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Giggles
2007-Oct-08, 06:43 PM
Hi all.

My question is a fairly simple one, and I don't think I've seen it asked or answered.

Let's say you were traveling through space at 98% the speed of light, perhaps upon a magical intergalactic space unicorn. You're traveling towards a star system a lightyear away from your starting position and, while riding, are observing the system with a wonderful telescope of wonderment.
What would you witness in regards to the motion of the astronomical bodies around the target star? Would they be moving absurdly fast due to the light delay between the star and your original position? Would you start seeing double, or triple, or sextruple copies of those objects?

antoniseb
2007-Oct-08, 06:56 PM
It would appear that the star system was coming toward you at 0.98c, and all of the light you see would be highly blue-shifted. 500nm visible light photons would be 10nm xrays. 25 micron far infrared light would be seen as visible.

You would only see one copy, but I can't tell you how quickly the planets would seem to be going.

grant hutchison
2007-Oct-08, 08:35 PM
The relativistic Doppler factor, η, is about equal to 10 when you look straight ahead at 0.98c. That's made up of a relativistic time dilation component (which slows everything down), and a classical Doppler component, which increases the frequency of signals coming from ahead.
(I think antoniseb has used the classical Doppler effect to come up with his blue-shift factor of 50. Time dilation undermines that by a factor of 5, giving us the final value of 10.)

So the star system ahead would appear speeded up by a factor of 10, as well as blueshifted by the same amount. It would also appear to be 10 times farther away, if you judged by its apparent size in your telescope, or if you attempted to triangulate it.

Grant Hutchison

Michael Noonan
2007-Oct-08, 08:53 PM
The relativistic Doppler factor, η, is about equal to 10 when you look straight ahead at 0.98c. That's made up of a relativistic time dilation component (which slows everything down), and a classical Doppler component, which increases the frequency of signals coming from ahead.
(I think antoniseb has used the classical Doppler effect to come up with his blue-shift factor of 50. Time dilation undermines that by a factor of 5, giving us the final value of 10.)

So the star system ahead would appear speeded up by a factor of 10, as well as blueshifted by the same amount. It would also appear to be 10 times farther away, if you judged by its apparent size in your telescope, or if you attempted to triangulate it.

Grant Hutchison

Wow, is this what sort of mathematics gets used on the other blue shifted set of stars we know about? On a much smaller scale of c naturally.

mugaliens
2007-Oct-12, 02:56 PM
It would appear that the star system was coming toward you at 0.98c, and all of the light you see would be highly blue-shifted. 500nm visible light photons would be 10nm xrays. 25 micron far infrared light would be seen as visible.

You would only see one copy, but I can't tell you how quickly the planets would seem to be going.

Faster than normal, as time for you has slowed.

Cougar
2007-Oct-12, 03:37 PM
Faster than normal, as time for you has slowed.
Relative to what? I hate these relativity mind-benders and probably should just stay out of it. Are you saying that the effect of time dilation is impervious to whether the relative velocity is toward or away from an observer?

I can finally see how the planets would appear (to the approaching spacecraft) to be orbiting faster, since the light coming from them has to travel less and less distance as you approach.

OH, darn, work calls.....

grant hutchison
2007-Oct-12, 04:06 PM
Faster than normal, as time for you has slowed.The second part of this seems to be wrong.
Time dilation makes time run slow in frames that are moving relative to you. So to you, aboard your spaceship, it's the stars that you are observing which "run slow".
But for stars up ahead the Doppler effect created by their approach velocity overwhelms the time dilation, and you see them run fast. (Meanwhile, you'll see stars astern running slow, because of time dilation and Doppler.)

Grant Hutchison

speedfreek
2007-Oct-12, 07:38 PM
The second part of this seems to be wrong.
Time dilation makes time run slow in frames that are moving relative to you. So to you, aboard your spaceship, it's the stars that you are observing which "run slow".
But for stars up ahead the Doppler effect created by their approach velocity overwhelms the time dilation, and you see them run fast. (Meanwhile, you'll see stars astern running slow, because of time dilation and Doppler.)

Grant Hutchison

This is where the problems occur with my understanding of the situation and I am unsure as to the mainstream view.

It seems that time-dilation is symmetrical between inertial frames of reference.

I was under the impression this meant that if you could observe your destination, you would see them "running slow" by the same amount that they see you "running slow".

You say that for stars up ahead the Doppler effect created by their approach velocity overwhelms the time dilation, and you see them run fast. If your spacecraft is moving at a constant speed and is therefore in an inertial frame of reference, why wouldn't an observer on one of those stars up ahead see you "running fast" too? After all, in that situation, both you and the star ahead of you can consider that you/they are at rest and it is the other that is moving.

How does this work with the twins paradox, if we consider one twin goes on a journey of 5 light years, turns around and comes back again, at say, 0.866c?

If the twin that makes the journey does indeed return having aged by half the amount that the twin that stayed at home has aged, where in the journey would they be able to observe the difference occuring? During the acceleration/deceleration/turnaround phases, when one of them is constantly changing reference frames?

And what if the whole journey is either acceleration, turnaround or deceleration? How does it work if one twin is constantly changing reference frames throughout that journey?

I'm not proposing anything here, I just strive to understand the mainstream view on all this.. :)

grant hutchison
2007-Oct-12, 07:50 PM
You say that for stars up ahead the Doppler effect created by their approach velocity overwhelms the time dilation, and you see them run fast. If your spacecraft is moving at a constant speed and is therefore in an inertial frame of reference, why wouldn't an observer on one of those stars up ahead see you "running fast" too? After all, in that situation, both you and the star ahead of you can consider that you/they are at rest and it is the other that is moving.Yep, that's eactly what would happen. Each would see the other running fast, and each would deduce (taking Doppler into consideration) that the other's clocks were running slow.

Easiest thing with the twin paradox is to chop it into three sections: coasting away from home, turning around, coasting towards home. During the "coasting away" phase, each observer sees the other's clocks running slow, and also deduces (after allowing for Doppler) that the other's clocks are running slow. During the "coasting towards" phase, each observer sees the other's clocks running fast, but deduces (after Doppler) that the other's clocks are running slow.
All the action happens during turnaround, when the traveller accelerates while the home twin just sits there. When the traveller shifts from one inertial frame ("coasting away") to another ("coasting towards"), this is accompanied by a shift in the traveller's perception of simultaneity. The simultaneity shift moves forward along the home twin's worldline, so that a short amount of the traveller's time equates to a long period of the home twin's time. That's where the paradox resolves, creating a mismatch in elapsed time between the accelerated twin and the unaccelerated twin.
But the travelling twin doesn't see this occur: light signals just arrive at longer intervals during the "coasting away" phase, and at shorter intervals during the "coasting towards" phase.

Performing the same journey under constant acceleration/deceleration doesn't change the outcome or the perceptions particularly: it just mixes the simultaneity shift in with the time dilation and Doppler so that everything is going on all the time.

Grant Hutchison

speedfreek
2007-Oct-12, 08:25 PM
All the action happens during turnaround, when the traveller accelerates while the home twin just sits there. When the traveller shifts from one inertial frame ("coasting away") to another ("coasting towards"), this is accompanied by a shift in the traveller's perception of simultaneity. The simultaneity shift moves forward along the home twin's worldline, so that a short amount of the traveller's time equates to a long period of the home twin's time. That's where the paradox resolves, creating a mismatch in elapsed time between the accelerated twin and the unaccelerated twin.

So can I take it that if they each could broadcast a picture of themselves holding up a watch, to the other, at regular intervals during the journey, they would each see the others watches running slower than their own by the same amount on the outward coasting part of the journey, faster than their own by the same amount on the return coasting part of the journey, but the twin at home would see travelling twin running slower than the travelling twin sees the twin at home, which would add up to the difference in their ages, during the acceleration/turnaround/deceleration phases (assuming several pictures were broadcast during these phases)?

And then for a journey that was all either acceleration, turnaround (which is acceleration) or deceleration (acceleration in the opposite direction), would the twin at home observe the travelling twin to be running slower than the traveller sees the twin that stayed at home, during the whole journey? Would they still see each other slowed on the outward phase and sped up on the return phase, but the twin at home always see the traveller running a little slower than the traveller sees them, which would add up to the age difference at the end?

grant hutchison
2007-Oct-12, 09:41 PM
Yep, I've I'm following it correctly, you've got it right. There's always perfect symmetry of observation, except during acceleration. The work of shifting the traveller's simultaneity forward along the home twin's world-line happens when the traveller's acceleration is directed towards the home twin (that is, when decelerating during the "away" phase and accelerating during the "towards" phase).

Grant Hutchison

grant hutchison
2007-Oct-12, 10:03 PM
If it helps, I still have some illustrations posted that I did for a previous discussion of the twin paradox. Each of them is a graph with time vertically and distance horizontally, showing the worldlines of the home twin and the travelling twin. The travelling twin does all the necessary acceleration instantaneously, so all we see are the "away" and "towards" coasting phases. The worldlines are marked off with equal intervals of proper time: twenty years for the home twin, twelve years for the traveller.

Here (http://www.ghutchison.pwp.blueyonder.co.uk/relativity/simultaneity3.jpg) are the simultaneity lines: blue horizontal lines for the home twin, green sloping lines for the traveller. If you follow the lines, you can see that after a year of his own time, the home twin finds that less than a year has elapsed for the traveller; but the traveller finds that after a year of her time, less than a year of the home twin's time has elapsed according to her own judgement of simultaneity. Symmetry!
When the traveller instaneously accelerates back towards the home twin at turnaround, her simultaneity line jumps forward along the home twin's world-line: more than twelve years of home-twin time elapse in an instant of traveller-time. There is the symmetry-breaker.

Here (http://www.ghutchison.pwp.blueyonder.co.uk/relativity/light1.jpg), in red, are light signals emitted every year by the home twin. The traveller receives only two of them during her six-year outbound leg, and receives the other eighteen during the six-year return: so she sees the home twin running slow by a factor of three when she's outbound, and running fast by a factor of three when she's inbound.
Meanwhile (http://www.ghutchison.pwp.blueyonder.co.uk/relativity/light2.jpg), the traveller's light signals are similarly distorted for the home twin: the six "away" signals take eighteen years to arrive at home, while the six signals emitted during the return trip are all crammed into the last two years of the home twin's worldline. The home twin sees the traveller run slow by a factor of three when outbound, and run fast by a factor of three when inbound. Symmetry!

Grant Hutchison

Fortunate
2007-Oct-12, 11:15 PM
It would also appear to be 10 times farther away, if you judged by its apparent size in your telescope, or if you attempted to triangulate it.

Grant Hutchison

I thought that it would appear closer.

grant hutchison
2007-Oct-12, 11:30 PM
I thought that it would appear closer.No. The universe certainly does shrink in your direction of travel, by Lorentz contraction, but the effects of light-travel time more than offset that when we consider what you actually see. At any given moment, as you approach a distant object, you're seeing light it emitted when it was farther away in your reference frame than it is at the moment you receive the light: so triangulation (by flying a couple of spaceships parallel to each other, for instance) will find that the object appears farther away than its current position. Relativistic aberration also causes objects ahead of you to appear to shrink in apparent diameter, compatible with that trangulated distance.

Grant Hutchison

speedfreek
2007-Oct-12, 11:42 PM
Thank you for those diagrams, Grant - very useful. :)

The reason I asked if the difference in "ageing" would be apparent only during the non-inertial phases of the journey is that I proposed this idea somewhere else and was informed that wasn't the mainstream view, but I was sceptical of that persons interpretation. I reasoned that if time-dilation is symmetrical between observers in inertial frames, that the difference in the twins ages must occur only during the non-inertial parts of the journey, and that would be when any difference would be observed by the other.

Further to your answer to the post by Fortunate, as the universe certainly does shrink in your direction of travel, by Lorentz contraction, but the effects of light-travel time more than offset that when we consider what you actually see, does this mean that if you were, as the person travelling, to subtract the effects of light travel time for your calculations, would you work out that the destination is indeed as close as the Lorentz contraction predicts it should be?

As an example, a journey of 5 light years at 0.866c should take you 5.77 years. But when we apply the Lorentz contraction we find that journey time is halved at that speed, so the journey would take you only 2.88 years. If you were to subtract the effects of light-travel time in your calculations, would you find that the destination looked to be only 2.5 light years away, and thus there would be no paradox in you arriving there in only 2.88 years when travelling at 86.6% of light speed?

Fortunate
2007-Oct-12, 11:46 PM
No. The universe certainly does shrink in your direction of travel, by Lorentz contraction, but the effects of light-travel time more than offset that when we consider what you actually see. At any given moment, as you approach a distant object, you're seeing light it emitted when it was farther away in your reference frame than it is at the moment you receive the light: so triangulation (by flying a couple of spaceships parallel to each other, for instance) will find that the object appears farther away than its current position. Relativistic aberration also causes objects ahead of you to appear to shrink in apparent diameter, compatible with that trangulated distance.

Grant Hutchison

grant hutchison
2007-Oct-13, 12:08 AM
I reasoned that if time-dilation is symmetrical between observers in inertial frames, that the difference in the twins ages must occur only during the non-inertial parts of the journey, and that would be when any difference would be observed by the other.You're right, the traveller and home twin are only ageing asymmetrically during the non-inertial portion of the traveller's journey. But they don't actually observe that effect, either at the time it occurs or afterwards: they simply see light signals "running slow" as they separate and "running fast" as they approach. In my extreme example of instantaneous acceleration, you should be able to see from the graphs that, although the travelling twin is instantaneously simultaneous with >12 years of the home twin's life, she never sees those twelve years suddenly shoot by: the light signals from home arrive at a steady pace during each coasting phase; slow, then fast.

Further to your answer to the post by Fortunate, as the universe certainly does shrink in your direction of travel, by Lorentz contraction, but the effects of light-travel time more than offset that when we consider what you actually see, does this mean that if you were, as the person travelling, to subtract the effects of light travel time for your calculations, would you work out that the destination is indeed as close as the Lorentz contraction predicts it should be?Yes. You'd otherwise be unable to explain how you got there so quickly, as you say.

(There's always a problem when discussing special relativity, to separate the optical effects of rapid travel from the coordinate transformations of relativity itself. I often find myself typing things like "The traveller will deduce" or "The traveller will calculate" when I'm talking about the coordinate transforms, and saving verbs like "see" and "observe" for the appearances after the optical effects have been added in. But it's kind of clunky and unnatural.)

Grant Hutchison

speedfreek
2007-Oct-13, 12:34 AM
Thank you again, Grant. It is nice to know that I am getting closer to comprehending it properly. :)

Fortunate
2007-Oct-13, 03:39 AM
The relativistic Doppler factor, η, is about equal to 10 when you look straight ahead at 0.98c. That's made up of a relativistic time dilation component (which slows everything down), and a classical Doppler component, which increases the frequency of signals coming from ahead.
(I think antoniseb has used the classical Doppler effect to come up with his blue-shift factor of 50. Time dilation undermines that by a factor of 5, giving us the final value of 10.)

So the star system ahead would appear speeded up by a factor of 10, as well as blueshifted by the same amount. It would also appear to be 10 times farther away, if you judged by its apparent size in your telescope, or if you attempted to triangulate it.

Grant Hutchison

My copy of Halliday and Resnick seems to indicate that the frequency would increase by a factor of (1 + v/c)/sqrt(1 - (v/c)2) for objects approaching each other. For v = .98c, this gives a factor of 9.95.

grant hutchison
2007-Oct-13, 12:33 PM
My copy of Halliday and Resnick seems to indicate that the frequency would increase by a factor of (1 + v/c)/sqrt(1 - (v/c)2) for objects approaching each other. For v = .98c, this gives a factor of 9.95.Yes, that's the formula.

Grant Hutchison

grant hutchison
2007-Oct-13, 10:41 PM
Thank you again, Grant. It is nice to know that I am getting closer to comprehending it properly. :)No worries.
Since Giggles seems never to have returned, I'd pretty have been talking to myself if it wasn't for you and Fortunate.:)

Grant Hutchison

speedfreek
2007-Oct-16, 08:34 PM
Grant, regarding my question about subtracting light-travel time (or aberration) from your calculations to determine the length contraction (and thus the distance) to your target, and finding that the shortened distance would indeed correspond to the time-dilation, meaning your shorter journey time would be consistent with the apparent distance to your destination...

If we, as travellers, subtract the light-travel time or aberration from our calculations, what do we see when looking backwards, behind the spaceship? Do we see the apparent distance from our departure point lengthened?

Or more simply, if you fly backwards at 86.6% of light speed, length contraction would put your destination (behind you) at half it's original distance. What would length contraction do to your view to the front, if you are flying backwards?

grant hutchison
2007-Oct-16, 09:03 PM
If we, as travellers, subtract the light-travel time or aberration from our calculations, what do we see when looking backwards, behind the spaceship? Do we see the apparent distance from our departure point lengthened?Lorentz contraction makes the Universe contract equally, fore and aft: if it halves the distance ahead, it halves the distance behind.
But light-travel effects stacked on top of that make objects ahead look farther away (as we've discussed) while objects behind look even closer than Lorentz makes them.
The relativistic Doppler factor for the view behind is just the inverse of the view ahead: if stuff looks ten times farther away ahead, it will look ten times closer behind.

I have diagrams :):
First, imagine that you are sitting at rest in the centre of a perfect circle of evenly-spaced stars. Here (http://www.ghutchison.pwp.blueyonder.co.uk/relativity/betazero.jpg) is a diagram, with you in the centre and your sightlines to the stars marked in yellow.
Now, imagine that you are merely passing through the same central point, but at 0.6 of lightspeed. The Universe is foreshortened, fore and aft, by a Lorentz contraction of 0.8: that's what you would figure if you allowed for light-travel effects.
But what you would see would be that the stars ahead had moved farther away by a factor of 2, while those behind had moved in to only half their original distance. The original circle of stars in the first diagram is distorted into an ellipse, with you, the travelling observer, at one focus of that ellipse. Here (http://www.ghutchison.pwp.blueyonder.co.uk/relativity/betapointsix.jpg) is a diagram of that situation, with the original circle of stars left in place for reference. Sightlines are colour red or blue according to whether there is blueshift or redshift. (The observer is travelling left to right.)
And here (http://www.ghutchison.pwp.blueyonder.co.uk/relativity/betapointnine.jpg) is the same situation with a velocity of 0.9 times lightspeed. Lorentz contraction is now around 0.4, fore and aft, but once light-travel is factored in, the stars ahead appear more than four times farther away than their rest-frame distance, and those behind have apparently moved more than four times closer.

Grant Hutchison

speedfreek
2007-Oct-16, 09:28 PM
That all makes sense to me, and again, thanks for the diagrams! :)

Would you indulge me with one last question (for now!)?

If one is travelling at, say 270,000 km/s (around 90% of c), and emits a light from the front and the rear, I am thinking that an observer sees the that light propagate at c in both directions from the point where the light was emitted. If we imagine we are viewing the ship side on and it is travelling to the right, when it emits the light in both directions we see the light move left at c, and right at c.

So we would see the light moving left at 300,000 km/s from the point the light was emitted, and we see the other light moving right at 300,000 km/s from that emission point. The ship is travelling right at 270,000 km/s, so we see the light moving right slowly moving away from the ship at 30,000 km/s...

You know what's coming don't you? :)

What would the ship see? Surely from it's POV it would see both lights moving away from it at c? How do we account for ships view if the observer sees the light in front only seemingly moving 10% of c faster than the ship, but sees the light behind moving away from the point of emission at c, and so moving away from the ship at 190% of c?

Where am I going wrong in trying to understand the ships view (after subtraction of light travel time or aberration from its calculations)? I could imagine that due to length contraction, the light in front (which to the observer outside is only moving away from the ship at 10% of c) would seem to the ship to be moving at c as its clock runs slower and the distance in front looks shorter, but how do we account for the ships view of the light that is receding behind it?

Where did I make the mistake, or what am I missing here? :confused:

grant hutchison
2007-Oct-16, 10:02 PM
You're missing the disagreement between "you" and "the ship" about simultaneity, which is right down at the roots of special relativity, driving the more familiar stuff about time dilation and length contraction (and resolving the twin "paradox", as we discussed earlier).
The two observers, in relative motion, can't agree about what distant events are simultaneous. In particular, the observers on the ship might judge a pair of events, positioned fore and aft, to be simultaneous for them, while you would measure the "aft" event preceding the "fore" event. (Note that this is not an issue relating to delays in signals transmitted by light; this is an actual shift in what is, or is not, simultaneous.)
So when the shipboard observers find that a pair of lightbeams, travelling fore and aft, are making journeys across equal distances in equal times, you (the "stationary" observer) tell them that they've messed up the synchronization of their distant clocks, and the journey times are not equal.
(They, meanwhile, observe that you have messed up your synchronization in a precisely symmetrical way. So when one observer is happy with a light-travel time experiment, the other says it's messed up by faulty synchronization.)

Grant Hutchison

speedfreek
2007-Oct-16, 11:38 PM
(Note that this is not an issue relating to delays in signals transmitted by light; this is an actual shift in what is, or is not, simultaneous.)

Ah-hah! I think that is where I have been going wrong. :doh: I appreciate both your answers and your rapid responses! Thanks yet again.

grant hutchison
2007-Oct-17, 12:03 AM
Ah-hah! I think that is where I have been going wrong.Well, if that is the case, you're certainly not alone. :)
It's common enough to think of the simultaneity shift as being some sort of minor consequence of signal delay, rather than a difference in what is real for each observer, after they have allowed for signal delay. (I edited the comment in as an afterthought, so I'm glad I thought to mention it.)

Grant Hutchison

zaphrentis
2007-Oct-22, 03:06 PM
Hi All,

Just for clarification might I ask whether the time shown on two identical synchronised indestructible hypothetical clocks with zero rest mass will still be the same if one was to travel (at light speed) for say 1 minute, bounce with perfect restitution off an immovable surface and then return (again at light speed) after another 1 minute.

Thanks.

grant hutchison
2007-Oct-22, 04:05 PM
Just for clarification might I ask whether the time shown on two identical synchronised indestructible hypothetical clocks with zero rest mass will still be the same if one was to travel (at light speed) for say 1 minute, bounce with perfect restitution off an immovable surface and then return (again at light speed) after another 1 minute.No. What you've got there is an (impossibly) extreme example of the twin paradox: two clocks separate and are brought together again, only one experiencing acceleration. The accelerated clock will show less elapsed time than the unaccelerated clock.

Grant Hutchison

zaphrentis
2007-Oct-22, 04:26 PM
Hi Grant,

Thanks for the response.

At what stage in the journey of the hypothetical clock does the aging (rate of ticking of the clock) slow below the rate of the stationary clock? Assuming that the 180 degree change of direction of constant velocity is instantaneous?

Steve

grant hutchison
2007-Oct-22, 04:51 PM
At what stage in the journey of the hypothetical clock does the aging (rate of ticking of the clock) slow below the rate of the stationary clock? Assuming that the 180 degree change of direction of constant velocity is instantaneous?The instantaneous turnaround is where the mismatch in clock times occurs. While two clocks are separating (or approaching each other) at constant velocity, each finds that the other is ticking too slowly. When the travelling clock makes its instantaneous turnaround, that instant of its own time is simultaneous with a broad stretch of the other clock's history: a whole chunk of the other clock's time shoots past in an instant for the traveller. The unaccelerated clock doesn't experience this effect.
So the unaccelerated clock finds that the traveller runs slow throughout the journey; the traveller finds that the unaccelerated clock mainly runs slow, but also skips through a big chunk of its own time very quickly. Hence the difference when they're reunited.

Grant Hutchison

zaphrentis
2007-Oct-22, 06:00 PM
Okay, let me try again.

Place an observer at a sufficient distance normal to the path travelled by the moving clock so that light emitted from either clock travels parallel towards the observer (not possible except in a thought experiment I admit).

Over the path travelled by the moving clock place at 1 light second intervals lights that will be obscured as the moving clock passes them, and at the surface that 'reflects' the clock place a light so that the moving clock breaks it when it bounces of the surface.

Let the stationary clock emit a flash of light every second, of the same duration as the obscurings and extinguishing produced by the moving clock.

Do you expect the flashes to be in sync with the obscurings?

If so where is the lost time for the moving clock?

grant hutchison
2007-Oct-22, 06:15 PM
Okay, let me try again.Certainly. Maybe it would help if I knew exactly what it was you were trying to do? :)

Let the stationary clock emit a flash of light every second, of the same duration as the obscurings and extinguishing produced by the moving clock.

Do you expect the flashes to be in sync with the obscurings?

If so where is the lost time for the moving clock?I do expect the flashes to keep time with the obscurings, assuming our distant observer is at rest relative to the unaccelerated clock (that is, they're in the same reference frame). All the obscurings are doing is counting off the travelling clock's journey as it would be timed by the unaccelerated clock. So our distant observer sees flashes and obscurations occurring at synchronized one-second intervals.
That tells us nothing about what sort of elapsed time the traveller is experiencing, however: the traveller finds that the lights along the flight-path are spaced at shorter intervals than one light-second, and passes them at intervals of less than one second according to his clock. So the traveller experiences a period of time that is shorter than the number of seconds the distant observer counts off in obscurations.
If the distant observer had asked the traveller to emit a flash at one-second intervals by the traveller's clock, the observer would see these flashes occurring less frequently than the obscurations caused by the traveller's passage, and would judge the traveller's clock to be running correspondingly slowly.

Grant Hutchison

zaphrentis
2007-Oct-22, 08:11 PM
I'm just trying to pin down where and hence the mechanism by which differential aging occurs.

I thought that earlier in the post you had mentioned that the aging slows during acceleration (both negative and positive acceleration producing a slowing) which is why I had hoped that you would have said yes in reply to my first post when I suggested a gedanken in which there was no period of acceleration, the 180 degree change of direction taking place instantaneously at constant speed (hence the zero rest mass).

I had not intended to follow this line of inquiry so far, being rather more interested in why light appears to decelerate into a gravitational field and accelerate out of it, which I assume to be due to light traveling at constant speed along the increased path due to the curvature of what might be called gravi-space (rather than space time) while being viewed in 'flat' 3D matter-space.

The slowing of time with increasing gravitational field strength is quite understandable if distances in gravi-space increase. What I am unclear about is what happens to gravi-space when a non-zero rest-mass object is accelerated through it that increases the path lengths along which the contained photons travel - resulting in a slowing of the passage of time they record as slowed processes.

My interest in this was stimulated by images of Einstein's rings followed by a struggle to understand 'why' null geodesics - something I can only visualise as being due to photons 'sniffing-out' the shortest path around the dimples in gravi-space.

Anyway, thank you for your assistance Grant.

grant hutchison
2007-Oct-22, 09:19 PM
I'm just trying to pin down where and hence the mechanism by which differential aging occurs.OK. You need to chase down the role of simultaneity in special relativity. It's the change in the slope of the simultaneity lines during acceleration that breaks the symmetry between the "stationary" and "travelling" clock.

I thought that earlier in the post you had mentioned that the aging slows during acceleration (both negative and positive acceleration producing a slowing) which is why I had hoped that you would have said yes in reply to my first post when I suggested a gedanken in which there was no period of acceleration, the 180 degree change of direction taking place instantaneously at constant speed (hence the zero rest mass).Ah, but there is acceleration: if an object changes from a positive velocity to a negative velocity, it accelerates. By making the change instantaneous, you made the acceleration infinite, is all. So the simultaneity shift is instantaneous.

You can model this without unfeasible acceleration, simply by handing off a clock setting between two travellers, one who recedes from the stationary clock, and one who approaches it.
Ann floats in space, and synchronizes her clock with Bob as he sweeps past her at close to lightspeed. As the distance between them widens, Bob and Ann communicate and find that each measures the other's clock as running slow, and that they differ about which distant events are simultaneous. Specifically, at a point in Bob's worldline that Ann considers simultaneous with her "now", Bob considers himself to be simultaneous with a point in Ann's worldline that she considers to be her past. The situation is exactly symmetrical, with Bob making similar observations about Ann.
So they separate, and at some point Bob encounters Carol, who is travelling in the opposite direction, towards Ann. Bob hands off his clock setting to Carol, who now carries it back towards Ann.
During Carol's approach, Ann observes Carol's clock to be running slow, and Carol observes Ann's clock to be running slow. Ann now observes that Carol's lines of simultaneity are directed into Ann's future (because Carol is approaching rather than receding).
When the clock setting returns to Ann, courtesy of Carol, Ann is unsurprised to discover that less time has elapsed on the travelling clocks: after all, she has observed them to be running slowly throughout the journey. Bob and Carol have likewise observed Ann to be running slowly throughout their separate journeys.
So where did the "missing" time go, to make Ann's time longer than that measured by the round-trip of the clock settings? It was lost during the handover between Bob and Carol, when a traveller simultaneous with Ann's past encountered a traveller simultaneous with Ann's future. The clock setting was handed between them, and therefore was never simultaneous with a big chunk of Ann's life which elapsed during the middle part of the journey.

The same thing will happen any time you hand off a clock setting between two inertial frames while observing a third. And all acceleration does is move you between one inertial frame and another: hence the simultaneity issues it generates.

Grant Hutchison

zaphrentis
2007-Oct-23, 04:37 PM
Hi Grant,

Thanks for the assistance.

Could you explain, though, how it is that the approaching clocks 'see' each other running slow - surely this is the opposite of what happens (unless they are accelerating towards each other which will reduce somewhat the apparently speeded up passage of time as recorded by the approaching clocks).

grant hutchison
2007-Oct-23, 05:52 PM
Could you explain, though, how it is that the approaching clocks 'see' each other running slow - I've tried to avoid using the word "see" throughout, so I apologize if it has crept in. What I've been describing has nothing to do with light-travel-time effects; the relativistic effects on clocks are what remains after light travel-time has been allowed for.
What one sees is compounded of the relativistic slowing of moving clocks, combined with light-travel-time effects. That combined effect is what is encapsulated in the relativistic Doppler factor, η, which I mentioned and described right at the start of this thread.
So observers approaching each other see each other's clocks run fast, because of light-travel-time effects; each deduces, after allowing for these effects, that the other's clock is running slow.

Grant Hutchison

zaphrentis
2007-Oct-23, 06:34 PM
Okay,

Just help me along a bit here, can you separate out for me what in what we have been discussing is 'real' and what is 'imaginary'?

By which I mean what is a permanent change - such as the difference in time between two initially synchronised clocks, one of which has experienced an acceleration before returning to the rest frame of the other clock - as opposed to an impermanent change such as the observed change of rate of clocks moving at constant speeds relative to the observer.

My reason for asking is that the former needs (for me at least, otherwise it is no better than a religion) an explanation beyond the - admittedly very good but still only theoretical - predictive mathematics, while the second does not.

If the process of accelerating a non-zero rest mass object causes all interactions between its component parts to take longer than those in a non-accelerating non-zero rest mass object then there must be some underlying physics that explains it - we can predict what will happen but sadly not only don't we know why but few seem to care.

grant hutchison
2007-Oct-23, 06:48 PM
Just help me along a bit here, can you separate out for me what in what we have been discussing is 'real' and what is 'imaginary'?Nothing is imaginary. The slowing of clocks, contraction of length and shifts in simultaneity are consequences of special relativity, one of the best tested theories in the world. The light-travel-time effects are clearly observable consequences of objects in motion: they are appearances, but none the less real for that. They have to be allowed for when observing the spectrum of distant galaxies, for instance.

If you want a deep philosophical explanation of why the Universe behaves that way, and not another way, then I'm sure I can't help you.
But you might consider that all our physics is in the same boat as relativity: we make observations about how the Universe behaves, we come up with some maths that describes it and allows us to make useful predictions; but we have no idea if our maths is telling us how the Universe actually does what it does.

Grant Hutchison

zaphrentis
2007-Oct-24, 10:06 AM
Sorry about that, imaginary was definitely not the word I should have used, it would have been better if I had said illusory, as in 'real'ly observed but unwindable in a classical relativity way.

What I was trying to get at in asking the question was where the difference (if any) lay between 'events that are purely due to the process of observing between frames' such as the actual operation of the GPS system in determining location and 'events that are due to an irreversible change' such as the different rates of the atomic clocks in the GPS system that depend on their positions within the earth's gravitational field.

I would take issue though with any contention that the why of how the (for the sake of a better word) universe behaves is in any way susceptible to philosophical explanation, deep or otherwise. Our continually improving ability (not mine, I don't have the maths ability unfortunately) to predict how the universe behaves should not have been accepted as the end of any quest to understand it, but rather as the tools to enable us to find out its currently hidden nature. Currently we seem to be promulgating the adulation of Michelangelo's chisel rather than his David.

grant hutchison
2007-Oct-24, 12:29 PM
What I was trying to get at in asking the question was where the difference (if any) lay between 'events that are purely due to the process of observing between frames' such as the actual operation of the GPS system in determining location and 'events that are due to an irreversible change' such as the different rates of the atomic clocks in the GPS system that depend on their positions within the earth's gravitational field.Well, I hope I answered that previously. The changes predicted and tested under SR are real changes: simultaneity shift, and the associated changes in clock rate and length, for moving objects.
Attendant upon movement are the light-travel-time effects, which determine when we receive signals, at what rate, and from what direction. I don't see these as being particularly illusory, though: that's just the way signals arrive when one is in rapid motion.
In the case of the twin "paradox", you might say that the total number of one-second signals emitted by one observer and received by the other during the course of the journey is determined by the simultaneity shifts of SR; the fact that we sample them at a faster rate when approaching (and a slower rate when receding) is determined by light-travel-time effects.

I would take issue though with any contention that the why of how the (for the sake of a better word) universe behaves is in any way susceptible to philosophical explanation, deep or otherwise.If not a topic of philosophy (or religion), then what?
Science builds mathematical models and just-so stories based on observation of the Universe in action. We judge its success by how well its predictions match other observations. It doesn't, and can't, say anything about "why" (though we often pretend it can).

Do you have a Third Way, neither philosophy nor science?

Grant Hutchison

John Mendenhall
2007-Oct-24, 02:53 PM
The instantaneous turnaround is where the mismatch in clock times occurs.

Interesting point. This instantaneous turnaround is something that obviously cannot happen in the real universe. It might be worthwhile to only consider thought experiments than can actually be performed, that is, that do not include impossible events, only physically difficult events (99% c, very high accelerations, very high gravity fields, etc.). Otherwise, we spend hours figuring out what is wrong with the question.

grant hutchison
2007-Oct-24, 03:33 PM
Interesting point. This instantaneous turnaround is something that obviously cannot happen in the real universe. It might be worthwhile to only consider thought experiments than can actually be performed ...Well, it can be performed by handing clock settings between two travellers moving at constant velocity in opposite directions, as I suggested.
But to generalize to "realistic" scenarios: the distant turnaround is where the mismatch in clock times occurs. An acceleration vector directed towards "home" makes the traveller's line of simultaneity sweep futureward along the stay-at-home's worldline. The larger the separation at turnaround, the greater the temporal displacement along the stay-at-home's worldline.
If turnaround is instantaneous, then that happens as an instantaneous jump for the traveller; for more conventional turnarounds, a long swathe of the stay-at-home's world-line elapses in a much shorter time period for the traveller.

Grant Hutchison