Absolute Velocity?

[Thabiguy]

Yes, but that's a statement about the events as measured against a reference frame. Earlier I thought you were making statements about what A "would see" as in "experience for himself".

Okay, let me be clear here. Here I write about what happens in A's non-inertial reference frame (the one in which A is always at rest):

For each observer, the other observer's clock will only be synchronized at three times: at the beginning of the journey, in the middle of the turn, and at the end of the journey.

During the outward leg, the other observer's clock will be behind and slower (thus getting increasingly more behind). During the return leg, the other observer's clock will be ahead and slower (getting increasingly less ahead, until they meet, at which point the clocks will be synchronized again).

And here I write about what A would "see" (as in, listening to the radio receiving time ticks from the other ship):

But note that that is what the observer observes (i.e. concludes from observations), not what he sees. What he would see is that radio ticks from the other ship would start arriving at a quicker pace after he makes the turn, and after the tick with the timestamp of the scheduled turn arrives, they will start coming even faster. But that is Doppler effect, not time dilation.

And I think that is what PS is asking about.

Well, it's possible, but I think he asked about what happens in the reference frame, not what can be heard on the radio. I think so because he talks of time dilation, and that is related to reference frames, not to what can be heard on the radio (for example, when another ship approaches at a relativistic speed, you hear the time ticks coming faster than yours, but the ship's clock is ticking slower than yours).

But there's no need to speculate, as PS can simply clarify what he asks about.

 

[Perpetual Student]

Originally Posted by Perpetual Student
How did that happen?
ans. It happened when the observer changed his reference frame by accelerating.


Originally Posted by Perpetual Student
Does this come about because of the acceleration?
ans. Yes. This caused a change of reference frame, thus a change in simultaneity. When this change happens over some timespan (as is the case in acceleration), the effect can be that the other clock "speeds up".


Originally Posted by Perpetual Student
If so, what is that transformation equation?
ans. You can find it here, but I would recommend you to read the whole article.

The article you show has no mention of acceleration, no mention of time speeding up and no "inverse" Lorentz transformation to acount for time speeeding up. Did you read it?

 

[RecoveringYuppy]

@Thaibiguy. Got ya. Lost you when you seemed to be claiming to be describing it solely from A's "point of view".

@P.S. Time never speeds up and there is no "inverse" Lorentz transformation. In your scenario the apparent "speed up" is due to the changing distance between A and B. There is a confusion between the time dilation and the time it takes the information about the dilation to reach a distant observer. It's easier to sort out if you imagine that there is an array of synchronized clocks, then think about what observers see as the ship passes each clock.

 

[Thabiguy]

The article you show has no mention of acceleration, no mention of time speeding up and no "inverse" Lorentz transformation to acount for time speeeding up. Did you read it?

Of course I did. And if you want someone to explain to you something that you don't understand, I would recommend that you choose a different tone.

Relativity of simultaneity means that two events that are simultaneous in one frame are not simultaneous in another. When you're asking the question, "What time does B's clock show in A's reference frame when A's clock shows T?", you have defined an event E - B's clock showing something - which in A's reference frame is simultaneous with the event "A's clock showing T", and you are asking about the time coordinate of this event in B's reference frame - that gives you the time that B's clock is showing.

When A changes velocity, his reference frame changes. This means that because of relativity of simultaneity, the original event E is no longer the one that is simultaneous with the event "A's clock showing T". Instead, it is some other event E', with a different time coordinate in B's reference frame. This means that A's change of velocity causes a change of the time that "A sees on B's clock". And this means that A's change of velocity over time (=acceleration) changes the rate of the time progression, as seen by A on B's clock.

In the mathematics of the Lorentz transformation (which I linked you to),

t' = (t - vx/c²) * γ

where γ = 1/sqrt(1-v²/c²).

Now, time dilation is slowing down of a clock in an inertial reference frame that moves with respect to the clock. You're examining Δt', compared to Δt. Because you're in a inertial reference frame, it follows that Δv = 0 and Δγ = 0 (and Δx = 0, as you're interested in the change of time between two events that are colocal in the clock's frame). From this, it simply follows that Δt' = Δt * γ. This is the relationship that you're probably familiar with. But you forgot to realize that this is only true for an inertial reference frame.

For a non-inertial, accelerating frame, Δv is no longer 0 and so the vx/c² term does not get cancelled out from the equation, as it did in the inertial case. This is the term that is relevant for the relativity of simultaneity, as it describes the change in time in response to a change in reference frame.

For this reason, it is no longer true that Δt' = Δt * γ, but the relationship becomes more complicated. The change in t' also depends on change in the term vx/c², the magnitude of which depends on x and Δv and is not constrained to always result in Δt' > Δt.

To sum up, time dilation as you understand it applies only to inertial frames. When the observer is accelerating, the observed clock can indeed "speed up". This is because of relativity of simultaneity.

 

[Perpetual Student]

@P.S. Time never speeds up and there is no "inverse" Lorentz transformation. In your scenario the apparent "speed up" is due to the changing distance between A and B. There is a confusion between the time dilation and the time it takes the information about the dilation to reach a distant observer. It's easier to sort out if you imagine that there is an array of synchronized clocks, then think about what observers see as the ship passes each clock.

Suppose the distance is not great. Say A and B move apart (accelerating simultaneously) for a few kilometers accelerating to, say, 10% the speed of light. They see each other's clocks slowing down. They then slow down, make u-turns and come together after repeating the exact acceleration and deceleration for the return trip. The whole trip is 10 kilometers for each (20 k. in total). Their clocks are now synchronized again. How does the "changing distance" you speak of work here? How does the time it takes the information to travel account for this? While A is looking at B he sees B's clock slow down during the outgoing trip, he also sees the clock slow down further during the return trip. I don't know what effect the accelerations have. When does B's clock speed up (actually or apparently, or however)? Of course the situation is totally symmetrical for B looking at A!

 

[Perpetual Student]

Of course I did. And if you want someone to explain to you something that you don't understand, I would recommend that you choose a different tone.

Relativity of simultaneity means that two events that are simultaneous in one frame are not simultaneous in another. When you're asking the question, "What time does B's clock show in A's reference frame when A's clock shows T?", you have defined an event E - B's clock showing something - which in A's reference frame is simultaneous with the event "A's clock showing T", and you are asking about the time coordinate of this event in B's reference frame - that gives you the time that B's clock is showing.

When A changes velocity, his reference frame changes. This means that because of relativity of simultaneity, the original event E is no longer the one that is simultaneous with the event "A's clock showing T". Instead, it is some other event E', with a different time coordinate in B's reference frame. This means that A's change of velocity causes a change of the time that "A sees on B's clock". And this means that A's change of velocity over time (=acceleration) changes the rate of the time progression, as seen by A on B's clock.

In the mathematics of the Lorentz transformation (which I linked you to),

t' = (t - vx/c²) * γ

where γ = 1/sqrt(1-v²/c²).

Now, time dilation is slowing down of a clock in an inertial reference frame that moves with respect to the clock. You're examining Δt', compared to Δt. Because you're in a inertial reference frame, it follows that Δv = 0 and Δγ = 0 (and Δx = 0, as you're interested in the change of time between two events that are colocal in the clock's frame). From this, it simply follows that Δt' = Δt * γ. This is the relationship that you're probably familiar with. But you forgot to realize that this is only true for an inertial reference frame.

For a non-inertial, accelerating frame, Δv is no longer 0 and so the vx/c² term does not get cancelled out from the equation, as it did in the inertial case. This is the term that is relevant for the relativity of simultaneity, as it describes the change in time in response to a change in reference frame.

For this reason, it is no longer true that Δt' = Δt * γ, but the relationship becomes more complicated. The change in t' also depends on change in the term vx/c², the magnitude of which depends on x and Δv and is not constrained to always result in Δt' > Δt.

To sum up, time dilation as you understand it applies only to inertial frames. When the observer is accelerating, the observed clock can indeed "speed up". This is because of relativity of simultaneity.

Sorry for my "tone" -- a manifestation of my frustration. I will read the above carefully. Thanks.

 

[69dodge]

I will rephrase and recast my question:

Two observers with synchronized clocks take off in opposite directions (both accelerate by equal amounts). They observe each others clocks slowing down. The relationship for this is dt* = dt(gamma) for both observers. They then simultaneously turn around and come together. In the end, if their clocks are again synchronized, then each observer saw the others clock speed up (constrict). How did that happen? Is there an inverse Lorenz transformation for this result? Does this come about because of the acceleration? If so, what is that transformation equation? Since it involves acceleration, is GR involved?

We don't need general relativity to figure out that the two clocks show the same time when they meet again. Special relativity suffices. But the way to solve a problem in special relativity is to pick one inertial reference frame and then use it consistently. It doesn't matter which one we pick. They'll all end up giving the same answer. But we can't switch reference frames in the middle of the problem.

Suppose we've picked an inertial reference frame, so that all talk of motion means motion relative to that frame. Then, a moving clock runs slower than a stationary one; no clock runs faster than a stationary one; and a clock's rate depends only on its speed, not on its acceleration.

We should not ascribe too much reality to statements such as "clock A is currently running slower than clock B." If we pick a different reference frame to do the calculations, the opposite statement might be 'true'. But when A and B meet again, either they will show the same time or they won't. The question whether they do is a real question, and the answer to that question doesn't depend on which reference frame is used in the calculations.

 

[RecoveringYuppy]

Suppose the distance is not great.

Didn't make any assumptions about the exact values before, so this doesn't change anything but the magntitude of the answer. Nothing qualitatively changes.

While A is looking at B he sees B's clock slow down during the outgoing trip, he also sees the clock slow down further during the return trip.

Could you clarify what you mean by "sees"? Are you referring to the time of arrival of broadcasts from the other ship? Or how ship A sees ships B's clocks compare to clocks that are near B? (and standing still compared to who?)

When does B's clock speed up (actually or apparently, or however)? Of course the situation is totally symmetrical for B looking at A!

I think I'll wait til I hear your answers to prior questions before tackling this. For the moment the answer is "when the velocities change".

 

[sol invictus]

Suppose the distance is not great. Say A and B move apart (accelerating simultaneously) for a few kilometers accelerating to, say, 10% the speed of light. They see each other's clocks slowing down. They then slow down, make u-turns and come together after repeating the exact acceleration and deceleration for the return trip. The whole trip is 10 kilometers for each (20 k. in total). Their clocks are now synchronized again. How does the "changing distance" you speak of work here? How does the time it takes the information to travel account for this? While A is looking at B he sees B's clock slow down during the outgoing trip, he also sees the clock slow down further during the return trip. I don't know what effect the accelerations have. When does B's clock speed up (actually or apparently, or however)? Of course the situation is totally symmetrical for B looking at A!

Thabiguy has done a very good job answering your questions, but sometimes a picture is worth a lot of words.

Take this diagram, double it (by reflecting it around the vertical line and matching along the line) so that there's a symmetrical traveling twin, extend those blue and red lines a bit so they intersect the other twin's worldline, and you have your answer.

If that doesn't do it, read the article it came from, which is very thorough.

 

[RecoveringYuppy]

@P.S. See if this description is the kind of answer you are looking for. If it is, see if Thabiguy [ETA: didn't mean to slight SI, he can obviously critique it too. His post just wasn't here when I first tried to post this last night] will critique it for you before trusting it. I've rounded the numbers a bit to keep things simple, but I think they are approximately right.

Scenario: Two ships A and B leave their home base at about .5c relative to home base, about a time factor of 1.15 relative to home base. That would be .8c relative to each other and a time factor of 1.67 between the two ships. The ships and the home base broadcast their local time at the same regular interval as measured in their own local time. Ships A and B are going to magically turn around instantly at some agreed upon distance and head back at .5c and return to home base at the time home base emits it's 100th pulse. And let's say that at home base the whole experiment takes 4 years.

First off, home base gets signals from both ships at exactly the same time over the whole experience. It receives about 88 of them from each ship. They arrive in pairs at exactly the same time. The first 44 come back spread evenly over the first 3 years. The last 44 come in spread evenly but at a faster pace over the last year of the experiment. Throughout this whole time, if the people on Earth look at the ships clock and compare it to clocks the ship is passing they will always see the ships clock is dilated by 1.15. The rest of the discrepancy in arrival time is due to the changing distance between Earth and the ship.

What ship A sees: Over the course of the whole trip it receives all the 100 pulses home base sends out and all the 88 pulses the other ship sends out. There are three phases to the trip:

During the first phase. Ship A looks at ship B and sees it receding at .8c. Ship A looks at ship B and compares ship B's clock to clocks near ship B that are stationary compared to ship A. Ship A sees ship B's clocks as dilated by a factor of 1.67. Due to the motion between ship B and ship A the "ticks" are arriving at ship A at an even slower pace, which I'm not going to figure out. The timestamp on these messages is always lower that our local clock and always falling behind.

Second phase. Ship A turns around and heads home at .5c relative to Earth. Ship A and B are now travelling the same direction at the same speed so there is no relative motion between them. Ship A looks at ship B and compares ship B's clock to clocks that are near B and stationary compared to ship A again. This time we find those same clocks are also stationary compared to ship B and there is no time dilation between the two ships. The "ticks" arriving at our ship from ship B are arriving at the same pace our clock is ticking. The timestamp on those messages is less than our timestamp because they are coming from a long way away but the difference is not changing.

Third phase. Ship B turns around and heads home at .5c relative to Earth. Ship A looks at ship B and compares ship B's clock to clocks near ship B that are stationary compared to ship A. Ship A sees ship B's clocks as dilated by a factor of 1.67, running slow compared to ship A's clock. Due to the motion between ship B and ship A the "ticks" are arriving at ship A at a faster pace, which I'm not going to figure out, but it's faster than our clock is ticking. The timestamp on these messages is always lower that our local clock but always catching up. It catches up exactly at home base.

 

[Perpetual Student]

Q. Could you clarify what you mean by "sees"? Are you referring to the time of arrival of broadcasts from the other ship? Or how ship A sees ships B's clocks compare to clocks that are near B? (and standing still compared to who?)

A. I should have been specific about this initially. I assume sophisticated observers. I intended to make the issue of the travel time for information and Doppler shifts irrelevant to my question by assuming that when A "sees" B's clock slow down he is taking into account the time for the information to reach his destination and any Doppler affects.

Thanks for all the responses from everyone; you have all provided me with a lot to work through. The question about what happens during the accelerations is obviously the key to the answer -- It's quite difficult to envision what's going on there.
In SR, the equations involve v, gamma, c, and x. There was no factor for acceleration -- "a' or "dv/dt." The following description (from Wiki.): "In a sense, during the U-turn the plane of simultaneity jumps from blue to red and very quickly sweeps over a large segment of the world line of the resting twin. The traveling twin reckons that there has been a jump discontinuity in the age of the resting twin." -- does not include a mathematical explanation. Is there a mathematical explanation to accompany the descriptive one in SR? Without mathematics involving acceleration, it would appear to be a lot of hand waving.
The GR explanation (a lot of complexity to digest -- but worth the effort) does feature the factor "a," which is to be expected and somewhat satisfying. The article mentions that the twin paradox was not really considered to be a paradox. Historically, was this "paradox" controversial before GR?

 

[Thabiguy]

To RecoveringYuppy: Not much to critique, just a clarification about the second phase:

By the time A turns around, it will find that B has also turned around (a long time ago, actually). So there is no time (in this particular experiment where the turn is instantaneous) when there is no relative motion and no time dilation between the two ships. But there is a second phase in which A will be receiving old ticks from B (from the time it was travelling outward), as you describe.

 

[RecoveringYuppy]

Yes. Darn, I was trying avoid description of spacially separated events but " Ship A and B are now travelling the same direction at the same speed so there is no relative motion between them." sure does that. What I had in mind was how the signal being received at A could appear to a naive observer aboard A.

 

[Thabiguy]

In SR, the equations involve v, gamma, c, and x. There was no factor for acceleration -- "a' or "dv/dt."

When you try to calculate instantaneous time dilation, you are looking for dt'/dt. When you differentiate the transformation equation with respect to time, dv/dt will appear on the right-hand side. So it is in there all right.

The instantaneous time dilation will depend on instantaneous velocity, acceleration and distance, and will generally not be constant over time (unlike the inertial case). Also, if you are looking for difference of time between two events, you would actually be looking for a finite integral of the instantaneous time dilation, which will get uglier still.

But in spite of the math being unpleasant, it's still only Lorentz transformations + calculus.

Is there a mathematical explanation to accompany the descriptive one in SR? Without mathematics involving acceleration, it would appear to be a lot of hand waving.

There is no hand waving. The Lorentz transformation is right there. It gives you everything you need. You are free to carry out the calculation to any degree that you like, until your complete mathematical satisfaction.

I tried to sketch for you how the solution would look like and why it would work, so that you could see it without going through the actual differentiation. But if that didn't help and you need the hardcore stuff, you can have it. If you don't want to do the unpleasant math (which I can sympathize with), maybe you can find something online. I tried to see if there's something handy online, but haven't found much. Perhaps you'll be more lucky.

Ultimately though, if you can't find the work already done or someone willing to do it for you, and you still demand a full mathematical analysis of your accelerating ships, I'm afraid you will just have to grit your teeth and do it yourself. I can at least offer to give some guidelines and even check if you've done it right.

The GR explanation (a lot of complexity to digest -- but worth the effort) does feature the factor "a," which is to be expected and somewhat satisfying.

SR is a special case of GR. So anything that can be solved in SR can also be solved in GR.

 

[Ziggurat]

In SR, the equations involve v, gamma, c, and x. There was no factor for acceleration -- "a' or "dv/dt."

The commonly used equations do not, because the commonly use equations are for transformations between two inertial reference frames. An accelerating frame is not innertial. This does not mean that special relativity cannot handle accelerating reference frames - it most certainly can, though things get uglier, and there is not much point in inflicting it on most students. It is a popular misconception (inspired by a common misunderstanding of the equivalence principle) that you need GR to handle acceleration, but this is not so.

Is there a mathematical explanation to accompany the descriptive one in SR?

Yes, there is. I can`t spell it out in detail now, but it is not hand waiving.

The GR explanation (a lot of complexity to digest -- but worth the effort) does feature the factor "a," which is to be expected and somewhat satisfying.

If it doesn`t have gravity, it`s not GR, but merely complex SR.

There`s an alternate version of the twin paradox involving the traveling twin going on a circular journey at constant speed with respect to earth, and constant acceleration as well. I like this version because you can pick whatever finite acceleration you want, and still ignore it completely in calculating your final answerfor how old they are when they meet up at the end (except to make sure he ends up in the right spot at the right time). You can also use different accelerations (for example, twice around a tight circle versus once around a larger circle) and still find the same answer, which only depends upon the speed of the traveling twin. It`s useful to illustrate how, even though acceleration is needed to break the symmetry between the twins, it isn`t the acceleration itself which causes the difference in times.

 

[GMB]

The appeal to "frame of reference" is a perpetual series of spoon-bender distractions. There is only one relevant "frame of reference" and that frame of reference is REALITY.

Always when we see the phrase "frame of reference" we are asked to take seriously appeals to pseudo-science since in all such cases no evidence is ever offered or yet even capable of being offered.

We need to follow the example of the Amazing Randi and chase out every last bit of voodoo from academic life and not merely from stage acts.

http://au.youtube.com/watch?v=QlfMsZwr8rc

"Relativity of simultaneity means that two events that are simultaneous in one frame are not simultaneous in another. "

This is rubbish and there is no such thing. Yet people are prattling on failing to distinguish between their version of what the theory says, and what would happen in reality. To interpret this theory as if it were revealed truth is to make a stooge of oneself. If something is simultaneous it is simultaneous IN REALITY and this reality is not dependent on this "frame of reference" sleight of hand.

In fact there can never be anything such as time dilation. Since there is no time to dilate. Time being merely a derived concept, derived from simultaneity and regular motion.

Its really very rude for people to prattle on in the vain when they know they have absolutely no evidence whatsoever for these spoon-bending absurdities. If two events are simultaneous they are simultaneous no matter what. No evidence exists otherwise or ever could exist otherwise. Everything that happens simultaneous happens at the same time everywhere. And in that last sentence the word time is a redundancy since time itself is derived from this simultaneity.

Note that spoon-bender-central could not make up its mind about the Dingle refutation. The official story about the clocks running slower than eachother was a committee decision and not something that the theory originally resolved. But our spoon-benders came out with at least 3 different answers anyway. Since we have had at least three versions of which clock ticks slower. The twins refutation IS the clocks refutation. They are the same refutation. They are the very same story. Notions to do with the aging of cells under acceleration are speculative. But notions to do with this same matter under different velocities are not. Since velocity is a relative concept. Hence it is ACCELERATION that we need to investigate as to its effect on this or that. Not velocity.

 

[Thabiguy]

Oh, he's back.

So much for physics in this thread.

 

[Ziggurat]

In fact there can never be anything such as time dilation. Since there is no time to dilate.

Except that it's been experimentally observed. Small fly in your ointment there, pal.

Its really very rude for people to prattle on in the vain

Oh the irony.

 

[RecoveringYuppy]

Oh, I missed a very amusing part of this thread.

 

[GMB]

"Except that it's been experimentally observed. Small fly in your ointment there, pal."

No it hasn't. There is no such thing as time dilation. Any effect so observed may be related to acceleration. But all this spoon-bender talk, to the contrary notwithstanding, it has not been observed with regard to velocity and never could be.

Lets hope that from now on, spoon-benders-united, can word things, so as to differentiate between reality and their own personal version of this failed and self-contradictory theory.

The problem is that the priesthood has been taken over by maths-boy 101. Its as if they chased out the natural philosophers and brought in Rainman. In fact the maths-barrier is the only reason the spoon-benders have been able to prevail in this area. In economics its the pull of parasitism. Here its the shield of advanced maths.

Being as the clocks and twins refutation are the self-same refutation spoon-benders everywhere might want to contemplate how three or more answers to the same refutation were possible. I'm not sure I want to know your explanations actually. It will be the same nonsense and "look over there!!!!" distractions.

But there can be no doubt about it that the refutation stands. At least it stands while any skeptics are still in the room while the spoon-bender goes to work. What we are not going to see is people scrambling over each-other to prove that this nonsense has been experimentally observed.

Directly below what do we see? A clarification of which answer to the clocks refutation spoon-benders are going to keep a united front on? I don't THINK!!!so. Well perhaps we see the evidence forthcoming for the wrong notion that time dilation has been experimentally verified? No thats not what we see directly below either.

Instead what we see is spoon-bender-central asking skeptics to leave the room. All skeptics must leave the room. Only the credulous and other spoon-benders need apply to stay on. Hurts their aura does this skepticism. Makes their powers weak.