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Assume that a space shuttle travels in a circular path having a radius $r_s$ at a tangential speed $v_s$ - a considerable fraction of the speed of light. There is a light clock inside the shuttle which can be easily seen by an inertial observer located at the center of the circular path. [See Figure 1.] (Remember that the path of the photon bouncing between the mirrors is perpendicular to the shuttle's motion direction as measured by the observer inside the shuttle.)

Figure 1. Spyglass and shuttle

The observer at the center of circular path looks through an spyglass at the shuttle, pursuing the motion of the shuttle by whirling his head around so that the light clock (shuttle) appears to be at rest with respect to the interior space (tubes) of the spyglass. In this case, the observer, indeed, has neutralized the relativistic motion of the light clock (shuttle) by the negligible rotation of his head/body, and he no longer detects an oblique (diagonal) path for the emitted photon pouncing between the mirrors, but rather detects the path of photon perpendicular to the shuttle's motion direction exactly similar to the path which is viewed by the observer inside the shuttle. [See Figure 2.]

Figure 2. Spyglass and light clock

How is it possible for the observer at the center of the circular path to measure any time dilation seeing that he cannot detect any diagonal path for the photon as long as he looks through an spyglass at the shuttle?

Before answering to this question, please be informed that:

1- The observer at the center of rotation is inertial because the radius of the observer ($r_c$) can be assumed to be infinitesimally small. That is to say, for a point observer, his centrifugal acceleration due to his rotation is calculated to be:

$$a_c=r_c\omega^2\approx0,$$ where $\omega$ is the angular velocity of the shuttle (spyglass).

2- The shuttle can also be taken account as an inertial system, if we assume that the magnitude of $r_s$ (radius of the circular path) is very great, and $\omega$ is very small such that:

$$v_s=r_s\omega\approx c {\space}{\space}{\space} and {\space}{\space}{\space} a_s=r_s\omega^2\approx 0.$$ For instance, assume $r_s=10^{25}{\space}m$ and $\omega=10^{-17}{\space}rad/s$.

3- According to 1 and 2, the effect of acceleration (equivalent gravitation) for the observer at the center of the circular path is negligible in the location of the shuttle considering the amendments of general relativity for the frequency of the light clock: [1]

$$\frac{\nu_c}{\nu_s}=\frac{T_s}{T_c}=1+\frac{a_cr_s}{c^2}=1+\frac{r_c\omega^2r_s}{c^2},$$ $$\lim_{r_c\to0}{\frac{T_s}{T_c}}=1.$$

By the way, I have discussed this problem in my book [2], however, I thought that someone may have an answer that differs from mine, thus I decided to explain it here.

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[1] R. Resnick, Introduction to Special Relativity, p. 213 (John Wiley and Sons, New York, 1968)

[2] M. Javanshiry, The Theory of Density: From the Effect of Pressure on Time Dilation to the Unified Mass-Charge Equation, Chap.1, Sec. 2, p. 10 (Nova Science Publishers, New York, 2017).

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  • $\begingroup$ It's not "Einstein's light clock." The light clock is a pedagogical device developed by other people decades later. Are you critiquing someone else's presentation, which includes all of these idiosyncratic elements, or is this your own presentation? There is no need to have the rocket ship going in a circle, and no need for the spyglass. Are the figures from a 1968 edition of Resnick, or are they ones you drew? $\endgroup$
    – user4552
    Jan 21, 2019 at 14:19
  • $\begingroup$ Whether or not Einstein himself had ever tried to introduce such a light clock, there are many books in which the mentioned device is called "Einstein's light clock" (google.com/…). $\endgroup$
    – Mohammad Javanshiry
    Jan 21, 2019 at 17:05
  • $\begingroup$ It is my question, and it is my own presentation. The figures belong to me, and I drew them all with my own cutesy hands! Would you please stick to the topic, and provide me with a clear answer?! Moreover, If the path is not a circle, the apparent size of the shuttle would change depending on whether the shuttle is approaching or receding. This alteration in size may confuse the observer who views the shuttle through a spyglass. However, if the distance is large enough, we can eliminate the circular path as well as the spyglass as you said. $\endgroup$
    – Mohammad Javanshiry
    Jan 21, 2019 at 17:09
  • $\begingroup$ If it's really the case that you can approximate everything by an inertial system, then you can approximate the trajectory of the shuttle by a tangent to its orbit at any point and you should get the same answer. Do you? Additionally I think treating the observer as inertial is clearly wrong: what paths do they observe for light in their rotating frame. $\endgroup$
    – user107153
    Jan 21, 2019 at 18:10
  • $\begingroup$ @tfb: I am not sure that I correctly understood the first three lines of your comment, but I think we get the same answer; don't we?! It is evident that Einstein himself believed that, e.g., on a rotating disk, clocks run slower and rulers are contracted by the traditional Lorenz factor. Indeed, their tangential velocities are the main agents in undergoing time dilation and length contraction regardless of the acceleration each element of the rotating disk experiences. Didn't Einstein approximate the trajectory of each rotating element by a tangent to its orbit at any point? $\endgroup$
    – Mohammad Javanshiry
    Jan 21, 2019 at 19:39

2 Answers 2

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The observer at the center of rotation is inertial because the radius of the observer can be assumed to be infinitesimally small.

The observer is not inertial, and more importantly for this problem, the observer’s coordinate system is not inertial.

Although the observer is not undergoing any linear acceleration, he is rotating. This can be detected by gyroscopes attached to the observer and his telescope. These gyroscopes will precess without an external force, which is a clear indication that the attached observer is not inertial.

The reference frame is also not inertial. If he tries to analyze the laws of physics in that reference frame then he will quickly find violations of the standard formulas. All of those violations will point to his frame being a rotating frame. One very easy test he can do is to look a little beyond the ship where in his frame massive objects are moving faster than c. This is a clear indication that the reference frame is not inertial.

How is it possible for the observer at the center of the circular path to measure any time dilation seeing that he cannot detect any diagonal path for the photon as long as he looks through an spyglass at the shuttle?

There is no one standard coordinate system for a rotating reference frame, but in all of them that I have seen time dilation depends on the product of $\omega$ and $r$. So although you can make $\omega$ arbitrarily small in your approach, you have done so by increasing $r$ such that $r\omega \approx c$ and therefore there is visible time dilation.

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  • $\begingroup$ I think Einstein himself considered the observer located at the center of a rotating disc as inertial: "... According to a result obtained in Section XII, it follows that the latter clock goes at a rate permanently slower than that of the clock at the centre of the circular disc, i.e. as observed from K. It is obvious that the same effect would be noted by an observer whom we will imagine sitting alongside his clock at the centre of the circular disc ..." gutenberg.org/files/5001/5001-h/5001-h.htm#pglicense [See Sec. XXIII] $\endgroup$
    – Mohammad Javanshiry
    May 26, 2019 at 15:26
  • $\begingroup$ The precession of the gyroscope would depend on $\omega$, and as I mentioned, we can consider a negligible value for $\omega$. (The precession is consequently negligible, that is.) In other words, the non-inertial observer located at the center of rotation tends to an inertial one if $\omega$ is small enough. $\endgroup$
    – Mohammad Javanshiry
    May 26, 2019 at 15:50
  • $\begingroup$ I see no claim that it is inertial, only that the time dilation of such a clock is 0. Since the time dilation depends on $r\omega$ this is correct. The question of time dilation is irrelevant to the question of inertia. Inertial clocks can be time dilated and non inertial clocks can be non time dilated $\endgroup$
    – Dale
    May 26, 2019 at 15:55
  • $\begingroup$ “the non-inertial observer located at the center of rotation tends to an inertial one” but the reference frame does not. Since the observer is making observations over a large volume of spacetime his requirements for sensitivity in his gyroscope also scale with $r$. There is therefore no radius at which $\omega$ becomes negligible. $\endgroup$
    – Dale
    May 26, 2019 at 15:58
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    $\begingroup$ I think you should ask that as another question. Or perhaps ask these questions on a discussion oriented forum. It seems that you want discussions more than answers, but this forum is not suited for that. $\endgroup$
    – Dale
    May 26, 2019 at 18:00
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How is it possible for the observer at the center of the circular path to measure any time dilation seeing that he cannot detect any diagonal path for the photon as long as he looks through an spyglass at the shuttle?

This procedure (looking at the clock) is not used to compare time intervals of stationary and moving clocks.

Inertiality or non inertiality has nothing to do with the topic.

The chart with the lightclock that you have mentioned roams from book to book, but apparently it confuses readers, because it usually lacks some details, namely synchronized clocks at point of departure and arrival.

The procedure for comparing time intervals is clearly described in the 1905 Einstein's work.

"The observer" does not work alone. The observer fills all space around him with clocks and synchronizes all these clocks by means of light. For example, a spherical wave travels through space and every that observer's assistant in immediate vicinity to a clock adjust it. After this synchronization procedure is finished, all clocks in his reference frame always show the same time, i.e. "the same time" is evenly distributed across the whole frame. Then this observer can measure rate of moving clock by means of comparing readings of "moving" clock with any clock of his frame successively with reading of moving clock at these moments, when they are in immediate vicinity. Sure, the moving clock will work slower than synchronized clock of the observer's frame.

Notion of observer is here

Also please look for Fig. 1-1

https://en.wikipedia.org/wiki/Spacetime

Also good picture is here (chapter time dilation)

http://www.pstcc.edu/departments/natural_behavioral_sciences/Web%20Physics/Chapter039.htm

There is animation for inertial observer.

enter image description here

Your case is not different. I have depicted only 4 synchronized clocks close to the circumference. The observer must compare readings of "stationary" clocks of his frame successively with reading of moving clock at these moments, when they are in immediate vicinity.

enter image description here

That is completely different story, how the moving object would appear visually . For example, you must take into consideration aberration of light. Beam of light that enters your telescope at right angle, had been emitted backward in the frame of the moving lightclock.

To be fair, I must say that if you observe a rotating source through a telescope, its apparent color will turn $\gamma$ times reddish.

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