In Einstein's 1907 paper, https://einsteinpapers.press.princeton.edu/vol2-trans/319 , he gives his exponential gravitational time dilation equation. Actually, he was working a special relativity problem (with accelerations and no gravitation) because he knew how to do that, and he was hoping the result would give him (via the equivalence principle) some help in his search for a gravitational theory. So Einstein's equation was actually a time dilation equation for accelerating clocks that are separated by a fixed distance. According to Einstein, for a pair of accelerating clocks separated by the distance "L" in the direction of the acceleration, the leading clock tics faster than the trailing clock by the factor R = exp(A*L), where "A" is the acceleration. In the iterations described below, I will limit myself to the case where "A" is constant during the acceleration. I will show that the exponential time dilation equation is incorrect. (I suspect that Einstein, and also physicists who came along later, didn't ever notice that the exponential equation is incorrect, because they never used it in the nonlinear range where its argument is large ... they only used it for very small arguments, where it is very nearly linear.) Suppose that the two clocks are initially inertial (unaccelerated) at time "t" = 0, and both read zero at that instant. Then, for t > 0, both clocks undergo a constant acceleration "A" (as determined by accelerometers attached to each of them, which control a rocket attached to each of them, so as to achieve the specified acceleration). Their separation remains constant at "L" during the acceleration. In all of my calculations below, I chose L = 7.520. Let "tau" be the duration of the acceleration. Therefore the reading on the leading clock, when the trailing clock reads "tau", is given by AC = tau * R = tau * exp(A*L) . I first take the case where the duration "tau" of the acceleration is equal to 1.0. I choose the magnitude of the acceleration "A" to be such that the velocity of the two clocks, after accelerating for a duration "tau" = 1.0, is 0.8660. The product of the constant "A" and the duration "tau" gives the rapidity "theta", which is related to the velocity by the equation v = tanh(theta), where tanh() is the hyperbolic tangent function, and is equal to tanh(theta) = { [exp(theta) - exp(-theta)] / [exp(theta) + exp(-theta)] } . So, for v = 0.8660, the rapidity theta = 1.3170. Since theta = A * tau, A = theta / tau = 1.3170 / 1.0 = 1.3170. So for the first case with tau = 1.0, we get R = exp(A*L) = exp( 1.3170 * 7.520 ) = exp(9.90384) = 20007. The reading on the leading clock, at the end of the acceleration at tau = 1.0, is AC = tau * R = tau * exp(A*L) = 1.0 * 20007 = 20007 = 2.0 * 10^4, where 10^4 is just 10 raised to the 4th power. So for the first calculation (with tau = 1.0), we have that at the end of the acceleration, the leading clock reads AC = 2.0 * 10^4. For the second case, we increase the acceleration "A" by a factor of 10, and decrease the duration "tau" by the factor 10 (which still, as required, results in the same speed change as in the first case). So we now have A = 13.170 and tau = 0.1. So R = exp(A*L) = exp( 13.170 * 7.520 ) = exp(99.0384) = 1.028 * 10^43 , and AC = tau * R = 0.1 * R = 1.028 * 10^42. Note that when we increased the acceleration by a factor of 10, the reading of the leading clock didn't increase by a factor of 10, it increased by ten raised to a power that increased by a factor of about ten. For the third case, we increase the acceleration "A" again by a factor of 10, and again decrease the duration "tau" by the factor 10 (which still, as required, results in the same speed change as in the first and second case). So we now have A = 131.70 and tau = 0.01. So R = exp(A*L) = exp( 131.70 * 7.520 ) = exp(990.38) = 1.31 * 10^430 , and AC = tau * R = 0.01 * R = 1.31 * 10^428. Note that, again, when we increased the acceleration by a factor of 10, the reading of the leading clock didn't increase by a factor of 10, it increased by ten raised to a power that increased by a factor of roughly ten. So we get the following table: tau -------- AC ----------------------- 1.0 ------ 2.0 * 10^4 0.1 ------ 1.0 * 10^42 0.01 ----- 1.0 * 10^428 Clearly, this iteration is NOT approaching a finite value for the leading clock's reading, as tau goes to zero. The leading clock's reading is clearly diverging as tau goes to zero. I.e., the leading clock's reading goes to infinity as tau goes to zero. So when we use this method to determine by how much the home twin's (her) age increases when the traveling twin (he) instantaneously changes his velocity by 0.866 when he reverses course at the turnaround, it tells us that the home twin gets INFINITELY older, which is not true. From the time dilation equation for an inertial observer (which the home twin IS), we KNOW that both she and he have a finite age at their reunion. Therefore the exponential gravitational time dilation equation CAN'T be correct.
No one has been able to exactly reproduce Einstein’s method used in the Special Theory of Relativity to obtain the exact same results. The reason is that it actually uses Galilean Relativity to solve the problem. It treats a ray of light as though it is a measuring rod or a javelin being thrown into the air and is being acted upon by unknown forces. You have to consider the physical meaning of epsilon. Epsilon is a co-moving frame that acts against all movement in the opposite direction, preventing the speed of light from being violated. It would be like trying to imagine some sphere closing in around the object which pushes against its progress in moving through space and time. It actually exists outside of space and time, because it is a pseudo force. You would have to convert epsilon into a real distance, since it exist outside of spacetime itself. The closing distance of spacetime is given by epsilon, so epsilon would have to be converted into the proper distance which is an equal and opposite opposing pseudo force. This can be done in the same manner tau is converted into the proper time, where it is assumed that x=x’, in sections 3 and 4 of the 1905 paper. Instead of multiplying by the beta factor, you would have to divide by the beta factor. The beta factor is the description of the closing spacetime around a moving object. The closing spacetime is the reciprocal of spacetime.
That formula is supposed to apply where one clock is "stationary" (inertial) and the other accelerating away from it. Isn't it? That's not the same as what you say, which is that the formula applies to two accelerating clocks.
No. Einstein's equation is for two clocks (separated by a constant distance "L") starting (from rest) at the same time, and accelerating with the same acceleration "A". P.S.: I haven't been able to post for several weeks, because I couldn't log in for some reason. Two weeks ago, Sasa gave me a new password that didn't work at first, but it DID work today ... hopefully that will continue to work.
