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BTW, you are welcome!

1. a is a parameter, as such , it can be >0 or <0 or [math]\infty[/math] (see the author's discussion) 2. a has dimensions of speed square, the author does not (initially) associate it with the speed of light, he could have written: If a>0 then [math]a=\sigma^2[/math] If a<0 then [math]a=-\sigma^2[/math] 2a. Speed of light (in vacuum ) is frame invariant Speed of gravitational waves (in anything) is frame invariant and equal to speed of light in vacuum. 2b. The equation of electromagnetic waves is what produces the identification.

It was a typo, I forgot to take away the [math]c^2[/math]. Shrug.

My calculations are correct. Yours are not. There was an extraneous [math]c^2[/math] [math]m_0 u du[/math] has dimensions of energy ([math]m_0 c^2[/math]), so you are wrong.

Not the "same calculation". Mine is correct, yours is wrong. In order to "prove relativity wrong" you need to understand relativity. You don't. Work is frame variant, you need to learn that, I just showed you the proof. The only way to disprove relativity is by constructing an experiment and RUNNING it. You are under the delusion that your INCORRECT CALCULATIONS "disprove" relativity. They don't since they only reflect your lack of knowledge. Yes, I know you can. RUN them. Come back when you have the result. For the time being, all you have is your basic misunderstandings.

And why is that a problem? Anyway, your calculations are incorrect. Here is how work transforms between frames: [math]W=dE[/math] where E represents the total energy [math]E=\frac{m_0c^2}{\sqrt{1-u^2/c^2}}[/math] wehere u is the speed of the onject of mass [math]m_0[/math] measured in frame F. So: [math]W=\frac{m_0c^2 u du}{\sqrt{1-u^2/c^2}^3}=m_0c^2 (u_xdu_x+u_ydu_y+u_zdu_z) \frac{1}{\sqrt{1-u^2/c^2}^3}[/math] Without any loss of generality, we can consider [math]u_y=u_z=0[/math] so: [math]W=m_0c^2 u_xdu_x\frac{1}{\sqrt{1-u_x^2/c^2}^3}[/math] If we want to transform the above into frame F' moving at speed V in the x direction we need to make the Lorentz transform for [math]u_x[/math]: [math]u_x=\frac{u'_x+V}{1+u'_xV/c^2}[/math] [math]du_x=\frac{(1-(u'_x/c)^2)(u'_x+V)du'_x}{(1+u'_xV/c^2)^2}[/math] [math]\frac{1}{\sqrt{1-u_x^2/c^2}}=\gamma(V)\gamma(u'_x)(1+\frac{u'_xV}{c^2})[/math] So: [math]W=m_0c^2 (u'_x+V) du'_x\gamma^3(V) \gamma(u'_x)[/math] If , on the other hand , V is in the y direction, then W transforms as: [math]W=m_0c^2 u'_xdu'_x\gamma^{-2}(V)\frac{1}{\sqrt{1+(u'_xV/c^2)^2-(u'_x/c)^2}}=W'\gamma^{-2}(V)\frac{1}{\sqrt{1+(u'_xV/c^2)^2-(u'_x/c)^2}}[/math] So, when V increases, the work W , decreases.

No need for your erroneous "program", relativity is one of the most tested theories (apart from QM) in MAINSTREAM science. Read here.

Yes, you are demonstrating that you do not understand the Lorentz transforms. This is typical of people that claim "relativity is wrong". Yep, work, energy, momentum, force are not frame invariant. You haven't learned that yet?

He wants to get the "altitude" adjusted to the tides. The answer is that it is being done but the errors are rather large. See here.

Funny article, I liked it.

I picked one claim, at random. Work (like energy) is frame dependent, so it MUST change with v. This is your first mistake. Work done by the force perpendicular on the direction of motion is NULL. So, it cannot "decrease as v increases". This is your second mistake. I did not bother reading past that. PS: I noticed that you "published" a "method for angle trisection". It would be good for your education to learn that this problem has been proven impossible to solve. That was in 1837, almost 200 years ago.

I understood perfectly, your statement is incorrect. Both the redlined one and the bluelined one.

Your first statement (the one I redlined earlier) was incorrect. You are trying to make it look as you agree with me. Your second statement (the one in blue) makes no sense whatsoever. So, no, we do not agree.

Err, no. The "front end", has a total proper elapsed time of [math]\tau_1=\int_0^T {\sqrt{1-(\frac{r_1 \omega}{c}})^2 dt}=T \sqrt{1-(\frac{r_1 \omega}{c}})^2[/math]. Here [math]T=1yr[/math] The "back end", has a total proper elapsed time of [math]\tau_2=T \sqrt{1-(\frac{r_2 \omega}{c}})^2[/math]. Since [math]r_1>r_2[/math] it follows that [math]\tau_1<\tau_2[/math], so your claim is FALSE. So "we know" that you are making erroneous statements, easily disproved by correct application of physics.

You know them with the given error. caveat emptor.

No, it is not. Janus explained this to you in words. I explained it with words AND math. In the general formula: [math]\displaystyle \frac{f_1}{f_2}= \sqrt{\frac{1-r_s/r_2-\frac{(v_2/c)^2}{1-r_s/r_2}-(\frac{r_2 \omega_2}{c})^2}{1-r_s/r_1-\frac{(v_1/c)^2}{1-r_s/r_1}-(\frac{r_1 \omega_1}{c})^2}}[/math] if you ignore the gravitational effect, you need to make [math]r_s=0[/math] If there is no radial movement between source and receiver, you need to make [math]v_1=v_2=0[/math]. When you do that you get: [math]\displaystyle \frac{f_1}{f_2}= \sqrt{\frac{1-(\frac{r_2 \omega_2}{c})^2}{1-(\frac{r_1 \omega_1}{c})^2}}[/math]

...for altitude determination.

The chart used by GPS is not flat, it is a fairly exact approximation of the geoid.

No, your question is answered precisely by the GPS functionality.

I am not sure I understand what you mean by "tidal correction" but all the corrections are listed here.

I misread your sentence, I missed that you have a period between "No, I don't think it is". and "In..... Anyways, there is no substantial difference between GPS and PR. In GPS, the frequency is altered at launch precisely in order to avoid the transmission of the frequency of the satellite clock to ground because this would require some very complicated processing by the ground-based device and would drive the price through the roof. So, in effect, GPS is a superset of PR, they are not different experiments.

Contrary to your claim, the frequency of the clock on the satellite IS changed. It is preset at launch.

This is false, the clock frequency is changed at launch to be 1,000000000000006 the clock frequency on the ground based GPS devices. For an exact description, see here. "The first two terms in Eq. (54) give rise to the “factory frequency offset”, which is applied to GPS clocks before launch in order to make them beat at a rate equal to that of reference clocks on earth’s surface. The last term in Eq. (54) is very small when the orbit eccentricity is small; when integrated over time these terms give rise to the so-called “” effect or “eccentricity effect”. In most of the following discussion we shall assume that eccentricity is very small." This part is correct. The GPS explanation is a superset of the Pound-Rebka explanation. There is only difference in gravitational potential and radial motion in the PR experiment. There is no difference in angular velocity.

No, it only proves that your understanding of relativity is wrong.

No, it does NOT create an "impossible situation" since this is exactly the situation solved in the case of GPS. Actually, GPS solves an even more complicated situation, where there is difference in the angular speed between the two "travelers" as you call them. GPS works EVEN when there is radial motion between the two travelers. I'll ask you one more time: do you have GPS in your car? If you do, does it work? You can prove that this is the case by answering the two questions I asked you.