# Viewing a (giant) clock at relativistic speed

1. Sep 13, 2014

### TJonline

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.

Last edited by a moderator: Sep 14, 2014
2. Sep 13, 2014

### 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.

3. Sep 13, 2014

### Staff: Mentor

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.

4. Sep 13, 2014

### TumblingDice

Thank you, Peter!

5. Sep 13, 2014

### Simon Bridge

Welcome to PF;

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.

6. Sep 14, 2014

### 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!

Last edited by a moderator: Sep 14, 2014
7. Sep 14, 2014

### ghwellsjr

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.

8. Sep 14, 2014

### pervect

Staff Emeritus
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.

9. Sep 14, 2014

### Simon Bridge

... 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.

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.

10. Sep 15, 2014

### quo

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.

11. Sep 15, 2014

### 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.

12. Sep 15, 2014

### 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?

13. Sep 15, 2014

### 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:

You may find more examples in other threads, too!

14. Sep 22, 2014

### 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]

15. Sep 24, 2014

### m4r35n357

16. Sep 24, 2014

### 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.

17. Sep 24, 2014

### pervect

Staff Emeritus
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.

18. Sep 25, 2014

### m4r35n357

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 in to the calculation. I hope that conveys my meaning better!

Last edited: Sep 25, 2014
19. Sep 25, 2014

### m4r35n357

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.

20. Sep 25, 2014

### TumblingDice

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?