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You can't teach hardened cranks. Whether you are nice, like Janus or not so much, like me.

This is not a problem with him not being able to relate to mainstream math. The real problem is with him accepting mainstream science. As evidence, see his response to Janus' excellent post. I call them the way they ARE.

Nope, see his response to Janus' excellent post.

It doesn't. Don't project your crackpoterry on the laws of nature. It is time that is affected, GR explains exactly how. I pffered to post the explanation but it involves understanding of basic GR, something that you obviously lack. Besides, you asked for a "touchy-feely" explanation. You can get the ratio of clock rates relatively easily by knowing GR. Start with the Schwarzschild solution to EFE: [math]ds^2=(1-r_s/r) (cdt)^2-\frac{dr^2}{1-r_s/r}-(rd\theta)^2[/math] Use the fact that [math]ds=c d\tau[/math] Then: [math](c d\tau)^2=(1-r_s/r) (cdt)^2-\frac{dr^2}{1-r_s/r}-(rd\theta)^2[/math] produces, by division with [math](cdt)^2[/math]: [math](\frac{d\tau}{dt})^2=(1-r_s/r)-\frac{(v/c)^2}{1-r_s/r}-(\frac{r \omega}{c})^2[/math] Write the above for two locations , at radial distances [math]r_1[/math] and [math]r_2[/math] and you get: [math](\frac{d\tau_1}{dt})^2=(1-r_s/r_1)-\frac{(v_1/c)^2}{1-r_s/r_1}-(\frac{r_1 \omega_1}{c})^2[/math] [math](\frac{d\tau_2}{dt})^2=(1-r_s/r_2)-\frac{(v_2/c)^2}{1-r_s/r_2}-(\frac{r_2 \omega_2}{c})^2[/math] Divide the two: [math](\frac{d\tau_2}{d \tau_1})^2=\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] where [math]v_i= \frac{d r_i}{dt}[/math] i=1,2. If the orbit is circular, then [math]v_i=0[/math] but this is not the usual case. The above equation relates the proper times of two clocks , located at different radial distances and moving at different angular speeds. It is the foundation of the calculations for setting up the frequencies of the GPS clocks (such that they are synched up). [math]\displaystyle \frac{d\tau_2}{d \tau_1}=\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] Since the above (exact) formula is rather complicated, the GPS calculations use a power series approximation, making use of the fact that [math]r_s<<r[/math]. The formula encapsulates the effects in: 1. gravitational potential difference , via the term in [math]\frac{r_s}{r_i}[/math] 2. radial speed difference (if it exists), via the term in [math]v_i[/math] 3. angular speed difference, via the term in [math]r_i \omega_i[/math] In general, the formula is expressed in terms of ratios of frequencies, not clock rates: [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] No, there is no "compression" Time passage is a complex function of the gravitational potential and of relative motion. Do you have GPS in your car, mobile phone? How do you think it works? Not through crank misconceptions, that much is sure.

This is exactly the situation solved by GPS. The clocks on the ground would disagree with the clocks in the satellites UNLESS the frequency of the satellite clocks is pre-adjusted at launch. This is exactly what happens and the frequency adjustment is done according to GR. I could post for you the exact derivation.

The absence of motion is judged with respect to the crystalline network of the material, not with respect to some arbitrary frame. There is always going to be motion wrt. an infinity of arbitrary frames, this is not what absence of motion is referred to.

Yes. Have you had any algebra classes? Because you seem not to know even the basics. How old are you?

Yes, you can. When you do that, something magic happens. The resultant speed is ....c!

All of them. You need to come to grips with this FACT.

Turns out that this is exactly the way things happen in reality. I am sorry it offends your "beliefs". You are confusing relative speed with closing speed. If you travel at [math]v_1[/math] and I travel at [math]v_2[/math] wrt the same reference frame , coming towards me, then our closing speed is [math]v_2+v_1[/math] If I travel at [math]v_1[/math] wrt a reference frame F and you travel at [math]v_2[/math] wrt me, then you are traveling at [math]\frac{v_1+v_2}{1+\frac{v_1v_2}{c^2}}[/math] wrt the frame F. Tough, this is the reality. This is not a coherent sentence.

Why do you think that? Turns out that , in SR speeds do not add like in Newtonian mechanics, so, if you are traveling at speed v wrt a reference frame and if you watch a light beam, that light beam travels at..... [math]\frac{c-v}{1-\frac{cv}{c^2}}=c![/math] with respect to you. From your post(s) you seem to believe that light is traveling at [math]c-v[/math], which would give you 1ms but your "belief" is trumped by the physical reality. It is clear that you have a great difficulty accepting the physical reality.

It has been MORE than 100 years that mainstream science has established that you cannot determine "absolute" speed. The diurnal strip variation is due to changes in temperature between day and night. This has also been known for about 70 years.

No, it cannot since there is no "moving" of the stripes. This has been known for over 100 years now.

There is no such thing as 'absolute velocity" There is no such thing as absolute time, time is frame-dependent. Your "theories" are fringe.

Err, false. c'=c.

Moderators, please add this one to the one already in Trash.

Riiiight!

Here. It is by far the best book on physics.

You will need to wait until you have learned calculus. Physics cannot be taught without it. The book is R.Feynman "Lectures on Physics". You can find it free on the internet but you will not be able to understand it until after you learned calculus and linear algebra.

OK, you are 15 . Have you started calculus? Do you know linear algebra? If, yes, I will recommend a book for you. It is the best book to understand physics.

What science did you study?

No, it didn't hurt me. What wouldn't hurt you is getting some education. This is the path of becoming a scientist.

For the time being, you are not a scientist. You will need to go to school and study science first. Based on your posts, you have not done this yet.

To illustrate the above: 1. If you execute a Sagnac experiment, you cannot ignore Earth rotation because this is precisely the cause of the Sagnac effect. 2. If you execute an MMX experiment, you can IGNORE the Earth rotation because its effect on the MMX experiment is NULL.

You got it. Finally