# I Time dilation of a rotating disk

1. Mar 11, 2017

### JonnyHilbert

Iv'e been recently interested in time dilation, but the relative time difference between two observers confuses me (i.e. that a high speed observer, and a stationary observer will each perceive the other's clock to run slow.)
I thought of the following experiment to help me understand, but i'm not sure if i'm correct about it.

Imagine four observers (A,B,C,D) and a disk rotating initially at rest.
Observer A is at the centre of the disk.
Observer B is at the outer edge of the disk.
Observer C is somewhere between A & B.
Observer D is outside the disk. (i.e. will not feel the effects of the rotation.)
The disk rotates to very close to the speed of light, maintains this speed for a while, then decelerates back to rest velocity.

I think that the oder of most time experienced to least time experienced (relative to the others) will go:
A = D > C > B

Am I right with this?

2. Mar 11, 2017

### PeroK

There is no such thing as a high-speed observer or a stationary observer.

Yes and no. This example involves accelerating observers. That changes things. Acceleration, unlike velocity, is not relative. In this case, the time dilation is not symmetrical and creates absolute differential ageing between the observers.

3. Mar 11, 2017

### Staff: Mentor

The rotating disk is one of the more tricky special relativity problems, so you probably do not want to take it on until you are comfortable with the much simpler case of ordinary straight-line constant speed motion. To get past the apparent paradox that both observers correctly find that the other clock is running slow you need to remember and apply relativity of simultaneity - it's behind most of the apparently paradoxical results of special relativity. (If you're not already familiar with it, google for "Einstein train simultaneity").

In the time dilation case: say that A and B are moving relative to one another at .87c, and as they pass one another they both set their clocks to 1:00 PM. Later (one hour later according to A) A looks at his clock and sees that it reads 2:00 PM. He's also watching B's clock through a telescope and after allowing for light travel time he works out that B's clock read 1:30 PM at the same time that his own clock read 2:00 and he correctly concludes that B's clock is running slow by a factor of two.

However, the events "A's clock reads 2:00 PM" and "B's clock reads 1:30 PM" are not simultaneous using the frame in which B is at rest and A is moving. In that frame, the events "A's clock reads 1:15 PM" and "B's clock reads 1:30 PM" are simultaneous and correctly concludes that A's clock is running slow by a factor of two.

4. Mar 11, 2017

### Jeronimus

Without claiming correctness, as i would imagine it, one could imagine the simpler case of a hexagon instead of a circle, where B would accelerate near instantaneously at the corners of the hexagon to remain on it.

Each time he accelerates near instantaneously at a corner, he would observe A's clock to be "jumping" up in time (run faster for a smoother acceleration case). While when he is travelling along the hexagon, not at any corner, it would be the classical case in which both observe each others' clocks to run slower.

One can easily see(or so i think), that by choosing a larger hexagon, it would take longer to reach the corners, and the times in which both observe each others' clocks to run slower would increase. Hence, given the same velocity B is moving along the hexagon (as observed by A), on a larger hexagon, B would observe A's clock counter to be moving slower in average than on a smaller hexagon.

You could then instead of a hexagon, use an octagon, dodecagon etc, until you get to the circle, given you have the math skills to compute this. Possibly it's doable with college math as well, most of which escaped my memory however.

In the final stage however, either A or B will have to accelerate into the current frame of the other. And by doing so, he will experience another "jump" on the clock counter as described above if we assume a near instantaneous acceleration.
And so the question on whose clock counter will be higher, will ALSO depend on who is going to accelerate into the other's frame in the final stage.

As for B observing A's clock, depending on the radius of the circle, considering above thoughts, my semi-educated guess would be that given a very large circle and a low velocity, he would observe A's clock counter to be moving slower by about the same factor as the usual case, where we consider them moving in a straight line.

As the radius becomes smaller and the velocity higher, B would observe A's clock to be ticking just as fast as his at some point. (note that one of the two has to accelerate into the other's frame still, so when one does, and they compare the clocks, the results might not be the same counts)

An even smaller radius and more velocity, B might see A's clock to be ticking faster than his.

Again, take this with a grain of salt. It might be completely false.

5. Mar 11, 2017

### PeroK

A bit like this, perhaps?

https://www.physicsforums.com/threa...ther-in-a-circular-orbit.896607/#post-5660815

6. Mar 11, 2017

### Jeronimus

7. Mar 11, 2017

### rootone

Both observers are stationary in relation to their own frame of reference.
Obviously something cannot be moving relative to itself. .
Both observers are moving in relation to the other observers frame.
There is no absolute reference frame in relation to which one observer is moving and the other is not.

8. Mar 11, 2017

### pervect

Staff Emeritus
The rotating disk is, as other posters have remarked, notoriously tricky. There's a pretty high chance that in thinking of time dilation, the OP is using a model based on the notion of "absolute time" from Newtonian physics with an extra mechanism on top, this added mechanism being time dilation.

Unfortunately, this model is known to be incompatible with special relativity, so if that is the OP's model, it simply won't work. And the OP would be correct in believing something is wrong.

The test that comes to mind for seeing if the OP believe in absolute time or not is their understanding of not the rotating disk, which is a more advanced problem, but the simpler problem of Einstein's train, which explores the relativity of simultaneity.

I've found in the past that using words like "absolute time" or "relativity of simultaneity" may not get through to some readers who aren't familiar with those terms. Before the name can have any meaning, the concept needs to be understood, and often the concept is not understood, so the names are not very useful.

I have found that trying to motivate people to study what would actually be productive rather than what they think will be productive is an uphill battle. But there's not much I can do other than recommend what I think is the best course of study.

As far as Einstein's train goes, there are numerous threads on the topic. Einstein's original words on the topic can be found at http://www.bartleby.com/173/9.html, in chapter 9 of one of his books on relativity. The chapter title is "The Relativity of SImultaneity". There is another article (not by Einstein) exploring some of the difficulties students have in understanding this concept that recommends a few changes. It's oriented towards teachers rather than students, however. I'll give the reference, and if the OP actually reads it, I'd appreciate feedback that they did read it, and on how useful it was to them. The paper can be found at https://arxiv.org/abs/physics/0207081, entitled "The challenge of overcoming deeply held student beliefs about the relativity of simultaneity", the author is Scherr et al.

9. Mar 12, 2017

### jartsa

That is correct. How do we know that? B and C are like traveling twins, from the twin paradox, because they are traveling. A and D are not traveling anywhere.

10. Mar 12, 2017

### jartsa

That is not quite correct. Just like inside a linearly accelerating spaceship, B sees A's clock ticking faster, and how much faster, that is proportional to potential difference between B and A, so it depends on distance and acceleration.

Now this potential difference between B and A, how do we calculate that? I think we cheat like this: According to A, B must be aging slower because of her velocity, so the potential difference must be the same as the ratio of the clock rates according to A.

For example, when B moves around at speed 0.87 c, then the potential difference Between B and A is 0.5 c2, or 0.5 if we set c to 1. So moving mass m from B to A takes energy 0.5 mc2.

11. Mar 12, 2017

### PeroK

I don't see the point of this. The time dilation is precisely the $\gamma$ factor due to the constant relative speed. There is no additional time dilation caused by acceleration.

12. Mar 12, 2017

### jartsa

Oh yes, in my scenario B must consider A not moving. Which is reasonable, because A's clock is not time dilated. Or is it reasonable? And why is A's clock not time dilated according to B, if A is moving according to B?

EDIT: Hey now I know what happened. I used a rotating reference frame, that rotates with the disk. IOW I used B's reference frame. That's why A is not moving in that frame.

Last edited: Mar 12, 2017
13. Mar 12, 2017

### Ibix

B isn't inertial, so you can't import intuition from inertial motion. You yourself described it as being like the twin paradox, and there are clear parallels.

It isn't straightforward to analyse what the orbiting observer sees, because they are only ever instantaneously at rest in any inertial frame. Working in A's rest frame, imagine A sends out two light pulses a second apart. Since the distance between A and B doesn’t change, they arrive at B a second apart. But B is moving, so the interval between the reception events is less than one second (crudely, $\Delta x$ and $\Delta y$ are non-zero; formally, you should use calculus or polar coordinates). Thus B sees less than one second between the pulses.

But if B emits pulses at one second (proper time) intervals these clearly are more than one second apart in A's coordinate time. Since the distance between A and B doesn’t change, the reception events must be more than one second of coordinate time apart. Since A's clock matches coordinate time in this frame, the flashes are slower than once a second.

So B sees A's clock ticking fast and A sees B's clock ticking slow. The "A emits" and "B emits" cases aren't symmetric because they aren't in similar states of motion.

14. Mar 12, 2017

### Jeronimus

So you are saying that if you took your spaceship, and accelerated to 0.999999c, moving away of your twin sister, measuring/calculating her clock to be almost standing still/ticking very very slowly, that by just commencing an infinitesimal small uniform acceleration back towards her, you would calculate her clock to be running faster than yours?

Not going to happen.

edit: Basically(as i imagine it), you have two vectors. One is the velocity vector responsible A and B observing each other's clock to be ticking slower. And you have the acceleration vector which is pointing towards A, keeping B on the circle. This acceleration towards A is responsible for increasing the tickrate of A's clock when observed by B, working against the former in a sense.

The acceleration vector towards A is smaller if the circle has a larger radius and or if the velocity is small. Now the exact details on how those work together, is for the math geeks.
What seems to be the case however, is that given any radius, there is an exact velocity for B at which he would measure A's clock to be running at the same pace as his. Increasing the velocity, he would measure A's clock to be running faster than his, while decreasing it, he would measure A's clock to be running slower.
This is the point where time dilation as a result of the speed between A and B and the effect of the acceleration towards A (switching of reference frames) cancel each other out exactly.

A would always measure B's clock to be running slower and always by a greater factor than B is measuring A's clock to be ticking slower. The factor would become almost the same for a huge circle with a very low velocity.

Another thing that APPEARS to be the case, is that no matter which velocity B would choose, if instead of going in a circle around A, he would choose to spiral into A, while de-accelerating in the process to become at rest just when he reaches A in the middle, he would end up with a lower clock count.
Hence, it appears that in this case, he would necessarily observe A's clock always running faster or to the least, most of his trip.
If this is not the case, i must be missing something fundamental, since from the perspective of A, B's clock would always tick slower, and since he is spiraling in to become at rest, i don't see any other solution.

Last edited: Mar 12, 2017
15. Mar 12, 2017

### jbriggs444

You are reasoning about the scenario based on intuitive guesses rather than actual calculations. The fact that the results are wrong is a good indication that the guesses are wrong.
That guess is wrong.

16. Mar 12, 2017

### Jeronimus

If you think i was making just intuitive guesses, then you clearly did not read what i wrote.

if you say so.

17. Mar 12, 2017

### Bartolomeo

You can always work it out by Transverse Doppler Effect

Received by rotating observer frequency will be:

$f_0 = \frac {f_s} {\gamma (1+ \frac v c \cos \theta_0) }$

$v=\omega r$

Obviously, for rotating observer photon always comes at oblique angle.

Angle of emission and angle of reception are tied by relativistic aberration formula.

$\cos {\theta_0} = \frac {\cos {\theta_s} - \frac v c} {1- \frac v c \cos \theta_s}$

Observer in the center launches photon at right angle to direction of motion of rotating observer, thus:

$f_0= \gamma (1 – \frac {v \cos \theta_s} {c}) f_s$

Since $\cos \theta_s = 0$

We get

$f_0= \gamma f_s$

i.e. photon is blueshifted gamma times at reception.

If rotating observer holds a mirror he reflects photon at oblique angle backward and photon comes back to centre redshifted i.e. at the same frequency it was released. So:

Clock in the centre A (and D) measures dilation of rotating clocks B and C.

Clock B and C measure, that clocks A and D run faster.

Clock B circles around clock A in xz plane. Clock A Is in the origin. Clock F is placed at point Y of y axis one million light years away from the origin (like a very distant star). Clock B will measure, that clocks A and F run faster at the same rate. Clock F will measure, that clock B is ticking slower.

Clocks B and G on the opposite sides of the rim of the disc will not measure any time dilation.

Imagine two clocks B and E, that are placed on the rims of two discs O and O’. The discs have the same radius. These discs rotate clockwise with the same angular velocity. At certain moment clocks B and E pass each other at points of closest approach. These clocks move in different directions at this instant with equal linear velocities. These clocks will not measure any dilation of each other.

http://mathpages.com/home/kmath587/kmath587.htm

“This is a stationary configuration (up to spatial isotropy), so we can definitely say the spatial distance travelled by the signal pulses is not changing. Classically it would follow that there was no Doppler effect, but the time dilation of special relativity implies that the proper time of the circling entity is reduced relative to the rest frame time coordinate of the central entity. As a result, the received signal will be either red-shifted or blue-shifted, depending on whether the transmitter or the receiver is moving in a circle.”

R.C. Jennison (Nature, 1964) considered ray path on rotating disc and concluded that light beams take different patches back and forth from the centre to the rim. Thus, by these patches rotating observers can determine their own rotation.

Some polemics appeared recently in Physica Scripta, starting with article by E. Zanchini "Correct Interpretation of two experiments on the Transverse Doppler Shift". Phys. Scr. 86 (2012). Followed by Chen et all, Kholmetskii et all.

Last edited: Mar 12, 2017
18. Mar 12, 2017

### Jonathan Scott

The situation when the system is rotating at a constant rate (usually referring to a "space station") can either be analysed using Special Relativity time dilation or by treating the acceleration like a gravitational field which is the gradient of a potential then relating the relative gravitational potential to the time dilation. These are alternatives, and must not be mixed together. Both methods give the same result, of course (at least for non-relativistic speeds).

Starting up and slowing down means that the fractional time dilation rate is less than it is for full speed (proportionally to $v^2$) but the overall effect is similar.

19. Mar 12, 2017

### Bartolomeo

Excerpt from: E. Zanchini "Correct Interpretation of two experiments on the Transverse Doppler Shift". Phys. Scr. 86 (2012)

Some discussion on the interpretation of the experimental results obtained in [1,2] are present in literature. The interpretation in terms of time dilatation was criticized by Essen [3], who stated that, while in general theory “the time dilatation can be a real effect”, in special relativity “the time dilatation is only apparent” and “the prediction based on the assumption that one clock actually goes slower than the other contradicts special theory” . Essen concluded that experiments [1,2] cannot be interpreted in framework of special relativity. A debate in literature followed: see the letter by Jennison [4] and the answer by Essen, published together with the letter.

[1] Hay H J, Schiffer J P, Cranshaw T E and Egelstaff P A 1960 Phys. Rev. Lett. 4 165-6
[2] Kundig W 1963 Phys. Rev. 129 2371 - 5
[3] Essen J 1964 Nature 202 787
[4] Jennison R C 1964 Nature 203 395-6

20. Mar 13, 2017

### jartsa

Look at the animation here: (the space-time diagram one)
https://en.wikipedia.org/wiki/Relativity_of_simultaneity
What does an accelerating observer observe far away at his left and what does he observe far away at his right? Let's say the observer is at the middle. What does the observer observe not so far away at his left and not so far away at his right?

(You are supposed to see that far away the line of simultaneity (horizontal green line) is moving fast, not so far away it is moving not so fast)
(then of course you are supposed to conclude that distance matters, and jartsa is right)

Last edited: Mar 13, 2017
21. Mar 13, 2017

### Jeronimus

I may have jumped the bullet with my statement there. While in theory you could move away at 0.999999c relative to your twin sister, and observe her clock move slower, in spite of you accelerating towards her, for sufficiently small accelerations, the further away you moved, the lower the acceleration you could use towards her would have to be.
You would never be able to accelerate back into your sister's frame that way in finite time as this would lead to a paradox.

I guess i will have to give this issue a few more thoughts.

22. Mar 13, 2017

### puzzled fish

I'll give you a hint. There's nothing strange or tricky about the rotating disk. Take an infinitely long strip of paper and roll it to a cylinder. Now treat this cylinder as a flat space ( which it is in its own right as a manifold ) familiar Minkowskian space-time and draw your lines. Repeat this process for the inner parts of your disk.

Last edited by a moderator: Mar 15, 2017
23. Mar 14, 2017

### Staff: Mentor

Can you elaborate on how the books you referenced support your contention about the disk?

Also, you're not thinking of a Tipler cylinder here are you?

Last edited: Mar 15, 2017
24. Mar 15, 2017

### puzzled fish

No, it's not a Tipler cylinder. For the circumrference of the disk, its rotating points trace spiral worldlines which are geodesics of a cylinder. This is the space-time of a rotating ring. Now we repeat for all the inner rings of the surface of our disk. There is a footnote no. 9 on page 14 ( Chapter 1.4 ) by Kip Thorne explaining that this space-time is defined only locally. I think, this definition coincides with the vanishing of acceleration due to the spirals being now geodesics.
Anyway, locally, it accounts for all the phenomena on a rotating disk: The watches running slowly as we move away from the center, the measuring of the circumference being greater than 2πr, etc.

25. Mar 17, 2017