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The picture shows an accelerating spaceship with two clocks inside it. It is so far away from all other bodys that gravity is of no importance.

Will the bottommost clock be slower than the topmost one? Or will both clocks have the same speed?

Spaceship with two clocks inside it.

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  • $\begingroup$ @lota I saw a video that explained theory of relativity, and one of the examples was this. It said the bottommost would be slower than the topmost clock. However, I've read that the slowing of time is due to the amount of gravitation. The lower you get in a gravitational field, the slower the time passes. Is this because the lower you are in the gravitational field, the stronger the gravitation? $\endgroup$ – Friend of Kim May 14 '14 at 9:14
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The bottom clock will run slower than the top clock.

The simple way to explain this is to appeal to the equivalence principle. This tells us that locally an acceleration is equivalent to a gravitational field. So if the rocket is accelerating at some acceleration $a$ this is equivalent to two clocks sitting stationary in a gravitational acceleration $a$.

We can use this to calculate the gravitational time dilation between the clocks because to a first approximation the relative time dilation is given by:

$$ \frac{\Delta t_{top}}{\Delta t_{bottom}} = \frac{1}{\sqrt{ 1 + \frac{2 \Delta\Phi}{c^2}}} $$

where $\Delta t_{top}$ is the time interval measured by the top clock, $\Delta t_{bottom}$ is the time interval measured by the bottom clock and $\Delta\Phi$ is the difference in the Newtonian gravitational potential. If the distance between the clocks is $h$, then the difference in the potential is simply:

$$ \Delta\Phi = ah $$

so:

$$ \frac{\Delta t_{top}}{\Delta t_{bottom}} = \frac{1}{\sqrt{ 1 + \frac{2ah}{c^2}}} $$

Let's do this calculation for an acceleration of $1g$ and a rocket length of $100$ m. We're taking the upward direction as positive, which means the acceleration is negative because it points down. The relative time is:

$$\begin{align} \frac{\Delta t_{top}}{\Delta t_{bottom}} &= \frac{1}{\sqrt{ 1 + \frac{2 \times -9.81 \times 100}{c^2}}} \\ &= 1.00000000000001 \end{align}$$

The ratio is possibly better written as $1 + 10^{-14}$ i.e. there are thirteen zeros after the decimal point. This is an extraordinarily small effect, but it can be measured. Indeed it was measured by the Pound-Rebka experiment.

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    $\begingroup$ I know we're not supposed to do this, so let the moderators feel free to delete this if they so wish, but anyway ; thank you for an absolutely supurb answer! I had never bothered to think about the Pound-Rebka experiment before, despite having heard of it often, however you motivate it so well. Maybe I'll study it (in detail) after-all! $\endgroup$ – Flint72 May 14 '14 at 10:23
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    $\begingroup$ @John Rennie: As far as I know, the value of gravitational acceleration differs depending on the altitude: Gravity decreases with altitude as one rises above the earth's surface because greater altitude means greater distance from the Earth's center.". Therefore clocks in the gravitational field experience different accelerations depending on altitude. In the case of the rocket, if we consider it a stiff body, the acceleration is exactly the same throughout its length. Don't you think it makes a difference? $\endgroup$ – bright magus May 14 '14 at 11:05
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    $\begingroup$ @brightmagus: the difference is (in the weak field approximation) down to the change in $\Phi$. The change in $\Phi$ is different in uniform and non-uniform fields so the time dilation per metre of vertical separation will be different. But provided you calculate $\Phi$ correctly you'll get the correct value for the time dilation. $\endgroup$ – John Rennie May 14 '14 at 11:09
  • $\begingroup$ Does it mean that different accelerations have nothing to do with the gravitational slowing of the clocks? $\endgroup$ – bright magus May 14 '14 at 11:13
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    $\begingroup$ The time dilation between two points is not dependant on the gravitational field being different at those two points, only on the gravitational potential being different. Wikipedia does not claim otherwise. $\endgroup$ – John Rennie May 14 '14 at 14:41
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Contrary to one of the answers given, if the distance between the clocks, as observed be either clock, remains constant, the two clocks cannot have the same proper acceleration (the acceleration according to an accelerometer attached to the clock); the clock 'in back' will have greater proper acceleration and, thus, will run slower than the clock 'in front'. This is a well known and uncontroversial result. See, for example, Rindler coordinates

If both clocks have the same proper acceleration, the clocks will run at the same rate but the distance between the clocks, according to either clock, will increase; the rocket would necessarily have to stretch.

From the above linked article:

It follows that if a rod is accelerated by some external force applied anywhere along its length, the elements of matter in various different places in the rod cannot all feel the same magnitude of acceleration if the rod is not to extend without bound and ultimately break.

Again, as far as I know, this is uncontroversial and quite easy to show.

As usual, a spacetime diagram is helpful:

enter image description here

Plotted are portions of the worldlines of two clocks with uniform proper acceleration to the right.

When the coordinate time is zero, both clocks are (momentarily) at rest in this frame and both clocks are (momentarily) synchronized with the coordinate time, i.e, both accelerated clocks read zero when the coordinate time is zero.

Now, according to either clock, the distance between the two clocks is constant (both clocks have constant Rindler spatial coordinate).

But, the accelerometer on clock A measures a greater acceleration than the accelerometer on clock B. This is clearly evident given that the curvature of clock A's worldline is greater.

And, indeed, we see that clock A is running more slowly than clock B.

To summarize these results as it relates to the OP's question, implicit in the question is the assumption that, according to either clock within the rocket, the distance between the clocks is constant.

What has been shown is that

(1) if the distance between the clocks, according to either clock, is constant

(2) and if the clocks are uniformly accelerating according to accelerometers attached to each clock

(3) then, the two accelerometers must read different accelerations.

This is a straightforward result from SR. Whether this is, in fact, the correct description of world is, of course, a matter of experiment. But, what SR predicts is unambiguous.


One final note: once acceleration is introduced into SR, one must be particularly careful about the concepts of acceleration and distance.

For example, there is proper acceleration, an invariant acceleration, and coordinate acceleration which is frame dependent. While there can be uniform proper acceleration, uniform coordinate acceleration is impossible.

Further, there are different notions of distance between observers with uniform proper acceleration. There are, for example, the notions of ruler distance and radar distance.

So, when thinking clearly about an acceleration thought experiment in SR, one must be careful to unambiguously specify the problem.

For example, if one says that two clocks have the same acceleration, it isn't clear if one means the same proper acceleration or the same coordinate acceleration. The answer one gets from SR crucially depends on the difference.

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  • $\begingroup$ Not so long ago it was "uncontroversial" here that you can consider accelerations in SR. But then I pointed out that this was achieved by a trick whereby two frames were frozen in time, and yet experiencing velocity (which requires time flow). Nice try ... (Or not so long ago black holes were - nearly - uncontroversial. And then came Hawking - again - and said they are actually ... grey) $\endgroup$ – bright magus May 15 '14 at 21:12
  • $\begingroup$ And by the way. Have you calculated this difference in accelerations? Is it the same as "the third" answer here shows? Does it coincide - for some reason - with the difference in gravitational potentials?? $\endgroup$ – bright magus May 15 '14 at 21:28
  • $\begingroup$ You must be aware of this? (Link from Rindler's link): "The distance between the spaceships does not undergo Lorentz contraction with respect to the distance at the start, because in S it is effectively defined to remain the same." This is really great. Just because the distance is "defined" to remain the same it must remain the same. What if we produced a rocket looking like two rockets connected with a rope but with permanent connection? The connection would not break then? How does the connection know if it is a permanent part of... $\endgroup$ – bright magus May 15 '14 at 21:48
  • $\begingroup$ ... the spaceship or not, and therefore if its length is "defined to remain the same" or not? $\endgroup$ – bright magus May 15 '14 at 21:50
  • $\begingroup$ I guess people playing with accelerations in SR should get serious one day. Obviously, one can rightfully argue that nothing, also acceleration, is transmitted instantaneously. But then the assumption of rigidity under acceleration is much closer to reality than assuming that clocks stop, and yet we can measure their instantaneous velocity at the same time. $\endgroup$ – bright magus May 15 '14 at 21:58
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The clock ahead is behind in time by an amount depending on the speed of the inertial frame of reference theyare in. If the acceleration is thought of as moving through faster and faster inertial frames of reference then the amount that the clock ahead in distance is behind in time increases as the speed of the inertial frames increases.

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If you consider your rocket a rigid body*, than at all times all points on and in it will experience exactly the same acceleration. Therefore there is no difference between the two clocks that might be the source of the difference of the rates of these clocks.

There is no time dilatation here due to equivalence principle saying that acceleration is equivalent to acceleration, since both clocks experience exactly the same acceleration. Gravitational time dilatation results from the difference in accelerations due to the difference in gravitational potentials between clocks' locations. The distance separation between the two clocks in the spaceship does not entail any difference in accelerations due to different distance to the "center of acceleration". They can be treated just like two clocks distance-separated on Earth but located at exactly the same altitude, and therefore experiencing exactly the same gravitational acceleration $g$ (which differs depending on altitude). Therefore the rates of the clocks will be exactly the same.

*I said "if ...", but Alfred Centauri opposed in his comment that nothing can be transmitted instantaneously, so there must be some differences in accelerations. Well ... yes, true... But then, let's think about it for a moment ...

Say, we are applying acceleration $a$ to the lowermost part of the rocket. If the spaceship is not perfectly rigid, the acceleration will be increasing (for a brief moment, or less) throughout its body, but finally the uppermost part of it will achieve the initial acceleration $a$. From this moment on the atoms of the whole body will be transmitting the initial value of the acceleration all the way to the top. One can say that this transmission will be "late", i.e. at any given instant atoms closer to the top will be transmitting the acceleration that the atoms closer to the bottom have experienced a fraction of a second earlier. Sure, but still, it does not matter at all, because the value of the acceleration will be uniform throughout as long as the acceleration applied at the bottom is constant. "Late" acceleration $a$ is still acceleration $a$. Therefore - safe for a brief moment until the whole rocket receives the acceleration $a$ - the whole rocket is experiencing exactly the same acceleration. (So we are back where we were at the beginning - no time dilatation between the clocks in the rocket.)

One last note: Should someone claim there are different phenomena due to acceleration in SR, and therefore accelerations throughout the spaceship do differ, one needs to remember that SR assumes inertial frames. There are claims, however, that it can be proven that accelerations are easily handled by SR. Well, I have shown in my answer here, how it is being achieved through mere sleigh of hand.

Should somebody still feel unconvinced, I will remind you, that SR says that within the moving frame no changes due to speed are showing. That all “strange” phenomena are visible by the outside observer only – through comparison. And here I'm going to hear the objection that this claim pertains to inertial frames only. Sure it does – and so does the whole Special Relativity. Which takes us back to the link above where I showed that one cannot derive equations for SR only to discard the axioms these equations are rooted in.

So, it is not that SR claims certain things about acceleration. It is simply certain people who claim SR claims these things.

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  • $\begingroup$ Indeed, but since you have given two different explanations, and one of them is wrong, some of the arguments are wrong. This is because the arguments aren't explained, only presented as facts. $\endgroup$ – Friend of Kim May 15 '14 at 14:07
  • $\begingroup$ You state that it is only the there isn't any gravitational potential difference, while @John Rennie says there is. This is why you end up with two different answers. Why do you think there isn't a gravitational potential difference? $\endgroup$ – Friend of Kim May 15 '14 at 14:17
  • $\begingroup$ Gravitational potential difference means a difference in acceleration. The further you are from the center of gravitation, the lower the acceleration due to this gravitation (see the link to Wikipedia in my answer). Take a look at Newton's equation: $g$ depends on $r$ (distance). Now, if we consider the rocket a rigid body, the acceleration is exactly the same in all points (except for some momentary stretches at the beginning). The whole rocket accelerates by exactly the same value. Otherwise it would get torn apart. Therefore there is no difference in "gravitational potential". $\endgroup$ – bright magus May 15 '14 at 14:23
  • $\begingroup$ Back to Einstein's equivalence principle. The lift comparison that assumes experiencing constant acceleration is equivalent to a person standing on Earth (or remaining on constant altitude), for whom $g$ doesn't change. Hence there is no gravitational time dilatation between clocks located at the same altitude. Dilatation occurs only for clocks located at different altitudes, where there is different acceleration and therefore different gravitational potential. $\endgroup$ – bright magus May 15 '14 at 14:37
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    $\begingroup$ I checked out your link and it's wrong too, for a similar reason: it starts by asserting a false statement and then, unsurprisingly, derives a contradiction from it. You should really be trying to understand the mistakes in your own thinking, rather than assuming the mistakes are Einstein's. $\endgroup$ – Nathaniel May 16 '14 at 23:13

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