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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

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7 Answers 7

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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.

Uniform acceleration

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.

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  • $\begingroup$ Even taking into account what you say it doesn't change anything because the difference in time dilation between rear and front due to the length contraction seen from the stationary reference frame is not gravitational time dilation. The gravitational time dilation is the one observed in the reference frame of the clocks, that it to say the change in simultaneity and not the one observed in the stationary reference frame which only takes into account the length contraction and which I wanted to neglect in my question and which Einstein did not take into account in his derivation. $\endgroup$
    – externo
    Commented Feb 14, 2023 at 9:57
  • $\begingroup$ To be more precise, the question asked approximates the gravitational field to the red and blue (not green) line assuming a constant g. This is the principle of local equivalence. $\endgroup$
    – externo
    Commented Feb 17, 2023 at 12:47
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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

Change in acceleration and time dilation

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.

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    $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Physics Meta, or in Physics Chat. Comments continuing discussion may be removed. $\endgroup$
    – Buzz
    Commented Feb 17, 2023 at 17:26
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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.

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  • $\begingroup$ Yet Einstein derived the gravitational time dilation from 2 clocks that accelerate in the same way. The two clocks in the gravitational field may have the same g too, they will not beat at the same rate because they will not have the same escape velocity. Imagine a virtual gravitational field in which g never varies, this would not prevent objects from falling in free fall faster and faster, as the escape velocity is the velocity of a faller from infinity. So time would get slower and slower too. In gh/c², g is constant, but this does not prevent gravit. time dilation at a height difference h. $\endgroup$
    – externo
    Commented Feb 14, 2023 at 9:37
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    $\begingroup$ What I'm saying is that the two clocks at different altitudes will not have the same g. In gh/c², g is not a constant. You can approximate it to a constant if you want, sure, but then, you must also approximate the escape velocities to be the same, as Ve, like g, varies with D, not h. And if you approximate the escape velocities to be the same, the clocks will beat at the same rate and your problem goes away. $\endgroup$ Commented Feb 14, 2023 at 14:04
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    $\begingroup$ Also, a virtual gravitational field in which g never varies presupposes that inertial frames don't exist, as you're postulating that the whole universe is subject to a uniform, non-zero g. I can't imagine D would have any meaning whatsoever in those circumstances, and I would think that they are way outside the scope of GR. Mathematically, I'm sure it's an interesting consideration, but you also have to be careful that your thought experiments don't become non-physical. $\endgroup$ Commented Feb 14, 2023 at 14:08
  • $\begingroup$ In the Pound-Rebka experiment the change is approximated as gh/c² with constant g, because between the top and bottom of the tower the variation of g can be neglected. If g is assumed to be constant, this does not prevent acceleration and thus an increase in escape velocity and thus an increase in time dilation. If we approximate both g and h as constant, it is no longer an approximation but an exact and meaningless result. $\endgroup$
    – externo
    Commented Feb 14, 2023 at 14:48
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    $\begingroup$ The Pound-Rebka, gh/c² approximation applies to Rindler observers. As it happens, that formula is numerically well-behaved in both g and and so subtractive cancellation is not an issue. The formula for time dilation in the Schwarzschild metric is incredibly not. (On the other hand, if you use Rindler observers for the gravitational clocks and Schwarzschild observers for the accelerated clocks, then no, the two sets of clocks won't be equivalent, like the other answers say.) $\endgroup$ Commented Feb 14, 2023 at 16:36
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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.

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  • $\begingroup$ I assume that g is the same and that the escape velocity is not the same, because even if g is the same everywhere a falling object accelerates and the faster it goes the more the escape velocity will increase. By neglecting the variation of g on small altitudes we still obtain a good approximate result. In the Pound-Rebka experiment the approximation is gh/c² with constant g , just as it is the case in Einstein's thought experiment. $\endgroup$
    – externo
    Commented Feb 15, 2023 at 13:27
  • $\begingroup$ If g is really the same everywhere, then the clocks will also run at the same rate, from the point of view of an observer in free-fall. In the Pound-Rebka experiment, the observer is not in free-fall, but located at one of the clocks, and therefore subject to the same gravitational acceleration. When the observer is located at one of the clocks, then they will not appear to run at the same rate, also not in the accelerated clocks scenario. Nullius in Verba's answer shows that extensively. $\endgroup$
    – fishinear
    Commented Feb 15, 2023 at 16:16
  • $\begingroup$ @externo "the faster it goes the more escape velocity will increase" What? No. $\endgroup$ Commented Feb 15, 2023 at 18:06
  • $\begingroup$ @externo No "of course". How does changing the current velocity change what the velocity would be if you dropped it in freefall from infinity? $\endgroup$ Commented Feb 15, 2023 at 18:27
  • $\begingroup$ You can imagine a gravitational field in which g saturates after a certain distance. It won't stop the objects from continuing to accelerate in their free fall, their acceleration will simply be constant instead of increasing, but it won't stop the escape velocity from continuing to increase as they fall. $\endgroup$
    – externo
    Commented Feb 15, 2023 at 18:50
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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.

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    $\begingroup$ Einstein derived the gravitational time dilation from a thought experiment where two clocks were accelarating. The situation must be equivalent and I find that they are not. $\endgroup$
    – externo
    Commented Feb 13, 2023 at 23:42
  • $\begingroup$ I don't know about that, but I do know that the equivalence to the escape velocity does work. $\endgroup$
    – Yukterez
    Commented Feb 13, 2023 at 23:48
  • $\begingroup$ Yes, because clocks at two different altitudes have two different escape velocities and therefore two different time dilations. On the other hand, if we accelerate two clocks in the same way they will have the same velocity and will not desynchronise. The situation is not the same and yet general relativity says that the situation is identical. $\endgroup$
    – externo
    Commented Feb 13, 2023 at 23:51
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    $\begingroup$ The equivalence principle only holds over small time spans and small length scales. Also in the equivalent scenario where you replace gravity with kinematic acceleration the clock corresponding to the lower clock has higher acceleration than the other clock corresponding to the higher one, not the same, so they will desynchronize as well. $\endgroup$
    – Yukterez
    Commented Feb 13, 2023 at 23:55
  • $\begingroup$ No, the clocks are accelerating in the same way. What happens is that by the time the light passes from one clock to the other the clocks have accelerated and the receiving frequency is not the same as the transmitting frequency. This is how Einstein derived the gravitational time dilation. en.wikipedia.org/wiki/Gravitational_redshift $\endgroup$
    – externo
    Commented Feb 14, 2023 at 0:01
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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.

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  • $\begingroup$ You have won. You are the only one who gave a correct answer. $\endgroup$
    – externo
    Commented Feb 16, 2023 at 20:31
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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 $g$.

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  • $\begingroup$ This is only an apparent equivalence, because clocks in the gravitational field get out of sync from all points of view, whereas accelerating clocks get out of sync to the same value from their own point of view only. $\endgroup$
    – externo
    Commented Feb 15, 2023 at 14:45
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    $\begingroup$ @externo That is simply false, on both counts. While we've only specifically assessed the discrepancy observed by clock 1, there will still be one from many other points of view (virtually any accelerating observer will see one). Meanwhile, all of the Schwarzschild observers you're referring to are accelerating. More pointedly: if you played the same game as I have here with an inertial observer in Schwarzschild (say, one following a radial geodesic), you'd find that they would see the two clocks at slightly different heights as ticking at the same rate with no first order correction. $\endgroup$
    – jawheele
    Commented Feb 15, 2023 at 15:58
  • $\begingroup$ If you think of a free-falling observer's point of view (not a Schwarzschild observer because Sch. obs. are not free-falling, they are at infinity) I can't imagine that. I do not believe that an observer in free fall could find that the two stationary clocks beat at the same rate. The difference in the rate of the clocks does not depend on their relative speed with respect to the faller but on their altitude only. $\endgroup$
    – externo
    Commented Feb 15, 2023 at 18:04
  • $\begingroup$ @externo Yes, I meant a free-falling observer's point of view-- this is what is meant by "inertial". Schwarzschild observers (at infinity or not) are indeed non-inertial. You may not believe the claim, but it's a simple consequence of the geometric fact that one can construct normal coordinates in a neighborhood of a point on the trajectory of one of the two clocks, wherein the metric is exactly Minkowski to first order. The rate at which you discern a given clock to progress depends heavily on your trajectory through spacetime, as demonstrated in my answer. $\endgroup$
    – jawheele
    Commented Feb 15, 2023 at 18:46
  • $\begingroup$ I don't see that. If you could complete your answer by showing it that would be a perfect answer. $\endgroup$
    – externo
    Commented Feb 15, 2023 at 20:18

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