocity v, as measured in the stationary frame of reference

starthaus Sep26-10 07:21 PM

Re: Relativistic bike
 
1 Attachment(s)
Quote:

Quote by pervect (Post 2900725)
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.

In the ground frame, each point on the rim is moving at a different speed. Therefore, each infinitesimal section of the rim is contracted differently. Certain sections are contracted more than others, depending on their relative speed wrt the ground. See attached, please.

Austin0 Sep27-10 03:52 AM

Re: Relativistic bike
 
Quote:

Quote by JesseM (Post 2902387)
They are stretched, relative to their length in the frame where the center of the wheel is at rest! Again just imagine a polygonal wheel where each straight segment lies flat on the track in succession. Each segment is at rest relative to the track when it's lying flat on it, so in the frame where the center is at rest and the track is moving backwards at high speed, each segment is moving backward at the same speed when it's lying on the track, so it's contracted by the same factor as the track in this frame. It's in this frame that the distance between the marks is the same as the circumference of the wheel. In the frame where the track is at rest, each segment is larger when it's at rest on the track, and thus the distance between marks is larger in this frame.

When looked at as a continuum wouldn't it be true that there is no finite interval of time when the segment is flat on the track and at rest with it??? It would only be the pivot point where this would be true. ANd contraction of a point is moot. That for any actual time interval all other points would have positive motion forward, yes???
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???

Passionflower Sep27-10 04:33 AM

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

DaleSpam Sep27-10 07:17 AM

Re: Relativistic bike
 
1 Attachment(s)
Quote:

Quote by psmitty (Post 2901930)
By all means, I would be very grateful.

Here you go. It has been a long time since I did this, so I am not sure that I remember it all. But I am sure it is just like riding a relativistic bike :smile:

Austin0 Sep27-10 09:26 AM

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??


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