# Clock hypothesis, gravity time dilation and Equivalence Principle

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BilboBaggins2

## Main Question or Discussion Point

1. The Clock Hypothesis states that the rate of a clock does not depend on its acceleration but only on its instantaneous velocity. This has been experimentally verified at very high accelerations.

2. A clock in a gravitational field experiences time dilation and runs slower that one not in a gravitational field.

The clock’s rate is a function, not of the gravitational acceleration (m/sec^2); but of the gravitational potential, which is the work done to move the clock from infinity, where the field strength is effectively zero, to the clock’s location. This definition of gravitational potential works fine when the gravitational field is caused by a body like earth or the sun, where the field strength falls to zero at infinity.

3. The Equivalence Principle states that the local effects of motion in a curved spacetime (gravitation) are indistinguishable from those of an accelerated observer in flat spacetime, without exception.

These statements seem to contradict themselves:

(1) says that acceleration has no effect on a clock's rate.

(2) says that gravity slows a clock’s rate.

(3) says that acceleration and gravity are essentially the same … which means that an accelerated clock is essentially the same as a clock in a gravitational field … which means that an accelerated clock will be slowed.

But (1) states that a clock’s rate is not affected by its being accelerated!

Question 1: Where, how and why am I going wrong?

I will be most grateful for an explanation and also for the name of any reference book which explains this in more detail. I have searched the web for answers but none seems to address the specific point.

Question 2: If acceleration does affect a clock's rate what is the formula for it?

I note there is no formula for the gravitational acceleration affecting a clock's rate - only a formula for the gravitational potential affecting a clock's rate.

(NB I completely understand that a moving clock runs slowly relative to a stationary observer due to its speed as given by the Special Relativity time dilation formula.)

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Gravitational acceleration does not affect clock rate, you said so yourself, it's the potential that causes the rate difference. This is also true in an accelerating rocket in the absence of gravity, the clock at the back is slower than the clock at the front even though the accelerations of the two clocks are (almost) equal. The reason is the same, difference in potential.

PeterDonis
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These statements seem to contradict themselves:

(1) says that acceleration has no effect on a clock's rate.

(2) says that gravity slows a clock’s rate.

(3) says that acceleration and gravity are essentially the same
You have left out some key factors in your statements. Including them removes the apparent contradiction. Here are restatements with the key factors included.

(1) says that proper acceleration has no effect on a clock's rate, as compared to an identical clock that is moving inertially and is momentarily at rest relative to the clock with proper acceleration.

(2) says that sitting at rest in a gravitational field slows a clock's rate, as compared to an identical clock far outside the gravitational field, which is at rest relative to the clock in the gravitational field.

(3) says that proper acceleration in free space, far away from all gravitating bodies, and sitting at rest in a gravitational field with the same proper acceleration, are essentially the same.

Dale
Mentor
Where, how and why am I going wrong?
I am surprised also. You already correctly stated and identified the resolution. The gravitational time dilation is a function of gravitational potential, not acceleration. So there is inherently no conflict with the clock hypothesis.

Let's say twins are inside an accelerating spaceship, Twin named Rear always stays at the rear, twin named Front always stays at the front.

After some time has passed, why does Front see an young Rear?

Maybe because he is looking to the distant past, because light travels a long time from the rear to the front. That seems like a plausible reason, right?

There are many question we can ask, and answer, like: If we force Front who is sitting at the front observing a young Rear at the rear, to move closer to Rear, what does he see then?

Funny, I just noticed that I don't know why Rear sees an old Front when Front is seeing a young Rear, because of the reason I gave above. I mean why is Rear reporting that a clock at the front seems to be many hours ahead of an identical clock at the rear?

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PeterDonis
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2019 Award
After some time has passed, why does Front see an young Rear?
Draw their respective hyperbolas on a spacetime diagram, and then draw the paths of light rays from Rear to Front (and from Front to Rear), and you will see why.

If we force Front who is sitting at the front observing a young Rear at the rear, to move closer to Rear, what does he see then?
A young Rear. When he finally reaches Rear, he will find that he has aged more than Rear has (assuming that they were the same age at some instant in the past when the ship was floating at rest in free space, under zero acceleration, and then the ship started accelerating after they had synchronized their clocks).

Hi. I would like to tell my thoughts about OP questions to be checked out.
(1) and (2) are OK. (1) attributes time dilation to speed of clocks in IF and (2) attributes same time dilation to gravitational potential in non inertial frame. (3) assures that (1) and (2) give the same result.
Acceleration might appear in explaining relation between IF and non IF, but it has no direct relation with time delation.

BilboBaggins2

From the answers it appears that the following statements are therefore correct:

Statement 1: An accelerating clock does not run slower (time dilation) due to its acceleration. The only effect on its rate is its velocity. An accelerating clock's rate is always equal to the rate of a co-moving clock which has the identical (instantaneous) velocity.

Statement 2: You cannot apply the Equivalence Principle to an accelerating clock - you cannot say an accelerating must run slower because it is effectively in a gravitational field.

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I think 1 is OK.
Statement 2: You cannot apply the Equivalence Principle to an accelerating clock.
Say the case of rotating disk, centrifuge gravity potential in rotating frame and tangent rotation speed in IF where clocks are in constant acceleration to the center, give the same result of time dilation.
Speed of accelerating clock in IF and gravity potential in rotating frame give equivalent explanation of time dilation here.

Dale
Mentor
An accelerating clock does not run slower (time dilation) due to its acceleration. The only effect on its rate is its velocity. An accelerating clock's rate is always equal to the rate of a co-moving clock which has the identical (instantaneous) velocity.
Pretty close. The one thing that I would add is an explicit reference to an inertial frame. E.g. "Its rate in a given inertial frame".

You cannot apply the Equivalence Principle to an accelerating clock - you cannot say an accelerating must run slower because it is effectively in a gravitational field
You most certainly can apply the equivalence principle. You just have to do it correctly. In fact, you can use the equivalence principle to derive (to first order) the gravitational time dilation.

The equivalence principle essentially states that locally the gravitational force is the same as the fictitious force in a uniformly accelerating reference frame.

Draw their respective hyperbolas on a spacetime diagram, and then draw the paths of light rays from Rear to Front (and from Front to Rear), and you will see why.
Rear - Front paths get longer as time passes. Longer in time and longer in space. And Front - Rear paths get shorter as time passes.

Now let's ask how does path of light being short explain Rear seeing Front's clock being many hours ahead? I mean can it explain the very large observed time differences?

It can not explain that. So it must the kinetic time dilation that explains most of the large time differences that Rear observes. By kinetic time dilation I mean Rear moving faster than Front, and Rear's clock running slower than Front's. (The contraction of the spaceship causes the speed difference, or the speed difference causes the contraction)

Statement 2: You cannot apply the Equivalence Principle to an accelerating clock - you cannot say an accelerating must run slower because it is effectively in a gravitational field.
In the case of constant acceleration or gravitation, say
1. Lattices of rockets are at rest in a IF. The pilots fire engines at the same time of the IF to get the same proper acceleration. SR explains the time dilation, the slower the more behind.
2. Lattices of rockets are at rest in constant gravitation field. GR explains time dilation by gravitation potential, the slower the lower.

Honestly saying, in 1 the lattice keeps expand in the direction of acceleration to rectangular one (ref. Bell's "paradox") like as in 2 the every story's ceiling expands.

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Janus
Staff Emeritus
Gold Member

From the answers it appears that the following statements are therefore correct:

Statement 1: An accelerating clock does not run slower (time dilation) due to its acceleration. The only effect on its rate is its velocity. An accelerating clock's rate is always equal to the rate of a co-moving clock which has the identical (instantaneous) velocity.

Statement 2: You cannot apply the Equivalence Principle to an accelerating clock - you cannot say an accelerating must run slower because it is effectively in a gravitational field.
Let's look at it this way:
You have two clocks at different heights in a gravity field. someone at the lower clock is watching the upper clock. Now between the time the second hand of the upper clock ticks off 1 second, it emits x waves of light at a given frequency. The person at the lower counts the same number of waves coming from the second hand as he sees it tick off one second. However, this light in traveling from the higher to lower clock gains energy, which it exhibits by increasing its frequency. Thus the person at the lower clock sees those x waves of light arrive closer together and thus also sees the upper clocks second hand take less than 1 sec to move tick off one sec. This continues to happen as long as he watches the upper clock and the upper clock slowly gets further and further and further ahead of his own. Someone at the upper clock watching the lower clock will see the opposite, the light waves climbing against gravity give up energy and shift to a lower frequency and so he sees the lower clock tick slower than his own and fall behind. This can't be attributed to the time delay caused by the light traveling from one clock to the other because that is fixed, and would only account for a fixed difference between the observed clock readings, not an ever increasing one.

Now let's switch to a constantly accelerating rocket with a clock in the nose and one in the Tail. Again, x waves of light leave the nose clock each time it ticks off one sec. That light travels toward the tail clock. Here we have to consider Doppler shift due to relative motion. Now while the instantaneous velocity of the two clocks is always the same, the velocity of the nose clock when it emits its light will be different from the velocity of the tail clock when it receives it, as it takes some non-zero time between emission and reception, during which the tail clock has changed velocity. The difference between The nose clock's velocity upon emission and the tail clock's velocity upon reception, produces a Doppler shift which results in an increase in the frequency seen by the tail clock observer. He again sees the nose clock tick faster than his own and getting further and further ahead as time goes by.
The reverse happens for the nose clock observer, he sees the tail clock tick slower than his and fall futher and further behind. Once again this can't be attributed to signal delay, but that, again would be fixed.

So this is how the equivalence principle applies to time dilation in these two scenarios.

Dale
But (1) states that a clock’s rate is not affected by its being accelerated!

Einstein seems to have made a distinction between pairs of two cases.

pair A) accelerating rocket with an observer inside and some observer on the surface of a planet "feeling gravity" (whatever feeling gravity means - a very unscientific expression IMO)

pair B) free floating in space compared to free falling in a gravity field (here, a gravity field as generated by a planet is not really exactly the same as free floating in space. The front gets accelerated more when observed from some far away inertial frame, hence an object would elongate, which Einstein knew of course, so i guess that he did not deem it that important for making the point)

Personally, i do not understand the distinction. I believe, that LOCALLY, in neither of those four cases could an "accelerometer" measure any acceleration. An observer free falling in a gravity field, floating in space, being on a planet surface or inside a rocket "accelerating", given the observer was infinitesimal small, would not "feel acceleration" at all.
Now i am not 100% certain about this, as i have to do the math first and ideally simulate it.

The idea is to place springs of various sizes inside the accelerating rocket. The largest would span all across the rocket, the front endpoint of the spring touching the front rocket endpoint. With the smallest spring's front endpoint getting ever closer to the observer at the front.
My assumption is, that the smaller the spring becomes, the less "acceleration" it will register relative to the larger springs. Hence, it would contract by a smaller percentage, approaching zero as the spring length approaches 0.

But if this was true(i might be wrong of course until i show the math), then i do not see why Einstein made the distinction.

Me and Ibix calculated the time difference between a clock at the front of a rocket and a clock at the back post acceleration in case of a near instantaneous acceleration towards 0.5c.

Unfortunately it is impossible to accelerate a rocket as shown in the diagrams. It would require for every part of the rocket along the x axis to be given a different acceleration "magically". Usually we accelerate the back and let the pressure moving at the speed of light push the rest. In the diagram however you can see, that the pressure would have to move faster than the speed of light.
That however is not a real issue, as you place "accelerators" all along the rocket, and pre-program them to go off at the right time. Still, you would not end up accelerating it perfectly as you cannot accelerate each particle of the rocket individually. Some would get pushed by others still.

But let's assume we could magically accelerate each particle of the rocket perfectly, as shown in the diagram for an instantaneous acceleration. In such a case, i believe(not certain yet) that no part would "feel" any acceleration. There would be no stress on the rocket's frame related to pressure.

But we can of course measure the time difference of clocks placed along the spring still. Meaning that the material at the front of the spring would be older vs the material at the back, post acceleration.
The closer the material is to the observer at the back of the rocket, the less of an age difference between the two points would occur, again, going towards zero age difference when approaching zero.

The acceleration profile shown in those diagrams is certainly not that of a free fall. As i mentioned above, a free fall would elongate any object if it was sufficiently elastic as the front gets accelerated faster than the back.
The acceleration profile as shown in the diagram would require you to cut a massive planet into two, then place a rocket between the two halves, such that the rocket back would be in a line with the planet surface. The front would be pointing towards the INSIDE, not up. Hence would be subject to less gravity.
Now you would have to keep the rocket in place while you move the planet's halves as far away as possible, the rocket back remaining in a line with the surfaces.

To accelerate the rocket perfectly, now you would start moving the two halves back towards the rocket. Then when close, you start moving the halves down, towards the rocket's front and away of the rocket to infinity.
This would give the rocket a perfect acceleration which could not be felt at all. The only way to tell would be to place clocks along the rocket and see them go out of sync. Locally however, you could not tell the difference.
I haven't done the math for this, but only inferred it from the acceleration profile the diagrams in the thread i pointed you to describe. So it is just an about. Getting the mass of the planets and the exact timing on when to pull the planet's halves down to infinity sounds like a job for some brainiacs. Certainly not me.

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Nugatory
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Personally, i do not understand the distinction. I believe, that LOCALLY, in neither of those four cases could an "accelerometer" measure any acceleration. An observer free falling in a gravity field, floating in space, being on a planet surface or inside a rocket "accelerating", given the observer was infinitesimal small, would not "feel acceleration" at all.

Now, I an not 100% certain...
It's a good thing that you aren't, because you are mistaken about two of the four cases. The observer in the accelerating rocket and on the surface of the planet are both undergoing proper acceleration, even if they are infinitesimally small. The tidal forces, which are the difference between the acceleration of various parts of their body, can be made arbitrarily small but that doesn't make the acceleration itself vanish.

It's a good thing that you aren't, because you are mistaken about two of the four cases. The observer in the accelerating rocket and on the surface of the planet are both undergoing proper acceleration, even if they are infinitesimally small. The tidal forces, which are the difference between the acceleration of various parts of their body, can be made arbitrarily small but that doesn't make the acceleration itself vanish.
Ok, but how does one prove that?

For example. If i imagine a spring with a mass on it, and consider this my accelerometer, then as i understand it, by design, this accelerometer works by checking the distance between the front and the back of the spring's body.
For the accelerometer, the spring in this case, to contract, it would require different parts of the spring's body to be in different inertial frames when observed by some far away observer in his rest frame.

But when we let the spring become infinitely small, both end and front are in the same rest frame. I do have troubles imagining an infinitely small spring, but if we were to assume that such a spring has both endpoints within a single frame at all times, then by design, the spring wouldn't register any acceleration at all.

So where does my thinking go wrong?

Dale
Mentor
But when we let the spring become infinitely small,
The term "locally" means sufficiently small that tidal effects are negligible. It does not mean infinitely small.

The two pairs of cases are distinct and non equivalent. Their distinctness is unrelated to tidal effects, as described by the equivalence principle.

In any case, the reading on a properly designed accelerometer does not depend on the size of the accelerometer. So in the limit that the size goes to 0 the reading is unchanged.

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The term "locally" means sufficiently small that tidal effects are negligible. It does not mean infinitely small.

The two pairs of cases are distinct and non equivalent. Their distinctness is unrelated to tidal effects, as described by the equivalence principle.

In any case, the reading on a properly designed accelerometer does not depend on the size of the accelerometer. So in the limit that the size goes to 0 the reading is unchanged.
Is a spring with a mass attached to it a properly designed accelerometer? If yes, then which spring's part reference frame would you use to measure the contraction of the spring while it is under acceleration as the various parts aren't in the same frame during the acceleration process, or at least at some point they are not.

In this https://images-fe.ssl-images-amazon.com/images/I/51Z0hryuRiL.jpg cover page of the famous text, the small lens plane can describe the small local surface of the apple almost correctly. This is similar to equivalence principle that any space time should be made flat locally. Apple surface corresponds to space time in GR and lens small surface corresponds to local IF.

We need many different lens planes in different place to describe all the surface of the apple. 2d surface of 3d body cannot be fully described by a single 2d flat plane geometry. Group of infinitely many and small local 2d surface flat planes could do it.

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The reverse happens for the nose clock observer, he sees the tail clock tick slower than his and fall futher and further behind. Once again this can't be attributed to signal delay, but that, again would be fixed.

1: Signal delay causes an apparent difference of reading of two clocks. (when signal from one clock reaches an observer late)
2: Small signal delay causes a small apparent difference of reading of two clocks.
3: Large signal delay causes a large apparent difference of reading of two clocks.
4: Increasing signal delay causes an increasing apparent difference of reading of two clocks.

Now the phrase "increasing apparent difference of reading of two clocks" is equivalent to phrase "one clock appears to tick slower than another clock"

So let's do a phrase substitution in the fourth statement. We get:
5: Increasing signal delay causes that one clock appears to tick slower than another clock.

Now I would even claim that increasing signal delay is the reason why one clock appears tick slower than another clock, when ever there exists an increasing signal delay.

Dale
Mentor
Is a spring with a mass attached to it a properly designed accelerometer?
Sure, under the right conditions.

If yes, then which spring's part reference frame would you use to measure the contraction of the spring while it is under acceleration as the various parts aren't in the same frame during the acceleration process, or at least at some point they are not.
You can use any reference frame. An object is not "in" just one frame. Every object is "in" all frames. It is moving in some and at rest in others, but it is in all of them so any frame can be used to analyze it.

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PeterDonis
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2019 Award
Rear - Front paths get longer as time passes. Longer in time and longer in space. And Front - Rear paths get shorter as time passes.
We are talking about the paths of light rays--they can't get "longer" or "shorter" since they are null.

The point is, a particular light ray leaves Rear at a particular proper time on Rear's worldline, and arrives at Front at a particular proper time on Front's worldline. If we consider successive light rays leaving Rear at equal proper time intervals on Rear's worldline, at what proper time intervals on Front's worldline do they arrive? And then consider the opposite case, of successive light rays leaving Front at equal proper time intervals on Front's worldline: at what proper time intervals on Rear's worldline do they arrive?

Answering those questions will explain the relative time dilation between Rear and Front.

it must the kinetic time dilation that explains most of the large time differences that Rear observes. By kinetic time dilation I mean Rear moving faster than Front, and Rear's clock running slower than Front's.
Rear and Front are at rest relative to each other, as seen in each of their momentarily comoving inertial frames (MCIFs), so there is no kinetic time dilation for either of them relative to the other.

To see this, draw the correct lines of simultaneity between Rear and Front's hyperbolas (hint: all of these lines pass through the origin of the diagram; their slopes increase with proper time along Rear's and Front's worldlines). You will find that the same line of simultaneity (i.e., with the same slope) matches the local line of simultaneity on both Rear's and Front's worldlines (i.e., that same line of simultaneity is orthogonal to the tangent vectors of both Rear's and Front's worldlines at the events of intersection).

Sure

You can use any reference frame. An object is not "in" just one frame. Every object is "in" all frames. It is moving in some and at rest in others, but it is in all of them so any frame can be used to analyze it.
well yes, what i meant is that not every part of the spring is in the same rest frame but in a different one during the acceleration process.

Dale
Mentor
well yes, what i meant is that not every part of the spring is in the same rest frame but in a different one during the acceleration process.
The correct thing to say is that not every part is at rest in the same frame. They are all in every frame, including each part's rest frame.

But so what? Why should that be a problem?

The correct thing to say is that not every part is at rest in the same frame. They are all in every frame, including each part's rest frame.

But so what? Why should that be a problem?
Maybe it isn't. I am currently in the process of figuring if measuring an object which does not have all its parts at rest within one frame is going to give me the proper results. I guess what i am looking for is the exact process to get the acceleration value a spring like accelerometer would display when not all of its parts are at rest in the same frame. Hence, i am looking for which spacetime locations of the spring's parts i am supposed to use in my formula computing the acceleration, such that i can check if my belief was right or not.