Time Dilation & Frame of Reference: Who's Time Slows Down?

[pervect]

In the real world situation the trajectories are mirror images of each other from the point of view of the original reference frames of the two observers.

This is not true in the example.

juju

I'm not sure why you say that. Initailly, a and b are receeding from each other with some velocity v. Regardless of who accelerates, after the turnaround point, a and b are approaching with some velocity v', which can be set to be equal to the inital velocity v by accelerating the correct amount.

Either a can accelerate or b can, it doesn't matter which one as far as the relative separation goes.

 

[geometer]

How's this? Relative velocity equally affects how each observer caclulates the expected time and space distortions ascribed to the other. When only one entity undergoes acceleration, that also affects those calculations but ONLY those made by the accelerating entity in assessing the distortions ascribed to the non-accelerating one. Nothing is concluded about states of motion, as they remain perfectly relative.

While the motion still remains relative, what is concluded is that the relationship has changed - it's not the same as it was before, and further, since the acclerated observer can tell he has been accelerated he can conclude that it is his motion which has changed.

 

[ostren]

...since the acclerated observer can tell he has been accelerated he can conclude that it is his motion which has changed.

I beg to differ: a gravitational field arose that (1) exactly countered the thrust of his engines and (2) caused the other twin/entity to accelerate.

 

[juju]

The formula for distant clock rate distortions, namely dt(1+gx)/gamma, is I believe a straightforward integration of SR's Lorentz transform. So sure, it's all the same, but no, there's no need to invoke GR.

Here you are referring, I believe to the observation of the stationary clocks as seen from the moving frame. This is a valid observation only if the observer in the moving frame is measuring the tick rate of the clocks in the stationary frame. It does not relate to what the clocks are actually measuring in the rest frame nor to the actual measurement of time in the moving frame.

juju

 

[juju]

No, I cannot accept that, unless you clarify further

In a situation with no endpoints defined,which means no distance is defined, then the situation is totally symmetrical.

Here you would just have an increasing or decreasing interval. In my opinion this case is unsolvable. Both observers see exactly the same thing from their point of view.

juju

 

[ostren]

...Here you are referring, I believe to the observation of the stationary clocks as seen ...

The formula refers to the assessment of a distant clock's tick rate as reckoned from a frame that FEELS acceleration. And you're right that it's entirely subjective and in no way says anything about what is experienced by the NATIVES of any frame. As always, the natives know of no distortions whatsoever.

 

[juju]

I'm not sure why you say that.

If you consider the starting point in the real world situation they both arrive at the end back at the starting point.

In the example, this is not true. They both arrive at the end some distance from the starting point.

The math works but I don't think the situational mapping is valid.

juju

 

[ostren]

In a situation with no endpoints defined,which means no distance is defined, then the situation is totally symmetrical.

Here you would just have an increasing or decreasing interval. In my opinion this case is unsolvable.

What?
If you are making an assertion about length contraction, and you definitely WERE (originally), then say exactly what SPAN is being contracted. It would help immensely if you restate from scratch your scenario and your resulting conclusions. I'd just love to bicker further!

 

[juju]

What?
If you are making an assertion about length contraction, and you definitely WERE (originally), then say exactly what SPAN is being contracted. It would help immensely if you restate from scratch your scenario and your resulting conclusions. I'd just love to bicker further!

If a distance is defined in the rest frame of the earth, and for all these cases it is, then this distance is contracted when measured by the moving observer (on his instruments) traveling this distance. Therefore, less time is measured by the moving observer on his instruments. He experiences less time.

If the situation was that two space ships were approaching each other with a relative velocity v in deep space with no reference frame defined other than the ones attached to each space ship, then the distance between them could not be defined as existing in a specific reference frame . It would exist in each reference frame simultaneously. In this case there would be no solution to the paradox.

juju

 

[geometer]

I beg to differ: a gravitational field arose that (1) exactly countered the thrust of his engines and (2) caused the other twin/entity to accelerate.

If a gravitational field arose that exactly countered the thrust of his engines, then he did NOT accelerate, so he has nothing to detect. But, now the other twin/entity can detect his acceleration and make the unequivocal statement that his state of motion relative to the first frame has changed.

 

[juju]

Hi,

this may help clarify my position. I hope so.

Consider a distance with fixed endpoints in the rest frame of observer A, with observer A near one end and observer Z at the far end.. Observer B is traveling parallel to this distance at a fixed velocity relative to the rest frame of observer A and at a known perpendicular distance.

When observer B crosses the perpendicular to A, they synchronize their clocks via light signals. At the same time observer A sends a light signal to observer Z to synchronize clocks with him.

When, B crosses the perpendicular to Z, what is the elapsed time relation between their clocks.

Here observer B measures less time on his clock than observers A and Z (who measure the same time) measure on their clocks.

According to my point of view this can be calculated directly from length contraction and velocity. There is no acceleration involved, and no round trip involved.

This occurs because the defining of the distance in a particular rest frame breaks the symmetry inherent in SR.

juju

 

[Doc Al]

This occurs because the defining of the distance in a particular rest frame breaks the symmetry inherent in SR.

Both frames measure the same "thing"--the time it takes B to travel from A to Z. Since the two frames are in relative motion, they will of course make different measurements.

But, since no acceleration is involved, the usual SR effects are completely symmetric. Each frame will say that clocks in the other frame are operating slow, etc.

Of course, this has little to do with the "twin paradox".

 

[juju]

Both frames measure the same "thing"--the time it takes B to travel from A to Z. Since the two frames are in relative motion, they will of course make different measurements.

But, since no acceleration is involved, the usual SR effects are completely symmetric. Each frame will say that clocks in the other frame are operating slow, etc.

Of course, this has little to do with the "twin paradox".

This has everything to do with the twin paradox and the breaking of symmetry. The distance traveled, the endpoints, are defined on the Earth's reference frame no matter what that distance is, Without this there is no symmetry breaking. Acceleration does not break the mathematical symmetry of SR, it just regauges it.

Another thing. You must understand the difference between time experienced (measured) in a specific frame,
and the time observed by counting the tick rate in some other frame. They are different concepts.

juju

 

[Doc Al]

This has everything to do with the twin paradox and the breaking of symmetry. The distance traveled, the endpoints, are defined on the Earth's reference frame no matter what that distance is, Without this there is no symmetry breaking. Acceleration does not break the mathematical symmetry of SR, it just regauges it.

Where's the "symmetry breaking"? In your example, the two frames see each other's clocks as being slow--total symmetry. In the "twin paradox" one twin really does age less--the symmetry is broken.

 

[juju]

Where's the "symmetry breaking"? In your example, the two frames see each other's clocks as being slow--total symmetry. In the "twin paradox" one twin really does age less--the symmetry is broken.

The symmetry breaking is in the FACT that the distance is defined in one reference frame not the other. This means the distance traveled is seen as different from the point of view of the two observers.

If the interval is undefined as being in one reference frame or the other, as in another example I have posted here, then no solution is possible.

juju

 

[ostren]

If a gravitational field arose that exactly countered the thrust of his engines, then he did NOT accelerate, so he has nothing to detect. But, now the other twin/entity can detect his acceleration and make the unequivocal statement that his state of motion relative to the first frame has changed.

If you want to deny the very essence of Relativity, then that's your prerogative. Each observer can rightly claim to be stock still in ANY scenario, bar none. All physical motion, bar none, is relative, with zero favoritism attached. No differentiation can be made as to what is "unequivocal" and what isn't, in terms of "state of motion". The proper conclusion is that each observer can claim his own proprietary world, in which HE is stock still: and the behavior of light rays corroborates that claim to utter perfection.

 

[geometer]

If you want to deny the very essence of Relativity, then that's your prerogative. Each observer can rightly claim to be stock still in ANY scenario, bar none. All physical motion, bar none, is relative, with zero favoritism attached. No differentiation can be made as to what is "unequivocal" and what isn't, in terms of "state of motion". The proper conclusion is that each observer can claim his own proprietary world, in which HE is stock still: and the behavior of light rays corroborates that claim to utter perfection.

Well, let me quote from an articleon the Twin Paradox in the "Encyclopedia of Physics, 2nd Edition," by Joseph Dreitlein of the University of Colorado, Department of Physics:

"...each twin can absolutely detect with accelerometers a different history of acceleration."

And - "Since in the inertial system in which A is at rest [the earthbound twin], the acceleration of B implies that v(t) is somewhere non-zero..."

Accelerations can be absolutely detected and an unequivocal conclusion drawn that the state of motion of the accelerated frame has changed.

 

[ostren]

Well, let me quote from an article on the Twin Paradox in the "Encyclopedia of Physics, 2nd Edition," by Joseph Dreitlein of the University of Colorado, Department of Physics:

"...each twin can absolutely detect with accelerometers a different history of acceleration."

And - "Since in the inertial system in which A is at rest [the earthbound twin], the acceleration of B implies that v(t) is somewhere non-zero..."

Accelerations can be absolutely detected and an unequivocal conclusion drawn that the state of motion of the accelerated frame has changed.

An "unequivocal conclusion drawn that the state of motion of the accelerated frame has changed"?? Dreitlein never said that. Why don't you quote him more fully, eh? What's the defintiion of "v(t)" for instance? The guy never voiced what YOU are claiming -- not in the passage you've cited, anyway!

Here's the truth:

Two spacecraft s, out in deep intergalactic space where gravitation is absent/negligible, drift away from one another at 50 MPH. Then one of them fires an exhaust jet and slows down. Now their relative speed is 40 MPH. Craft A can "make the unequivocal statement that his state of motion has changed" (your words) vis-a-vis Craft B. Yet Craft B can also "make the unequivocal statement that his state of motion has changed" vis-a-vis Craft A. Ok, you might say, but that's AFTER the acceleration... what about DURING the acceleration, you might ponder. During, it's the same old hackneyed tale: the one that fired his engine remained stock still in His World for the duration, and the other guy remained stock still in His World for the duration. A spurious gravitational force arose which nullified the one's engine thrust, is all. How you define the phrase "state of motion relative to" might be at the heart of this delightful quibble.

If you think you have a valid authority, quote it more completely. I don't see any accord with your assertions in the cited passage.

 

[geometer]

Ok - Here are the full quotes, and the quoted definition of v(t).

With respect to V(t) Dreitlein says - " The time t [refering to the differential in the integral of the time dilation equation] is the synchronized time read by clocks in a chosen inertial frame, and v(t) records the history of the velocity of the moving clock measured in this frame."

The full quotes from my previous post:

"To mitigate the force of the paradox, it might be remarked that each twin can absolutely detect with accelerometers a different history of acceleration."

"Since in the inertial reference system in which A is at rest, the acceleration of B implies that v(t) is somewhere non-zero, the clock B will necessarily measure a shorter lapse of time between D and R [Departure and Return] than A whose velocity
v(t) is always zero; see Eq 1."

Note that in the last quote Dreitlein uses v(t) for both B's velocity and A's velocity.

My conclusion is a restatement of Dreitlien's assertion that "...the acceleration of B implies that v(t) is somewhere non-zero..." In combination with his statement that A and B can absolutely detect accelerations, this certainly seems to me an unequivocal statement that due to the acceleration of B we are able to say that its state of motion of has changed.

 

[ostren]

Thanks, I VERY MUCH appreciate your efforts. but there's still not enough of the passage to see what he's driving at. His references to a "v(t)" are simply the relative velocity of the clock that is seen to be moving, from A's vantage. From B's vantage, there is a v(t) as well, describing A's velocity. So where's the beef? Regardless of accelerations, v(t) will be non-zero simply by virtue of the fact that B departed (and later returned)! So where's the beef? A's velocity is not zero in any absolute sense! Why does the bozo say, "clock B will necessarily measure a shorter lapse of time between D and R [Departure and Return] than A, whose velocity v(t) is always zero"?

Can you scan the article and post the image? I'll tear it to pieces, I swear. Why does the bozo say, "To mitigate the force of the paradox, it might be remarked that each twin can absolutely detect with accelerometers a different history of acceleration"? Mitigation?? Force of the paradox?? There are only four fundamental forces and the "Force of the Paradox" is not one of them!

The best approach is to realize that the Twin Paradox is never -- I repeat, NEVER -- resolved by concluding that one twin "really truly" moved, while the homebound twin "really truly" stayed stationary... that would be some childish nonsense! Post the whole article.

And you'll see in http://www.sysmatrix.net/~kavs/kjs/addend4.html that the spacebound twin can be adjudged stock still throughout; and there's still no contradiction.

 

[ostren]

...My conclusion is a restatement of Dreitlien's assertion that "...the acceleration of B implies that v(t) is somewhere non-zero..." In combination with his statement that A and B can absolutely detect accelerations, this certainly seems to me an unequivocal statement that due to the acceleration of B we are able to say that its state of motion of has changed.

SCENARIO. One clock (A) is plummeting to Earth from the heavens above; while another clock (B) is some distance below it, on board a rocket ship that has left the launch pad and is now using controlled burns to stay hovering motionless just above the ground. The accelerometer says that B is the one undergoing acceleration. Is clock B moving? is its "state of motion" changing, hmmm??

 

[ostren]

Your analysis is fine, but realize that you still invoke acceleration ... she must change inertial frames. That's acceleration.

Sure, it's the same thing essentially, BUT the math is considerably simpler the way I laid it out http://www.sysmatrix.net/~kavs/kjs/addend4.html. If conventional accelerations were used then they must be finite in duration, and the clock reading assessments would then vary all over the gamut, depending on where exactly in the journey the accelerations are started and stopped... isn't that right??

Maybe there's a way to incorporate (hypothetical) "instantaneous" accelerations and arrive at the same figures, but I am not aware of any, Can you supply such a treatise?

 

[Doc Al]

Each observer can rightly claim to be stock still in ANY scenario, bar none. All physical motion, bar none, is relative, with zero favoritism attached. No differentiation can be made as to what is "unequivocal" and what isn't, in terms of "state of motion". The proper conclusion is that each observer can claim his own proprietary world, in which HE is stock still: and the behavior of light rays corroborates that claim to utter perfection.

It's not clear what you are talking about. In special relativity, the topic of this thread, all inertial frames are equally acceptable. But an accelerated frame can certainly be unambiguously distinguished.

Of course general relativity allows one to take the viewpoint of any frame, regardless of acceleration.

 

[Doc Al]

Sure, it's the same thing essentially, BUT the math is considerably simpler the way I laid it out http://www.sysmatrix.net/~kavs/kjs/addend4.html. If conventional accelerations were used then they must be finite in duration, and the clock reading assessments would then vary all over the gamut, depending on where exactly in the journey the accelerations are started and stopped... isn't that right??

The kind of analysis you provide in your "humble treatment" is standard fare. If I'm not mistaken, N. David Mermin gives a similar treatment of the twin paradox in his classic book "Space and Time in Special Relativity".

 

[ostren]

.. But an accelerated frame can certainly be unambiguously distinguished.

That's a balk at my assertions, "Each observer can rightly claim to be stock still in ANY scenario, bar none. All physical motion, bar none, is relative, with zero favoritism attached. No differentiation can be made as to what is "unequivocal" and what isn't, in terms of "state of motion". The proper conclusion is that each observer can claim his own proprietary world, in which HE is stock still: and the behavior of light rays corroborates that claim to utter perfection."

The keyword is "motion"; look for it in this thread and you'll see it's central. When an entity undergoes acceleration and a G-force felt, then yes, his frame is "unambiguously distinguished"; but nothing is garnered about his motion. Symmetry with an unaccelerated frame is broken, aye. But the fact remains that under Rel*a*tiv*ity, each observer rightly claims to be stock still in his own proprietary frame/platform, ALL the time.

Simple spinning is not considered motion, for the purposes of the above truisms.

 

[ostren]

The kind of analysis you provide in your "humble treatment" is standard fare. If I'm not mistaken, N. David Mermin gives a similar treatment of the twin paradox in his classic book "Space and Time in Special Relativity".

Is that your admission that no, you cannot supply such a treatise; in answer to my:

Maybe there's a way to incorporate (hypothetical) "instantaneous" accelerations and arrive at the same figures, but I am not aware of any, Can you supply such a treatise?

I'll take it as such.

As for ND Mermin, his book isn't in my library -- and anyway, your operative words were "if I'm not mistaken..". Maybe I can get hold of a copy. Hey, I never claimed that my humble treatment is unique... but I certainly didn't plagiarize! It's standard fare because it's right, yo! My treatment is also most eloquent and concise.

 

[kawikdx225]

That's a balk at my assertions, "Each observer can rightly claim to be stock still in ANY scenario, bar none. All physical motion, bar none, is relative, with zero favoritism attached. No differentiation can be made as to what is "unequivocal" and what isn't, in terms of "state of motion". The proper conclusion is that each observer can claim his own proprietary world, in which HE is stock still: and the behavior of light rays corroborates that claim to utter perfection."

The keyword is "motion"; look for it in this thread and you'll see it's central. When an entity undergoes acceleration and a G-force felt, then yes, his frame is "unambiguously distinguished"; but nothing is garnered about his motion. Symmetry with an unaccelerated frame is broken, aye. But the fact remains that under Rel*a*tiv*ity, each observer rightly claims to be stock still in his own proprietary frame/platform, ALL the time.

Simple spinning is not considered motion, for the purposes of the above truisms.

Are you implying that after the twin paradox experiment both twins will be the same age??

We know the traveling twin will be younger so how do you explain this?

 

[kawikdx225]

SCENARIO. One clock (A) is plummeting to Earth from the heavens above; while another clock (B) is some distance below it, on board a rocket ship that has left the launch pad and is now using controlled burns to stay hovering motionless just above the ground. The accelerometer says that B is the one undergoing acceleration. Is clock B moving? is its "state of motion" changing, hmmm??

Your accelerometer is broken. If you are hovering motionless above the ground then your accelerometer will measure 0m/s²

You ask: Is clock B moving
answer: with respect to what?

You ask: is it's state of motion changing
answer: NO, the accelerometer should read 0, if it reads anything other than 0 then it's broken.

 

[ostren]

Are you implying that after the twin paradox experiment both twins will be the same age??

Of course I'm not!

We know the traveling twin will be younger so how do you explain this?

I explain it http://www.sysmatrix.net/~kavs/kjs/addend4.html

 

[ostren]

Your accelerometer is broken. If you are hovering motionless above the ground then your accelerometer will measure 0m/s²

Well folks start throwing around words like accelerometer, of which I have no direct knowledge, but I've got to assume that such a device merely measures the G-force present and from that infers acceleration. There is no other possible configuration of such a device that comes to mind: it measures the G-force. So, my clock B would feel such a force, yet my clock A would most definitely NOT.

You ask: Is clock B moving?
You ask: is it's state of motion changing?

Those were rhetorical questions addressed to geometer to get him to back off his latest claims implying that unequivocal statements can be asserted about "moving" and "state of motion", based on accelerometer readings.

You'd need to review the back-and-fro between geometer and me, this thread, to appreciate exactly the context.