Viewing a (giant) clock at relativistic speed

[TJonline]

Can anyone answer this question?

What you would see while watching a stationary clock that you are moving directly away from at a constant relativistic velocity? Yes, any practical clock would immediately shrink to a tiny point an instant after you passed it. So, either assume a really huge clock (or alternatively a fast-rotating planet that you could watch while moving along the line defined by its poles). I'm really only interested in a precise formulation of the time that the clock's hands would read to you the observer as you move. As you move away from it (or towards it for that matter), the clock should show a component of slowing due to a permanent time loss due to relativistic effects. It should also show a recoverable component of slowing while moving away (that you could regain as an apparent quickening while moving back towards it) due simply to your growing or shrinking distance from it and regardless of your speed.

 

[TumblingDice]

If you are only interested in what your eyes would see while moving away from the clock at relative velocity near c, the doppler effect would drastically red-shift the light reaching you, and the clock would seem to almost stop. The slow time of the clock you observe has nothing to do with relativistic transformations. Only because you are receding and the light is not reaching you as 'timely' as it's leaving the clock.

 

[PeterDonis]

The slow time of the clock you observe has nothing to do with relativistic transformations.

Actually, in part it does. The relativistic Doppler equation is different from the non-relativistic one, meaning that at relativistic speeds, the observed Doppler shift is only partly due to the increase in light travel time due to the observer moving away from the light source (which is what the non-relativistic Doppler equation describes); the other part of the observed relativistic Doppler shift is due to relativistic time dilation.

 

[TumblingDice]

Actually, in part it does. The relativistic Doppler equation is different...

Thank you, Peter!

 

[Simon Bridge]

Welcome to PF;
in addition to TumblingDice ...

The key to understanding relativity is to be very careful about how you talk about it.
For instance, the question is posed in a way that implies some absolute reference frame so that it makes sense to talk about some "you" moving wrt some "stationary" clock.
There is no absolute stationary.
It makes equal sense to say that the clock is moving away from you... and it is usually easier to think about.

The question also talks about a "recoverable" or "permanent" "time loss" - none of these things make any sense either. There is no "lost" time to recover - there is only time lived through. But you can talk about the amount of time measured by one clock in comparison to the time measured on another clock.

Putting the clock on a rapidly rotating planet does not simplify things because this introduces a non-inertial frame (it's rotating, and it has gravity). But these sorts of thought experiments are common - the usual simplifying assumption is that "you" have a very powerful telescope or the clock is very bright (picture a digital display).

It is usually unhelpful to talk about the observer of interest being "you". That introduces all sorts of baggage - like the intuitive feeling of time passing - that can mess up the concepts. As a kind-of self discipline, we usually talk about some other observer - usually more than one.

As TumblingDice points out - the main effects that concern you are the doppler shift and time dilation.
Doppler shift is likely to dominate what you physically see with your eyes (+telescope etc).
Time dilation will always slow the moving clock.
The doppler shift changes the rate that the information on the clock reaches you... if you imagine a small blinking light on the clock, then the blinking speeds up (compared with your clock) when it is approaching, and slow down when it is retreating.

Most texts on special relativity tend to fudge over the doppler effect part - what "you see" in those descriptions implies that you have taken all other effects into consideration.

If you were to pass someone with a clock, the behavior you see for their clock is the same behavior they see for your clock - doppler shift, time dilation, the works. You passing them, as long as the frames are inertial, is the same as them passing you.

If you decide to turn around and go back for another look, you'll discover, when you get back, that the other person has lived through more time than you. This is called the Twin's "Paradox". It occurs because one frame is no longer inertial.

 

[TJonline]

Thanks for the responses!

Yes, I'm aware of the danger of assuming anything being stationary or absolute (much less absolutely stationary!) when discussing relativity and I really meant only to consider an (ideally massless) observer's view of the (ideally massless) clock's hands as a function of the observer's (implied and ideally massless) pocketwatch that he carries with him.

In terms of recoverable, permanent and lost, I only meant the apparent time as read by the observer's view of the clock relative to his own pocketwatch.

Actually, I wasn't putting the clock ON a planet (rotating at, say, one revolution per minute), I was asking that you consider the clock to BE a rotating planet (pole facing toward you) so as to avoid your needing to build a too large clock in this thought experiment.

As for the paragraphs that follow, I'm aware of those factors.

In addition to the Doppler color shift, there would also be (as I understand it), an 'apparent' brightening with fast motion towards the clock and an 'apparent' darkening while quickly moving away, there being greater or fewer numbers of photons arriving at the observer's eye per unit of time. Let's ignore that. Let's also ignore gravity. Let's also ignore Doppler color shift.

Edit by mentor:Deleted inappropriate link for the second time

I'm just not certain how to apply the wavelength components to the observer's perception of the clock's time. Can anyone help me with that? Maybe I should put this in the homework section. Thanks!

 

[ghwellsjr]

...Let's also ignore Doppler color shift.

Can anyone come up with a precise formulation of the current clock time viewed by the observer as a function of the observer's pocketwatch, their relative velocity and their relative distance?

It might be a good idea to ignore Doppler color shift but Doppler also applies to all other repetitive operations such as the tickings of clocks or the motions of their hands. Please look up in wikipedia the article on the Relatistic Doppler Effect for the correct formulation to answer your question.

 

[pervect]

Can anyone come up with a precise formulation of the current clock time viewed by the observer as a function of the observer's pocketwatch, their relative velocity and their relative distance?

The current time viewed by the observer, assuming no gravitational effects (i.e. flat Minkowskii space-time) can be written as a function of the observer's pocketwatch time and the relative velocity only.

Let t_o be the visually observed time on the clock

Let $\tau$ be the time on the observers pocketwatch.

Let the clocks be syncrhonized in such a manner that when t_0 = 0, $\tau$ is also zero, and that when t=0 and $\tau$ = 0, the distance between the oberver and the clock is zero.

Let v be the relative velocity between the observer and the clock, with a positive v meaning the observer is approaching the clock.

Let $\beta = v/c$, so that $\beta$ is the normalized velocity relative to the speed of light.

Then we can write:

$$t_o= \sqrt{\frac{1+\beta}{1-\beta}} \tau$$

As other posters have noted, this will be familiar as the relativistic doppler shift formula. I know you said that you were not specifically interested in doppler shift, but the formula for doppler shift turns out to be identical to the formula you are interested in. There is a reason for this.

Imagine that your clock hands "tick" and move once every millisecond. Imagine that the clock also emits an electromagnetic wave of 1 khz, a wave with a period of 1 ms. Then the EM wave will be in some particular phase (say it is at a peak) i whenever the clock moves. Counting the number of peaks of the EM wave will thus be equivalent to counting the number of clock ticks. Thus the transform of the EM wave's period will be the same as the transform of the clocks period.

The period of all EM waves are multiplied by the same number, independent of frequency, and this same number multiplie the visual appearance of the observed clock. This number is typically called the doppler shift.

If you dig up a copy of Bondi's "Relativity and Common Sense" (I've seen some E-copies online, I can't say whether or not they are legitimate), you can find a further discussion of how you can derive the above formula from some very simple assumptions.

 

[Simon Bridge]

In addition to the Doppler color shift, there would also be (as I understand it), an 'apparent' brightening with fast motion towards the clock and an 'apparent' darkening while quickly moving away, there being greater or fewer numbers of photons arriving at the observer's eye per unit of time. Let's ignore that. Let's also ignore gravity. Let's also ignore Doppler color shift.

... though, as observed above, the Doppler frequency shift must be taken into account to calculate what time the retreating clock reads to the observer.

As before probably best tot think of a digital clock - the numerals change at regular intervals. This interval defines a frequency that the numerals change.

I'm just not certain how to apply the wavelength components to the observer's perception of the clock's time. Can anyone help me with that? Maybe I should put this in the homework section.

Even wikipedia can help with that ;)

http://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Motion_along_the_line_of_sight
The derivation includes time dilation and Doppler effects - even though it talks about it in terms of a color change, you can easily get the time interval shift from $f=1/\Delta t$ ... then it is a matter of counting off time intervals since the observer's clock synchronized with the retreating clock. You should be able to manage that yourself.

... also, if you scroll up a bit from that section you'll see a useful analogy.

Aside: I see the planet-as-clock approach, you are thinking that a point on the equator acts like the end of a big hand? I wonder why everyone seems to think of analog clocks for these things?

Instead: put a bright spot or a beacon of some kind at the equator and have the rotation axis perpendicular to the velocity. The observers sees a series of pulses as the planet rotates - though the signal still has to climb out of a gravity well. Astronomy has lots of examples of large objects that work as a clock in this or other ways. However, this sort of detail is just distracting for your purposes.

 

[quo]

Can anyone answer this question?

What you would see while watching a stationary clock that you are moving directly away from at a constant relativistic velocity? Yes, any practical clock would immediately shrink to a tiny point an instant after you passed it. So, either assume a really huge clock (or alternatively a fast-rotating planet that you could watch while moving along the line defined by its poles). I'm really only interested in a precise formulation of the time that the clock's hands would read to you the observer as you move. As you move away from it (or towards it for that matter), the clock should show a component of slowing due to a permanent time loss due to relativistic effects. It should also show a recoverable component of slowing while moving away (that you could regain as an apparent quickening while moving back towards it) due simply to your growing or shrinking distance from it and regardless of your speed.

The stationary clock doesn't slow down.
Only a moving clocks slow down, but the Doppler efect is symmetric, therefore there is no possibility to resolve who moves: you or the clock.

a stationary clock and a moving observer:
1 - v/c - the Doppler factor for a moving observer,
but a clock of the moving observer slows down, therefore he observes higher frequency - gamma times:
(1 - v/c) * gamma = ...


A stationary observer and a moving clock:
1/(1+v) - it is a Doppler for a moving source;
but a frequency of the moving source slows down additionaly, and gamma times, therefore:

1/(1+v/c) / gamma = ...

the both results should be equal.

 

[TJonline]

Thanks again to all responders! I could press the Thanks button to everyone's response, but I guess I'll save myself the trouble and just thank you all at once (unless there is a good practical reason to use the Thanks button).

I finished my model and as far as I could tell, it was correct. If I started the observer moving away from the clock, the observed second hand would slow relative to that of the pocketwatch, more so as the relative velocity increased, until the observed second hand stopped completely when the observer moved at the velocity of light relative to the clock, as expected. If I then stopped the observer some distance away from the clock, the observed time offset became constant, as expected. If I moved the observer all the way back to the clock, then only the constant relativistic component of time lost would remain, as expected.

My only remaining question was, would your intuition (or a more rigorous measure) of the proportion of relativistic time lost compared to the non-relativistic time lost agree with my model.

And then I broke my model somehow such that when the observer's velocity relative to the clock approached the speed of light, the clock started running backwards. Just a bug that I need to ferret out. I've implemented my model in LabView. I can wrap up the model into a Windows executable and share it with you guys when I'm finished, if anyone expresses an interest and the Mentor doesn't delete my link to it as inappropriate (?).

Again, thanks for all the input. I'll be... back.

 

[TJonline]

I'm currently struggling with the model. Assume the model begins at t=0 (according to the pocketwatch) with the observer right at the clock and moving at a constant velocity of 0.5 m/s away from the clock. Assume also that the speed of light is 0.5 m/s. At t=1s (since the observer is moving at the speed of light) the model must subtract 1s from the observed clock time (clock time appears to have stopped completely). OK so far. But at t=1s, the observer is now 0.5m away from the clock, so my current model subtracts ANOTHER second from the clock (due to the time for light leaving the clock to transit that distance). I know that it's incorrect to subtract 2s from the observer's view of the clock (the clock has apparently run backwards by 1s). I'm stymied as to how to reconcile the relativistic and non-relativistic time offsets to the model for that moment. Any suggestions?

 

[TumblingDice]

I can think of two possible cause of problems. First, I wouldn't test the model with the observer moving at the speed of light, because relativity doesn't allow that, and the math in your model shouldn't work/allow that either. Also, it sounds as though your model is double-counting a classical model's measurement, as you only mentioned distance and time (no references to relativistic calculations).

The reason you're double counting is because you begin by adding 1s for your receding motion, and then add another for the light to reach you. But the light is traveling while you are - it doesn't wait for you before it begins to travel, so you wouldn't want to add your 1s and the light travel at 1s, too. But with that said, I think you want to use the Relativistic Doppler calc in your model to reach what I *think* you really want to accomplish. Here's a link to a thread that specifically addresses what one would "see" in the manner you're describing, and it uses clocks, too! George provides a good post #5 that covers exactly what you want to calculate about observing a clock:

www.physicsforums.com/showthread.php?t=761982

You may find more examples in other threads, too!

 

[TJonline]

Hi again,
Finished the model. Checked it by comparing its results with a few relativistic there and back scenarios. Was even able to determine that a Shockwave demo at a PBS Einstein-oriented website is currently providing erroneous results whereas a scenario provided in text is basically accurate. Thanks again for all the help and comments.

Lessee... a there and back scenario is modeled as follows:
I choose a number of increments n.
dt is my time increment = (e * dist) / (n * v)
e is the reciprocal of the Lorentz Factor = sqrt( 1 - (v/c)^2 )
dist is the rest frame distance to target
v is a constant (less than c)
The traveler travels a shorter than rest frame distance (e * dist), due to length contraction while traveling.
vel is the velocity of the traveler/observer = v * [0..n-1] concatenated with -v * [n..2n-1] (positive velocity outgoing, negative velocity incoming for the round trip)

Each pocketwatch increment = dt * [0..2n-1]
f is the observed ticks/sec of the resting clock = (1 - vel/c) / sqrt( 1 - (vel/c)^2 )
Each apparent clock increment = f * dt * [0..2n-1]

 

[m4r35n357]

Can anyone answer this question?

What you would see while watching a stationary clock that you are moving directly away from at a constant relativistic velocity? Yes, any practical clock would immediately shrink to a tiny point an instant after you passed it. So, either assume a really huge clock (or alternatively a fast-rotating planet that you could watch while moving along the line defined by its poles). I'm really only interested in a precise formulation of the time that the clock's hands would read to you the observer as you move. As you move away from it (or towards it for that matter), the clock should show a component of slowing due to a permanent time loss due to relativistic effects. It should also show a recoverable component of slowing while moving away (that you could regain as an apparent quickening while moving back towards it) due simply to your growing or shrinking distance from it and regardless of your speed.

 

[m4r35n357]

I am wondering if you have had you question answered as clearly as it might have been. The time you will see on a huge clock is the time it is showing in its own reference frame, minus the light delay time to your eyes. It's that simple, your relative speed makes no difference whatsoever!

Of course you can integrate the ticks using the Doppler formula given above, but that is a roundabout way of doing it in my opinion.

 

[pervect]

I am wondering if you have had you question answered as clearly as it might have been. The time you will see on a huge clock is the time it is showing in its own reference frame, minus the light delay time to your eyes. It's that simple, your relative speed makes no difference whatsoever!

Of course you can integrate the ticks using the Doppler formula given above, but that is a roundabout way of doing it in my opinion.

I'm not sure if the OP is still around or what he was really up to, but since we haven't heard from him, I assume he thinks he's happy.

I'm not sure your answer is actually equivalent to the doppler formula, because of the relativity of simultaneity. It is true that the time difference visually seen on the big clock between an observer in your reference frame co-located with you and a second observer in your reference frame co-located with the big clock is equal to the propagation delay in your reference frame, that you see an earlier time on the clock (visually) than an observer in your reference frame who is colocated with the clock. But that is not precisely what you said, it may or may not be what you intended, I can't really tell. It's not the same as what you said because the concept of "now" is different in your reference frame and in the clocks reference frame.

 

[m4r35n357]

I'm not sure your answer is actually equivalent to the doppler formula, because of the relativity of simultaneity. It is true that the time difference visually seen on the big clock between an observer in your reference frame co-located with you and a second observer in your reference frame co-located with the big clock is equal to the propagation delay in your reference frame, that you see an earlier time on the clock (visually) than an observer in your reference frame who is colocated with the clock. But that is not precisely what you said, it may or may not be what you intended, I can't really tell. It's not the same as what you said because the concept of "now" is different in your reference frame and in the clocks reference frame.

I am talking about a MUCH simpler scenario than you describe above, in the sense that my reference frame doesn't enter into the calculations at all! Let's see if I can be clearer. I set off from the clock as it reads t = 0 and travel at 0.6c (observing rods & clocks through a porthole) until I reach the 3ly milestone. I notice the clock on the milestone reads 5y. My colleague looks at the huge clock through a telescope. I say to him "does the big clock read 5 - 3 = 2 years?". He says "yes".
At the same time, I notice an observer located at the milestone with a telescope. I say to him "does the big clock read 5 - 3 = 2 years?". He says "yes".
So, all I am saying is the light signal (which could be a radio transmission of a photograph of the clock face) that reaches me at a distance of 3ly is the one that reaches the stationary observer at 3ly which is the one that left the clock when it was reading 2 years, because the light takes 3 years to get to both me and the stationary observer at that location. I notice that my wristwatch reads 4y but that does not enter into the calculation. I hope that conveys my meaning better!

 

[m4r35n357]

I'm not sure your answer is actually equivalent to the doppler formula, because of the relativity of simultaneity.

Well on a spacetime diagram like the one with the red lines in http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_vase.html
(look at "Terence sends pulses to Stella"), my light delay analysis agrees with the doppler analysis, and there is no discussion of simultaneity in that scenario. I used a similar diagram to confirm my figures.

 

[TumblingDice]

I am talking about a MUCH simpler scenario than you describe above, in the sense that my reference frame doesn't enter into the calculations at all! Let's see if I can be clearer. I set off from the clock as it reads t = 0 and travel at 0.6c (observing rods & clocks through a porthole) until I reach the 3ly milestone.

Might there be a problem nailing down a 3ly milestone between both observers? I'm wondering if your simpler scenario requires a more complex set of prerequisites to maintain its integrity. Would the observers be required to mark this milestone in advance so both can agree on the same location?

Also, the OP specifically asked about observing while moving:

I'm really only interested in a precise formulation of the time that the clock's hands would read to you the observer as you move.

The additional posts you've contributed don't embrace or address this.

 

[ghwellsjr]

I am talking about a MUCH simpler scenario than you describe above, in the sense that my reference frame doesn't enter into the calculations at all! Let's see if I can be clearer. I set off from the clock as it reads t = 0 and travel at 0.6c (observing rods & clocks through a porthole) until I reach the 3ly milestone. I notice the clock on the milestone reads 5y. My colleague looks at the huge clock through a telescope. I say to him "does the big clock read 5 - 3 = 2 years?". He says "yes".
At the same time, I notice an observer located at the milestone with a telescope. I say to him "does the big clock read 5 - 3 = 2 years?". He says "yes".
So, all I am saying is the light signal (which could be a radio transmission of a photograph of the clock face) that reaches me at a distance of 3ly is the one that reaches the stationary observer at 3ly which is the one that left the clock when it was reading 2 years, because the light takes 3 years to get to both me and the stationary observer at that location. I notice that my wristwatch reads 4y but that does not enter into the calculation. I hope that conveys my meaning better!

Your analysis is correct but it doesn't answer the OP's question as expounded in post #6:

...I really meant only to consider an...observer's view of the...clock's hands as a function of the observer's...pocketwatch that he carries with him.

You haven't provided the requested function as pervect did in post #8.

 

[m4r35n357]

Might there be a problem nailing down a 3ly milestone between both observers? I'm wondering if your simpler scenario requires a more complex set of prerequisites to maintain its integrity. Would the observers be required to mark this milestone in advance so both can agree on the same location?

Also, the OP specifically asked about observing while moving:

The additional posts you've contributed don't embrace or address this.

Yes they do; I am talking about observing while moving (ie. I never said that I stopped, just asked a stationary observer at the event)! Was there something I said that implied otherwise? As to the formulation, if we call the time shown by the hands on the clock T and the reception event is (t, x), then T = t - x/c. I was using c = 1ly/y in my earlier posts.
I'm not sure how much more information I can give, let me know what you need me to say.
Ah OK, I was answering the OP, not Pervect's, and the OP does not ask for the formulation in terms of the traveller's proper time, so I gave it in terms of coordinate time.

 

[m4r35n357]

Your analysis is correct but it doesn't answer the OP's question as expounded in post #6:You haven't provided the requested function as pervect did in post #8.

Agreed; I answered post #1.

 

[ghwellsjr]

Ah OK, I was answering the OP, not Pervect's, and the OP does not ask for the formulation in terms of the traveller's proper time, so I gave it in terms of coordinate time.

He didn't specifically ask in post #1 what he wanted so maybe we could say you provided an answer that fit what he originally asked but as I pointed out, he expounded on his request in post #6. He wants to know what time the traveler sees on the "stationary clock" that he left behind as a function of the Proper Time on his wristwatch that he carries with him, not the Coordinate Times on additional "stationary clocks" that have been previously synchronized.

 

[m4r35n357]

Yep, you are right, no complaints from me, I must admit I wanted you guys to look over my reasoning too ;)

 

[TumblingDice]

He didn't specifically ask in post #1 what he wanted so maybe we could say you provided an answer that fit what he originally asked

@jhwellsjr - The last thing I want to do is get in the middle here. I'm looking towards you for your quality diagnosis I always enjoy. And yet I'm still looking at the OP and seeing conflicts:

I'm really only interested in a precise formulation of the time that the clock's hands would read to you the observer as you move. As you move away from it (or towards it for that matter), the clock should show a component of slowing due to a permanent time loss due to relativistic effects. It should also show a recoverable component of slowing while moving away (that you could regain as an apparent quickening while moving back towards it) due simply to your growing or shrinking distance from it and regardless of your speed.

I don't see how m4r35n357 posts address this movement, or the relativistic effects that relative motion contribute to the query regarding "the observer as you move". I mentioned the 3ly milestone in my last post - not sure if that was appropriate - and wondering why you're agreeing with m4r35n357:

Your analysis is correct...

I thought I had my feet on the ground until m4r35n357 added these follow-up posts. @PeterDonis added a tweak to my first reply with his post #3 to my post #2. I thought my post #2 was pretty much what m4 has been posting, and Peter's correction remains valid to me.

Apologies if my ability to understand is getting in the way here. Many TIA for helping me see through the smoke!

 

[m4r35n357]

I don't see how m4r35n357 posts address this movement, or the relativistic effects that relative motion contribute to the query regarding "the observer as you move". I mentioned the 3ly milestone in my last post - not sure if that was appropriate - and wondering why you're agreeing with m4r35n357:

None taken ;) I think I get where you think I am wrong. If I understand your #2, you were talking about the rate of the clock, whereas I am talking about the reading on the clock. The observed rate does depend on speed via the doppler effect, whereas the reading only depends on the separation between two events, and NOT on motion. Of course the two must remain consistent, which I think everyone here agrees is the case (otherwise I'm sure someone will chip in). Does this help?

 

[ghwellsjr]

Hi again,
Finished the model. Checked it by comparing its results with a few relativistic there and back scenarios. Was even able to determine that a Shockwave demo at a PBS Einstein-oriented website is currently providing erroneous results whereas a scenario provided in text is basically accurate. Thanks again for all the help and comments.

Lessee... a there and back scenario is modeled as follows:
I choose a number of increments n.
dt is my time increment = (e * dist) / (n * v)
e is the reciprocal of the Lorentz Factor = sqrt( 1 - (v/c)^2 )
dist is the rest frame distance to target
v is a constant (less than c)
The traveler travels a shorter than rest frame distance (e * dist), due to length contraction while traveling.
vel is the velocity of the traveler/observer = v * [0..n-1] concatenated with -v * [n..2n-1] (positive velocity outgoing, negative velocity incoming for the round trip)

Each pocketwatch increment = dt * [0..2n-1]
f is the observed ticks/sec of the resting clock = (1 - vel/c) / sqrt( 1 - (vel/c)^2 )
Each apparent clock increment = f * dt * [0..2n-1]

This looks pretty much correct except I don't understand why you attached the bracketed part that I have set in bold above.

I think you should be aware that your equation for f is the same as the one the pervect gave you in post #8. Keep in mind that pervect defined β as v/c but with v in the opposite direction from your vel. So we can take your formulation and derive pervect's:

(1 - vel/c) / sqrt( 1 - (vel/c)^2 ) = (1+β)/√(1-β²) = √(1+β)²/√(1-β²) = √(1+β)²/√((1-β)(1+β)) = √(1+β)/√(1-β)

Can you provide a link to the Shockwave demo?

 

[TumblingDice]

I think I get where you think I am wrong. If I understand your #2, you were talking about the rate of the clock, whereas I am talking about the reading on the clock.

I was also talking about the reading on the clock in my #2. But that's not the issue I have regarding the posts you've added...

It seemed to me that the OP had not only received the answers s/he was looking for, but had also expressed thanks and reported back on progress made. And then, two days later you posted:

I am wondering if you have had you question answered as clearly as it might have been. The time you will see on a huge clock is the time it is showing in its own reference frame, minus the light delay time to your eyes. It's that simple, your relative speed makes no difference whatsoever!

Was it really "that simple" to "answer the question as clearly as it might have been"...? If you had posted that as a first responder, would the OP have been as pleased as the thread played out?

It bothers me a tad when a plan comes together for the OP and then posts are added that derail the success. Yet, in all honesty, I was the member that ended up confused, because everything made sense to me until your posts. I have a ton of respect for pervect, and he seemed a bit confused with your perspective (or politely suggested, it was how you tried to express what you meant to write).

I've twice mentioned your "3 light year" milestone, and if you and your colleague would have a problem agreeing where that was. Maybe you can hand wave around that - neither @ghwellsjr nor @PeterDonis have replied, so I'm taking their silence as an indication that answers have been given and it's up to me to understand better.

Does this help?

(see above)

 

[ghwellsjr]

I've twice mentioned your "3 light year" milestone, and if you and your colleague would have a problem agreeing where that was. Maybe you can hand wave around that - neither @ghwellsjr nor @PeterDonis have replied, so I'm taking their silence as an indication that answers have been given and it's up to me to understand better.

Does this spacetime help?

The thick blue line represents the "stationary" clock. The traveler is the red line moving away from the clcok at 0.6c. The black line is the 3 ly milestone. The clock at the milestone has been previously synchronized to the "stationary" clock and they both read the same value as the Coordinate Time. So when the traveler gets to the 3 ly milestone, he sees that its clock reads 5 years and since it is 3 ly from the "stationary" clock and he knows that the image of the "stationary" clock is traveling at c, it took 3 years to get to him so he subtracts 3 from 5 to determine that the time he sees on the "stationary" is 2 years.

In general, all the traveler has to do is subtract the distance marker on each milestone from the time on the clock at each milestone and he knows the time that he will see on the "stationary" clock. The point of this is that it works even if the traveler is not traveling at a constant speed. For example, suppose he travels for 1 year of his Proper Time at 0.943c and then slows down to 0.081c for 2 more years, he will arrive at the same 3 ly milestone at the Coordinate Time of 5 years so he can determine that he will see the "stationary" clock reading 2 years at that time. Note that his wristwatch reads 3 years in this circumstance. Here is another spacetime diagram to illustrate this new scenario:

 

[TumblingDice]

@ghwellsjr - Thank you for your reply. I never had a problem using a 3 light year milestone - only that this was not the same as the OP asked, and would be an added prerquisite to the setup that mr4 referred to as simpler, without mentioning thie significance it would add to the preparation.

I am talking about a MUCH simpler scenario than you describe above, in the sense that my reference frame doesn't enter into the calculations at all! Let's see if I can be clearer. I set off from the clock as it reads t = 0 and travel at 0.6c (observing rods & clocks through a porthole) until I reach the 3ly milestone.

Might there be a problem nailing down a 3ly milestone between both observers? I'm wondering if your simpler scenario requires a more complex set of prerequisites to maintain its integrity. Would the observers be required to mark this milestone in advance so both can agree on the same location?

ghwellsjr wrote:

The clock at the milestone has been previously synchronized to the "stationary" clock and they both read the same value as the Coordinate Time.

The OP has been derailed. The simpler scenario is a different animal than the OP. Now you have replied by supporting why the "supposedly simpler" scenario is accurate. My exceptions have not been about accuracy of the scenario. My concern is that it's more confusing than helpful, providing the OP with an answer to a question that's different than that being asked, and leaving out important requirements, too. It certainly isn't simpler if it requires locating and synchronizing two clocks three light years apart in advance.

This may be a case of posting without reading enough to understand the thread's evolution first. George, I'm really sorry that I've taken up your time with the space diagram and all, and I do appreciate the effort! This is just getting further off topic the harder I try, so maybe best for me to let it go...

<sigh>

 

[m4r35n357]

I have recently wrestled with the precisely the question in the title and the OP myself, so I answered them as they were both written, with the answer I would have liked to read myself to the same question. OK I slipped up quoting the actual message, but tripped up over the new interface (the quote is there in the previous message). I just wanted to share a moment of clarity with a fellow learner. I still think I answered #1 and the thread title more accurately than the following discussion (again, speaking from the viewpoint of a learner).

 

[ghwellsjr]

@ghwellsjr - Thank you for your reply. I never had a problem using a 3 light year milestone - only that this was not the same as the OP asked, and would be an added prerquisite to the setup that mr4 referred to as simpler, without mentioning thie significance it would add to the preparation.
...
The OP has been derailed. The simpler scenario is a different animal than the OP. Now you have replied by supporting why the "supposedly simpler" scenario is accurate. My exceptions have not been about accuracy of the scenario. My concern is that it's more confusing than helpful, providing the OP with an answer to a question that's different than that being asked, and leaving out important requirements, too. It certainly isn't simpler if it requires locating and synchronizing two clocks three light years apart in advance.

This may be a case of posting without reading enough to understand the thread's evolution first. George, I'm really sorry that I've taken up your time with the space diagram and all, and I do appreciate the effort! This is just getting further off topic the harder I try, so maybe best for me to let it go...

<sigh>

It would be awfully boring for me if people didn't ask for spacetime diagrams, either directly or indirectly, and I appreciate the feedback.

But the OP did ask in his first post about a solution related to distance and unrelated to speed which none of us except mr4 addressed:

It should also show a recoverable component of slowing while moving away (that you could regain as an apparent quickening while moving back towards it) due simply to your growing or shrinking distance from it and regardless of your speed.

And if you look at his description of his program in post #14 that he is concerned about distance:

dist is the rest frame distance to target

To me, the problem is that the OP didn't say precisely what he wanted in his first post and he asked about issues that appeared off base (the recoverable component of time). This happens all the time with people who are trying to learn SR and the variation of responses in this thread is typical. I just wish the OP would come back and provide more feedback because without that, we are all trying our best to interpret what we think he wants.

 

[TJonline]

The algorithm that I presented above agreed exactly with the there and back scenario given below (please don't delete link as inappropriate, mysterious Mentor, and if you do, please say why you consider it inappropriate). The difference between clock and pocketwatch when the observer finally gets back to the clock and comes to an instantaneous stop were exactly the same. It might be erroneous to assume that what the clock reads to the observer during the journey is thus also correct, but I don't think so.

http://en.wikipedia.org/wiki/Twin_paradox

4 light years one way as measured in clock (rest) frame, v=0.8 c (instantaneous acceleration), clock and pocketwatch time upon return: 10 and 6 years

It also agreed pretty well with the one in the text in the page below. The error I attribute to approximation errors in the data they provided:

http://www.pbs.org/wgbh/nova/einstein/hotsciencetwin/

25 light years one way, v=0.9999 c, clock and pocketwatch time upon return - algorithm: 50 and 0.704 years, in text: 50 and 0.5 years

The algorithm agreed not at all well with the shockwave demo on the same page.

4.2 light years one way (Proxima Centauri), v=0.5 c, clock and pocketwatch time upon return - algorithm: 16.811 and 14.486 years, shockwave: 47.30 and 40.96 years. Be sure to subtract off the two fellows' starting ages (assume both were just born at the start of one's journey). Maybe that demo assumed something other than instantaneous acceleration. Who knows?

Thanks.