Biking at near the speed of light

In summary: 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... 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.cbI see. You are saying that the bike would be sliding on the ground because the wheels are spinning less fast than the ground. I can see how that could be confusing. I think it would be helpful to think about this in terms of the frames of reference of the bike and the ground. From the bike's frame of reference, the wheels are moving at
  • #36
phinds said:
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.
 
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  • #37
phinds said:
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.
 
  • #38
phinds said:
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 [itex]2\pi r[/itex], 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 [itex]2\pi r[/itex], 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 [itex]2\pi r[/itex]. 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.
 
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  • #39
DrGreg said:
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 [itex]2\pi r[/itex], 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)
 
  • #40
Nugatory said:
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.
 
  • #41
liometopum said:
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.
 
  • #42
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.
 
  • #43
liometopum said:
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.
 
  • #44
liometopum said:
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.

liometopum said:
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
 
  • #45
A.T. said:
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.
 

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