Mike_Fontenot
Registered Senior Member
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.
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.
