# Relativistic bike

1. Sep 25, 2010

### psmitty

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?

2. Sep 25, 2010

### Hurkyl

Staff Emeritus
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.

3. Sep 25, 2010

### Staff: Mentor

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

4. Sep 25, 2010

### pervect

Staff Emeritus
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.

5. Sep 25, 2010

### JesseM

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:

6. Sep 26, 2010

### Austin0

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???

Last edited: Sep 27, 2010
7. Sep 26, 2010

### Austin0

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

8. Sep 26, 2010

### Drakkith

Staff Emeritus
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?

9. Sep 26, 2010

### Austin0

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.

10. Sep 26, 2010

### psmitty

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

By all means, I would be very grateful.

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.

Last edited by a moderator: May 4, 2017
11. Sep 26, 2010

### pervect

Staff Emeritus
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.

12. Sep 26, 2010

### JesseM

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.

13. Sep 26, 2010

### yuiop

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

14. Sep 26, 2010

### starthaus

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.

#### Attached Files:

• ###### RelativisticRollingObjectsSig18.pdf
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Last edited: Sep 26, 2010
15. Sep 27, 2010

### Austin0

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???

16. Sep 27, 2010

### Passionflower

It does not have to be a circle that moves along:

[URL]http://mathworld.wolfram.com/images/gifs/roll3gon.gif[/URL]
[URL]http://mathworld.wolfram.com/images/gifs/roll4gon.gif[/URL]

http://mathworld.wolfram.com/Roulette.html

Last edited by a moderator: Apr 25, 2017
17. Sep 27, 2010

### Staff: Mentor

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

#### Attached Files:

• ###### RelativisticRolling.nb
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18. Sep 27, 2010

### 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??

19. Sep 27, 2010

### 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).

20. Oct 10, 2010

### 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)?

21. Oct 11, 2010

### 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.

Last edited: Oct 11, 2010
22. Oct 11, 2010

### 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.

23. Oct 11, 2010

### 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.

24. Oct 11, 2010

### kamenjar

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

25. Oct 11, 2010

### 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, http://www.spacetimetravel.org/tompkins/node7.html" [Broken] 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?

Last edited by a moderator: May 5, 2017