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Say a circle is rotating about its centre at a relativistic speed.

If it were not rotating in your reference frame, its circumference divided by its diameter would be $\pi$.

Now, because it is rotating, let us say we have been using little sticks all along to measure circumference and diameter. We would need a ratio of $\pi:1$ sticks to build the static circle. However, these sticks change when the circle is spinning because of length contraction.

So, the sticks along the diameter become thicker, but not shorter. The sticks along the circumference, however, become smaller, and not thicker. Therefore, if a circle is rotating in a reference frame, $\frac{circumference}{diameter}$ is not $\pi$. This is an example of relativistic geometry.

My question is: how does area change? I believe the area becomes larger. Here is my reasoning:

The rotating circle has a larger circumference than before, therefore the 'new $\pi$' is greater than the 'old $\pi$'. And, because this circle is spinning around its centre, it is still a circle to you even when its spinning. Therefore, area is still $\pi r^2$; but this time, $\pi$ is larger, therefore the new area is larger.

I can also think of a counterexample:

Imagine using a bunch of smaller circles within the larger circle to estimate its area. When the circle begins spinning, these circles shrink because of length contraction. Therefore, these circles become smaller, and the overall area is smaller. However, this may also be interpreted as changing the metric of the space by requiring more sticks to create the diameter of the circles, contradicting the argument that the circle gets smaller and actually agreeing that it gets larger.

Which is correct? Why?

EDIT: I chose to distinguish my question from others because the reasoning I offer is not something explored in-depth in other questions/answers and is a part of the question that I believe enhances it.


marked as duplicate by Aaron Stevens, John Rennie special-relativity Oct 18 '18 at 5:05

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.

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    $\begingroup$ The linked question identifies this as the Ehrenfest paradox. $\endgroup$ – rob Oct 18 '18 at 4:55