Clocks in special relativity

Question

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.

Answer 1

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.

Answer 2

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

Answer 3

Don't they contradict each other?

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.

Answer 4

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.

Answer 5

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}}} $

Answer 6

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.

Answer 7

I will use this reference Chapter 3 Time dilation from the classical wave equation from Understanding Relativistic Quantum Field Theory by Hans de Vries

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.

EDIT ADD :
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

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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)