Undergrad Clock hypothesis, gravity time dilation and Equivalence Principle

[Dale]

I am currently in the process of figuring ...

There is a huge gulf between this and claiming that you believe that all four scenarios are the same and have no proper acceleration. If you do not have the math figured out to justify such a conclusion and if you have not read a peer reviewed paper justifying that conclusion then it is a completely baseless claim that does not belong on PF anywhere, and particularly not as a thread hijack in somebody else's thread.

 

[jartsa]

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.

Let me defend signal delay as one of the reasons for different apparent clock rates just a little bit more:

In those explanations that are explanations in an inertial frame there is an increasing signal delay.

And in those explanations that are explanations in the non-inertial frame of the spaceship there is no increasing signal delay. And the explanation in that frame is gravitational time dilation.

It is unlikely that an astronaut would give gravitational time dilation as the explanation, so his explanation is an explanation in an inertial frame, in which case the explanation involves an increasing signal delay.

 

[PeterDonis]

The spaceship accelerated to high speed, and then it stopped accelerating - maybe I forgot to mention that.

Yes, you did. But if this is the case, then your scenario is underspecified, because you have not given the stopping condition: how each spaceship knows when to stop accelerating. Or, equivalently, you have not said whether the spaceships end up at rest relative to each other after they have both stopped accelerating, or whether they are moving relative to each other after they have both stopped accelerating.

it's just a boost from one frame to another frame

This would appear to indicate that the spaceships should be at rest relative to each other after they have stopped accelerating. But you should state that explicitly.

Assuming that is the case, then the following will be true:

(1) The two spaceships must both stop accelerating when they have the same velocity relative to their original launch points.

(2) The events at which the two spaceships stop accelerating will be simultaneous, in the inertial frame in which they are both at rest after they have stopped accelerating. (Note that this implies that these events will not be simultaneous in the original inertial frame, the one in which the launch points are at rest--the front spaceship stops later than the rear spaceship in this frame.)

(3) The clock readings on the two spaceships will be different when they stop accelerating (even though, as above, the two stopping events are simultaneous in the final inertial frame). The front clock will read an earlier time than the rear clock (assuming they started out synchronized before the period of accelerating), by an amount that depends on the time spent accelerating, or, equivalently, on the final velocity (more precisely, the final gamma factor) relative to the original launch points.

 

[jartsa]

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?

Oh I just noticed that I need to draw some marks on the world lines, and corrcet placement of the marks takes care of time dilation of the clocks, even differential time dilation of clocks that accelerate slightly differently. Well, the drawing does not 'stop working' at high speeds then.

 

[Raymond Potvin]

Hi Guys,

What about using the mind experiment of the light clock to understand the way light travels in an accelerated one? We know it will take more time for light to travel between the mirrors of a clock considered to be moving compared to one that is considered to be at rest, so if we assume it is accelerating, we can simply conclude it will take more and more time. Now, if we have two clocks on the floor of the ship, and if we put one higher than the other, it will suffer the same acceleration so it will suffer the same time dilation for the observer at rest, which is different from what happens when we do the same thing with two clock on earth, because with gravitation, the acceleration varies with height. Nevertheless, the light exchanged between two detectors placed at different heights on the ship should still appear blue shifted when it travels downwards, and red shifted when it travels upward, because the relative speeds of the two detectors will not be the same at emission and at detection. So we seem to have two different phenomenon here: one related to the way light travels both ways in an accelerated light clock and one related to the way it travels one way in an accelerated ship. The two phenomenon are linked though, because we can replace the two detectors by two mirrors and get a light clock that will give the observer at rest the same time dilation data.

 

[Janus]

Hi Guys,

What about using the mind experiment of the light clock to understand the way light travels in an accelerated one? We know it will take more time for light to travel between the mirrors of a clock considered to be moving compared to one that is considered to be at rest, so if we assume it is accelerating, we can simply conclude it will take more and more time. Now, if we have two clocks on the floor of the ship, and if we put one higher than the other, it will suffer the same acceleration so it will suffer the same time dilation for the observer at rest,

If the distance between the two clocks is constant as measured from the ship, then it won't be according to our observer "at rest". As the ship increases speed relative to this observer, the ship must also be undergoing an increasing Length contraction and the distance between the two clocks decreases with time. But this means that the lead clock can't be moving as fast at any given moment as the trailing clock, and thus according to the rest observer, not exhibiting the same time dilation factor.

 

[jartsa]

Hi Guys,

What about using the mind experiment of the light clock to understand the way light travels in an accelerated one? We know it will take more time for light to travel between the mirrors of a clock considered to be moving compared to one that is considered to be at rest, so if we assume it is accelerating, we can simply conclude it will take more and more time. Now, if we have two clocks on the floor of the ship, and if we put one higher than the other, it will suffer the same acceleration so it will suffer the same time dilation for the observer at rest, which is different from what happens when we do the same thing with two clock on earth, because with gravitation, the acceleration varies with height. Nevertheless, the light exchanged between two detectors placed at different heights on the ship should still appear blue shifted when it travels downwards, and red shifted when it travels upward, because the relative speeds of the two detectors will not be the same at emission and at detection. So we seem to have two different phenomenon here: one related to the way light travels both ways in an accelerated light clock and one related to the way it travels one way in an accelerated ship. The two phenomenon are linked though, because we can replace the two detectors by two mirrors and get a light clock that will give the observer at rest the same time dilation data.

Let's make 100 short pulses of light bounce between the floor and the ceiling of an static spaceship. The light pulses are distributed evenly.
At any moment 50 pulses are heading down and 50 pulses are heading up.

What happens when the spaceship starts to accelerate?

Observer at rest sees the number of pulses that are going forwards increasing. Let's say that at some moment 60 pulses are heading up and 40 pulses are heading down according to said observer.

And at that same moment an observer at the nose of the spaceship is saying that the clock consisting of all those pulses is 10 pulses late.

 

[Raymond Potvin]

If the distance between the two clocks is constant as measured from the ship, then it won't be according to our observer "at rest". As the ship increases speed relative to this observer, the ship must also be undergoing an increasing Length contraction and the distance between the two clocks decreases with time. But this means that the lead clock can't be moving as fast at any given moment as the trailing clock, and thus according to the rest observer, not exhibiting the same time dilation factor.

Hi Janus,
Length contraction is needed to explain the null result of the MM experiment, but it doesn't change the behavior of light clocks. The horizontal arm of the MMx is a light clock, and light takes more time between the mirrors if the apparatus is moving wrt another apparatus at rest. Light would only take more time if the apparatus did not contract. Since it is a speed phenomenon, contraction should happen to the moving ship as a whole, so both clocks should contract at the same rate, and the distance between them too, which means that all the clocks aboard should exhibit the same time dilation factor whatever their original length.

 

[Raymond Potvin]

Let's make 100 short pulses of light bounce between the floor and the ceiling of an static spaceship. The light pulses are distributed evenly.
At any moment 50 pulses are heading down and 50 pulses are heading up.

What happens when the spaceship starts to accelerate?

Observer at rest sees the number of pulses that are going forwards increasing. Let's say that at some moment 60 pulses are heading up and 40 pulses are heading down according to said observer.

And at that same moment an observer at the nose of the spaceship is saying that the clock consisting of all those pulses is 10 pulses late.

Hi Jartsa,
If your observer is at rest behind the ship for instance, what he will observe is a composite increasing redshift if the ship starts to accelerate: the one from standard doppler effect, plus the one from time dilation. What your observer inside the ship will observe is only doppler effect: if he is with the clock at the bottom of the ship for instance, the clock at the top will appear blue shifted, because he will have had the time to get speed during the time the light from the top clock will have taken to reach him.

 

[Dale]

Since it is a speed phenomenon, contraction should happen to the moving ship as a whole, so both clocks should contract at the same rate,

But as @Janus pointed out the clocks don’t move at the same speed. There isn’t a single speed of the “ship as a whole”.

 

[jartsa]

Hi Jartsa,
If your observer is at rest behind the ship for instance, what he will observe is a composite increasing redshift if the ship starts to accelerate: the one from standard doppler effect, plus the one from time dilation.

Yes. But my observer is only interested in counting how many pulses are heading up and how many are heading down.By the way, anything that delays a light pulse can be used to change how many pulses are heading up and how many ate heading down. For example a block of glass that we put at the path of upwards heading pulses can be used. Two observers see a redshift and a blueshift when the glass is being inserted ... No actually both observers see a redshift, and then a blueshift back to normal. When a block of glass is moved from one light path to the other light path, then two observers on the two paths see a redshift and a blueshift, and then a blueshift and a redshift back to normal. Increasing delay - redshift. Decreasing delay - blueshift. Constant delay - no shift.

 

[Raymond Potvin]

Yes. But my observer is only interested in counting how many pulses are heading up and how many are heading down.

You're right then. Do you see a problem with that?

But as @Janus pointed out the clocks don’t move at the same speed. There isn’t a single speed of the “ship as a whole”.

Hi Dale,
Janus said that the clocks would have different speeds because the ship was contracting constantly, so let's assume that the two ends would be contracting at the same rate towards one another. It means that the rear end would be accelerating a bit faster than the front one, so it would suffer a bit more time dilation wrt an observer at rest, but both clocks would be getting closer to one another with time at the same rate, so they should both appear to be dilated the same wrt one another, which is still a bit different from what happens with two clocks placed at two different heights on earth.

 

[Janus]

You're right then. Do you see a problem with that?

Hi Dale,
Janus said that the clocks would have different speeds because the ship was contracting constantly, so let's assume that the two ends would be contracting at the same rate towards one another. It means that the rear end would be accelerating a bit faster than the front one, so it would suffer a bit more time dilation wrt an observer at rest, but both clocks would be getting closer to one another with time at the same rate, so they should both appear to be dilated the same wrt one another, which is still a bit different from what happens with two clocks placed at two different heights on earth.

Your confusing the two frames. In the "rest" frame or inertial frame that the rocket is accelerating with respect to, the lead and trailing clocks have different accelerations and thus different velocities with respect to the rest frame and thus have a different time dilation than each other as measured from this frame.

The apparent closing speed between the two clocks has no bearing on this, nor can it be applied to the ship frame, because in that frame, the distance between the clocks remains a constant. Reciprocal time dilation applies between two clocks when the two clocks are in relative motion with respect to each other as measured by the clocks, when they compare their respective tick rates, It does not apply according to a third frame for which both clocks are moving. That frame will only measure time dilation in each clock according to its relative motion with respect to that frame.

 

[Raymond Potvin]

Your confusing the two frames. In the "rest" frame or inertial frame that the rocket is accelerating with respect to, the lead and trailing clocks have different accelerations and thus different velocities with respect to the rest frame and thus have a different time dilation than each other as measured from this frame.

You seem to consider that the two clocks would not be accelerating at the same time, maybe that the one closer to the rear would accelerate sooner than the one at the top. If that's what you mean, can you explain to me the underlying mechanism.

 

[Janus]

You seem to consider that the two clocks would not be accelerating at the same time, maybe that the one closer to the rear would accelerate sooner than the one at the top. If that's what you mean, can you explain to me the underlying mechanism.

Imagine The accelerating rocket is 10m long as measured in its own frame. It is under an acceleration such that the length remains 10 m as the rocket measures its own length.
You have an inertial frame which, during one moment (t1=0), the rocket has an instantaneous velocity of 0.( the rocket doesn't start accelerating at this point,as it has already been accelerating up till now.). At that moment, this frame measures the rocket to be 10 m long. The rocket's velocity is increasing with time from this instant on. So let's say that at time t2= 0+T after this moment, the rear of the rocket is moving at 0.866c relative to this frame, and the rear of the rocket has traveled a distance of x since t1. But now the rocket is no longer 10 m long a measured in our inertial frame but closer to 5 meters long. This means that the front of the rocket is X+5 m from where the rear of the was at t1. But it was 10 m from the rear of the rocket at t1. So between t1 an t2 the rear of the rocket has traveled a distance of X and the front a distance of X-5. In the same amount of time the front has traveled a shorter distance than the rear. This means that it had to have a lower average velocity as measured from the inertial frame and an lower average acceleration. This means that that a clock at the front will show less time dilation than one in the rear or the time T measured in the inertial frame.

 

[Dale]

both clocks would be getting closer to one another with time at the same rate, so they should both appear to be dilated the same wrt one another,

No. That is not how time dilation works. It does not depend on the rate that the clocks are getting closer to one another. It depends on the speed in a given inertial frame. They are not traveling at the same speed in any inertial frame. And you cannot naively apply rules from inertial frames to a non inertial frame.

 

[Raymond Potvin]

This means that that a clock at the front will show less time dilation than one in the rear or the time T measured in the inertial frame.

Firstly, if the contraction happens equally along the ship, it would add some speed to the rear end and subtract the same amount to the front one. That's what I was saying in my previous message, and that would effectively give less time dilation to the rear clock for the observer at rest. Now what I'm trying to do is test the equivalence principle inside the ship. If inertial acceleration is equivalent to gravitational acceleration, the clock at the front should suffer less time dilation than the one at the rear for an observer aboard the ship, and it doesn't seem to be the case.

 

[Janus]

Firstly, if the contraction happens equally along the ship, it would add some speed to the rear end and subtract the same amount to the front one. That's what I was saying in my previous message, and that would effectively give less time dilation to the rear clock for the observer at rest. Now what I'm trying to do is test the equivalence principle inside the ship. If inertial acceleration is equivalent to gravitational acceleration, the clock at the front should suffer less time dilation than the one at the rear for an observer aboard the ship, and it doesn't seem to be the case.

But that is exactly the case. If you are at the rear of the clock watching the front clock, you would see it running fast, and if you were in the front, you would see the rear clock running slow. The larger the distance between the clocks, the greater the tick rate difference. This is exactly equivalent to two clocks at different heights in a gravity field.

 

[Dale]

add some speed to the rear end ... that would effectively give less time dilation to the rear clock

You have this backwards

If inertial acceleration is equivalent to gravitational acceleration, the clock at the front should suffer less time dilation than the one at the rear for an observer aboard the ship, and it doesn't seem to be the case.

Inertial acceleration is indeed equivalent to gravitational acceleration. Locally (no tidal effects), the clock at the lower potential runs slower by an amount determined by the difference in the potential.

 

[Raymond Potvin]

But that is exactly the case. If you are at the rear of the clock watching the front clock, you would see it running fast, and if you were in the front, you would see the rear clock running slow. The larger the distance between the clocks, the greater the tick rate difference. This is exactly equivalent to two clocks at different heights in a gravity field.

We would see the other clock running at a different pace, but would it really do so or would it only be a kind of doppler effect due to the fact that the speed of our clock increases in the time light takes to reach it? When two clocks are in relative motion, those two effects add up to give what we call the relativistic doppler effect. Don't they add up when we adjust the signal of the GPS clocks from the ground?

 

[jartsa]

You're right then. Do you see a problem with that?

No problem. The time dilation of that longitudinal light clock consists of two time delays. We may call the time delays Doppler shifts too. A Doppler shift is a quite large effect, while time dilation is quite small effect, at low speeds, and so is length contraction. How small effect is then such time dilation that is caused by length contraction. It's a very very very small effect at low speeds.

 

[Raymond Potvin]

You have this backwards

Right! Thanks! :0)

Inertial acceleration is indeed equivalent to gravitational acceleration. Locally (no tidal effects), the clock at the lower potential runs slower by an amount determined by the difference in the potential.

I still don't see where the potential difference lies. For an observer inside the ship, the two clocks are accelerated at the same rate, whereas in a gravitational field, they are not...

...wait a minute, I think I get something: time dilation applies to moving clocks, and those in the ship are accelerating, so they are necessary dilated more when they receive the light than when they send it, which means that they always see the other clock as less dilated. Is that it?

If so, it's still a bit different than with gravitational acceleration, where it is always the higher clock that is less dilated.

 

[Dale]

I still don't see where the potential difference lies.

Locally the PE = mgh, so the potential is gh. It doesn’t matter if g is the gravitational acceleration or the fictitious force’s acceleration in a non inertial reference frame. Either way, the time dilation depends on gh, not just g.

 

[Raymond Potvin]

If you don't mind of course, I prefer to study the way light would travel instead of studying the math. In my previous message, I was saying that the two clocks in the ship had to be more dilated at detection than at emission, and you did not comment. Do you agree?

 

[vanhees71]

How do you study the way light would travel without math? An "antimathematical" attitude towards physics is not helping in understanding the subject!

 

[Raymond Potvin]

Hi Vanhees,

I finally understood relativity out of a simulation, so I prefer those to maths. I'm actually learning JavaScript to be able to make them myself. I'll try to make one about that accelerating ship, meanwhile, here is <off topic content deleted by moderator>

 

[Dale]

I prefer to study the way light would travel instead of studying the math.

If you are uncomfortable with math like PE = mgh then you have no business studying relativity.

I was saying that the two clocks in the ship had to be more dilated at detection than at emission, and you did not comment. Do you agree?

It is neither correct nor relevant.

 

[weirdoguy]

so I prefer those to maths.

You can't learn properly physics without the maths. Period. It's like studying japanese poetry without knowing japanese.

 

[vanhees71]

It's a contradiction! You cannot simulate anything that you haven't fully understood. There's no simulation without math!

 

[PeterDonis]

You cannot simulate anything that you haven't fully understood. There's no simulation without math!

Exactly. A simulation is not an alternative to math. It's just one way of doing the math.