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Let us say you are in an inertial reference frame with a circular planar disk. If you take your meter measuring rods (or perhaps tape measure) you can find the diameter and circumference of the disk. If you divide the circumference by the diameter, you will get exactly $\pi$. Now you start rotating the disk. A book I have claims that now the ratio will be different (lets call it $\pi_\circ$ to avoid ambiguity.) This has caused to reevaluate my understanding of special relativity (this is a set up to general relativity, but the problem itself only requires the special theory.) My problem is that, sure, the measuring rods (or tape) would contract on the circumference, but wouldn't the circumference contract a similar amount, therefore giving a measurement of $\pi$ in every reference frame according to your own measuring rods. Of course, this reasoning goes on to make me question how I understood all of special relativity up this point.

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Can you please tell me how the diameter and circumference of a circular disc changes during rotation? –  Dws_kool Jun 26 at 12:50
    
This entry of the usenet Physics FAQ is related to this question: math.ucr.edu/home/baez/physics/Relativity/SR/rigid_disk.html The Rigid Rotating Disk in Relativity –  b_jonas Jun 26 at 13:05
    
The answer is in your question. You, and your meter stick, are in an inertial frame. No contraction of the meter stick. –  garyp Jun 26 at 13:50
    
Related: physics.stackexchange.com/q/8659/2451 and links therein. See also Ehrenfest paradox on Wikipedia. –  Qmechanic Jun 30 at 23:15

2 Answers 2

Instead of a disc, let's use a ring. Now remember it is a ring I refer to when I tell you to make a measurement. If you sit inside a ring, you aren't moving. Like sitting inside a hula hoop which someone is rotating around you without touching you by applying tangential forces on the hula hoop.

Your metre sticks are always in a state of rest w.r.t. you or it defeats the purpose, now only the length of circumference shortens for you, but not your metre sticks. Hence when you measure the circumference using metre sticks, the circumference is shorter than what you would have had if it was in a state of rest w.r.t. you, hence the value $\pi_o<\pi$.

To measure the circumference, sit on each diameter one by one at an instant , and measure $rd\theta$ and add them all, and what you will get will be shorter than before as your metre sticks are still metre sticks as they aren't moving w.r.t. you, but the arc length has reduced because it is moving w.r.t. you.

The length of the diameter doesn't change, because there is no component of motion along the radial direction.

Now imagine a disc, a disc is just infinitely many rings, so in each ring this is happening.

Now, for clarity carefully again see the derivation of lorentz transformation.

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Actually the issue is much more subtle (in spite of the existence of several naive statements written in popular books like Landau's one). The point is that there is no standard reference frame at rest with the rotating disk or ring. In other words, there is no way to synchronize all clocks at rest with a rotating ring or disk using Einstein's prescription. It is a theorem of differential geometry actually... en.wikipedia.org/wiki/Born_coordinates –  Valter Moretti Jun 26 at 17:28
    
V. Moretti: "**there is no standard reference frame at rest with the rotating disk or ring**" -- The word "rest" can (and should) be used more subtly: The constituents of a rotating ring or disk are not at rest to each other, or anyone, but (merely, chronometrically) rigid to each other. The reference frame to which they belong is not a "rest frame" (a.k.a. "inertial frame"), but a "rotating frame". And the ratios of ping durations between them are not necessarily to be taken as ratios of distances. –  user12262 Jun 26 at 18:03
    
@V.Moretti I haven't studied differential geometry completely, but I would really like to learn what you've said in your comment. I hope you can explain it further, or perhaps I will ask a separate question for it. –  Iota Jun 26 at 18:07
    
@user12262 I did not refer to the problem of rigidity but just the notion of extended rest frame (obviously non-inertial in this case) like that of Rindler coordinates for instance. See the link I pointed out. –  Valter Moretti Jun 26 at 18:09
    
@V. Moretti: "I did not refer to the problem of rigidity but just the notion of extended rest frame. [...] See the link I pointed out." -- The indicated link is to en.wikipedia.org/wiki/Born_coordinates and there's no mentioning on that page of "extended rest frame", nor only of "rest frame", nor even only of "rest" at all. I like to point out that this absence is sensible, as far as I understand the meaning of "(mutual) rest" (in distinction to "(mutual) rigidity"). –  user12262 Jun 26 at 18:23

Let us say you are in an inertial reference frame with a circular planar disk. If you take your meter measuring rods (or perhaps tape measure) you can find the diameter and circumference of the disk.

Then the distance ratios between constituents of the disk edge can certainly be determined;
and a midpoint of the disk may be identified (also in terms of distance ratios, involving constituents of the disk edge).

If you divide the circumference by the diameter, you will get exactly $\pi$.

Of course, the number $\pi$ is correspondingly defined (by polygonal approximation) in the first place.

Now you start rotating the disk.

Consequently, ideally: the constituents of the disk are now moving along circles (with respect to the given inertial reference frame, arond a common center), with constant duration $T$ of the rotational period (as measured by members of the given inertial reference frame). The constituents of the disk edge are moving at constant speed along a circle of the same diameter $2 R$ as had been determined earlier of the disk while it had been at rest.

The disk constituents (including the center of rotation) are therefore no longer at rest to each other; they remain (merely) rigid to each other, in the chronometric sense that any two disk constituents $A$ and $B$ keep measuring constant ratios of ping durations between each other,

$$\tau_{ABA} / \tau_{BAB} = \text{constant}.$$

The values of these ratios are generally different from $1$, and different from each other. In particular, for the center of rotation, $C$, and any constituent on the edge of the disk, $E$, the ratio of mutual ping durations is

$$\tau_{ECE} / \tau_{CEC} = \sqrt{ 1 - \beta_E^2 } = \sqrt{ 1 - \left( \frac{2~\pi~R}{c~T} \right)^2 }.$$

Further, the ratio of ping durations $\tau_{EAE} / \tau_{ECE}$ attains its maximum value if disk constituent $A$ is on the disk edge moving antipodal to $E$; and the corresponding maximum value is less than 2. In fact

$$\text{Max}[ \tau_{EAE} / \tau_{ECE} ] \approx 2 - \beta_E^2 + \frac{13}{12} \beta_E^4 - \frac{541}{360} \beta_E^6 + ...$$

Now, in order to adapt the polygonal approximation of $\pi$ to these mutually rigid disk constituents, and especially to those which make up the disk edge, we may consider some large number ($N$) of them distributed (as "marks") evenly on the disk edge, i.e. such that the ping durations between adjacent marks are equal.

Let $E$ and $F$ be two such adjacent marks on the disk edge. Then we may consider (for instance) the number

$$2~\pi^C_{\beta} := N ~ \tau_{EFE} / \tau_{ECE}.$$

The number $\tau_{ECE} / \tau_{EFE}$ also represents the number of counts obtained by $E$ of successive pings (signal round trips) to $F$ and back during just one ping to the rotational center $C$ and back.

In order to evaluate this ratio, for given numbers $N$ and $\beta_E$, we can consider separately the number $\#EFE_T$ of successive pings of $E$ to $F$ and back which $E$ is counting in the course of one full turn around the center $C$. Of course, this number can also be unambiguously determined in terms of quantities which were entirely measured by members of the given inertial frame (who were passed by the marks on the disk edge). Namely, setting

$$x := 2~\pi~R / N,$$

$$x + t_{EF} ~c ~\beta = c~t_{EF}; \, \, \, \, \, t_{EF} = \frac{x}{c} \frac{1}{1 - \beta},$$

$$x = t_{FE} ~c ~\beta + c~t_{FE}; \, \, \, \, \, t_{FE} = \frac{x}{c} \frac{1}{1 + \beta},$$

$$t_{EFE} = t_{EF} + t_{FE} = \frac{2~x}{c} \frac{1}{1 - \beta^2},$$ and

$$\#EFE_T = T / t_{EFE} = \frac{2~\pi~R}{c~\beta}~c~\frac{1 - \beta^2}{2~x} = N~\frac{1 - \beta^2}{2~\beta}.$$

The other relevant value, $\#ECE_T$, which $E$ can obtain directly by counting (and which can be likewise unambiguously evaluated from the perspective of the inertial frame) is the ratio between some (large) number of successive pings to $C$ and back and the corresponding (also large) number of successive full turns. Obviously obtains:

$$\#ECE_T = \frac{2~\pi~R}{c~\beta} / \frac{2~R}{c} = \frac{\pi}{\beta}.$$

Inserting these two values in the calculation of $2~\pi^C_{\beta}$:

$$2~\pi^C_{\beta} := N ~ \tau_{EFE} / \tau_{ECE} = N ~ \frac{\#ECE_T}{\#EFE_T} = N ~ \frac{\pi}{\beta} \times \frac{1}{N}~\frac{2~\beta}{1 - \beta^2} = \frac{2~\pi}{1 - \beta^2} \gt 2~\pi.$$

Comparing instead to the "antipodal maximum":

$$\pi^A_{\beta} := N ~ \tau_{EFE} / \tau_{ECE} \times 1 / \text{Max}[ \tau_{EAE} / \tau_{ECE} ] = \frac{2~\pi^C_{\beta}}{\text{Max}[ \tau_{EAE} / \tau_{ECE} ]} \gt \pi^C_{\beta} \gt \pi.$$

For both these cases therefore (with notation as used in the question): $$\pi_o > \pi.$$

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