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In writings concerning time dilation and GPS (incl. on PSE) one can find statements such as

When viewed from the surface of the Earth, the clocks on the satellites appear to be ticking faster than identical clocks on the ground.

To me, this statement and its variants are in need of explanation ...

Apparently it is thereby assumed that the mentioned clocks (for concreteness let's refer to one of the clocks on the satellites as clock $\mathfrak B$, and to one of the clocks on the ground as clock $\mathfrak C$) are each characterized by their particular tick rates; accordingly $\nu_{\mathfrak B}^{\,}$ and $\nu_{\mathfrak C}^{\,}.$

And further, the prescription of those clocks being "identical" is to be understood that all those clocks have equal tick rates, in particular $$\nu_{\mathfrak B}^{\,} = \nu_{\mathfrak C}^{\,}.$$

But then, what exactly is meant by:
"clock $\mathfrak B$ appearing to be ticking faster than clock $\mathfrak C$, as viewed by someone else (say $\mathbf A$)"
??

Would this phrase in fact be referring to a comparison of certain rates of $\mathbf A$, namely $\mathbf A$'s rate of receiving the tick signals issued by clock $\mathfrak B$ being greater than $\mathbf A$'s rate of receiving the tick signals issued by clock $\mathfrak C$; symbolically: $$\nu_A^{(\circledR \, \mathfrak B)} > \nu_A^{(\circledR \, \mathfrak C)}$$ ?

Qmechanic
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user12262
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  • The question is a little unclear. Your quote at the beginning talks about clock $B$ being viewed from location $C$. Where are you getting location $A$ ? – RC_23 Sep 29 '21 at 03:40
  • @RC_23: "[...] Your quote at the beginning talks about [...]" -- As far as I understand, the quote refers to three kinds of subjects or protagonists or participants: some constituents of the surface of the Earth, some clocks on the satellites, and some clocks on the ground. The constituents of the surface of the Earth and the clocks on the ground need not necessarily be disjoint, I suppose. "clock $B$" -- I prefer and suggest to denote plain participants (such as "material points") with plain font capital letters; while using special font for [contd.] – user12262 Sep 29 '21 at 06:18
  • ... using a special font for denoting additional structure which is attributed to participants; such as clock $\mathfrak B$ referring to some particular identifiable participant, say $B$, together with the ordered set $\mathcal B$ denoting $B$'s tick indications, and together with some particular assignment $$ t_{\mathfrak B} : \mathcal B \rightarrow \mathbb Z $$ by which $B$'s tick indications are enumerated. – user12262 Sep 29 '21 at 06:18

3 Answers3

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The quoted sentence is probably intended to mean what you suggest: the two clocks emit ticks, which are received somewhere, and the ratio between the time-averaged rates at which the ticks are received isn't $1$.

This omits details about the nature of the tick signals (light, sound, etc), and where the receiver (your $A$) is located, but the ratio of the rates turns out to be independent of those details, as long as the whole system including the signals and receiver is in a quasi-steady state. (That condition is intended to rule out silly situations like bouncing the signal from the satellite off of an ever-receding mirror, or using slower-than-light signals that get slower and slower as time goes on.)

benrg
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The statements are misleading at best and possibly nonsense. In SR all (ideal) clocks measure time at the same rate in their respective rest frames. The phenomenon referred to as time dilation arises because the planes of simultaneity of two references frames in motion relative to each other are tilted, so a level plane in one frame, across which it is the same time everywhere, is a tilted slice through time in the other frame, and vice versa.

To take a concrete example, suppose you start walking down a corridor at 1m/s beginning at time t=0. Upon the wall where you start is a clock identical to your own, also showing t'=0. Along the corridor at 5m intervals are other clocks identical to you own and ticking at exactly the same rate as your own. However, each of the clocks placed along the corridor has been set to be 1 second ahead of its nearer neighbour.

After you have walked for 5 seconds you reach the first clock along. Your clock reads t=5, but the clock on the wall, having been set 1 second ahead, reads t'=6. When you reach the next clock 5 seconds later, your reads t=10s but the adjacent clock being two seconds ahead reads t'=12s. At the next clock, yours reads t=15s and it reads t'=18s. And so on down the corridor. All the clocks- yours and those one the wall- are ticking at exactly the same rate, but the cumulative time on your clock gets progressively further behind the time shown on the walls clocks as you move down the corridor- in other words, your clock seems to running slow. It is doing nothing of the sort, it is simply that the wall clocks were already reading progressively later times at the moment you set off.

Exactly the same effect is the cause of relativistic time dilation. When you start to move relative to another frame at t=0, it is t=0 everywhere in your frame. In the other frame, however, it is only t'=0 where you are- further ahead in your direction of travel it is already a progressively later time. Your clock and all the clocks in the other frame are measuring time at the same rate, but since you are moving between successive clocks in the other frame, each of which started off ahead of the previous one, the time shown on your clock is progressively less than the readings on the clocks you pass.

So you will see that a statement that the moving clock runs slower than the stationary one is wrong, and shows a complete misunderstanding of what is happening. It is also a source of endless confusion and questions about 'how can both clocks be running slower than the other?'

Marco Ocram
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  • Marco Ocram: "[...] In SR all clocks tick at the same rate in their respective rest frames." -- This strongly disagrees with my understanding; I've just set up the question "How to call the "constant factor $K$" in Gourgoulhon's definition of "ideal clocks"? And: May distinct ideal clocks have unequal values $K$?" (PSE/q/670030) as opportunity to resolve our disagreement. The rest of your answer I find less disagreeable, but also less relevant to my OP question. – user12262 Oct 05 '21 at 18:46
  • Hi- my language was sloppy. Clearly an atomic clock ticks at a different rate than a seconds pendulum. I meant to say that all clocks measure time passing at the same rate. I will edit my answer accordingly. – Marco Ocram Oct 05 '21 at 19:59
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    Indeed, I often wonder whether mentioning clocks in connection with SR is partly the cause of the confusion associated with the subject. What we really mean is the passage of time, which happens without the presence of clocks. – Marco Ocram Oct 05 '21 at 20:02
  • Marco Ocram: "Clearly an atomic clock ticks at a different rate than a seconds pendulum." -- Alright: this matches my understanding of "tick rate" as $1/K$, in line with Gourgoulhon's terminology. (Btw., this easily generalizes to non-ticking clocks; incl. forensic clocks, geological, biological, astro-physical clocks...) "{ You edited } In SR all (ideal) clocks measure time at the same rate in their respective rest frames." -- There you use "rate" apparently in a different sense, or perhaps sloppily, or possibly even non-sensically. Also: [contd.] – user12262 Oct 05 '21 at 21:01
  • Also: we can have clocks which aren't members of any inertial system (if that's what you mean by a "rest frame"); e.g. the clocks of my OP question. "whether mentioning clocks in connection with SR is partly the cause of the confusion" -- I appreciate the sentiment, but I'm more consequent/radical: aiming to establish geometry/kinematics/frames, durations, distances, speeds etc. without involving any coordinates, parametrizations, manifolds, clock readings. It can and must be only subsequently determined which parametrizations make a "good clock" of any given world line, and which don't. – user12262 Oct 05 '21 at 21:09
  • It is a challenge to put these nuanced points into words. What I was trying to convey is the idea that a second is a second at every point in every interval reference frame, and to make that point in contrast to those who take the view that time dilation means a slowing down of time in moving reference frames. The point I am trying to make is that time does not slow down in moving reference frames- the phenomenon of time dilation is a consequence of the relativity of simultaneity. – Marco Ocram Oct 05 '21 at 21:24
  • It seems to me that there is no absolute distance or absolute time- the only absolute is the ratio of the two expressed as the speed of light. – Marco Ocram Oct 05 '21 at 21:28
  • If you have three identical clocks ticking at the same rate and initially separated and at rest to each other, and then you move one of the clocks between the other two, the elapsed time for the trip shown on the moved clock is less than the difference in the readings on the two stationary clocks at the start and end of the trip. You can say that the moving clock slowed down, or you could say that the difference arose because the two stationary clocks were out of synch in the frame of the moving clock. The first explanation seems entirely untenable for the following reason... – Marco Ocram Oct 05 '21 at 21:33
  • ...namely that you can consider the moving clock to be simultaneously moving in relation to n other reference frames each moving at a different speed to all the rest. In each of those frames the degree of time dilation associated with the moving clock will be different- clearly it is impossible for the moving clock to slow down by n different rates simultaneously. – Marco Ocram Oct 05 '21 at 21:38
  • Marco Ocram: Take your own advice! -- Disregard clocks and work coordinate-free!: In a flat region, consider principal participants $A$, $B$, $P$ and auxiliaries $H$, $J$ such that for the spacetime intervals between suitable coincidence events holds: $$ s^2[ \varepsilon_{AP}, \varepsilon_{BH} ] + s^2[ \varepsilon_{BH}, \varepsilon_{BJ} ] = 0, \ s^2[ \varepsilon_{AP}, \varepsilon_{BJ} ] = 0, \ \sqrt{ s^2[ \varepsilon_{BJ}, \varepsilon_{BP} ] } = \sqrt{ s^2[ \varepsilon_{BJ}, \varepsilon_{BH} ] } + \sqrt{ s^2[ \varepsilon_{BH}, \varepsilon_{BP} ] }.$$ Then: [contd.] – user12262 Oct 05 '21 at 22:00
  • Then $$s^2[ \varepsilon_{AP}, \varepsilon_{BP} ] = s^2[ \varepsilon_{BH}, \varepsilon_{BP} ] \times (1 - (s^2[ \varepsilon_{BH}, \varepsilon_{BJ} ] / s^2[ \varepsilon_{BH}, \varepsilon_{BP} ])).$$

    Why?: Because "flat" means "vanishing Cayley-Menger determinant".

     – user12262 Oct 05 '21 at 22:01
  • ha! I'm afraid that on principle I refuse to hold conversations about physical concepts using the language of mathematics. If you would care to explain what your symbols mean, I would be delighted to hear what you have to say. – Marco Ocram Oct 05 '21 at 22:10
  • Marco Ocram: "ha! I'm afraid that on principle I refuse to hold conversations about physical concepts using the language of mathematics." -- Well, what do you think you're doing yourself when writing of "ratio" and "difference", for instance ? (just to quote some obvious cases from your above comments). "If you would care to explain what your symbols mean, [...]" -- Ha! ... – user12262 Oct 06 '21 at 06:31
  • Touché. All I have to say is x cos tan pi sigma square root. – Marco Ocram Oct 06 '21 at 07:04
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The main principle in relativity theory is that anyone will see physical processes in their local vicinity proceeding like normal. Any time you have to observe something far away, the rate of those processes may be different from the far away perspective, because of relative motion or gravitational fields (which are a type of relative motion).

The far away observation could be accomplished in any number of ways, e.g. through a telescope, or through the far away clock sending regular signals that the other observer receives. Does that help?

RC_23
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  • RC_23: "Does that help?" -- Thanks, but not a lot, to be honest and kind. The purpose of my question was rather to provide proper explicit translations to statements which are less ... straightforward. E.g. above: If the title phrase "viewed by $A$, $\mathfrak B$ ticks faster than $\mathfrak C$" is not meant as "$\nu_{\mathfrak B}^{,} > \nu_{\mathfrak C}^{,}.$ can be and has been properly measured by those directly involved; and everyone understands and agrees; incl. $A$" then the title phrase requires and admits translation in which certain rates of $\mathbf A$ feature explicitly. – user12262 Sep 30 '21 at 21:40
  • Referring to another perhaps better known example: Instead of following the habit since 1905 of referring to "the length of the train in the frame of the railway station platform" and to end up concluding: "In the frame of the railway station platform, the train appears to be contracted by $\gamma$.", I prefer and I'd suggest ...............: "The train is longer than the railway station platform, by a factor $\gamma$." – user12262 Sep 30 '21 at 21:41
  • So your question is about the semantics or terminology used, and not about the underlying physical concepts? – RC_23 Oct 01 '21 at 04:45
  • RC_23: "So your question is about the semantics or terminology used, " -- Explaining a phrase (e.g. a phrase which refers to a physical concept, setup, methodology) necessarily involves the semantics or terminology used. Accordingly, my question had from the outset, and still has, the label [terminology], along with other appliable labels. "and not about the underlying physical concepts? " -- Explaining a phrase necessarily also involves the content of that phrase; e.g. the physical concepts to which it alludes. Ideally, the relevant physical concepts are thereby even further clarified. – user12262 Oct 01 '21 at 06:10