# Does acceleration cause time dilation?

They are equivalent: neither acceleration nor gravity cause time dilation.
You can't be serious can you? Its a well known fact of general relativity that gravitational time dilation occurs in both accelerated frames and in gravitational fields. Einstein himself proved this and it was observed back as far as the early 60's. The GPS system depends on it.

Pete

As in my previous thread, I'm trying to eliminate the velocity aspect and only focus on gravity versus accelertion and their effect on time dilation.
When analysing a complex issue you should start with the simplest possible thought experiment and make it more general when you have a better handle on the issue.

Comparing a model of a planet with gravitation and rotation to a turntable is not the simplest model. It is better to compare a model of a planet with gravitation and no rotation with with a rotaing turntable with gravitation. That way, you can isolate and compare gravitaion effects with rotation effects.

This would mean that the GR "equivalency principle" for gravity and acceleration doesn't apply to time dilation?
The equivalence principle does apply to time dilation, but the issues are subtle.

That wasn't my intention, I was only interested in the near 1 g of gravitational field strength at the equator. The speed was only mentioned so I could show another example of a clock moving at the same speed and experiencing near 1 g of acceleration.
The answer is that the clock #1 on the Earth experiences additional time dilation due to gravitational potential (not acceleration) relative to clock #2 on the spacestation.

OK, but doesn't acceleration on clock #2 at the perimeter of the space station have the same component of time dilation due to acceleration as clock #1 does due to gravitational potential?
The acceleration on clock #2 at the perimeter of the space station contibutes no component of time dilation to clock #2. The proof, is in this experiment in the FAQ of this forum (Experimental basis of Special Relativity) titled "The Clock Postulate" http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html#Clock_Hypothesis

This is where all the hand waveing stops. We cannot argue with what is actually measured in experiments. The FAQ item states:

" The clock hypothesis states that the tick rate of a clock when measured in an inertial frame depends only upon its velocity relative to that frame, and is independent of its acceleration or higher derivatives. The experiment of Bailey et al. referenced above stored muons in a magnetic storage ring and measured their lifetime. While being stored in the ring they were subject to a proper acceleration of approximately $$10^{18}g$$ where (1 g = 9.8 m/s2). The observed agreement between the lifetime of the stored muons with that of muons with the same energy moving inertially confirms the clock hypothesis for accelerations of that magnitude."

In other words, the time dilation of muons experiencing 10,000,000,000,000,000,000 times the surface gravitational acceleration of the Earth, did not experience any additional time dilation above the time dilation accounted for by the instantaneous linear velocity.

However illogical or unreasonable the result of the experiment seems, that is the facts of the case.

Dale
Mentor
You can't be serious can you? Its a well known fact of general relativity that gravitational time dilation occurs in both accelerated frames and in gravitational fields. Einstein himself proved this and it was observed back as far as the early 60's. The GPS system depends on it.
I am 100% serious. Speed differences cause time dilation in SR. In GR gravitational potential differences also cause time dilation. In neither case is the cause of time dilation the acceleration.

The GPS satellites are at a higher gravitational potential than the ground, similaraly with the Pound-Rebka experiment you refer to. Similarly in an accelerating reference frame.

Look at the expression for gravitational time dilation in the Swartzschild metric: $$\frac{t_g}{t_f} = \sqrt{1 - \frac{GM}{Rc^2}}$$ which is clearly only a function of the gravitational potential, $$\frac{GM}{R}$$, and not only a function of the gravitational acceleration, $$\frac{GM}{R^2}$$. You find a similar conclusion in a uniform field or accelerating rocket.

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Fredrik
Staff Emeritus
Gold Member
A clock in free fall at ground level ticks at the same rate as a clock on the satellite, right? Doesn't that contradict your claim that gravitational potential causes time dilation?

I am 100% serious. Speed differences cause time dilation in SR. In GR gravitational potential differences also cause time dilation. In neither case is the cause of time dilation the acceleration.
I like this line of thinking however I prefer to say that velocity (not necessarily speed) differences and non Euclidean spacetimes ('gravitational potential' is nice but how would you define 'gravitational potential' in GR?) cause time dilation. And needless to say that time dilation is always a comparison between two or more clocks, as there is no absolute notion of time.

A clock in free fall at ground level ticks at the same rate as a clock on the satellite, right? Doesn't that contradict your claim that gravitational potential causes time dilation?
Your statement is a little vague. A clock dropped dropped vertically from a couple of meters above the ground is briefly if freefall and so is a clock that is orbiting a couple of meters above the ground around some planet with no atmosphere.

Assuming you meant the latter and assuming a non rotating planet and assuming a non rotating satellite that is orbiting much higher up, then the clock rates will not necessarily be the same. The lower orbiting clock is at a lower gravitational potential (higher gravitational time dilation) and a higher orbital velocity and so it also has greater kinetic time dilation.

A object in free fall "feels no gravity" but is still subject to gravitational time dilation. The "feels no gravity" part just means that an accelerometer attached to the object would not measure any acceleration and in a free falling closed lab for example, it would appear as if there is no gravity as far as the occupents are concerned (if we ignore some tidal effects that are hard to detect in a small volume).

I like this line of thinking however I prefer to say that velocity (not necessarily speed) differences and non Euclidean spacetimes ..
I believe that non-Euclidean spacetimes are neccesary but not sufficient for gravitational time dilation. For instance Minkowski spacetime will not produce gravitational time dilation but is a non-Euclidean spacetime.
...'gravitational potential' is nice but how would you define 'gravitational potential' in GR..
The term gravitatational potential is more precisely defined in the plural, i.e.gravitatational potentials in general relativity is defined as the components of the metric tensor. They are referred to as a set of ten independant gravitational potentials. The metric tensor can then be thought of as a tensor potential (or potential tensor).

Pete

Fredrik
Staff Emeritus
Gold Member
Your statement is a little vague. A clock dropped dropped vertically from a couple of meters above the ground is briefly if freefall and so is a clock that is orbiting a couple of meters above the ground around some planet with no atmosphere.

Assuming you meant the latter and assuming a non rotating planet and assuming a non rotating satellite that is orbiting much higher up, then the clock rates will not necessarily be the same. The lower orbiting clock is at a lower gravitational potential (higher gravitational time dilation) and a higher orbital velocity and so it also has greater kinetic time dilation.

A object in free fall "feels no gravity" but is still subject to gravitational time dilation. The "feels no gravity" part just means that an accelerometer attached to the object would not measure any acceleration and in a free falling closed lab for example, it would appear as if there is no gravity as far as the occupents are concerned (if we ignore some tidal effects that are hard to detect in a small volume).
I meant that any clock in free fall should tick at the same rate as any other, but I have to admit that I don't fully understand gravitational dime dilation and I also haven't really thought this through, so maybe I'm way off. Doesn't any clock in free fall define a "default ticking rate" that we compare everything else to? (I guess you have already answered that with a no). I'm going to have to start thinking about this now.

I am 100% serious. Speed differences cause time dilation in SR. In GR gravitational potential differences also cause time dilation. In neither case is the cause of time dilation the acceleration.
That is incorrect. No offense my frien but you keep confusing the clock hypothesis with the phenomena of gravitational time dilation. The clock hypothesis refers to the notion that the rate at which a clock runs as measured locally depends only on the speed of the clock as measured locallyand not the acceleration of the clock. It does not refer to the relative rates at which two different clocks run which are not locally compared but which are seperated by a finite distance, either in a higher gravitational potential or higher up in the acclerated frame. I can then be shown, by utilizing the clock hypothesis, that the clock at z = 0 will run at a different rate than an identical clock at z = h, even though they will run at identical rates when compared locally. Each clock having, for all practical purposes, the same acceleration (even though the acceleration makes no difference). The reason? Consider an observer at rest in frame S' which is a frame of reference which is accelerating with respect to an inertial frame S. Let the origin of S and S' be coincident at t = 0. Let light be emitted from z = 0 where the coordinate clock is located, in the +z direction which is the direction in which the frame is accelerating. The light will arrive at z = h where there is an identical clock of identical construction. The light takes a finite amount of time to reach the clock as observed from both S and S'. Observers at rest in S will determine that the observer at rest at z = h in S' will be moving with respect to S with a speed v as measured in S. Since the observer at h has a clock whose rate does not depend on acceleration but merely on speed then that observer will detect a redshift in the frequency of light. Now look at this from observers at rest in S'. The clock located at z = 0 is not moving relative to the observer at z = h. Since the equivalence principle tells us that this same thing will happen in a uniform gravitational field, or in a region of a gravitational field in which the field is uniform for all practical purposes. The later has been measured in the lab with positive results. The ratio of the the two clocks are a function of $\gamma$ where, for the accelerating frame of refernce, of acceleration a, that $\gamma = (1+ az/c^2)$. For the gravitational field $\gamma = (1+ gz/c^2)$. The equivalence principle is evident here in the relation a = g.

Funny thing! Nobody has explained they physics up until now!
The GPS satellites are at a higher gravitational potential than the ground, similaraly with the Pound-Rebka experiment you refer to. Similarly in an accelerating reference frame.
I['m confused. Elsewhere you told me that Therefore gravity does not cause time dilation.. Please clarify for me.
Look at the expression for gravitational time dilation in the Swartzschild metric: $$\frac{t_g}{t_f} = \sqrt{1 - \frac{GM}{Rc^2}}$$ which is clearly only a function of the gravitational potential, $$\frac{GM}{R}$$, and not only a function of the gravitational acceleration, $$\frac{GM}{R^2}$$. You find a similar conclusion in a uniform field or accelerating rocket.
Why are you using that metric? The week form of the equivalence principle states that a uniformly accelerating frame of reference is equivalent to a uniform gravitational field. It doesn't say that any accelerating frame of reference is equivalent to the Earth's gravitational field.

Best wishes

Pete

I meant that any clock in free fall should tick at the same rate as any other, but I have to admit that I don't fully understand gravitational dime dilation and I also haven't really thought this through, so maybe I'm way off. Doesn't any clock in free fall define a "default ticking rate" that we compare everything else to? (I guess you have already answered that with a no). I'm going to have to start thinking about this now.
It is a complex and subtle issue and I am not 100% convinced I am right and there seems to quite a lot of difference in opinion in this forum. The confusion comes about I think because the equivalence principle is usually stated in terms of acceleration but I beleive the equal acceleration is coincidental because as I have shown in my other posts we can construct thought experiments where the acceleration is equal but the time dilation is different and equally we show situations where the acceleration is different but the time dilation is equal.

Another example is to consider what happens when a clock is lowered down a long mine shaft to the centre of a non rotating earth. (Assume it does not melt). If we have a small hollow spherical cavity at the center the acceleration of gravity at the centre of the planet is zero according to both Newton and Einstein. Now if you assume zero acceleration = zero time dilation you would get the wrong answer. The clock at the centre runs slower than a clock at the surface of the gravitational body. So why is the clock at the centre running slower than the clcok at the surface, when the clock at the centre has zero acceleration and zero velocity? The answer is that the gravitational potential at the centre of the Earth is not zero and is less than the potential at the surface which is why objects fall towards the centre. For a given gravitational potential there is an equivalent escape velocity. The escape velocity at the center of the Earth is not zero. The escape velocity is the "effective velocity" and you can think of the time dilation as kinetic time dilation proportional to that that "effective velocity". It is a bit like an object sitting on a table experiences acceleration even though it is not actually moving anywhere. Just by sitting there it has an effective veocity which is numerically the escape velocity which is a function of gravitational potential.

Just for info, the Newtonian acceleration inside a body of even density is not proportional to GM/R^2 but:

$$\frac{GMR_{inside}}{R_{surface}^3}$$

where the variable R(inside)<=R(surface) and R(surface) is constant.
At least I think that is right, but I could not find a handy reference to it. At the surface R(inside) = R(surface and the expression becomes equal to the regular GM/R^2.

Einstein's first derivation of gravitational time dilation is found in Jahrbuch der Radioaktivität und Elektronik (1907). Note that Einstein stated the clock hypothesis in the section labeled Space and time in a uniformly accelerated reference system. After stating the equivalence principle and just prioer to his derivation of the gravitational time dilation effect he wrote
First of all, we have to bear in mind that a specific effect of acceleration on the rate of the clocks in $\Sigma$ need not be taken into account, since they would have to be of order $\gamma^2$.
I would hazard to guess that Einstein later accepted that there is no effect at all on the acceleration of an ideal clock. With this assumption in mind Eistein derived the gravitational time dilation relation. So its erroneous to hold that the clock hypothesis implies that gravitational time dilation doesn't happen. It is a misuse of that hypothesis since this dilation effect has to do, not with the rate a single clock runs, but at the rates between two different clocks which are seperated in an accelerating frame.

Pete

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rcgldr
Homework Helper
Ok, one thing that is confusing me. Why would altitude in an accelerating rocket in open space (assume the ideal case that there is zero gravitational effects in the region the rocket is traveling through) cause any difference in time dilation effects.

It's my belief that once in a stable accelerating state, every molecule of the rocket and all attached components are moving at the exact same velocity, and experience the exact same acceleration (otherwise there would be continuous compression or expansion) (ignoring the spent fuel and the flow of unspent fuel here).

Given this, then how could the positioning of two clocks within such a rocket experience any difference in time dilation?

Ok, one thing that is confusing me. Why would altitude in an accelerating rocket in open space (assume the ideal case that there is zero gravitational effects in the region the rocket is traveling through) cause any difference in time dilation effects.
I explained the reason for the time dilation above. In essece the farther up in the rocket the longer it takes the light, as observed from an observer in an inertial frame of reference, to go from tale of the rocket to the nose of the rocket. The greater the speed the greater the redshift in the light due to the greater speed of the nose.
It's my belief that once in a stable accelerating state, every molecule of the rocket and all attached components are moving at the exact same velocity, and experience the exact same acceleration (otherwise there would be continuous compression or expansion) (ignoring the spent fuel and the flow of unspent fuel here).
That is correct.
Given this, then how could the positioning of two clocks within such a rocket experience any difference in time dilation?
Did you read the explaination of the derivation that I provided above? Was it unclear>

Fredrik
Staff Emeritus
Gold Member
Why would altitude in an accelerating rocket in open space [...] cause any difference in time dilation effects.
...
how could the positioning of two clocks within such a rocket experience any difference in time dilation?
I didn't read pmb_phy's explanation so I can't comment on that, but one way to understand this is to just draw the world lines of the front and rear of the rocket in a space-time diagram corresponding to an inertial frame (or just imagine doing it). The two world lines will not be identical in the inertial frame. If they were, the rocket would remain the same length in the inertial frame even though it's being Lorentz contracted more and more as the speed increases. So the world line of the rear must be curved more than world line of the front, making it look like the rear is "catching up" with the front. In the inertial frame, the rear is always moving faster than the front.

Now consider the fact that what a clock really measures is the integral of $\sqrt{dt^2-dx^2}$ along its world line. The contribution from the dx displacements make the total smaller, and the dx displacements are bigger on the world line of the rear, and that means less proper time. So the clock in the rear runs slower than the clock in the front.

rcgldr
Homework Helper
accelerating rocked two clocks
My intended observer would also be inside the rocket accelerating and moving at the same speed as the rocket. It this observer going to see a difference between the two clock rates?

To elminate the ever changing velocity issue, then go back to the rotating space station, clock #1 at the perimeter on one side of the space station, clock #2 on the perimeter on the opposite side of the space station. The observer is at the center of the space station. Does the observer see both clocks running at the same rate? If the observer is next to clock #1, does clock #2 "above" appear to be running at a different rate?

Ok, one thing that is confusing me. Why would altitude in an accelerating rocket in open space (assume the ideal case that there is zero gravitational effects in the region the rocket is traveling through) cause any difference in time dilation effects.

It's my belief that once in a stable accelerating state, every molecule of the rocket and all attached components are moving at the exact same velocity, and experience the exact same acceleration (otherwise there would be continuous compression or expansion) (ignoring the spent fuel and the flow of unspent fuel here).

Given this, then how could the positioning of two clocks within such a rocket experience any difference in time dilation?
A diagram illustrating the point made by Fredrik, that a continuously accelerating rocket undergoes continuous length contraction is shown here. http://www.mathpages.com/home/kmath422/kmath422.htm

Fredrik
Staff Emeritus
Gold Member
Does the observer see both clocks running at the same rate?
Yes. They have the same speed, so he sees the same time dilation effect on both of them.
If the observer is next to clock #1, does clock #2 "above" appear to be running at a different rate?
Yes. If the speed relative to the center of the space station is v, clock #2 is moving with speed 2v/(1+v2) relative to clock #1. So he sees a time dilation corresponding to that speed.

....
Yes. If the speed relative to the center of the space station is v, clock #2 is moving with speed 2v/(1+v2) relative to clock #1. So he sees a time dilation corresponding to that speed.
I don't think is quite right. Both clocks are effectively at the same altitude and moving at the same speed so they will be running at the same rate and see each other to be running at the same rate.

Dale
Mentor
The ratio of the the two clocks are a function of $\gamma$ where, for the accelerating frame of refernce, of acceleration a, that $\gamma^2 = (1+ az/c^2)$. For the gravitational field $\gamma^2 = (1+ gz/c^2)$. The equivalence principle is evident here in the relation a = g.
don't you see that we are in agreement here? As you have derived and as I have stated several times: the gravitational time dilation is a function of the gravitational potential, gz, rather than just a function of the gravity, g. "What we have here is failure to communicate"
Why are you using that metric?
because jeff's question was specifically set up as an experiment with a sea level clock and a space clock, so swarzschild is the appropriate metric
The week form of the equivalence principle states that a uniformly accelerating frame of reference is equivalent to a uniform gravitational field. It doesn't say that any accelerating frame of reference is equivalent to the Earth's gravitational field.
yes, I know. And I pointed that out to Jeff too

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Fredrik
Staff Emeritus
Gold Member
I don't think is quite right. Both clocks are effectively at the same altitude and moving at the same speed so they will be running at the same rate and see each other to be running at the same rate.
I disagree. This is a simple SR problem. The effect that makes two clocks on a rocket tick at different rates isn't present here (in this case the two world lines look identical in the inertial frame, so the rear isn't catching up with the front, or anything like that), but there is still plenty of time dilation due to the velocity difference.

I also think that even though the equivalence principle can be useful sometimes, it's usually just a cause of confusion. It was invented as a criterion that candidate theories of gravity must satisfy to be taken seriously. For example if you find a theory of gravity that doesn't predict that the ceiling ages faster than the floor, as in SR (on an accelerating rocket), the theory would be dismissed. There may be times when the equivalence principle can help you quickly find the correct result in SR from a known result in GR, but I think those times are rare. It's probably more useful when we want to push an SR calculation over to GR.

It's important to note that there are no situations when the equivalence principle must be used. The alternative theories have already been dismissed, and we're left with a theory that does predict, all by itself, that the ceiling ages faster than the floor. So we should only use it when it significantly simplifies the calculations.

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Fredrik
Staff Emeritus
Gold Member
The Equivalence principle says:

-The ceiling in a house on earth ages faster as the floor,
-The ceiling in a linear accelerating rocket also ages faster as the floor

They do so by the same amount if both g and h(eight) are the same.
If the windows are blindfolded then you can't tell the difference. You
can't tell in which room you are by comparing the clocks on the floor
and at the ceiling.
These are some very good comments, obviously, but I didn't realize when I read them the first time that they suggest an answer to a question that (I believe) was left unresolved in the discussions in recent threads: What is a homogeneous gravitational field in GR?

A "homogeneous gravitational field" must be a metric that somehow causes this effect.

That statement is rather vague and needs to be made more formal. Unfortunately I haven't figured out how to do that yet.

I don't think is quite right. Both clocks are effectively at the same altitude and moving at the same speed so they will be running at the same rate and see each other to be running at the same rate.

I disagree. This is a simple SR problem. The effect that makes two clocks on a rocket tick at different rates isn't present here (in this case the two world lines look identical in the inertial frame, so the rear isn't catching up with the front, or anything like that), but there is still plenty of time dilation due to the velocity difference.

I also think that even though the equivalence principle can be useful sometimes, it's usually just a cause of confusion. It was invented as a criterion that candidate theories of gravity must satisfy to be taken seriously. For example if you find a theory of gravity that doesn't predict that the ceiling ages faster than the floor, as in SR (on an accelerating rocket), the theory would be dismissed. There may be times when the equivalence principle can help you quickly find the correct result in SR from a known result in GR, but I think those times are rare. It's probably more useful when we want to push an SR calculation over to GR.

It's important to note that there are no situations when the equivalence principle must be used. The alternative theories have already been dismissed, and we're left with a theory that does predict, all by itself, that the ceiling ages faster than the floor. So we should only use it when it significantly simplifies the calculations.
I do not disagree with anything you say in this post, but that is not the situation I was talking about. I agree that the ceiling ages faster than the floor. When asked if clocks A and B would both run at the same rate as seen from the centre (C) we both said yes they would. In this case the perimeter where A and B are located represents the floor, while the centre represents the ceiling. Clocks A and B both run slower than clock C.

When asked if if an observer located by clock A on the perimeter, would see clock A as running at a different rate to clock B (also located on the perimeter), the answer should also be yes, because both clocks A and B are on the floor and not moving relative to each other.

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My intended observer would also be inside the rocket accelerating and moving at the same speed as the rocket. It this observer going to see a difference between the two clock rates?
As I mentioned above ...look at this from observers at rest in S'. The clock located at z = 0 is not moving relative to the observer at z = h. Since the equivalence principle tells us that this same thing will happen in a uniform gravitational field, .. So we have already addressed this above. The answer is that even though the clocks are at rest with respect to each other there is a change in the light as it moves between them.
To elminate the ever changing velocity issue, then go back to the rotating space station, clock #1 at the perimeter on one side of the space station, clock #2 on the perimeter on the opposite side of the space station. The observer is at the center of the space station. Does the observer see both clocks running at the same rate? If the observer is next to clock #1, does clock #2 "above" appear to be running at a different rate?
No. Since these clocks are at the same gravitational potential they will run at the same rate.

Pete

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don't you see that we are in agreement here?
No. In fact I can't seem to figure out what it is you're saying. What you wrote here
As you have derived and as I have stated several times: the gravitational time dilation is a function of the gravitational potential, gz, rather than just a function of the gravity, g.
is inconsistent with what you wrote above, i.e.
They are equivalent: neither acceleration nor gravity cause time dilation.
What do you mean in this last statement when you use the term time dilation. Do you mean something different than gravitational time dilation? If so then I see the communication problem.

Pete

I think all Dalespam is trying to say is that gravitational time dilation is caused by gravitational potential and not gravitational acceleration.

An example I showed earlier showed that an observer in small hollow spherical cavity at the centre of the Earth, experiences no gravitational acceleration, yet is subject to gravitational time dilation due to the non zero gravitational potential there.

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