# I Two twins moving towards each other in a circular orbit

1. Dec 10, 2016

### Battlemage!

So, it seems there's a new twin paradox thread every day, but I don't think I've seen this particular situation looked at.

The two twins move towards each other each in a circular orbit, and at one point they get close enough to each other that they can compare clocks directly (but their circles do not cross each other: they meet on a line tangent to both circles).

Now, from the symmetry of the situation it appears to me that each twin will see the other as having the exact same age every time they meet at the intersection point.

But if each twin looks at the other at a different point along their respective orbits, it seems to me that whether or not the twin will see the other as having a slow clock (or a fast one) will depend on their respective positions, and in particular whether or not they are approaching each other or moving away from each other along their orbital paths.

My question is two fold:

(a) What would a space time diagram of this look like, since the motion for the twins is two dimensional (would we need a three dimensional space time diagram, or some form of parameterization?). Furthermore, if one twin is considered at rest, the motion through space for the other twin seems like it is going to be weird enough without throwing in the added complication of time. I'm guessing each twin will see the other as following an elliptical path?

(b) Since this is circular motion for both twins, and hence acceleration, would there be any need to bring general relativity into it, and if there is no need, would it be simpler or more complex to do so if it's possible?

2. Dec 10, 2016

### Staff: Mentor

Yes.

Also yes, and as with the ordinary version the twin paradox you can imagine that they're both carrying identically designed strobe lights that flash once every second in their rest frames, and they can see and count the flashes from the other one. The number of flashes received between meetings will be equal to the number of flashes sent.

You need three dimensions: either a three-d model or a two-d perspective drawing works. Draw it from the point of view of an inertial observer and the worldliness of both ships will be helixes, like a coil spring oriented along the time axis.

A frame in which a twin is at rest is not an inertial frame, and loses the symmetry of the situation. Don't go there, it's just making an easy problem difficult.

There is no need to bring in general relativity because we're still in flat spacetime, no gravity. Special relativity handles acceleration just fine, it's only when gravity (the real thing, not the pseudo-gravity caused by acceleration and the equivalence principle) is involved that you need GR.

As an aside..... You can make the acceleration arbitrarily small by making the radius of the circle arbitrarily large.

3. Dec 10, 2016

### Ibix

It's all flat space. You can easily draw a spacetime diagram for it - the paths are helices with the axis parallel to some time-like direction.

You can't really use intuitions from the classic twin paradox about what each twin will see because they are continuously accelerating - so you have to be continuously correcting for the simultaneity change if you want to use the instantaneously co-moving inertial frame. Perhaps better to use radar coordinates?

4. Dec 10, 2016

### Battlemage!

Thanks both of you. So it's a helix world line. Interesting. That actually makes perfect sense now that I think about it, if x and y were parallel with the floor and t was vertical (of course that kind of graph is really Euclidian, isn't it? It doesn't have the asymptote we see in two-dimensional space time diagrams to represent the speed of light).

But if I wanted to look at it from that perspective, with one observer at rest, would I be dealing with things like fictitious forces? If I understand you correctly, since there is no gravity (no tidal effects in particular, I assume), there would be no need for general relativity. Could I approach the perspective of choosing one twin to be at rest the same way they do in Classical Mechanics intro courses with problems like, for example, a kid in a merry-go-round?

I tried looking that up and couldn't find anything. This is similar to polar coordinates, I presume?

5. Dec 10, 2016

### Ibix

6. Dec 10, 2016

### Staff: Mentor

Yes, there would be fictitious forces and other related effects. However, the bigger problem is actually before that. It is to properly define what you mean by "that perspective, with one observer at rest".

When you talk about inertial observers there is a standard and fairly natural coordinate system defined for a given observer. With non inertial observers there is no single natural system. You have to specify the coordinate system in detail.

7. Dec 10, 2016

### Battlemage!

Thanks. That PDF is going to take some time to absorb!

Since my "at rest" frame would be in circular motion, wouldn't any transformation equation to another frame actually change over time? By that I mean, x wouldn't always be parallel to x', for example. They'd have to be related by some function of the angle between them at any given instant, right? I assume there would be a lot more to it than merely figuring that part out, though.

Would it even be linear? I'm not sure about that, since any coordinate transformation equation would change based on angle between axes, and that angle would change based on time (so you'd have a derivative dθ/dt somewhere in there, possibly multiplied by θ or r- I don't know, I haven't thought it out in detail [in the process of trying to do that]).

Although I could imagine a situation where one coordinate system rotates with respect to the tangent line of the circle in such a way that the axes remain parallel to each other; i.e., the x and x' axes are NOT always perpendicular to or tangent to the circle, but rather only at two specific points (the x' axes of the S' coordinate system would be tangent to the circle of orbit at the top and bottom, but perpendicular at the nearest point to the S coordinate system and the furthest point).

Anyway, I think my aimless rambling has accomplished it's purpose. I know the shape of the world lines now ;) (just a matter of how distorted they would be if we actually did an accurate space time diagram). Any thoughts you want to add are welcomed. I enjoy exploring this stuff and finding new ways of looking at it.

8. Dec 10, 2016

### PeroK

Perhaps you should progress to the triplet paradox!

9. Dec 10, 2016

### vanhees71

The rotational reference frame is a nice example for a non-inertial reference frame that can be calculated analytically. Here is a problem+solution to evaluate the motion of a free particle in such a reference frame. You get the same "fictitious forces" (I prefer the name "inertial forces") as in non-relativistic mechanics, i.e., Coriolis and centrifugal force:

http://th.physik.uni-frankfurt.de/~hees/art-ws16/index.html
Sheet 4/Problem 4

10. Dec 10, 2016

### Mister T

On the usual 2-d spacetime diagram light lines are drawn as diagonal lines at 45°. On these 3-d spacetime diagrams it's a light cone. The pitch of the helices that are the worldlines of the twins cannot reach or exceed 45° as that's the upper speed limit. That's a restriction imposed by the non-Euclidean geometry of spacetime. At least I think that's the right way to say it.

Note that this version of the twin paradox can be carried out, and actually is carried out IIRC, at places like the LHC and Fermi Lab when particles are sent in opposite directions around the ring.

11. Dec 10, 2016

### vanhees71

12. Dec 10, 2016

### Staff: Mentor

Yes, x and y parallel to the floor and t vertical is how these space-time diagrams are usually drawn. We're representing a non-Euclidean spacetime in a Euclidean space, but that's no different than the ordinary two-dimensional space-time diagrams you're accustomed to seeing; these are also drawn on a Euclidean plane even though the spacetime is Minkowski. The only problem is that ruler distances between points on the sheet of paper ($\Delta{s}^2=\Delta{x}^2+\Delta{y}^2$) don't match the Minkowski distance ($\Delta{s}^2=\Delta{x}^2-\Delta{t}^2$) between the corresponding events in spacetime, and once we know this we aren't surprised by it.

The pitch of the helix corresponds to the orbital speed of the object; the shallower the pitch the less time passes on each orbit. If the object is continuously accelerating (which of course implies that it is somehow constrained to move on its circular path - think object at the end of a rope, not planet in free fall orbit) then the pitch of the spiral is getting shallower. Asymptotic approach to the speed of light means that the pitch asymptotically approaches one turn per unit of time (choose your time unit so that your clock ticks once in the time it takes a light signal to travel a distance $2\pi{R}$ where $R$ is the radius of the circular path) but never quite gets there.

13. Dec 10, 2016

### Staff: Mentor

I also prefer "inertial forces".

14. Jan 8, 2017

### PeroK

I've been meaning to look at this for some time. Imagine that in A's inertial reference frame: B is moving in a circular orbit of radius $R$ with speed $v$. Here's my analysis by treating the circular orbit as a large number of linear segments with an equal number of instantaneous changes of direction:

In A's frame the situation is clear and there is a time dilation of $\frac{\Delta t_A}{\Delta t_B} = \gamma$.

We approximate B's motion as $n$ segments of length $d \approx \frac{2\pi R}{n}$ with $n$ instantaneous changes of direction of $\theta = \frac{2\pi }{n}$. We imagine a series of clocks, synchronised in A's frame, placed round the orbit at each vertex.

Now, we analyse the situation in B's frame. Imagine B passes the first clock at $t_A = 0$ and $t_B = 0$. B then changes direction. The second clock is now straight ahead, a distance $D = \frac{d}{\gamma}$ in B's frame. But, by the relativity of simultaneity, this clock already reads $t_0 = +\frac{vd}{c^2}$ in B's frame.

B takes $\Delta t_B = \frac{D}{v} = \frac{d}{v \gamma}$ to reach this second clock.

During this time, as A's clocks are time dilated, the second clock only advances $t_1 = \frac{d}{v \gamma^2}$

When B reaches the second clock it reads:

$\Delta t_A = t_0 + t_1 = \frac{vd}{c^2} + \frac{d}{v \gamma^2} = \frac{d}{v}(\frac{v^2}{c^2} + \frac{1}{\gamma^2}) = \frac{d}{v}$

So, now we have (as expected): $\frac{\Delta t_A}{\Delta t_B} = \gamma$

Alternatively, we can use a single clock at A, at the centre of the orbit. At each change of direction, A moves a distance $x = R \sin \theta$ ahead of B in the direction of B's motion. For small $\theta$ this gives $x = R\theta$. Also, the distance to the second vertex is $d = R\theta$, and we get the same equations as before to compare the times on A's and B's clocks between a pair of vertices:

$\Delta t_B = \frac{R\theta}{v \gamma}$

$\Delta t_A = \frac{R\theta v}{c^2} + \frac{R\theta}{v \gamma^2} = \frac{R\theta}{v}$

In summary, for a circular orbit it is a case of a simple (asymmetric) time dilation with the orbiting clock running slower, as expected.

15. Jan 8, 2017

### Bartolomeo

Amount of time dilation depends only on linear speed of an observer. It does not depend on acceleration. They can observe each other’s clock rate in distant points by means of measuring frequency. I also think each of them will always see, that clock of another twin runs at the same rate as his own. Hence, in this case, they will never see any time dilation of the twin brother.

We can try to put things in order the following way:

Let us there is a set of observers who are in motion at a high velocity, V, around a circle with a large diameter. Let’s assume that an observer is also located in the center of the circle. We will further assume that each of the observers, including the one in the center, is in possession of a clock, a tube (let it be very thin one), and a lamp with a green monochromatic light.

Let’s imagine that the velocity of the observers at the circumference is such that the observer in the center sees a slowed rate for each of the clocks moving around the circle due to the Transverse Doppler effect, and perceives the color of the lamps hurtling around the circle with the observers as yellow.

What the observers positioned around the circle would see in the center of the circle?

Let’s recall what happens if an observer moves in a transverse direction with respect to a monochromatic emission. Due to aberration, the light flux falls on the observer not at a right angle relative to its direction of movement, but rather at an oblique angle to this direction. An emission spectral line shift toward short waves simultaneously occurs.

The observers at the circumference will see an accelerated clock rate, while the lamp color will be perceived as blue. The observers in motion at the circumference will never see a yellow lamp or clocks that have slowed rates, since the light beams from the central observer’s lamp are propagated along the circle’s radii. If the observers at the circumference direct their gaze into the circle perpendicular to the direction of their movement (strictly toward the center of rotation), they will then not see either the lamp or the observer in the center due to the light aberration – because of the aberration, the light in fact falls on the observers moving around the circle not perpendicular to their direction of movement, but rather at a certain angle in front. If an observer in motion uses a tube that is kept perpendicular to the direction of its movement, the light emanating from the central observer’s lamp will not enter the tube and will be absorbed by the tube’s walls. To see the central observer, an observer at the circumference must look at an oblique angle to the direction of its movement (or orient the tube at this angle). Only then will it see the central observer, notice that the color of the lamp is blue, and that the clocks are running at an increased rate. The observer can find the oblique angle experimentally, or can calculate it, knowing the V velocity at which it is hurtling around the circle at right angles to the beams.

The fact that the color of the lamp will be perceived as blue is not difficult to explain if one assumes that the central observer sent a green-colored light pulse to a mirror that one of the observers around the circle is holding in its hands in such a way that the pulse reflected from it is returned to the central observer. I.e., the mirror must be oriented along a normal line to the center of the circle. To the central observer, the returned pulse appears to be green in color, since the transverse movement of a mirror in the special theory of relativity does not change the color of a beam reflected from it. But if a pulse received from an observer in motion is green, that means the observer in motion perceived it as shifted into the short-wave (blue) region.

The observers moving around the circle are not justified in explaining the blue color of the lamp by way of the Doppler effect caused by the movement of the central observer (in fact, only they themselves, not the central observer, are in motion, during which they move at right angles to the beam). They can only explain the blue color by way of the fact that, in moving at right angles to the green beam, they experience time dilation and perceive the green beam as blue due to the slowing of all the processes in their rotating “laboratory”.

And what will a co-moving inertial observer (an observer of a co-moving inertial frame of reference) see, who at some point in time ends up near one of the observers hurtling around the circle?

This will depend upon what velocity it ascribes to itself. Indeed, an observer of a co-moving inertial frame of reference is entitled to assume both a state of proper rest and a state or proper motion. If it regards itself to be at rest, it will then direct its gaze perpendicular to the direction of movement of the central observer within its own frame of reference, and will see a yellow lamp and a slowed clock rate. If, however, like an observer at the circumference, it regards itself to be in motion at a velocity of V at a right angle to a beam, it will be looking in the same direction in which the observer at the circumference is looking, and like the latter, of course, will also see a blue lamp. And finally, if for some reason, it ascribes a velocity equaling ½V to itself, then by calculating the necessary angle and directing the tube at this angle, it will see a green lamp and a normal central observer clock rate.

This direction of gaze (of a tube) and the rate variability of the observed moving clocks noticed in this direction corresponds to a co-moving observer’s recognition of its own motion relative to a certain frame of reference at the same velocity of ½V at which the central observer is moving within this system, but in the opposite direction. Since the co-moving and central observers are in motion at identical velocities within this frame of reference (albeit in different directions), they will then experience an identical time dilation. The mutual compensation of identical slowing will result in the fact that the clock rate and lamp color of the central observer in motion within this frame of reference at a velocity of ½V is perceived as normal by a co-moving observer in motion in the other direction at a velocity ½V.

In the physical sense, the possibility of an inertial observer of a co-moving frame of reference observing the dilatability of time in the center of a circle is explained by the fact that the observer can receive the emission given off by a central lamp not only at the circumference, but also outside its confines.

The following fact is interesting. If an observer in motion around a circle receives a light pulse given off by a moving source from a diametrically opposed point of a circle, it then would not notice a transverse Doppler effect, even though the source and the observer are in motion relative to one another in this instance. Let’s assume, for example, that a light pulse proceeds from a green light source at the circumference, which, bypassing the center of the circle, reaches a diametrically opposed point of the circle. This might be a pulse that was sent either by an observer at the circumference or by a co-moving observer. Another observer is in motion around the circle at the moment that the pulse intersects it. What pulse color will it see? It is not difficult to understand that it will see a green pulse. But in fact, a green light pulse emitted from a point of a circle will appear yellow to a central observer past which this pulse proceeds due to the time dilation of the source moving around the circle that emitted this pulse. On the other hand, the light wave frequency of a yellow pulse emitted from the center of a circle and arriving at an observer in motion at the circumference will be perceived by the latter at shifted into the short-wave region due to time dilation itself, and the pulse will take on a green color for this observer.

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

16. Jan 8, 2017

### Ibix

Clearly they will see time dilation - and in fact will be able to see it directly witht his setup - since (as you say in your first sentence) time dilation depends on the speed.

Clearly not. You can't change a physical result (the light frequency received) by changing your mind about what your current state of motion is.

17. Jan 8, 2017

### Bartolomeo

Let's don't rush. At the page the author requests to resolve a very, very simple (it seems) task about time dilation. The task is in the chapter “The Transverse Doppler Effect”
http://spiff.rit.edu/classes/phys200/lectures/doppler/doppler.html

Q: Joe has an ordinary red laser pointer which
emits red light, with a wavelength of
about 650 nm. He points the laser directly
up into the night sky.
Aliens flying past the Earth at v = 0.9c
look straight down at Joe and his laser.

What wavelength do they measure?

I would be glad to know – what do you think about measured wavelength (or frequency)? If proper frequency of Joe's laser pointer is f, what frequency the Aliens will measure?

18. Jan 8, 2017

### Ibix

I'm not rushing. I'm pointing out that, no matter how hard I really, really, believe that I'm stationary, or really, really believe that I'm moving at 0.99c, the Sun does not change colour, not for me nor anyone else.

The relevant formula is given in a large red box directly above the question.

19. Jan 8, 2017

### Bartolomeo

That formula does not work in this case.
It is clearly stated, that the Aliens MOVE in Joe's reference frame. The Aliens will see nothing. Nothing at all, because of aberration. Photon was released at right angle to direction of Aliens in Joe's reference frame. It approaches the Aliens at oblique angle. Thus, the Aliens have to take aberration of light into account and turn their gaze into front at oblique angle. The source will appear "in the front" of them. The "faster" the Aliens move in Joe's reference frame, the sharper angle of gaze will be. In this case the Aliens will see, that laser pointer's frequency increased and if laser pointer was "red" it will become "green" or even "blue". Thus, according to their observations, Joe's clock runs faster.
Good?
You can easily check it. Look at the Transverse Doppler Effect https://en.wikipedia.org/wiki/Relativistic_Doppler_effect.
Particularly motion in arbitrary direction. Angles of emission and reception are always tied with relativistic aberration formula. Just add different values.
Yes, we can say that the Aliens are at rest and Joe is in motion. Then the Aliens will look straight down, but Joe must tilt his laser pointer backward to direction of his motion.The Aliens will see redshift (dilation) of Joe's clock, but Joe will see blueshift (acceleration) if the Aliens will send him the beam back.
If you consider yourself "at rest", you look straight down. If you think you are in motion, so as to evaluate clock rate of the source, you have to direct your gaze into front at corresponding to your relative speed angle.
Look at the very last sentence in Wikipedia article. It means, that if Joe and Aliens ascribe themselves half of relative speed each, they will see neither dilation nor acceleration. The same frequency. Joe tilts laser pointer back and the Aliens look into front at equal angles.

Last edited: Jan 8, 2017
20. Jan 8, 2017

### jartsa

I guess a wrong formula is given. Because: A photon moving up hits the bow of the spaceship that is moving horizontally. The higher the speed of the spaceship, the more energetic the collision, and the higher the observed frequency.