0
$\begingroup$

I have been running this through my head for a few days and I understand I am probably missing something very simple here. If a wheel is spinning the outer edge will travel at a speed that is greater than the rotational speed of the center of the wheel as there is more distance to cover to keep up with the center. Should this same concept not apply to all speeds, including the speed of light. If the center is clocked at a rotation that equals the speed of light would the edge be clocked at a speed that is faster? I know this cannot be tested in a practical sense but is there a mathematical answer to this or have I, as I said earlier, missed something fundamental here?

$\endgroup$

marked as duplicate by Norbert Schuch, John Rennie, Danu, ACuriousMind, Sebastian Riese Dec 11 '15 at 15:48

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ One fundamental thing that is relevant is that the speeds have to be "clocked" by a certain observer, and to fully answer your question, the position of that observer must be chosen before it is even meaningful to talk about measuring speeds. Another complication is that on the one hand we have to ignore the fact that any physical disk than can be imagined for this though experiment would destroy itself before getting anywhere close to those speed, and yet we need to pay attention to the forces and deformations in order to imagine how the edge of the disk would be accelerated. $\endgroup$ – Todd Wilcox Dec 11 '15 at 4:46
  • $\begingroup$ Related and this seems to indicate that it would be impossible to set any disk into motion as you describe, as it would take infinite energy to accelerate it to such a speed, even if the disk would survive the process. $\endgroup$ – Todd Wilcox Dec 11 '15 at 4:56
  • $\begingroup$ This was more of a thought experiment in my head more than anything else. In my head it excluded the ideas of energy and distortion and assumed that an observer could clock the speeds at both the central area of the disk and the outer edge. It would assume the observer would be positioned at the flat side of the disk to be able to observe the entire surface. Under those conditions it would seem reasonable, from my thinking, that if the center were "clocked" at the speed of light the outer edge would be moving faster. I understand I may be over-simplifying things to satisfy things, though. $\endgroup$ – Kevin Duffield Dec 11 '15 at 5:09
  • $\begingroup$ After careful consideration, I'm commenting to explain why I've decided to downvote this question. Even though it's common to ignore certain physical realities in thought experiments, like treating something as a point mass, whenever we do that it must be done carefully and meaningfully. I don't think this question acheives that. Instead, this seems to be asking, "If we ignore some laws of physics, can we break another law of physics?" Even if we could answer this question definitively, we would not understand actual physics any better for it. $\endgroup$ – Todd Wilcox Dec 11 '15 at 14:55
2
$\begingroup$

The short answer: Not in a physical system . In physical systems nothing can go faster than the velocity of light . The constraints of Lorenz invariance have been validated experimentally innumerable times. The limit of the velocity of light to any motion comes out mathematically from Lorenz invariance.

A physical disk , if enough energy is supplied, [since Lorenz invariance transforms energy to an inertial mass the higher the speed the more energy needed] can approach the speed of light only at its periphery: approach but never reach.

$\endgroup$
2
$\begingroup$

To paraphrase the accepted answer here, the disk becomes subject to non-Euclidean geometry and therefore the ratio of the radius of a point on the disk and the circumference travelled by that point is less than 2$\pi$.

That means in the limit case the entire disk, or rather each chosen point on the disk, is rotating at the same speed of light, even though the various points have different radii, because the normal way (using Euclidean geometry) of calculating the path length traveled by a given point on the disk in a given amount of time doesn't apply.

Put more simply, at relativistic speeds, the difference in speed between the edge of a disk and a point near its center is lower than classical mechanics would predict. As the innermost moving point on the disk (note the very center does not move) approaches the speed of light, the difference in speed between the outermost and innermost points approaches zero.

In reality, such a disk in such motion is not possible.

$\endgroup$
  • $\begingroup$ As alluded to in my comment to the question, even though this answer does add some value in pointing out that we can't rely on Euclidean geometry when working with relativity, it still doesn't sit right with me as a complete answer, and I think that's because the question has insurmountable problems. I'll leave it here in case it is edifying to anyone (it has a few upvotes) and resist the unusual temptation to downvote my own answer. $\endgroup$ – Todd Wilcox Dec 11 '15 at 14:58
0
$\begingroup$

This type of question is complicated because you are talking about dynamics of a continually accelerating body in a relativistic manner, but it's easy to show that the speed of light can not be surpassed without needing to get into the nasty details.

Let's simplify the problem and look at just a single particle in a pseudorelativistic way. A particle in the disc is moving at a constant speed which we can adjust, and just look at the tangential velocity for a second. We will find that it would still require infinite energy to rotate that single particle s.t. it's speed approached the speed of light. The rotational kinetic energy is going to be.

$T_{rotational}={1\over2} I\omega^2 = {1 \over 2} mv^2 $

$\omega={v\over r}$

$I=mr^2$

The energy of a single particle rotating in the disc is going to be proportional to the. mass m, defined as $m = \sqrt{{m_0}\over{1-{v^2\over{}c^2}}}$

Which will tend to infinity as speed increases, ergo the energy scales to infinity at high c. Now scaling this up its easy to see that each particle has at least this much energy. This idea gets much trickier once you go relativistic in angular momentum, with the introduction of the stress–energy tensor and a much more rigorous treatment is required.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.