'Gravitational' time dilation using Newtonian physics alone

Question

I assume most people are familiar with the thought experiment for gravitational time dilation. First the time dilation effect is demonstrated for two observers in an accelerating rocket, one at the top and one at the bottom. The one at the bottom sends light signals to the one at the top. Since the observer at the top is accelerating away from the light signals, he receives them at a slower rate than than they are emitted. It is then argued that time runs at different rates at the top and bottom of the rocket, and the equivalence principle is invoked to extend the argument to a gravitational field.

My question is this: consider the same experiment, but analysed entirely using Newtonian physics with absolute space and time. I know that there should be no time dilation effect predicted. Where exactly in the above argument is the absolute space and time of Newtonian physics rejected?

Short question: Why can't we argue (using the standard thought experiment) in Newtonian physics that clocks at different ends of an accelerating rocket tick at different rates?

Answer 1

The key difference in Newtonian physics is that the speed of light (or whatever signals are being sent between the observers) are dependent on the velocity of the emitter.

Let's suppose that the observer at the bottom of the rocket has a position $z_B(t) = \frac{1}{2}g t^2$ while the observer at the top has a position of $z_T(t) = \frac{1}{2} g t^2 + h.$ Suppose that your signals travel at a speed $v_s$ relative to their emitter. At time $t = 0$, the person at the bottom of the rocket emits a signal, which travels up to the top of the rocket and is received at time $t_1$. At time $t = \Delta \tau_B$, a second pulse is emitted at the bottom of the rocket, and this second pulse is received at time $t_1 + \Delta \tau_T$.

At time $t = 0$, the emitter is moving with velocity $v = 0$, and so the signal travels at speed $v_s$ to the top of the rocket. This implies that $$ z_T(t_1) - z_B(0) = v_s t_1. $$ At time $\Delta \tau_B$, however, the rocket (and the observers) are travelling at a velocity $g \Delta \tau_B$, and so the signals will travel at speed $v_s + g \Delta \tau_B$; this implies that $$ z_T(t_1 + \Delta \tau_T) - z_B(\Delta \tau_B) = (v_s + g \Delta \tau_B)(t_1 + \Delta \tau_T - \Delta \tau_B) $$ In the limit that $\Delta \tau_A$ and $\Delta \tau_B$ are "small", so that we only care about linear terms in these quantities, these equations become \begin{align} \frac{1}{2} g t_1^2 + h &= v_s t_1\\ \frac{1}{2} g (t_1^2 + 2 t_1 \Delta \tau_T) + h &\approx v_s (t_1 + \Delta \tau_T - \Delta \tau_B) + g t_1 \Delta \tau_T \end{align} and combining these two equations leads us to the conclusion that $\Delta \tau_T = \Delta \tau_B$ (at least to linear order in these quantities.)

Note that the last term in the second equation ($g t_1 \Delta \tau_T$) arises from the assumption that the speed of the signal depends on the speed of the emitter. If you instead assume that the speed of the signal is still $v_s$, you find that $\Delta \tau_T \neq \Delta \tau_B$ to linear order. This is exactly what is done in the standard thought experiment you describe.