General Relativity passes the Ratio's Test

May 26, 2020
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It seems the author and the researcher still don't know that Einstein's relativity has already been disproved both theoretically and experimentally.

The fatal error of relativity is the redefinition of time which is no longer the time measured with physical clocks, as shown in the following:

We know physical time T has a relationship with the theoretical time t of both Newtonian mechanics and Einstein's special relativity: T = tf/k where f is the frequency of the clock and k is a calibration constant.

In Newtonian mechanics, since the theoretical time t is the absolute Galilean time and thus the frequency f is a frame independent constant. We can set k = f to make T = tf/k = tf/f = t, which proves our physical time T is the absolute theoretical Galilean time t, and also confirms that our physical time is absolute too.

Now let’s look at special relativity. We would like to use the simultaneity of events measured with both physical time T and relativistic time t in different inertial reference frame through Lorentz Transformation to verify whether they have the same property.

If you have a clock (clock 1) with you and watch my clock (clock 2) in motion and both clocks are set to be synchronized to show the same physical time T relative to your inertial reference frame, you will see your clock time: T1 = tf1/k1 = T and my clock time: T2 = tf2/k2 = T, where t is relativistic time of your frame, f1 and f2 are the frequencies of clock 1 and clock 2 respectively observed in your inertial reference frame, k1 and k2 are calibration constants of the clocks. The two events (Clock1, T1=T, x1=0, y1=0, z1=0, t1=t) and (Clock2, T2=T, x2=vt, y2=0, z2=0, t2=t) are simultaneous measured with both relativistic time t and clock time T in your reference frame. When these two clocks are observed by me in the moving inertial reference frame, according to special relativity, we can use Lorentz Transformation to get the events in my frame (x', y', z', t'): (clock1, T1', x1'=-vt1', y1'=0, z1'=0, t1') and (clock2, T2', x2'=0, y2'=0, z2'=0, t2'), where

t1' = r(t1-vx1/c^2) = r(t-0) = rt
t2' = r(t2-vx2/c^2) = r(t-tv^2/c^2) = rt/r^2 = t/r
T1' = t1'f1'/k1 = (rt)(f1/r)/k1 = tf1/k1 = T1 = T
T2' = t2'f2'/k2 = (t/r)(rf2)/k2 = tf2/k2 = T2 = T

in which r = 1/sqrt(1-v^2/c^2).

That is, no matter observed from which inertial reference frame, the two events are still simultaneous measured with physical time T i.e. the two clocks are always synchronized measured with clock time T i.e. clock time T is absolute, but not synchronized measured with relativistic time t'. In real observations, we can only see clock time T but not relativistic time. Therefore, clock time is our physical time and absolute, totally different from relativistic time in Lorentz Transformation and thus relativistic time is a fake time without physical meaning. The change of the reference frame only makes changes of the relativistic time from t to t' and the relativistic frequency from f to f', which cancel each other in the formula: T= tf/k to make the physical time T unchanged. This proves that even in special relativity our physical time is still absolute. Therefore, special relativity based on the fake relativistic time is wrong.

For more details, you can read the journal paper here:
Apr 22, 2020
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In the Buffon's Needle exhibit that the article links to, I don't get why the range of ϴ is from 0 to π. If the needle is horizontal, ϴ = 0 and sin (ϴ) = 0. Rotate the needle 90° ccw, then ϴ = 90°, sin (ϴ) = 1, and (½)sin ϴ = ½. So far, so good. Rotate another 90°, and the line is horizontal again, and sin (ϴ) = 0 again. But there the value of ϴ would be 180°. So where does π come from? Or, in what units is this measured as π? And in that case, in what square units is the area of the rectangle π/2?