Biking at near the speed of light

[phinds]

Traveling near c and standing still? This is not a good way to put it.

I didn't say it WAS standing still, I said it will APPEAR to be standing still.

Time dilation means that clocks on the bike run slower in the ground frame, than clocks on the ground. But the clocks on the wheel bottom still run at the same rate as the clocks on the ground. And that is the whole point of the OP:

How can the RPM at the wheel hub be reduced due to time dilation, but the tangential velocity at the wheel bottom still cancel the linear velocity of the bike, so the wheel bottom is stationary relative to the ground?

And the answer to this, is the distortion along the circumference shown in the links a posted above.

OK, I did get it wrong about the bottom of the wheel being stationary in the ground frame. My bad.

The point I was getting at (and missing the bottom of the wheel thing) is that things moving at high speed relative to you look slow from your perspective but I agree that's not what the OP was getting at.

 

[A.T.]

things moving at high speed relative to you look slow from your perspective

Things moving fast are fast and look fast. What is slow is the rate of clocks attached to these fast things.

 

[phinds]

Things moving fast are fast and look fast. What is slow is the rate of clocks attached to these fast things.

So you are saying that to a remote observer an object falling into a black hole does NOT appear to slow down as it approaches the event horizon? It's the same thing.

EDIT: Or, I suppose, perhaps you believe that objects falling into a black hole are not moving fast?

 

[phinds]

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?

Actually, I'm talking about the whole thing. Although I realize that the top of the wheel is moving even closer to light speed, the whole thing still appears to a remote observer to be moving slowly, just like an object under gravitational time dilation that is falling into a black hole.

 

[phinds]

Things moving fast are fast and look fast. What is slow is the rate of clocks attached to these fast things.

When you say the clocks are slow, what are you comparing them to? Certainly they are NOT slow in the frame of reference of the object they are with, so you must mean that they appear slow to a remote observer. I agree. So does the object they are with.

 

[A.T.]

So you are saying that to a remote observer an object falling into a black hole does NOT appear to slow down as it approaches the event horizon? It's the same thing.

No, it's not. We are talking about relative movement in flat space time here. There is no gravitational time dilation involved here.

 

[A.T.]

When you say the clocks are slow, what are you comparing them to?

Moving clocks run slow compared to resting clocks. That has nothing to do with "remoteness". The clocks can pass very close to each other.

 

[DrGreg]

Although I realize that the top of the wheel is moving even closer to light speed, the whole thing still appears to a remote observer to be moving slowly

The speed of the bicycle relative to the ground is already specified to be 0.99c, so you can't say it is going slower than that. That is also the speed of the ground relative to the bicycle and therefore the speed of the wheel rim relative to the wheel centre.

[STRIKE]You are failing to take account of length contraction. According to the observer riding the bicycle, the circumference of the wheel is length contracted and a lot less than 2\pi r, and therefore the angular velocity of the wheel (revs per second) required the make the rim move at 0.99c must be a lot faster than you would expect from Newtonian physics. This counteracts the dilation you refer to above.[/STRIKE]

EDIT: I got that the wrong way round. In the frame of the biker the circumference still has to be 2\pi r, but that value is length-contracted from the "rest length of rubber in the tyre", which is much more than that. The rubber that touches the ground is at rest relative to the ground and so the effective length of rubber that touches the ground over one revolution of the wheel is more than 2\pi r. So to summarise:

In the biker frame the wheel's angular velocity is what Newton would expect.

In the ground frame the wheel's angular velocity is slower than Newton would expect, but this is explained by the rest-length of rubber in the tyre being longer than expected.

 

[jartsa]

The speed of the bicycle relative to the ground is already specified to be 0.99c, so you can't say it is going slower than that. That is also the speed of the ground relative to the bicycle and therefore the speed of the wheel rim relative to the wheel centre.

You are failing to take account of length contraction. According to the observer riding the bicycle, the circumference of the wheel is length contracted and a lot less than 2\pi r, and therefore the angular velocity of the wheel (revs per second) required the make the rim move at 0.99c must be a lot faster than you would expect from Newtonian physics. This counteracts the dilation you refer to above.

The height of the wheel is the same in the bike frame and in the road frame. In the bike frame the width of the wheel is the same as the height of the wheel. But in the road frame that width is contracted.

(I think we have been considering a wheel that does not contract in the bike frame: spokes sturdier than rim)

 

[liometopum]

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 .

Great post! I question, though, the last part:

The top of the tire may have a velocity of .99c relative to the rider, but it is not going anywhere; it is not moving translationally relative to the rider. There is no change in the distance between the two. The top of the tire does not move.

Rotational velocity and translational velocity are different. The use of the relativistic velocity addition equation to add rotational velocity to translational velocity is thus questionable.

 

[Nugatory]

Great post! I question, though, the last part:

The top of the tire may have a velocity of .99c relative to the rider, but it is not going anywhere; it is not moving translationally relative to the rider. There is no change in the distance between the two. The top of the tire does not move.

Rotational velocity and translational velocity are different. The use of the relativistic velocity addition equation to add rotational velocity to translational velocity is thus questionable.

Perhaps I should have said "the piece of rubber that is at the top of the tire right now". That piece of rubber is, right now, moving at .99c relative to the rider, and this is a straightforward movement in the same direction as the rider. Yes, if we wait a moment that piece of rubber will be moving in a different direction, but that just means that we have a different velocity to plug into the velocity addition formula a moment later.

 

[liometopum]

I understand what you are saying (and you say and know it well!). But the velocities are different, like apples and oranges to use the old cliche. I don't think you can use the relativistic velocity addition equation here. The velocities are not of the same type.

 

[Nugatory]

The velocities are not of the same type.

Why not? Velocity is defined to be $\frac{d\vec{r}}{dt}$, where $\vec{r}$ is the position vector. That definition works just fine for things moving in circles like the wheel, in straight lines like bicyclist and the road, or in arbitrarily complicated trajectories.

 

[A.T.]

it is not moving translationally relative to the rider. There is no change in the distance between the two.

Transnational velocity doesn't require change in distance.

Rotational velocity and translational velocity are different.

Yes, and the top of the wheel has both in the frame of the bike:

radius * angular_velocity = linear_velocity

 

[liometopum]

radius * angular_velocity = linear_velocity

That equation gives tangential velocity, not linear velocity. More specifically, it is the tangential velocity of a spinning object at rest (not moving translationally relative to the observer) in your reference frame.

Which leads to maybe the most critical part of this idea. Translational velocity changes the apparent tangential velocity. That is, a spinning object moving relative to an observer appears to have a lower rim speed.