# Biking at near the speed of light

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1. Mar 23, 2014

### Cosmobrain

Biking at near the speed of light

If a biker is going at 0.99c, an observer standing still would notice his clocking moving slower.Fine. But then how would the wheels of the bicycle match the ground? Would the bike hover over the ground? thanks

Last edited by a moderator: Mar 23, 2014
2. Mar 23, 2014

### Staff: Mentor

The velocity of the wheels, according to the stationary observer, varies from 0c at the point where they touch the ground, to faster than the bike is moving at the apex of the wheel's arc.

Last edited by a moderator: Mar 23, 2014
3. Mar 23, 2014

### Cosmobrain

Sorry but that didn't explain much to me. I don't know what you're talking about.

I am going to attempt to answer my own question. Perhaps because of the effects of length contraction, even with a slower spin of the wheel, the tire can match the ground and not hover over. I'm not quite sure

cb

4. Mar 23, 2014

### phinds

Why do you think the tires would hover as opposed to touching the ground? I just can't figure any basis for even HAVING that thought, so please share yours.

5. Mar 23, 2014

### Cosmobrain

sorry, that's not what I meant. What I'm saying is that the bike would seem to be sliding on the ground, because of time dilation making the wheels spin less fast.

cb

6. Mar 23, 2014

### phinds

Ah, I see. You have some misconception that the wheels would exhibit time dilation in a way differently from the bike and rider, which is not the case. The whole thing would actually appear pretty much motionless, depending on how close to c we are talking about. Far from looking like it was sliding over the ground, it would appear to be amazing that the guy could keep the bike upright while going so slow

7. Mar 23, 2014

### Staff: Mentor

You have two frames to consider this from. The ground, and the bike. From the Bike's frame, each point on the tread of the tire is moving at a steady speed but with a constantly changing direction. However, from the ground's frame of reference this is not so.

The part of the wheel that is in contact with the ground is stationary with respect to the ground. As the tire rotates, that particular point increases with velocity until it reaches maximum at the very top of the wheel. This is because the top of the wheel has to move faster than the bike in order to get in front of the bike. Remember that the part of the wheel in contact with the ground is applying a force in order to keep the bike moving against friction, air resistance, etc. It is only when the bike begins to skid or when it is "burning rubber" that the point of contact is no longer stationary with respect to the ground.

8. Mar 23, 2014

### Staff: Mentor

Not true. Different parts of the wheel are moving at different velocities with respect to the ground, and the wheels would have to spin very, very quickly indeed to keep up with how fast the bike is moving.

9. Mar 23, 2014

### craigi

You can't treat a wheel with SR, because every point on it is accelerating. You need GR for that and I doubt that a bicycle is a good introduction to the subject.

Last edited: Mar 23, 2014
10. Mar 23, 2014

### Staff: Mentor

First, think about the easy situation where the speed is nowhere near the speed of light, just somebody pedaling their bicycle down the road at the everyday speed of 20 km/hr. I'm going to describe this situation in detail because it's not clear from your response to Drakkith that you understand this case completely - if you do, and I'm telling you something that you already know, I apologize.

How fast is the bottom of the wheel, the part that's touching the ground, moving relative to the ground? It's not moving relative to the ground at all - if it were moving relative to the ground, the rubber would be rubbing against the ground, causing tire-squeal noises and smoke and leaving a black skidmark on the ground.

How fast is the the center of the wheel moving relative to the ground? Well, the center of the wheel is pretty obviously moving at the same speed as the bicycle because they aren't moving relative to one another. So if the bicycle is moving at 20 km/hr, that's the speed of the mounting point for the axle and therefore for the center of the wheel.

There's no contradiction here because the wheel itself is turning; if it's turning clockwise, the bottom of the wheel is moving from right to left and the top of the wheel is moving from left to right relative to the center of the wheel, which is moving at the same speed as the bicycle.

So in the easy non-relativistic case, the bottom of the wheel and the road are moving at 0 km/hr, the center of the wheel and the bicycle are both moving at 20 km/hr, and the top of the wheel is moving at 40 km/hr, all relative to the road. Relative to the bicycle, the bottom of the wheel and the road are moving at -20 km/hr, the center of the wheel and the bicycle are moving at 0 km/hr, and the top of the wheel is moving at 20 km/hr.

STOP HERE AND SATISFY YOURSELF THAT THE ABOVE MAKES SENSE BEFORE PROCEEDING (and as I said above, if this part is obvious and you already understand it, I apologize).

Now, what happens as the speed of the bicycle becomes relativistic, say .99c as you suggest?

As the far as the bicycle rider is concerned, the bottom of the wheel and the ground are both moving at -.99c relative to him; they're moving at the same speed so there's no skidding/smoking/skid marks being laid down on the road. The center of the wheel is at rest relative to him, and the top of the wheel is moving at a speed of .99c relative to him.

Relative to someone standing at the side of the road? The bottom of the wheel is still at rest relative to the road and this observer. The center of the wheel and the bicycle are moving at .99c. And the top of the wheel is moving at...? Well, it's moving at .99c relative to the rider, and the rider is moving at .99c relative to the road, so we have to use the relativistic velocity addition formula (google will find it if you're not already familiar with it) to calculate that the top of the wheel is moving at .99995c relative to the ground.

(It would be a good exercise to calculate the tension in the spokes of the wheel as a result of the centrifugal force at these speeds. Make the reasonable assumptions that the wheel has a mass of 5 kg, a diameter of one meter, and there are 100 wire spokes in the wheel, and you will conclude that this is not a situation in which you should trust your intuition).

Last edited: Mar 24, 2014
11. Mar 23, 2014

Staff Emeritus
Why do people keep saying that? SR handles accelerations just fine. It's gravity that it doesn't handle.

12. Mar 23, 2014

### Staff: Mentor

That's not true, although it's one of the more common misconceptions and repeated so often by so many pop-sci treatments that you can be forgiven for believing it up to now.

In fact SR works just fine for accelerations. The limitation of SR is that it only applies to flat spacetimes, which is to say no gravitational fields.

Last edited: Mar 23, 2014
13. Mar 23, 2014

### Staff: Mentor

Different points on the circumference of the wheel are moving at different speeds relative to the ground (I just did a huge long post on this) so there is no one time dilation formula that you can apply to all points on the circumference. In particular, the point that is touching the ground is always at rest relative to the ground.

14. Mar 23, 2014

### jartsa

The maximum velocity of a bike is k * c.

I have always thought k is 0.5.

But if we consider the maximum velocity of a caterpillar, that kind of seems to be 0.5 c. (a piece of caterpillar track moves half of the time, is still half of the time)

So maximum speed of a bike may be less than 0.5 c.

EDIT: k is actually 1. It's 1 for bikes and caterpillars, and also for the average velocity of a person that does interval training: runs very fast 1 minute according to his clock, and then rests 1 minute according to his clock.

Last edited: Mar 23, 2014
15. Mar 23, 2014

### phinds

Yes but my point was that time dilation would make the entire thing appear to be standing still, or very close. That isn't changed by the fact that the top of the wheel is moving faster than the bike.

16. Mar 23, 2014

### Staff: Mentor

Imagine the bicycle is on a treadmill, so that the bicycle is at rest and the ground under it is moving. This is just looking at the things from an inertial frame in which the bicycle is at rest, instead of one in which the road is at rest.

Why should the speed of the bicycle be constrained? Bicycle at rest, bottom edge of wheel and road can move at any speed less than $c$ in the backwards direction relative to the bicycle, and the top of the wheel moves in the forward direction with the same speed.

17. Mar 23, 2014

### jartsa

Ok. Good reasoning.

18. Mar 24, 2014

### Staff: Mentor

But it wouldn't appear to be standing still. If you could watch an object moving at that velocity, the wheels would be turning extremely fast. Are you talking about the rest of the bike and the biker?

19. Mar 24, 2014

### bahamagreen

I think you are all missing the point of the Op's question...

Cosmobrain, are you thinking of the spinning wheel as a clock, and expecting the period of this clock to increase... and so expecting the wheel to slow its rotation, the discrepancy being this slowing of rotation with increasing approach to c?
As in, at the limit, the wheel approaches zero rotation, yet the bike speed approaches c, hence the "hovering" or sliding translation of a non-rotating wheel?

I'm curious to see how this goes.

20. Mar 24, 2014

### Staff: Mentor

I'm pretty sure we understand the OP's question. The answer just isn't easy. We aren't dealing with an object moving in a constant linear manner, but a much more complicated scenario involving not only multiple reference frames, but accelerating frames at that. And all of them are attached to each other.