Two twins moving towards each other in a circular orbit

[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?

 

[Nugatory]

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.

Yes.

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.

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.

(aWhat 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?).

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.

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?

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.

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?

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.

 

[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?

 

[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).

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.

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?

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?

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

 

[Ibix]

Radar coordinates: https://arxiv.org/abs/gr-qc/0104077

 

[Dale]

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?

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.

 

[Battlemage!]

Radar coordinates: https://arxiv.org/abs/gr-qc/0104077

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

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.

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.

 

[PeroK]

Any thoughts you want to add are welcomed. I enjoy exploring this stuff and finding new ways of looking at it.

Perhaps you should progress to the triplet paradox!

 

[vanhees71]

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.

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

 

[Mister T]

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).

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.

 

[vanhees71]

The time-dilation effect was indeed measured with heavy ions at GSI with very high accuracy:

https://www.tu-darmstadt.de/vorbeischauen/aktuell/news_archive/news_details_en_103104.en.jsp
http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.113.120405

 

[Nugatory]

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).

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.

 

[Dale]

You get the same "fictitious forces" (I prefer the name "inertial forces")

I also prefer "inertial forces".

 

[PeroK]

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?

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.

 

[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

 

[Ibix]

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.

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.

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.

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.

 

[Bartolomeo]

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.

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?

 

[Ibix]

Let's don't rush.

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.

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”

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

 

[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.

 

[jartsa]

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

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.

 

[Bartolomeo]

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.

Yes, that's right.

 

[Ibix]

That formula does not work in this case.

Let's do some analysis and see what happens. Let Joe be at some point on the y-axis shining his laser along it. The aliens pass through the beam at $x=0$ moving in the $+x$ direction. Let's assume their ship is sufficiently long to receive two successive wave crests. The first one arrives at $t=0$ and the second at $t=1/f$. What time do the aliens measure between wavecrests? The first one is obviously $t'=0$ and the second arrives at $t'=\gamma/f$. That gives a frequency of $f'=f/\gamma$, as expected.

You are correct that the question is sloppily worded because the aliens are only looking straight down in Joe's frame. In the context of a page about the transverse Doppler effect, though, I think it is clear what is meant.

Of interest: the emission events for the two wavecrests occur at $(t,x,y)=(-T,0,Y)$ and $(-T+1/f,0,Y)$, where $cT=Y$. Transforming those into the aliens' frame gives us $(t',x',y')=(-\gamma T,\gamma vT,Y)$ and $(-\gamma(T-1/f),\gamma v(T-1/f),Y)$. So the aliens will certainly attribute their frequency measurement to a mixture of the movement and consequent time dilation of the source, whereas Joe attributes it entirely to time dilation on the part of the receiver. But the fact remains that the received frequency will be about 2.3 times the frequency printed on Joe's laser pointer.

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.

No. It means pretty much what I said in my last paragraph. Depending on the frame in which you do the analysis, your analysis process will be different because the state of motion you assign to source and receiver will be different. However, physical invariants such as "was the alien looking in the right direction?" and "what did the alien's spectrometer read?" will necessarily come out the same. In the symmetric case mentioned in the Wikipedia article, it means that the difference in emitted and received frequencies is attributed entirely to the classical Doppler effect, whereas in any other frame you need to factor in a time dilation calculation.

 

[Ibix]

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.

The aliens will see a blue shift, but won't agree that the light was red when emitted (correcting only for classical Doppler). In the end, they'll agree that they receive light at a lower frequency than that printed on the laser pointer.

 

[jartsa]

The aliens will see a blue shift, but won't agree that the light was red when emitted (correcting only for classical Doppler). In the end, they'll agree that they receive light at a lower frequency than that printed on the laser pointer.

Well, I think they receive gamma times higher frequency than what is printed on the laser pointer. I used four-vectors to find that out! IOW I looked at the energy-momentum boosting formula here:
http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/vec4.html#c2
... and it seems to me that if x-momentum is zero, then E' = γ E

So the energy of a photon or the energy of a light pulse is multiplied by gamma. And frequency is proportional to energy.

EDIT: And of course aliens say that the pointer emits at lower frequency than what is printed on the pointer, because the pointer is moving.

 

[Bartolomeo]

No. It means pretty much what I said in my last paragraph. Depending on the frame in which you do the analysis, your analysis process will be different because the state of motion you assign to source and receiver will be different. However, physical invariants such as "was the alien looking in the right direction?" and "what did the alien's spectrometer read?" will necessarily come out the same. In the symmetric case mentioned in the Wikipedia article, it means that the difference in emitted and received frequencies is attributed entirely to the classical Doppler effect, whereas in any other frame you need to factor in a time dilation calculation.

The Aliens don’t do any analysis. They find the angle experimentally. They turn their tube at different angles until they see Joe’s light. If the Aliens are “at rest and look straight down” and Joe is “in motion and tilts his pointer backward”, green photon turns red. If the Aliens are “in motion and look into front” and Joe is “at rest and hold the pointer straight up”, green photon turns blue.
Champeney and Moon time dilation test (1965). Absorber and source were placed on the opposite sides of rotating disc. No frequency shift was detected.
Kholmetskii at all time dilation test. Absorber rotates - blueshift. Absorber is in the center - redshift.

 

[Bartolomeo]

EDIT: And of course aliens say that the pointer emits at lower frequency than what is printed on the pointer, because the pointer is moving.

The Aliens conduct spectrum analysis of Joe’s radiation. For example, they see Balmer series and realize, that received frequency is higher than proper one. They even know how much it is higher.

There is a mark on the laser pointer: „this laser pointer releases frequency f(s)“. In special relativity we call it „proper frequency“ which is equivalent to „proper time“. The aliens see that received frequency is not f(s), not f(s)/gamma but f(s)*gamma. Do you want to say, that the aliens think that the „proper frequency“ was not proper? Ok, the Aliens move relative to Joe and hold a mirror. Joe is „at rest“ and holds pointer (which is an observation tube too) straight up. Joe releases green photon. Photon reflects from the mirror and comes back to Joe. Color is green again. What color of photon was at the mirror?

You maybe think how to fit all this mess into the SR. That‘s easy! Both Aliens and Joe have tube – pointers. Aliens look straigt down, Joe tilts pointer backward. Aliens measure dilaton.

Then Joe forgets, that he was „in motion“ and changes direction of his tube - pointer into „straight up“. So as Joe would see the beam the Aliens also change direction of their tube – pointer „backward“. Now Joe measures dilation.

Thanks God, everything is allright, and the both measure dilation.

Please look at this diagram. https://en.wikipedia.org/wiki/Time_dilation#/media/File:Time_dilation02.gif

Synchronized clocks – it is inertial Einsteinian reference frame. Single clock moves in this reference frame. How many oscillations moving clock makes during time of travel? How many every synchronized in the same time? Does moving red clock „sees“ dilation or acceleration of time in the reference frame? What has to be done, so as red clock would meaure dilation of any of green clock?

Well, it has to leave „green“ reference frame and introduce it‘s own „red“, in which it is „at rest“. Now the red clock considers itself not in motion, but at rest. Now every moving single green clock dilates. This is the special relativity. Joe is at rest, Aliens are at rest too, everyone is at rest and nobody is in motion, since motion is „undetectable“ and is „the same as rest“. Everyone „sees“ only dilation and contraction.

Page 4, chapter Time Dilation

http://isites.harvard.edu/fs/docs/icb.topic455971.files/l09.pdf

 

[Bartolomeo]

By the way. Imagine that red moving clock is the Aliens. Row of green synchronized clock are identical brothers Joes. Clockfaces of green clocks are higlighted in green monochromatic light. Aliens compare their own single clock rate with the time in the „Joes green“ reference frame. They see that the set of clocks runs faster at gamma. What color of clockfaces they will see? If they look straight down, it will be red. But since they move, they have to look into front. Clockfaces will be blue. Frequency increases at gamma too. Now set of clock (time in reference frame) runs faster and every single clock too.

 

[jartsa]

Ok, the Aliens move relative to Joe and hold a mirror. Joe is „at rest“ and holds pointer (which is an observation tube too) straight up. Joe releases green photon. Photon reflects from the mirror and comes back to Joe. Color is green again. What color of photon was at the mirror?

The photon is a green photon in Joe's frame all the time, and the photon is a blue photon in the mirror frame all the time.

In the mirror frame the blue photon is produced by a device that emits wave crests at slow pace, the wave crests move away from the device at slow pace, because the device is moving, chasing the crests. The crests are placed close to each other, in the mirror frame. And that's why it's a blue photon in the mirror frame: Wave crest distance is a blue photon wave crest distance, in the mirror frame.

(for a less incorrect version of that replace "photon" by "light pulse")

If a light bulb is moving very fast, then it is chasing almost all the photons it emits, so almost all photons it emits are blue shifted. I mean the bulb moves so fast that there's a lot of relativistic beaming occurring.

 

[Ibix]

Well, I think they receive gamma times higher frequency than what is printed on the laser pointer. I used four-vectors to find that out!

Agreed. I made a mistake in my analysis above and didn't account for the motion of the receiver. Serves me right for trying to cure insomnia with physics...

 

[Ibix]

The Aliens don’t do any analysis. They find the angle experimentally. They turn their tube at different angles until they see Joe’s light. If the Aliens are “at rest and look straight down” and Joe is “in motion and tilts his pointer backward”, green photon turns red. If the Aliens are “in motion and look into front” and Joe is “at rest and hold the pointer straight up”, green photon turns blue.

I think this is true, but it's worth pointing out that these are two separate experiments. They are done at different times, one when the aliens see Joe moving purely transversely and one when Joe sees the aliens moving purely transversely. If you are aware of that it hasn't come across to me in your writing. As @jartsa says in his last post, the photons don't change frequency depending on the thinking; rather the frequency of the light received by Joe from the aliens and vice versa is time dependent and two observations made at different times do, indeed, have different frequencies.

Again, say that in Joe's frame the aliens travel along the x-axis and pass through the origin at t=0, and Joe lies on the y-axis at y=Y. Let Joe continually illuminate the alien ship with his laser. The light Joe emitted at t=-Y/c is the light Joe considers to have been emitted parallel to the y-axis (he has to lead a moving target) while the light he emitted at t=t'=0 is what the aliens consider to have been emitted parallel to the y-axis (he's a moving source aiming at the spatial origin). It's this difference in times that drives the difference in received frequency in your two cases.

And, dragging this back on topic before Dale yells at me again, this is all irrelevant to the original case of circular motion compared to an inertial observer on the axis, because there's a time symmetry there that isn't present in the linear motion being discussed since #15.

 

[Bartolomeo]

The photon is a green photon in Joe's frame all the time, and the photon is a blue photon in the mirror frame all the time..

At least we agree, that if we say that „the aliens move“ or „the mirror move“ a photon of „green“ proper frequency appears to be „blue“ at reception. It can be only blue, not red. Blue means time acceleration.
We also agree, that if we say, that „the source move“,
SR claims, however, that photon of proper frequency „green“ can only be observed as „red“ at reception, i.e. clock dilates.

In the mirror frame the blue photon is produced by a device that emits wave crests at slow pace, the wave crests move away from the device at slow pace, because the device is moving, chasing the crests. The crests are placed close to each other, in the mirror frame. And that's why it's a blue photon in the mirror frame: Wave crest distance is a blue photon wave crest distance, in the mirror frame.
.

Your jump into another frame. In that frame Joe moves but not Aliens. It is a different story. This effect is not transverse, but longtitudal in this frame since photon does not move at normal between patches of motion of source and receiver. It takes long path. This interpretation says nothing about clock rate. But yes, there maybe this interpretation of observations. Either Transverse, or longtitudal, this is a matter of taste.

 

[Bartolomeo]

I think this is true, but it's worth pointing out that these are two separate experiments. They are done at different times, one when the aliens see Joe moving purely transversely and one when Joe sees the aliens moving purely transversely.

It is exactly what Special Relativity does – conducts two separate experiments and they lead to the same outcome. I have already noted that in previous post. Reciprocity of observations takes place when we change frame and conduct another experiment. The task about Aliens means, that it is not that easy and each observer has to take into account „opinion“ of other one. They have decide who at which relative velocity moves relatively to each other. They cannot admit state of proper rest simultaneously.

Again, say that in Joe's frame the aliens travel along the x-axis and pass through the origin at t=0, and Joe lies on the y-axis at y=Y. Let Joe continually illuminate the alien ship with his laser. The light Joe emitted at t=-Y/c is the light Joe considers to have been emitted parallel to the y-axis (he has to lead a moving target) while the light he emitted at t=t'=0 is what the aliens consider to have been emitted parallel to the y-axis (he's a moving source aiming at the spatial origin). It's this difference in times that drives the difference in received frequency in your two cases.

Light received when they are at the closest approach. That means the Aliens are in motion and Joe is at rest (Aliens see blueshift, since their own clock dilates).

Light emitted when they are at the closest approach. That means the Aliens are at rest and Joe is in motion (Aliens see redshift, since their own clock tick as usual and moving clock dilates)

„ simple way of expressing this is to point out that the null frequency shift occurs for the pulse that travels the shortest distance from emitter to receiver, and this pulse is obviously neither emitted nor received at the point when the emitter and receiver are at their point of closest approach” http://mathpages.com/home/kmath587/kmath587.htm
That means, Joe and Aliens admit motion at „equal“ velocities (simplified 0,45 c and 0,45 c) and their clocks dilate at the same magnitude, i.e. they see the same clock rate. They tilt their tubes at equal angles backward and into front respectively.

What if the Aliens admit that they move with velocity 0,1 c and Joe with 0,8 c?

But which observation is the "correct" one? First? Second? Third? Fourth? Only the second one?

Do you still think that it doesn‘t matter what the Aliens think about their own state of motion?

this is all irrelevant to the original case of circular motion compared to an inertial observer on the axis, because there's a time symmetry there that isn't present in the linear motion being discussed since #15.

This animation is on the first page of google: Please look at the last episode.
For rotating observer accumulated amount of time dilation solely depends on linear velocity. If two observers rotate with the same LINEAR velocity, accumulated time dilation amount will always be the same and their clock will always show the same time. Direction of rotation does not affect it.