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Not by much, you only need to know how to apply the Schwazschild solution. The answer is a one line simple expression.

To fix the ideas , assume that event A has coordinates [math](x_A,t_A)[/math] and event B coordinates [math](x_B,t_B)[/math], both measured in the SAME frame F. Also assume that , for example, [math]t_A>t_B[/math] and [math]x_A>x_B[/math] From the perspective of frame F' , moving wrt. F at speed v: [math]t'_A=\gamma(v)(t_A-\frac{vx_A}{c^2})[/math] [math]t'_B=\gamma(v)(t_B-\frac{vx_B}{c^2})[/math] so, [math]t'_A-t'_B=\gamma(v)((t_A-t_B)-\frac{v(x_A-x_B)}{c^2})[/math] We can see that the sign of the expression [math]t'_A-t'_B[/math] depends not ONLY of the sign of the time difference [math]t_A-t_B[/math] but also of the spatial separation of the two events [math]x_A-x_B[/math]. If the events (A,B) are "timelike separated", then [math]x_A-x_B<c(t_A-t_B)[/math] and [math]t'_A-t'_B=(t_A-t_B)\gamma(1-\frac{uv}{c^2})[/math] (the order of events is maintained, [math]u=\frac{x_A-x_B}{t_A-t_B}[/math]) If the events are "lightlike" , then [math]x_A-x_B=c(t_A-t_B)[/math] and [math]t'_A-t'_B=(t_A-t_B) \gamma(v) (1-\frac{v}{c})[/math] (order of events is still maintained) If the events (A,B) are "spacelike separated", then , it is possible that [math]x_A-x_B>c(t_A-t_B)[/math] and [math]t'_A-t'_B=(t_A-t_B)\gamma(1-\frac{uv}{c^2})[/math] (the order of events MAY be reversed, since [math]u=\frac{x_A-x_B}{t_A-t_B}>c [/math] can induce [math]1-\frac{uv}{c^2}<0[/math])

But it is neither necessary nor useful. By contrast, the calculation in the inertial frame is useful and necessary, this is how it is done in all the published papers.

This is extremely tough and unnnecessary. The standard treatment is done from the POV of the inertial frame attached to the center. They don't, [math] c \pm v [/math] is the closing speed , not the speed of the signal.

Spyman (and pzkpfw) explained to you why you are wrong. Put his words into math and , hopefully, if you do the math right, you will realize why, despite your insistence, you are wrong. I asked you to do that before. In the frame of the box, your statement is false. In any frame in motion wrt. the box, your statement is false. You made two false statements, taken together, they do not make a correct statement.

The above is , of course, incorrect. The error is in your "experiment". Since you are making assertions without doing any computations you will have a hard time figuring out the error. Try putting your assertions in mathematical formalism and you'll find your error (or I 'll point it out to you).

It isn't if you are looking at the periphery of the rotating disc, as explained. The inertial frame is the one attached to the center of the disc. When the calculations are done from the perspective of that frame, everything is very clear, not to mention easy to calculate. No.

The modern re-enactments of the Sagnac experiments use Fiber Optics Gyroscopes (FOGs) or Ring Laser Gyroscopes (RLGs). The precision of the experiments is much increased over the original experiments and the principle is exactly the same: [math]t_{CW}-t_{CCW}=\frac{2 \pi R}{c-\omega R}-\frac{2 \pi R}{c+\omega R}=\frac{4 \pi R^2 \omega}{c^2-\omega^2R^2}[/math] The principle is different from Sagnac (Sagnac has a Newtonian/Galilean explanation), Hafele-Keating has a GR explanation.

From the POV of an inertial observer situated in the center of the disc (non-rotating), the equations of motion are: 1. In the clockwork direction (see the mathpages drawing): [math]ct_{CW}=2 \pi R+\omega R t_{CW}[/math] (the wavefront "chases" the detector) 2. In the counter clockwork direction: [math]ct_{CCW}+\omega R t_{CCW}=2 \pi R[/math] (the wavefront "runs into" the detector) The above resolve to: 1a. [math]t_{CW}=\frac{2 \pi R}{c-\omega R}[/math] 2a. [math]t_{CCW}=\frac{2 \pi R}{c+\omega R}[/math] Hence, the [math]c \pm \omega R[/math]

In the (inertial) frame of the launch pad the distance between the rockets stays unchanged (by the way the problem was constructed). The string , on the other hand, is subject to Lorentz contraction. This means that the rockets tend to stretch the string, breaking it.

I suggested a few fringe journals for you to submit, they are the most likely to publish the stuff you wrote. Your stuff doesn't stand a prayer in a reputable journal. There is nothing there, there, as already pointed out. So, the best thing for you is to abandon the "paper".

I asked you a question, you didn't answer: from whose perspective is the redshift being measured? Let me make it easier for you to answer: -for [math]\theta=\frac{\pi}{2}[/math] [math]f_0=f \gamma[/math] i.e. [math]f=\frac{f_0}{\gamma}[/math] Yes, the formula clearly says [math]f_o=f_s \gamma (1-\beta \theta_s)[/math] as pointed out several times. [math]f_0[/math] represents the proper frequency (the frequency in the frame of the emitter). See Moller, "Theory of Relativity", top of page 62, formula (90). The confusion may be caused by the fact that this notation is opposite to the one used by wiki.You seem not to understand that blueshift from the perspective of one frame means redshift from the perspective of the other frame, as in [math]f'=f \gamma <=> f=\frac{f'}{\gamma}[/math].

No, you should try reading the book, it is very good. From whose perspective? think about it.

Incorrect, at [math]\theta>\theta_{critical}[/math], the blueshift changes into redshift, as I said. Incorrect as well, at [math]\theta=\frac{\pi}{2}[/math] light is redshifted, not blueshifted. In the notation used by the formula [math]f_0=f \gamma (1-\beta cos \theta)[/math] , [math]f_0[/math] represents the proper frequency (the frequency in the frame of the emitter). See Moller, "Theory of Relativity", top of page 62, formula (90). The confusion may be caused by the fact that this notation is opposite to the one used by wiki.

The angle where blueshift changes into redshift. The point was that it is smaller than [math]\frac{\pi}{2}[/math].

Small correction: [math]\frac{f_o}{f_s}= \gamma (1-\beta cos \theta_s )[/math] Based on the above, redshift occurs for all: [math]\frac{f_o}{f_s}<1[/math] meaning for all [math]\theta>\theta_{critical}=arccos \frac{1-\sqrt{1-\beta^2}}{\beta}[/math]

I am glad that you are getting independent confirmation of the things I've been telling you.

correct

Scientific research, peer reviewed and published in Nature shows that black or white..... it makes NO difference. Surprise!

Slower, I had the square root inverted by mistake: The derivation is based on the metric for rotating frames: [math](cdt')^2=(1-\frac{\omega^2 R^2}{c^2})(cdt)^2-(dx^2+dy^2+dz^2+2 \omega (xdy-ydx)dt)[/math] for the particular case [math]dx=dy=dt=0[/math] Particles in a particle accelerator (cyclotron or synchrotron) exhibit time dilation according to the rule: [math]\Delta t'=\Delta t \sqrt{1-(\omega R/c)^2}[/math] where: [math]\Delta t'[/math] is the elapsed time in the frame attached to the (accelerated) particle [math]\Delta t[/math] is the elapsed time in the frame of the lab, so, clearly [math]\Delta t' < \Delta t[/math], i.e. the clock in the rotating frame ticks slower than the one in the (inertial) lab, meaning that , in the frame of the lab, the plasma state persists longer than in the frame of the moving particles, exactly as in the muon experiment: [math]\Delta t=\frac{\Delta t'}{ \sqrt{1-(\omega R/c)^2}}[/math] [math]\omega[/math] is the angular speed of the (plasma) particle [math]R[/math] is the radius of the trajectory of the plasma jet

You have it right.

Yes, I know the answer, I never make statements I don't already know the answer. The reason I asked is that taking a class would be a much better way of spending your time than making fringe posts and writing fringe "papers". So why didn't you take the class? You might learn physics (relativity) the proper way...

Have you considered taking a class?

Nope. If it were, there wouldn't be a separate term for it. You have been talking all along about it without knowing what it is. Also no. Try to learn the basics first, before you go into more complex issues.

Particles in a particle accelerator (cyclotron or synchrotron) exhibit time dilation according to the rule: [math]\Delta t'=\frac{\Delta t}{\sqrt{1-(\omega R/c)^2}}[/math] where: [math]\Delta t'[/math] is the elapsed time in the frame attached to the (accelerated) particle [math]\Delta t[/math] is the elapsed time in the frame of the lab, so, clearly [math]\Delta t' > \Delta t[/math] [math]\omega[/math] is the angular speed [math]R[/math] is the radius of the trajectory