Clocks in special relativity

One book on special relativity says:

Any observer at rest relative to his own timepiece will see that other clocks moving with respect to him run fast - the greater their speed, the faster they are.

Other book says:

Observers measure any clock to run slow if it moves relative to them.

Don't they contradict each other? If yes - who's right? If no - why are they both right? I assume it is a very newbie question, but relativity is one of the topics where you recheck every statement again and again, so I want to be sure.

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Maybe the statement about faster clocks included the Doppler effect, or something like that. It can be hard to understand special relativity because, in practice, observation is limited by the time taken for the light to travel, but the theory ignores this, assuming that observations are back-calculated. – Alan Rominger Jun 12 '14 at 15:11
Which book claims that they run faster? – George G Jun 12 '14 at 15:13
@user50381 could you give the title and page number? – Danu Jun 12 '14 at 15:21
@Danu - "What Is Relativity? L. D. Landau, G. B. Rumer" (p. 47). See preview of the page: books.google.com/… – user50381 Jun 12 '14 at 15:23
insti.physics.sunysb.edu/~siegel/sr.html "The analog in Euclidean geometry is that a board at an angle looks shorter than one standing up. So if you try to fit a board through a window, from the window's point of view the board looks like it will fit through better. But from the board's point of view it is the window that looks shorter, so the board will not fit. Here is a "paradox", and we didn't even need relativity! Needless to say, the solution is just as simple. " – Count Iblis Jun 12 '14 at 18:15

In the first book you linked to in the comments, I think the author is trying to say that if an observer sets their watch by a clock in a train station, then gets in a train and travels to another train station, they will find that their watch is slow compared to the station's clock.

This is true because in order to travel from one station to another, the rider must accelerate. During the acceleration, the clock in the station will appear to tick much faster, and during the constant velocity part of the trip (if there is one) the clock in the station will tick more slowly.

These lecture notes do a good job of explaining how acceleration works in SR.

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Hmm, very interesting. It is unclear that they mean acceleration here. They could've said explicitly - "other clocks accelerating with respect to him" instead of "other clocks moving with respect to him". I would definitely have a look at those lectures. Thanks. – user50381 Jun 12 '14 at 16:06
Yeah, the wording is not clear, but on the next page, there is a little discussion about acceleration, so I think this must be what the author meant. – George G Jun 12 '14 at 16:08
@GeorgeG - I don't think it has anything to do with acceleration. In Landau's book he describes a passenger in a train which compares his own clock with the stations' clock. In his particular example: if the time it took for the train to get from station $A$ to station $B$ is 1 hour (according to the clock on station $B$) then according to the clock of the passenger it took him about 36 minutes to get from $A$ to $B$. Thus, for the passenger, the clocks on the stations run faster (because when he gets to station $B$ he sees that the clock there is 24 min ahead of his own clock). – cth Sep 11 '15 at 14:25
Equations for SR take no account for acceleration. – bright magus Sep 16 '15 at 12:43
Inertial frames don't accelerate, but there is nothing wrong with talking about how an accelerated object appears to one or more inertial observers. – George G Sep 18 '15 at 14:38

Here's a basic, basic and rather heuristic explanation. When looking at velocity through spacetime with respect to the time of the person doing the traveling, (the person who left the train station's frame), we can use something called proper velocity. Proper velocity is the distance you travel as measured in the train stations frame, (the frame you will eventually return to), divided by the time in your moving frame, (called the proper time, $\tau$).

The magnitude of your proper velocity with respect to a given frame is fixed at c, the speed of light. If you're sitting at rest in the frame, you're hurtling through the time dimension at the speed of light. As you begin to increase your speed through space, your velocity through time with respect to the train station's fixed frame slows down. The entire time you're moving at this increased speed through space, the train station is moving through its dimension of time more quickly than you are. Here's an important point:

Acceleration matters in that you change your speed through time relative to the train station, but the interval that you remain at the new speed matters as much if not more. This was detailed in a fairly recent AJP article[2].

As for your original question... let's see. As usual in physics paradoxes, everyone is correct. Here's my Cinton-esque interpretation of the two statements above. It depends on what the word 'observer' means. In the first statement, the observer is the traveler. You can tell because he's at rest with respect to his own clock and concerned about the clock back at the train station. This comment fairly screams 'proper time' and 'proper velocity' once you've seen enough of these articles/books. Furthermore, L&L are bigtime into proper spacetime velocity which you'll see written as a two-vector a little further on in the text. I don't have my copy sitting right here, but they do tend to focus on proper velocity.

The observer in the second statement is the usual special relativity 'observer' used in most texts. He stays at the train station and measures everything with respect to his frame's distance and his frame's time. If he had a way to magically look at your clock, then yes, it would be moving slower than his. He can verify for certain that your clock ran slowly if you return.

I hope this helps as it rambled a bit and didn't include much of the underlying math at all. If you'd like to get into that aspect of it, or if you have any other questions, please let me know.

References

1. Geometrization of the Relativistic Velocity Addition Formula, Robert W. Brehme, Citation: Am. J. Phys. 37, 360 (1969); doi: 10.1119/1.1975576, View online: http://dx.doi.org/10.1119/1.1975576

2. Zero time dilation in an accelerating rocket, Ronald P. Gruber and Richard H. Price, Citation: Am. J. Phys. 65, 979 (1997); doi: 10.1119/1.18700 View online: http://dx.doi.org/10.1119/1.18700

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Notice that the first citation says "other clocks moving with respect to him run fast". Obviously, you can switch the moving frames in SR, however, if you want it all to make any sense, you also have to switch the "direction" of time dilatation. Otherwise it all looses all appearance of science and becomes just magic tricks. So, if you are right the authors of this book are just ... magicians. – bright magus Jun 12 '14 at 19:19
If I had a nickel for every time someone called one of my special rel explanations magic, I'd have 10 cents so far in the last year :) L&L are in fact at the caliber of what some might call magic, in the Arthur Clarke sense of the word, in their conciseness and elegance. The 'direction' of time dilation is a twin paradox misdirection. The attribute that determines whose clock runs slow is what frame is 'started from' and 'returned to'. L&L refer to proper velocity, i.e. moving frame. The second ref. almost certainly refers to laboratory frame. – dolphus333 Jun 12 '14 at 20:15
Whoops! @bright magus, my bad! I hadn't realized the book referred to wasn't L&L, but L&R? Looks like an interesting book, and thanks to the post's author for the cool reference! I'm still holding to the proper velocity frame though. – dolphus333 Jun 12 '14 at 20:19
What you are saying boils down to SR being bogus. Looks like Einstein's magic is not working anymore. Fine with me. Pity though other gods are taking over apparently. Not much room left for old good physics. What can one do? Magic is easier than reality, though not necessarily more fascinating. – bright magus Jun 12 '14 at 20:33
Actually, what I'm saying isn't as clear as I'd hoped it might be. I need to take more time to say it more precisely, which I hope to be able to do in the near future. Thanks for pointing out one of the blocks to clarity. – dolphus333 Jun 12 '14 at 21:00

Well, if both statements are interpreted sympathetically (and both are so short and improperly phrased that they can use a lot of sympathetic interpretation) then they are arguably consistent with each other and are referring to the same fairly simple experimental situation, described from opposite perspectives:

We have two participants, say $A$ and $B$, who are and remain at rest to each other, and another participant, $J$, who moved from $A$ to $B$; uniformly, with speed $\beta~c$. (This short description is sufficient to describe the setup unambiguously.)

Corresponding to these three participants in this setup there are three durations of particular relevance:

• the duration of $A$ from $A$'s (own) indication of having been left by $J$ until $A$'s (own) indication simultaneous to $B$'s indication of having been met by $J$; symbolically: $\tau A[ \circ_J, \circledS B_J ]$,

• the duration of $B$ from $B$'s (own) indication simultaneous to $A$'s indication of having been left by $J$ until $B$'s (own) indication of having been met by $J$; symbolically: $\tau B[ \circledS A_J, \circ_J ]$, and

• the duration of $J$ from $J$'s (own) indication of having been left by $A$ until $J$'s (own) indication of having been met by $B$; symbolically: $\tau J[ \circ_A, \circ_B ]$.

Obviously (due to $A$ and $B$ being at rest to each other)

$$\tau A[ \circ_J, \circledS B_J ] = \tau B[ \circledS A_J, \circ_J ];$$

and it is not difficult to derive (by appealing to the notions of "mutual rest" and "duration" and "speed", as defined within the theory of relativity) that

$$\tau J[ \circ_A, \circ_B ] = \sqrt{1 - \beta^2} \times \tau A[ \circ_J, \circledS B_J ];$$

and therefore (due to $0 \lt \beta^2 \lt 1$)

$$\tau J[ \circ_A, \circ_B ] \lt \tau A[ \circ_J, \circledS B_J ].$$

The suggested interpretation of the first statement is then to identify $J$ as "any observer (incl. his timepiece)" and $A$ and $B$ as the "other clocks";
while the suggested interpretation of the second statement is to identify $A$ and $B$ as the "observers" and $J$ as "any clock".

There's one more "subtlety" to note:
Earlier in the section "Clocks and Rulers Play Tricks" of Landau/Rumer's brochure (namely in the second paragraph of that section) it is pointed out:

But the watchmaker assured the traveller that his clock is perfectly alright.
[My translation from a German edition of Landau/Rumer's brochure, which I happen to have available at the moment.]

Therefore:

1. All clocks considered in Landau/Rumer's examples are (arguably) "running at equal rates"; there aren't some "running slow(er)" and/or others "running fast(er)",
but instead, corresponding to the inequality shown above, it could be said more correctly that $P$'s duration (or "run") was shorter than the corresponding durations (or "runs") of $A$ and $B$. And

2. It can be noted that the equations and the inequality shown above (incl. their derivation) are only concerned with comparing durations, not "rates" or "readings". These relations are independent of the "rates" of the various clocks being equal and "alright (in comparison to each other)", or not. Instead, these relations are useful for determining in the first place whether the "rates" of different clocks remained equal (as any watchmaker may have readily promised), or not, especially if the clocks to be compared were moving with respect to each other.

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The reason these statements are consistent becomes clear if we quote from the Landau & Rumer book a little more extensively:

Ahead of us is a very long railway line with Einstein's train moving along it. At a distance of 864,000,000 kilometers from each other there are two stations. At its speed of 240,000 kilometers per second, Einstein's train needs an hour to cover this distance.

There is a clock at each of these stations. A passenger boards the train at the first station and before its departure sets his watch by the station clock. On arriving at the second station, he notices with astonishment that his watch is slow.

The watchmaker had assured the passenger that his watch was in perfect order.

What has been going on?

[explanation of how these effects work in relativity, which is the standard textbook material]

So any clock in motion will run slow compared to a clock at rest. But doesn't this result contradict the principle of the relativity of motion, which was our starting point? Doesn't this mean that the clock that goes faster than any other is in a state of absolute rest? No, because we compared the watch in the train with the clocks at the stations in completely unequal conditions. We used not two but three timepieces! The traveler compared his watch with two different clocks at two different stations.

In other words, the crucial point (as cth notes in a comment) is that we are comparing the time elapsed on the passenger's watch to the difference in times between a pair of clocks, where the clocks in the pair are at rest relative to one another, and have been synchronised in their rest frame. (When I say they are synchronised, I mean that they have done the following thing: a light signal is fired from the first clock to the second, which reflects the signal back immediately. The first clock then sends the second clock a letter in which it specifies the times that its face indicated when it sent and received the light signal: call these $\tau_1$ and $\tau_2$ (so we might have $\tau_1$ = 17:00 and $\tau_2$ = 17:10, for example). The second clock then sets its timeface to read $\frac{1}{2}(\tau_2 - \tau_1) + \Delta \tau$, where $\Delta \tau$ is the time it has recorded elapsing since the reflection event. In other words, it sets its timeface such that the reflection event is assigned time $\frac{1}{2}(\tau_2 - \tau_1)$ (so in the example, it gets set such that the reflection event is assigned time 17:05. This is generally known as the Einstein-Poincaré synchrony convention.)

Had the passenger instead sought to calculate the time intervals between the ticks of one clock (let's say the clock at the first station) using his watch, by figuring out which events on his watch are simultaneous with those ticks, then he would have determined that the tick at 17:00:00 is separated from the tick at 17:00:01 by some interval larger than 1 second; that is, supposing the first tick is simultaneous with his watch reading $t$, the second tick will turn out to be simultaneous with his watch reading $t + 1 + \epsilon$, for some positive $\epsilon$. (You could actually calculate the numbers here, but I'm in a hurry and can't be bothered.) But had someone located at the station throughout (let's say, the station manager) sought to do the same thing, they would have reached the conclusion that the ticks on the passenger's watch are separated by intervals of greater than 1 second. That is, the station manager judges that the ticks on the passenger's watch are simultaneous with events on the station clock separated by more than 1 second. There's no contradiction here either: the fact that they reach "opposite" conclusions merely makes vivid the fact that the passenger and the station manager disagree about which events on the passenger's worldline are simultaneous with which events on the station manager's worldline.

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The Special Relativity Theory says that the moving clock is slower. It results from the the transformation equation for time that shows time dilatation:

$$\Delta t' = \Delta t \gamma$$

where $\Delta t'$ is time measured in a reference frame considered stationary, and $\Delta t$ is measured in a reference frame considered moving (with respect to the stationary one) and $\gamma>0$*.

As you can see, whatever time period you choose as $\Delta t$, $\Delta t'$ will be always greater, because $\Delta t$ will be multiplied by $\gamma$. This means that the stationary clock will always measure larger number of seconds for a given number of seconds measured by the moving clock, and therefore the moving clock will always be the slower one according to SR.

*$\gamma = \frac{1}{\sqrt{1 - \tfrac{v^2}{c^2}}}$

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The statement is

"Any observer at rest relative to his own timepiece will see that other clocks moving with respect to him run fast - the greater their speed, the faster they are".

According to one source, Don Koks (a physicist text book author)this statement is true..... on the condition A orbits B.

http://math.ucr.edu/home/baez/physics/Relativity/SR/movingClocks.html

Let A very closely orbit B. The orbit is so close that A is almost touching B; therefore time delays in signal exchanges can be neglected. The faster A orbits B, the FASTER B's clock runs in A's view. Conversely, the SLOWER A's clock runs in B's view.

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3.1 Signal propagation: The bouncing photon clock The classical wave equation tells us that the propagation is on the light cone, and the propagation speed is c. With this as a starting point we will show that we should expect that physical processes which move progress slower as they do when at rest.

With the help of a Bouncing photon clock and beautiful pictures (Figure 3.1: Bouncing photon clock, at rest (left), moving (right))
he analyses two configurations (bounce vertical or horizontal irt the motion) to show that...
His explanation is so simple, and the graphics are so clear, that one deserves to consult that document.

to ease the interpretation I'will post that figure along with some words:

situation A
on top left: a clock A at rest, distance between faces is always L, even when in motion
on top right: the same clock A in motion with v velocity to the right
”tick” and the ”tock” are equal in duration, being 'tick' the time interval for the photon to move from the top to the bottom mirrors and 'tock' idem from bottom to top;

The photon moves on the diagonals (the arrows) with the speed of light c The vertical component of the speed which determines the duration of the ticks is thus $\sqrt{c^2-v^2}$ and the duration of the ticks for a distance 2L between the mirrors becomes:
$T_{tick}+T_{tock}=\frac{2L}{\sqrt{c^2-v^2}}=2\gamma L/c$

situation B - The two bottom images

In the case, where the photon bounces horizontal we get an asymmetry. The photon moving along with the mirrors in the same direction takes more time to go from one mirror to the other as the photon moving in the opposite direction as the mirrors. The times$T_{tick}$ and $T_{tock}$ are different.
$T_{tick}+T_{tock}=\frac{\lambda/L}{c-v}+\frac{\lambda/L}{c+v}=2\gamma L/c$

However, in both cases the total time for the tick plus the tock is $=2\gamma L/c$,compared with a total time of $=2L/c$ for a clock at rest. In both cases the clock runs slower by a factor $\gamma$. The factor $\gamma$ which determines the time dilation.

Notice the agreement with the text of Einstein(1905)
Note: a longer time interval, as compared to the one at rest, to perform the same TickTack, correspond a slower clock rate.

To avoid incorrect interpretations I will decompose the bottom right image in two, corresponding to the same mirror B at time $t_0$ (photon moving to the right) and time $t_0+\delta t$ (photon moving to the left, after reflection) and

The document deserves a closer attention, finally, it explains the Twin paradox

Closer observation of figure 3.1 shows that the wave length of light changes when .... the Doppler effect on the photons ... ... Even more interesting are the diagonal wavefronts

(this video 'Theory of relativity explained in 7 mins' at 1'30" can clear your mind, I hope)

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Helder Velez: "Chapter 3 [...] by Hans de Vries" -- There: Sect. 3.1: "[...] bouncing between two mirrors [...] The clock-rate becomes higher if the distance between the two mirrors decreases [...]" -- There's neither a value of clock-rate nor of distance to be attributed to a pair of mirrors not at rest to each other. Instead, for mirrors at rest to each other plainly: $$L=\text{distance } := \frac{c}{2 \text{ clock-rate}}.$$ "the duration of the ticks [...] $T_{tick}$, $T_{tock}$" -- These denote apparently not durations of the mirrors, but durations of members of certain "frames". – user12262 Sep 16 '15 at 9:48
Continued (2/2): Sect. 3.2: "Figure 3.2 shows the cases we will calculate here. [...] The stripes on the lines are an indication of the clock ticks." -- The numbers of "stripes" (white gaps) shown in the two parts of "Figure 3.2" are apparently unequal for corresponding lines representing "Twin A": $$36 \text{ vs. } 9 + 9 = 18,$$ while corresponding lines representing "Twin B" show in both parts $15$ "stripes". But surely the number of "clock ticks" indicated by "Twin A" should be a constant for both cases, too. – user12262 Sep 16 '15 at 9:48
@user12262 To avoid misinterpretations I posted a more detailed answer, (here, I will not discuss the twins). I suspect that your interpretation of the image at bottom right is not correct and so I decomposed it in two, to stress that at any time the distance between the mirrors is L. – Helder Velez Sep 16 '15 at 11:50
Helder Velez: "I posted a more detailed answer [...] your interpretation of the image [...]" -- The image which I mentioned in my above comment, and which I interpreted by counting "stripes" (which I understood to mean white gaps in certain lines) is Figure 3.2 of the document you linked in your answer. (It's not any one of the images you've added recently to your answer.) Concerning my other complaint: I like that there's mentioning of "duration(s)" at all. But [... contd.] – user12262 Sep 16 '15 at 20:08
But this terminology should be used even more consistently (rather than falling back to the unspecific word "time"); and it should be used more explicitly, case by case, as "duration of some particular participant, from some particular indication of this participant, until some (other) particular indication of this (same) participant. This would help to recognize that section 3.1 is actually concerned with the comparison of durations (being ""longer", or "shorter", or "equally long"); not comparison of proper rates (being or running "slower", or "faster", or "equally fast"). – user12262 Sep 16 '15 at 20:35

protected by Qmechanic♦Sep 16 '15 at 12:12

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