Gravitational time dilation using an accelerating light-clock

  • #1
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Main Question or Discussion Point

I've been trying to understand gravitational time dilation by considering a light-clock of length ##l## undergoing an equivalent acceleration ##a## from rest along the direction of the bouncing light pulse.

I find that the time ##t## that the light pulse takes to travel to the forward receding mirror and then back to the approaching back mirror is given by
$$t \approx \frac{2l}{c}-\frac{al^2}{c^3}$$
Is this result obviously wrong as it implies that the light clock is taking less time to tick in a gravitational field than the time ##2l/c## that it takes without any gravitational field?
 

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  • #2
Janus
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For gravitational time dilation, or it equivalent due to acceleration, You really need to compare two clocks.
It isn't the local acceleration or gravity that is directly responsible for the time dilation, but the difference in potential between the two clocks.
You can use Doppler shift to visual this:
You have two clocks accelerating such that, according to them, the distance between them remains constant.
They are separated by along a line parallel to their acceleration.
You have observers by each clock.
Light leaves the lead clock heading for the trailing clock. Now this light leaves when the pair of clocks have a certain velocity, but arrives at the trailing clock some time later, at which time, the accelerating clocks have changed their velocities. Thus the trailing clock has a different velocity when it receives the light than the lead clock had when the light left it. This difference in velocities between transmission and reception results in the trailing observer seeing a blue shift in the light coming from the lead clock. He will see the lead clock ticking faster. Since, according to him, there has been no change in distance between the lead clock and himself to cause this, he is force to conclude that the lead clock is indeed running fast.
If consider the lead observer's view, he also sees a Doppler shift in the light coming from the trailing clock, only he sees a red shift, and concludes that the trailing clock runs slow.

The same kind of thing happens in a gravity field. If you put one clock a fixed distance above another, light leaving the upper clock is seen as blue shifted by the lower clock, and light leaving the lower clock is seen as red shifted by the upper clock. Again there is no changing distance between them to account for this, so both clocks have to conclude that the lower clock runs slower than the higher one.

Sometimes you'll hear it said that clocks in a stronger gravity run slower than ones in a weaker one. This is misleading. If you put a clock at sea level and one on a mountain top, the clock at sea level runs slower. But this is not because gravity is stronger at sea level. In fact, if you could "smooth out" the gravity such that there was no decrease in gravity as you went from sea level to mountain top, not only would the clock on the mountain top still run faster than sea level clock, but the difference in the tick rates would be greater than if gravity dropped off as one gained altitude.
 
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  • #3
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Ok - Gravitational time dilation is due to a difference in gravitational potential and not the local field. Though I guess one could say that the field is itself an infinitesimal difference in gravitational potential with respect to distance.
 
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  • #4
PeroK
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Ok - Gravitational time dilation is due to a difference in gravitational potential and not the local field.
Acceleration does not "cause" time dilation. Imagine two clocks are passing you, both at speed ##v##.

The first clock is travelling at a constant speed ##v##.

The second clock is accelerating rapidly and reaches the speed ##v## just as it passes you.

The time dilation factor for both clocks is the same at the instant they pass you. The acceleration makes no difference: only the speed relative to you matters.

The time dilation of the accelerating clock will vary over time in accordance with its speed relative to you:
$$d\tau = dt\sqrt{1 - \frac{v(t)^2}{c^2}}$$
Where I've emphasised that ##v## is a function of time. There is no additional "gravitational" factor for the acceleration of the clock.
 
  • #5
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I was trying to understand the accelerating lightclock example given by
At 4:57 he describes an upward accelerating lightclock running slow because the top mirror runs away from the rising photon faster than the bottom mirror catches up with the falling photon. I thought in an accelerating rocket both mirrors would accelerate upwards at the same rate. Am I wrong?
 
  • #6
pervect
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I was trying to understand the accelerating lightclock example given by ....

At 4:57 he describes an upward accelerating lightclock running slow because the top mirror runs away from the rising photon faster than the bottom mirror catches up with the falling photon. I thought in an accelerating rocket both mirrors would accelerate upwards at the same rate. Am I wrong?
My best guess is that his explanation given in the you-tube video is just wrong, though perhaps there is some interpretation of his words that I don't see that saves it.

The answer previously given to you by Janus seems to me to be on the correct track. Since you're interested in the topic, it might be time to start looking at textbook references, though I don't have any specific suggestions at the moment.

The most promising explanation in wiki is not under gravitational time dilation, but under Rindler coordinates, the current as of this posting version is https://en.wikipedia.org/w/index.php?title=Rindler_coordinates&oldid=950444078

wiki said:
A more general derivation of the transformation formulas is given, when the corresponding Fermi–Walker tetrad is formulated from which the Fermi coordinates or Proper coordinates can be derived.[19] Depending on the choice of origin of these coordinates, one can derive the metric, the time dilation between the time at the origin ... and the coordinate light speed ...
This approach may be more mathematical than what you're looking for. Unfortunately, most of the textbook discussions I can think of do involve a similar idea of working out the transformation equations between accelerated and inertial coordinates, and finding the metrc, but if one is not already familiar with the very concept of a metric, this sort of explanation would be hard to follow, even if one was comfortable with the fairly complicated mathematical manipulations required.

I'm sure there are less complex textbook derivations out there, which ones are recommendable as accurate and convincing is another question.
 
  • #7
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I was trying to understand the accelerating lightclock example given by
At 4:57 he describes an upward accelerating lightclock running slow because the top mirror runs away from the rising photon faster than the bottom mirror catches up with the falling photon. I thought in an accelerating rocket both mirrors would accelerate upwards at the same rate. Am I wrong?

Let's consider one cycle of an upwards accelerating light clock. The cycle starts from the floor. It takes some extra time for the light to reach the ceiling, because the ceiling is moving to the same direction as the light. Then the light returns to the floor. That happens in extra short time, because the floor is moving towards the light.

Extra time and extra short time cancel out almost completely at low constant speeds of light clocks.

But our light clock moves faster during the light's return phase. So because of that the cycle lasts an extra short time.


If the cycle started from the ceiling then the cycle would last an extra long time.


Does the above sound correct?


A small clarification: As motion of the floor towards the light causes the light to reach the floor extra fast, then extra velocity of floor causes the light to reach the floor extra extra fast. Two "extras". The extra velocity of the floor is caused by the acceleration of the floor.
 
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  • #8
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Does the above sound correct?
That is a correct description of what happens when we use a frame in which a light clock is moving at a constant velocity. Using the frame in which the light clock is at rest of course the time is the same on both legs and there is no time dilation. However,here we're talking about constant acceleration not constant velocity, and that's a bit trickier; there is time dilation even using a frame in which one or the other end of the light clock is at rest, which you haven't yet explained.

The key is that only is the ceiling "moving away" (using any inertial frame in which the ceiling is not at rest) from the light but also, because of the acceleration, the speed with which it is "moving away" is changing while the light signal is in flight. As @pervect says above, the math here gets difficult (part of the difficulty is that "constant acceleration" is itself a very tricky concept). However, two things that may help developing an intuition for the problem are:
1) Draw a spacetime diagram showing the worldlines of the two accelerating ends of the clock and the path followed by the light signals between them. Contrast with the worldlines when the two ends are moving at a constant velocity. The tricky part here will be correctly drawing the accelerated worldlines and choosinga scale that makes the effect we're looking for clear.
2) Spend some time understanding Bell's Spaceship Paradox (google for it) to get a better understanding of what "constant acceleration" means in these problem.
 
  • #9
When you have accelerated light clock big enough, that you have to include different potential along light path, you have limited choice of clocks which are at rest with them (to eliminate different speed efects and to have enough time to complete at least one tick) to compare . Good choice are much smaller local clocks in the same accelerating frame.
This is exactly situation which jartsa or janus described (agree).
When you want to choose for comparison clocks which are not accelerated, they are not at rest with the big clocks during one their tick (big clocks).
When you make the big accelerated light clock smaller to be at rest during their whole tick with non accelerating clocks, the effects from the acceleration will be negligible and you will see no effect from acceleration.
 
  • #10
….. I thought in an accelerating rocket both mirrors would accelerate upwards at the same rate. Am I wrong?
Accelerating rocket as solid body accelerates as hyperbolically accelerated frame (Rindler frame). Proper acceleration varies along length of the rocket (at nose is lowest at the bottom is highest).
Also external observer at rest see different acceleration of parts of the rocket (see that rocket is contracting, so all parts can not accelerate at the same rate)
 
  • #11
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Proper acceleration varies along length of the rocket (at nose is lowest at the bottom is highest).
Also external observer at rest see different acceleration of parts of the rocket
Yes, and this is why we need to be so careful in problems are phrased in terms of "constant acceleration". The question in this thread ("how do acceleration and the equivalence principle produce an effect equivalent to gravitational time dilation?") can be answered without digging into these complexities, but they're there just below the surface.
 
  • #12
pervect
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The point I'd like to make is that if you have a short light clock, at the origin of an accelerated system, it ticks normally.

That's not a very mathematically precise description, but it captures the essence. I suppose I'll mention that "short light clock" is mathematically equivalent to "using a linear approximation", but I won't go into further details to make the whole statement more mathematically precise.

The vidoe's descrition doesn't really imply this behavior by my viewing, so I'd say it's basically wrong. At best, it's misleading.

What you need to understand gravitatioanl time dilation is not one light clock, but rather two or more light clocks, one at the origin of the accelerated system, and one located elsewhere at a constant distance. Then you need a way to compare their rates. This is essentially what Janus described in an earlier post.

It turns out that it is both necessary and sufficient that the round-trip travel time of light between the two light clocks be constant for the distance to be constant. This makes the comparison process pretty easy, the propagation delays simply don't need to be accounted for in detail to compare the rates.

Thus, a correct description of the behavior of the light clocks is that the one at the origin of the accelerated system ticks normally, one "above" the one ticks faster, so that light from the origin, when it reaches the higher clock, is seen as blue shifted, and one "below" the origin ticks more slowly, so that light coming from the origin is seen as blue-shifted.
 
  • #13
…... The question in this thread ("how do acceleration and the equivalence principle produce an effect equivalent to gravitational time dilation?") can be answered without digging into these complexities, but they're there just below the surface.
Yes, there is a lot of ways how to derive the time dilation equation. I made one, which is really simple and is based on pure geometry of Minkowski diagram. Here it is:
deriving of time dilatation.jpg
 
  • #14
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I understand that gravitational time dilation is not caused by proper acceleration itself (the clock hypothesis) - can one think about it in the following manner?

Imagine that there is a clock fixed at a distance ##r## from a body with mass ##M##.

Imagine that I am an inertial observer at rest at a very large distance from the gravitating body.

As an inertial observer I freely fall to a distance ##r## from the body in such a way that my total energy = kinetic energy + potential energy = 0. Thus my kinetic energy at distance ##r## is given by:
$$\frac{1}{2}mv^2=\frac{GMm}{r}\tag{1}$$
In my inertial frame the clock will pass me at velocity ##-v## so that an interval of its proper time ##d\tau## is related to an interval of my time ##dt## by the relativistic time dilation formula:
$$d\tau=dt\Big(1-\frac{v^2}{c^2}\Big)^{1/2}\tag{2}$$
Substituting the squared velocity ##v^2## from Eqn ##(1)## into Eqn ##(2)## we find the gravitation time dilation formula
$$d\tau=dt\Big(1-\frac{2GM}{r c^2}\Big)^{1/2}\tag{2}$$

Is this a reasonable way to understand the gravitational time dilation of a clock in a gravitational potential with respect to an inertial observer at zero gravitational potential?
 
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  • #15
pervect
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I understand that gravitational time dilation is not caused by proper acceleration itself (the clock hypothesis) - can one think about it in the following manner?

Imagine that there is a clock fixed at a distance ##r## from a body with mass ##M##.
You've just put yourself in to the realm of general relativity with this statement. You'd be better off to stick with an accelerating observer in flat space-time.

Imagine that I am an inertial observer at rest at a very large distance from the gravitating body.
Unfortunately, "inertial observers" and "kinetic and potential energy" don't really exist as such in GR. So, your logic as written is just not correct :(.

You were on the right track earlier by considering the flat space-time case though you had a few important details wrong. Primarily, you seemed to be trying to consider one light clock, but the phenomenon you were interested in involves comparing two different light clocks. So from my point of view, you appeared to be floundering. (I'm not sure if that's how you see the situation).

Unfortunately, the current post makes me think you've just moved your floundering from shallow watters (Special relativity) to deeper waters (general relativity) :(.
 
  • #16
PeterDonis
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"inertial observers" and "kinetic and potential energy" don't really exist as such in GR.
In a general spacetime, they don't, but this is actually one of the special cases (a stationary spacetime) where they do. There are indeed issues with what @jcap is doing, but that actually isn't one of them. See my responses to @jcap below.

my kinetic energy at distance ##r## is given by
Careful. You're assuming that you can use the non-relativistic approximation for kinetic energy. How good that approximation is depends on how massive and compact your gravitating body is. It will work all right for, say, the Earth (as long as you don't need too many decimal places of accuracy), but if the body were, say, a neutron star, your formula would result in significant error in estimating ##v## in the general case.

As it turns out, you've gotten lucky, because the error involved in using the non-relativistic formula for kinetic energy here is exactly compensated for by another error, which is to treat ##r## as actual radial distance. In fact, in the coordinates in which the gravitational time dilation formula is the one you give, ##r## is not radial distance but "areal radius", i.e., ##r## is defined such that the surface area of the 2-sphere at radial coordinate ##r## is ##4 \pi r^2##. Again, the difference between this and the actual radial distance is small for a body like the Earth, but it is not for much more compact bodies.

The simplest way to obtain the exact correct formula for ##v## in this case is to observe that, as you've set up the scenario, the infalling object is falling inward at exactly escape velocity (since it started from rest at infinity), and escape velocity is ##v_{e} = \sqrt{2 G M / r}##. As it turns out, this formula, which is exact in Newtonian physics with ##r## interpreted as radial distance, remains exactly correct for the Schwarzschild solution in GR with ##r## interpreted as the areal radius. Then you just plug this ##v## into the SR relative motion time dilation formula, and it gives you a result which is identical to the GR gravitational time dilation formula.

Is this a reasonable way to understand the gravitational time dilation of a clock in a gravitational potential with respect to an inertial observer at zero gravitational potential?
By itself, not really, no. But it could be part of a reasonable way to understand it.

The problem with taking your explanation as it stands is twofold. First, you are implicitly equating the clock rate of the infalling observer with the clock rate of the inertial observer at infinity. That would be wrong even in SR, since they are in relative motion.

Second, the time dilation due to relative motion is symmetric: the infalling observer sees the "hovering" observer's clock running slow, and the "hovering" observer sees the infalling observer's clock running slow. But gravitational time dilation is asymmetric: both the "hovering" observer and the inertial observer at rest at infinity agree that the latter's clock is running faster, and they can confirm this by exchanging round-trip light signals.

In fact, exchanging round-trip light signals between the two observers is the simplest way to understand the gravitational time dilation in this case: it's just a result of gravitational redshift/blueshift. The observer at lower altitude emits light signals of a certain frequency; those light signals are redshifted when they reach the observer at higher altitude. The latter interprets this as the former's clock running slow compared to his own.

Conversely, the observer at higher altitude emits light signals of a certain frequency; those light signals are blueshifted when they reach the observer at lower altitude. The latter interprets this as the former's clock running fast compared to his own.

As for where gravitational redshift/blueshift comes from, that's a necessary consequence of conservation of energy, as Einstein pointed out. The infalling observer can be used as part of a scenario to demonstrate this. Suppose the observer comes to rest at the hovering observer's altitude, and stores all his kinetic energy in a big battery. That battery is then used to power a laser that emits a light signal back outward to the inertial observer at infinity. If there is any energy left in the light signal when it reaches infinity, we would have a perpetual motion machine--the infalling observer had zero kinetic energy when he started, so there should be zero excess energy left at the end. That means the light signal has to redshift as it travels upward, since that's how light loses energy.

All of this can be thought of in terms of "gravitational potential energy" and "kinetic energy" (the light, just like free-falling observers, can be thought of as gaining potential energy and losing kinetic energy as it rises, or the reverse as it falls). But the validity of all of this depends on the spacetime in question being of a very special kind, a stationary spacetime. Basically, this is a spacetime which has a family of observers (of which our "hovering" observer is one) who never see any change in the spacetime geometry in their vicinity--in other words, they can view the spacetime as "not changing with time". Their worldlines in turn can be taken to mark "points in space", so that we can say we have a "space" that remains the same with "time". And that is what we need for concepts like "gravitational potential energy" and "kinetic energy" to be well-defined (note that "kinetic energy" here thus has to mean specifically "kinetic energy relative to a hovering observer co-located with you").
 

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