User:Albert Gartinger/sandbox

Transverse Doppler effect
Suppose that a receiver, moving in a straight line, is at its closest point to the source. It would appear that the classical analysis predicts that the receiver detects no Doppler shift. Due to subtleties in the analysis, that expectation is not necessarily true. Nevertheless, when appropriately defined, transverse Doppler shift is a relativistic effect that has no classical analog. The subtleties are these: Fig. 3-7a. If a receiver is moving in a straight line, what is the frequency measurement when the receiver is at its closest approach to the source?

Fig. 3-7b. If a source is moving in a straight line, what is the frequency measurement when the receiver sees the source as being closest to it?

Fig. 3-7a1. If receiver is moving in a circle around the source, what frequency does the receiver measure?

Fig. 3-7b1. If the source is moving in a circle around the receiver, what frequency does the receiver measure?

In scenario (a), when the receiver is at its closest to the source, due to aberration of light the apparent position of the source appears to be displaced towards the direction of motion of the observer at relativistic aberration angle. The point of closest approach is frame-independent and represents the moment where there is no change in distance versus time (i.e. dr/dt = 0 where r is the distance between receiver and source) and hence no longitudinal Doppler shift. The source emits light of frequency f', but also the receiver has a time-dilated clock. In frame S, the receiver is therefore measures blueshift of frequency of light
 * $$f = f' \gamma = f' / \sqrt { 1 - \beta ^2  }$$

Scenario (b) is best analyzed from S, the frame of the receiver. The illustration shows the receiver being illuminated by light from when the source was closest to the receiver, even though the source has moved on. Because the source's clocks are time dilated, and since dr/dt was equal to zero at this point, the light from the source, emitted from this closest point, is redshifted with frequency
 * $$f = f' / \gamma = f' \sqrt { 1 - \beta ^2 }$$

Scenarios (a1) and (b1) can be analyzed by simple time dilation arguments. In (a1), the receiver observes light from the source as being blueshifted by a factor of $$\gamma$$, and in (b1), the light is redshifted. The only seeming complication is that the orbiting objects are in accelerated motion. However, if an inertial observer looks at an accelerating clock, only the clock's instantaneous speed is important when computing time dilation. (The converse, however, is not true.) Most reports of transverse Doppler shift refer to the effect as a redshift and analyze the effect in terms of scenarios (b) or (b1).

Twin paradox as seen by Moving observer


The picture demonstrates a "stationary" reference frame (lattice of Einstein - synchronized clocks). The clock-faces are highlighted with green monochromatic light. A non - inertial observer ascribes to himself state of motion by means of turning his eyes (telescope) into front of direction of his motion at relativistic aberration angle sin(aplha) = v/c.

Since he "moves" himself, he sees, that clock-faces turned blue. He explains blueshift of frequency by dilation of his own clock. When he goes around the last clock he always see, that this clock is highlighted with blue color.

Since he moves in the "stationary" reference frame, he changes spatial position himself, hence, he doesn't need any second clock in his frame. If he compares his readings of his clocks successively with synchronized clocks he will see, that clocks "at rest" run faster - the row of clock as a whole and each individual clock (blueshift)

If he ascribes himself state of rest and will look at right angle to direction of his motion, he will see, that clokcfaces turned red, but the row of clocks still runs faster, i.e. time in the reference frame ticks faster. When he goes around the last clock, the light will go out. If he understandы, that during rotation only he can be moving, he will turn his telescope into front and will see blue light of the clock.

This way moving observer always sees, that his time flows more slowly, than time in the stationary reference frame.

Terrell rotation????


I believe there are some issues with observations by eyes of "moving" and eyes by "stationary" observers,

The Terrell’s article clearly observes the sphere by eyes of "stationary" observer. However, James Terrell writes in his 1959 article : “ Of course, it would make no difference if the camera, were, instead, considered to move at high speed past the stationary object". The problem is that the same picture taken by this Terrell's "moving" camera (if we consider it moving) is taken when the camera is already far, far away from the sphere. His "moving" camera takes the picture of the sphere "from behind", plus, we have to take into consideration Lorentz - contraction of the film.

So as to take "correct" picture of the sphere by "moving" camera, the picture must be taken exactly at that moment, when the camera crosses point of closest approach to the sphere. At least good photographer aims his camera straight into front of the object, unless his client wants to take a picture of his ears or nape.

We can imagine a camera which is at rest relative to the sphere. It is clear, taken by this camera image will be not distorted. If a "moving" film slides directly over the stationary film, clearly, due to Lorentz - contraction of the film the sphere will appear gamma times elongated.

Let’s work it out in the sphere’s frame. Let’s assume, that the sphere is at rest in the origin. When moving camera is at closest approach to the sphere, it crosses Y – axis and its aperture briefly opens. At this moment (of snapshot) photons from the left and right edges of the sphere simultaneously pass through the pinhole and travel further towards the film. They hit the film simultaneously (in the sphere's frame) some later.

It is clear, that these photons were emitted simultaneously in the sphere’s frame, because distances from the left and right edge to the pinhole are the same (sides AO and BO).

Please note that patches of photons from the edges of the sphere towards pinhole and from pinhole to the film form similar triangles AOB and A'OB'. Due to aberration of light image will be a bit shifted, but due to Lorentz – contraction of the photo emulsion, after developing the film the sphere will appear gamma times stretched and you will be able to see the left and right edge. There won't even be any smell of rotation.

Yes, some cameraman can shout with foam at his mouth that he was "at rest" and speculate about simultaneity. But more thoughtful and flexible cameraman can also explain the stretching of the photograph by his own movement relative to the sphere and contraction of the film in his camera. Especially if he agrees with the sphere, which of them is at rest and who is moving.

Absence of the transverse Doppler effect in the Champeney and Moon time dilation experiment for detector and source on the same circular orbit


For that special case when source and receiver move in the same frame at equal and opposite velocities, so their clocks dilate at the same magnitude, I have prepared this diagram for this case. Light emitted backward at angle alpha (so as to go through the center) comes to the center of circumference at right angle (redshifts) travels further and approaches opposite side of the rim at angle alpha from the front and blueshifts again. So, it appears in the same color when was emitted from the opposite side of the rim. This diagram is based on the primary and secondary sources: D C Champeney and P B Moon article, R C Jennison (Ray path in rotating frame) in Nature, Kevin S. Brown (Mathpages)

Einstein 1905 quote
Should someone wishes me to provide quotes, in support of scenario a I would like to quote celebrated A. Einstein’s work “On the Electrodynamic of Moving Bodies”, &7, Theory of Doppler Principle and Aberration:

From the equation for $$\omega$$ it follows that if an observer is moving with velocity $$v$$ relatively to an infinitely distant source of light of frequency $$\nu$$, in such a way that the connecting line “source-observer” makes the angle $$\phi$$ with the velocity of the observer referred to a system of co-ordinates which is at rest relatively to the source of light, the frequency $$\nu'$$ of the light perceived by the observer is given by the equation

$$\nu'=\nu \frac {1-cos \phi \cdot v/c}{\sqrt {1-v^2/c^2}}$$

Mr. Einstein clearly speaks "an observer is moving with velocity $$v$$" "with the velocity of the observer referred to a system of co-ordinates which is at rest relatively to the source of light"

That exactly what my diagram case a demonstrates. It is clear, that at the moment when "the connecting line “source-observer” makes the angle $$\pi/2$$ with the velocity of the observer" makes angle that formula reduces to:

$$\nu'=\frac {\nu}{\sqrt {1-v^2/c^2}}$$

That exactly what my diagram demonstrates.