Why don't two accelerated clocks behave like two clocks in a gravitational field? If we immerse two clocks in a gravitational field at different altitudes (with the approximation that both heights share the same g for equivalence to be true), the falling and Schwarzschild observers will see them running at different rates (they have slighty different escape velocity).
On the other hand, if we accelerate two distant clocks in the same way, the inertial observers will see them running at the same rate because they move at the same velocity.
So it seems that the two phenomena are not equivalent.
Answer I choose : https://physics.stackexchange.com/a/750456/346390
 A: You seem to be running afoul of subtractive cancellation.
It's true that two clocks at different altitudes in the same gravitational field will tick at slightly different rates, but that's because you can't really approximate them to the same g. I presume you're approximating g because you're considering that altitude << radius, but when you're comparing (i.e. subtracting) two very similar quantities, you have to be very careful with the level of approximation you're dealing with.
The two clocks in the gravity field will tick at slightly different rates, but then, to obtain an equivalent situation with the two distant clocks, you also have to accelerate them by slightly different amounts such that they have slightly different final velocities. You will then find that the difference between the two distant clocks is pretty much the same as the difference between the two clocks in the gravity field.
A: 
two clocks in a gravitational field at different altitudes (with the approximation that both heights share the same g), the Schwarzschild observer will see them running at different rates (they have slighty different escape velocity)

$$g = -\frac{GM}{r^2}\hat{r}$$ and $$v_{esc}=\sqrt{\frac{2GM}{r}}$$
You are claiming that $g$ is the same for both objects since $r$ is approximately the same, but if $r$ is the same then $v_{esc}$ is also the same for both objects. You don't get to pick different $r$ when calculating $g$ vs. $v_{esc}$ if you want the results to make sense relative to each other.
A: The gravitational time dilation is not related to the gravitational acceleration but to the escape velocity vₑ=c√(rₛ/r) at the given height, just plug the escape velocity at your height into the gammafactor and you get the gravitational time dilation relative to a field free observer at infinity.
Since the escape velocity at smaller r is higher, the time difference between the higher and the lower clock is equivalent to the difference between two clocks that fly in circles in flat space, every time the faster and slower ones meet to compare their clocks their proper times have the same ratio as in the other example with gravity.
A: In strict physics terminology, the strong equivalence principle is a statement about observers in freefall, not accelerating observers. Also, the weak equivalence principle is the statement that all nearby things in freefall have the same acceleration relative to any given local inertial frame. So strictly speaking neither of these principles can be invoked to study either clocks held at fixed locations above a planet or clocks in accelerating rockets.
Having said said, there is a less formal notion of equivalence which asserts that effects of gravitation (such as clocks going at different rates at different places) are like effects of acceleration and this is what the question is asking about. The equivalence here is between what is observed at fixed locations above a planet, and what is observed inside the rockets, by observers sitting in the rockets. It has nothing to do with inertial observers.
At observer sitting in an accelerating rocket will indeed find that light waves emitted from the top of the rocket arrive at him with a higher frequency; it is a Doppler effect owing to his upwards motion. After some more analysis he will also conclude that the clock at the front of rocket is running faster than the one he keeps next to him. This is owing to changing lines of simultaneity as the rocket gets faster relative to any given inertial frame, and the associated changing relative velocity between top and bottom of rocket.
I am going to omit the proof of the above. It can be done by standard tools of special relativity, chiefly simultaneity and Doppler effect, and diagrams can be useful. But I will mention another neat observation. This is that a static spacetime is, locally at every event, to first approximation flat (Minkowski metric) and to second approximation Rindler metric! That is, you can always expand the actual metric by aligning axes with the local direction of gravitational acceleration (in whatever coordinates you started with) and then express the space- and time-dependence to low order in displacements. This is much quicker than all the algebra I see in other answers here, and it shows immediately the equivalence between the two cases, such as between Schwarzschild and Rindler, and therefore between effects of gravitation and effects of acceleration. You can even get a horizon and associated observations this way.
A: Acceleration in special relativity works a bit differently from the Newtonian version. It takes a while to build up a new set of intuitions.
A uniformly accelerating particle moves along a hyperbola in spacetime. Consider the following spacetime diagram. Time goes up the page, the spatial direction in which the particle is accelerating is shown horizontally. The dotted diagonals show the light cone at the origin. The red line shows a uniformly accelerating particle.

At intervals along the red line we draw unit lengths along local time and space axes (black lines) in the instantaneous reference frame of the accelerating particle. As the velocity changes, the time dilation and length contraction change the lengths and orientations of the axes. Another particle that maintains a constant distance from the red particle in the accelerating reference frame is shown as a green line. It is also a hyperbola, with the same $45^\circ$ asymptotes passing throught the origin. This, too, is accelerating uniformly, but with a smaller acceleration. The acceleration is one over the distance from the hyperbola to the origin. So these particles maintain a constant separation, in the accelerating frame of reference, but are nevertheless accelerating at different rates!
If we want a particle accelerating at the same rate, we need to translate the red line horizontally, in the spacetime diagram. This gives us the blue line. The distance between them is constant in the stationary frame of reference, but is changing in the accelerating frame of reference, first shrinking, reaching a minimum when it passes through the stationary reference frame, and then increasing.
The analogous situation to two clocks stationary at the bottom and top of a tower in a gravitational field, a constant distance apart in the accelerated frame, is the red and green lines. Even though the separation remains constant, their clocks tick at different rates. The tick marks are further apart on the green line than the red line. (Draw a paralellogram on each pair of unit time/space axes - the black lines - to see this.)
The situation of two particles with the same acceleration is that of the red and blue lines, considered in the stationary frame. A stationary observer can draw horizontal lines across the page, representing clock ticks, and a distant (stationary) observer will see them moving at the same velocity at the events where each tick intersects red and blue lines, and hence the same time dilation.
But this is for the distant stationary observer. For the accelerating observer, these ticks do not occur at the same time, and the other particle is not stationary.
If you want to know more, this uniformly accelerated coordinate system is usually referred to as Rindler coordinates in the literature.
A: 
the Schwarzschild observer will see … two distant clocks … So it seems that the two phenomena are not equivalent.

Indeed, they are not equivalent as described.
This does not violate the equivalence principle. The equivalence principle is strictly local, and the Schwarzschild observer is at infinity. You cannot get more not local than infinity. To use the equivalence principle you must use nearby clocks and observers, not distant ones.
EDIT: to address the question in a situation where the equivalence principle does apply we can use local observers as follows.
First, we start with the usual Schwarzschild metric $$ ds^2=-c^2 d\tau^2=-\left(1-\frac{2GM}{c^2 r} \right) c^2 dt^2 + \left(1-\frac{2GM}{c^2 r} \right)^{-1} dr^2 + r^2 d\theta^2+ \sin(\theta)^2 r^2 d\phi^2 $$ Then for a stationary clock the the time dilation is $$\gamma = \frac{dt}{d\tau}=\left(1-\frac{2GM}{c^2 r} \right)^{-1/2}$$ The worldline is $x^\mu = (\gamma\tau,r_0,\theta_0,\phi_0)$
The four-velocity is $$u^\mu=\frac{Dx^\mu}{D\tau} = \left(\frac{1}{\sqrt{1-\frac{2GM}{c^2 r}}},0,0,0 \right)$$ The four-acceleration is $$a^\mu=\frac{Du^\mu}{D\tau}=\left(0, \frac{GM}{r^2},0,0 \right)$$ This gives a proper acceleration of $$a=\sqrt{a_\mu a^\mu} = \frac{c^2 R}{2\sqrt{r^3(r-R)}}$$ where we have introduced the Schwarzschild radius $R=\frac{2GM}{c^2}$ to simplify the notation.
Now, to describe two clocks that are nearby we change $r \to r+h$ and do a series expansion to first order in $h$. This is the local case when $h$ is small. This gives a proper acceleration of $$a(h)=\frac{c^2 R}{2\sqrt{r^3(r-R)}} - \frac{c^2R(4r-3R)}{2\sqrt{r^5(r-R)^3}}h + O(h^2)$$ Since $R<r$ we know that $4r-3R$ is positive and so for any $0<h$ we have $a(h)$ is a lower proper acceleration than $a(0)$. This means that there is no $h$ small enough that $a(h)=a(0)$, so the approximation that both heights share the same $a$ is problematic. It only holds in the limit as $h \to 0$ and $a(h)$ is first-order in $h$.
However, all is not lost. What we can do is we can write the same series expansion for $\gamma$ and see if $\gamma(h)$ goes to $0$ faster or slower than $a(h)$ in the limit as $h \to 0$. So we find $$\gamma(h)=\sqrt{\frac{r}{r-R}}-\frac{R}{2\sqrt{r(r-R)^3}}h+O(h^2)$$ With this we can calculate the relative change in gravitational acceleration and time dilation as a function of h as $$\Delta a=1-\frac{a(h)}{a(0)}$$$$\Delta \gamma = 1-\frac{\gamma(h)}{\gamma(0)}$$ Which we can then plot

Note that for all $r$ the gravitational time dilation $\gamma$ goes to zero much faster than the gravitational acceleration $a$ does. So the approximation that the gravitational acceleration is the same necessarily implies that the gravitational time dilation is also the same. They are both non-zero for any finite $h$, but they are both first-order in small $h$ with $\Delta\gamma$ always being smaller than $\Delta a$.
For a numerical sense of the scale, if we use $M$ and $r$ for the earth with $h=1 \mathrm{\ m}$ and carry out all calculations to high precision then $a(0)=9.8199492 \mathrm{\ m \ s^{-2}}$ and $a(h)=9.8199461 \mathrm{\ m \ s^{-2}}$ for a fractional difference of $\Delta a = 3.14 \ 10^{-7}$. Similarly $\gamma(0)=1.00000000069610706$ and $\gamma(h)=1.00000000069610695$ for a fractional difference of $\Delta \gamma = 1.09 \ 10^{-16}$. So at the surface of the earth, time dilation vanishes about 9 orders of magnitude faster than the gravitational acceleration does for small $h$.
Thus the equivalence principle holds. If the gravitational acceleration is the same for both clocks in the gravitational case then the gravitational time dilation is also the same for both clocks in the gravitational case. This is equivalent to the non-gravitational acceleration case where identical accelerations also leads to identical time dilations.
A: What's missing from this analysis is a critical assessment of what it means for two clocks to be running at different rates, in particular from the perspective of one of the clocks. You are correct that two clocks identically accelerating would be seen to be running at the same rate by an inertial observer. This is irrelevant, however, as you also correctly point out in the comments to Nullius in Verba's answer that the apparent discrepancy arises in the rest frame of one of the clocks, not of some arbitrary inertial observer. Your mistake in that comment was assuming that the distance between the clocks being constant means that one sees the other as ticking at a the same rate-- this does not follow because the clock's frame is not inertial.
So, how do we assess the rate at which one clock sees the other tick, given that the frame is non-inertial? We must use the fundamental procedure by which simultaneity is established in special relativity: the Einstein synchronization condition. Let us work in the initial instantaneous rest frame of the first clock, which passes $x=0$ at $t=0$, and call the positions of the clocks $x_1(t)$ and $x_2(t)$. Let's neglect fancy high-speed corrections to the trajectories: to first order, both clocks move along parabolas
$$x_1(t) = t^2, \qquad x_2(t) = t^2 + \delta$$ (I've set the acceleration to $2$, in appropriate units with $c = 1$, for simplicity), with $\delta$ the separation between the clocks, and the coordinate times are the same as each clock's proper time. This is justified since, for sufficiently small $\delta$, the following synchronization check happens very quickly and so the clocks do not have time to accelerate to relativistic speeds. Assume that both clocks read $0$ at $t=0$, so they are initially synchronized. The question is: do they remain synchronized? We answer this by finding a pair of points respectively on the trajectories of $x_1$ and $x_2$ which $x_1$ deems simultaneous, and checking whether the two trajectories have had the same amount of proper time pass up to these points since $t = 0$.
Clock 1 emits a light pulse at $t = 0$ which clock 2 reflects at some point $p_r = (t_r, x_r)$, so that it returns to clock 1 at some point $p_f = (t_f, x_f)$. The Einstein synchronization condition indicates that clock 1 will deem the reflection point $p_r$ as being simultaneous with the point on its trajectory that occurred at half the time between the emission time $0$ and the final return time $t_f$, so we seek to compare $t_r$ with $\frac{t_f}{2}$.
We solve for $t_r$ by setting $t = x_2(t) = t^2 + \delta$, which is a simple quadratic with solution
$$t_r = \frac{1 - \sqrt{1-4 \delta}}{2} = \frac{1-(1-2 \delta - \frac{(4 \delta)^2}{8} + O(\delta^3))}{2} = \delta + \delta^2 + O(\delta^3)$$
We solve for $t_f$ by setting $t^2 = x_1(t) = x_2(t_r)-(t-t_r) = 2t_r - t $ (the latter two expressions are the position of the reflected light pulse at time $t$). This is again a quadratic with solution
$$t_f = \frac{\sqrt{1+8 t_r}-1}{2}.$$
While $t_f \to 2t_r$ in the limit of $\delta \to 0$, saying the clocks approach ticking the same rate as their separation goes to $0$, in general $t_f$ is not equal to $\frac{t_r}{2}$. Let's look at the first order correction:
$$\frac{t_f}{2t_r} = \frac{\sqrt{1+8 t_r}-1}{4t_r} = \frac{\sqrt{1+8\delta + 8\delta^2 }-1 + O(\delta^3)}{4\delta + 4 \delta^2 + O(\delta^3)} = \frac{4\delta + 4\delta^2 - \frac{(8 \delta + 8 \delta^2)^2}{8} + O(\delta^3)}{4\delta + 4 \delta^2 + O(\delta^3)} \\ = \frac{4 \delta - 4 \delta^2 + O(\delta^3)}{ 4 \delta + 4 \delta^2 + O(\delta^3)} = \frac{1 - \delta}{ 1 + \delta} + O(\delta^2) = 1-2\delta + O(\delta^2)$$
So, all of this means that clock 1 says the time is $1-2\delta$ times what clock 2 says it is in the limit of a small displacement $\delta$, i.e. clock 1 runs slower, when comparing points that clock 1 considers simultaneous. This is precisely as expected from the equivalent gravitational field scenario, and moreover the redshift is $z = 2\delta$, as claimed here (replacing $g = 2$ and setting $c = 1$ as we have). It's not too much trouble to rework the above with $x_1(t) = \frac{g}{2}t^2$ and $x_2(t) = \frac{g}{2}t^2 + \delta$ and find the redshift is indeed proportional to the acceleration $g$.
