# I Moving clocks

1. Jul 7, 2017

### Grimble

I am starting a new thread as my last one was unceremoniously hijacked by those who should know better and then closed by moderators when boundaries were crossed in the hijacked thread. I am not complaining as it was an interesting diversion, but now, if acceptable, I would like to return to the question of how this diagram, courtesy of 'Ebeb' from #124 in thread "Proper (and coordinate) times re the Twin paradox", works.
For when I shave it with Ockham's Razor I see some significant differences

Reducing the time frame to 1 second the facts are:
• after 1 second each light will have travelled 1 light second.
• measured by the observer with each clock, the light in their clock will have reached their mirror
• Relative to clock A, clock B will have travelled 0.6 light seconds, the light in B will have travelled 1 light second and will have reached point (0.6,0.8)
• The light in clock B arrives at the mirror in clock B after travelling 1 light second measured within clock B, yet, relative to clock A, it has travelled 1.25 light seconds taking 1.25 seconds to arrive at the mirror.
• Relative to clock A, the light has travelled the extra distance 0.6 light seconds along the x axis to point (0.6,0.8) and 0.75 light seconds with clock B (in clock A's frame) by the time it has reached the mirror.
• So between two fixed events; the generation of the light and the arrival at mirror B, we have two different times. 1 second within the resting light (B's frame) and 1.25 seconds with the moving light in A's frame.
• This is what I see as time dilation - just as the moving clocks distance travelled 0.75 light seconds is length contracted to 0.6 light seconds in the resting frames: that is clock A's resting frame and clock B's resting frame.
• The moving clock's time units are longer (dilated) so that measurement of the clock runs slower - each tick takes longer can only be equated to the clock running slower but there are still the same number of ticks counted for the resting clock. i.e. the clock still reads the same which ever frame it is measured in.
• Between the same two events the moving clock ticks slower and travels further for each tick - time dilation - yet ticks the same number of ticks - as shewn by the clock reading. Or to put it another way the moving clock travels further and measures more time.

2. Jul 7, 2017

### Staff: Mentor

What does this mean? I understand @Ebeb's plot (and it seems correct), although it is very cluttered looking, but I have no idea what this second plot is supposed to represent.

Last edited: Jul 7, 2017
3. Jul 7, 2017

### Ibix

Both clock A and clock B are 1ls long "transverse" light clocks, and clock B is doing 0.6c in the +x direction. The diagram is a snapshot of the x-z plane of the rest frame of clock A, constructed at t=1.25s.

Mostly.

Note that the x and z scales aren't the same. The green arrow represents the path of the light pulse in clock A, but inexplicably terminates it at t=1s. The purple arrow represents the path of a light pulse in clock B over the full 1.25s. The red arrow doesn't represent anything. The green curve is just a 1ls circle centered on the origin.

@Grimble - it would be interesting to see where you think the light pulses are at 1.25s. Your diagram does not accurately represent that.

4. Jul 7, 2017

### Staff: Mentor

Thanks @Ibix. If that is correct then the second plot and the first plot have little to do with each other. I don't know why @Grimble chose to present them as related or bring Occhams razor into the discussion.

@Grimble, you have previously indicated displeasure at the fact that many discussions with you become bogged down in semantics and never get to the heart of your question. This OP is a great example of why that happens. These two diagrams are nearly unrelated. Occhams razor is irrelevant to this thread. And reading over it again I still don't know what your question is.

5. Jul 7, 2017

Staff Emeritus
That would be a Grimble Grumble?

6. Jul 7, 2017

### Ibix

One more observation: I think that Grimble is trying to draw a slice through Ebeb's Minkowski diagram in the frame of the A clock, and the Occam's razor is simply a play on words. If so, it's rather unhelpful because that kind of thing is where confusion creeps in. I think his complaint is that the result is not consistent with the Minkowski diagram.

@Grimble - am I right?

Assuming I am, the problem is that you're presenting a snapshot of the position of the clocks in A's rest frame at t=1.25s but not representing the light pulses in the same manner. In a snapshot like that they should be points, not arrows. If you want to record the path followed by the pulses then you should make sure that you show the path over the same time period, which you aren't doing - your green arrow covers t=0 to t=1, while your purple arrow covers t=0 to t=1.25, and your red arrow doesn’t really represent anything in this frame.

7. Jul 8, 2017

### Grimble

OK, I will try and explain (and I apologise for referring to Occam's razor ).
I was attempting to draw a simple diagram shewing 2 light clocks (with the lights along the z axis) moving apart along the x/x' axis at 0.6c.

I have tried to draw this as a simple 2 dimensional slice through time (Galilean transformation?), no time axis just the x and z dimensions.
When clock B is 0.6 light seconds from clock A the lights in the clocks will have reached the mirrors; 1 second will have passed, (time physically measured by each clock)

What the diagram is intended to shew is that the light in clock B will have travelled 1.25 light seconds from the base of clock A to the mirror in clock B; i.e. relative to clock A.

All true even using classical Newtonian mechanics where the speed of the light would be 1.25c, which is OK using classical mechanics.
The difference using relativistic mechanics is that that light, in clock B, must take 1.25 seconds to reach the mirror in B, as it must be measured to travel at 'c'.
(that is, after all, the only real difference introduced by Special Relativity - Einstein's 2nd Postulate)

For me this shews exactly where time dilation comes from:
measured from A, after 1 second the light in B will have travelled 1 light second to (0,6.0.8) and will arrive at mirror B (0.75,1.0) after 1.25 seconds giving time dilation from 1 to 1.25(γ=1.25) and length contraction 0.75 to 0.6 (1/γ = 0.8)

8. Jul 8, 2017

### Ibix

@Grimble - am I understanding you right? Your diagram is an attempt at a non-relativistic diagram of a scenario that depends explicitly on the behaviour of light in different frames?

You realise that the fact that that's entirely self-contradictory is what led us to relativity theory? And that's why no-one can understand your diagram? It is internally inconsistent because there is no frame in which light behaves as you describe.
Not quite. Time dilation arises from two things in this context: First, viewed in the frame of the green clock it takes light longer to reach the other mirror of the red clock. Second, that someone co-moving with the red clock must see it ticking every two seconds the same as the green clock in its frame. Ignoring the latter gets you an ether theory.

Last edited: Jul 8, 2017
9. Jul 8, 2017

### Staff: Mentor

Time in which reference frame? I assume the clock at rest's frame.

In a simple slice through time the light pulses will be dots, not lines.

No, the light will have just reached the moving mirror, but it will have reached and reflected and started heading back for the resting clock.

1 second will have been measured for the moving clock, but 1.25 seconds will have been measured for the resting clock.

Yes

Yes

10. Jul 8, 2017

### Grimble

OK.
Classic Newtonian mechanics. Two light clocks moving apart at 0.6c for 1 second. Absolute time. The lights in the two clocks have each travelled 1 light second to its mirror. From the perspective of clock A, the light in clock B has travelled 1.25 light seconds from the base of clock A to the mirror in clock B at a speed of 1.25c. (impossible I know)

The difference in relativistic mechanics is c, the speed of light.

From a relativistic perspective the light would take 1.25 seconds to reach the mirror in B. Which seems obvious to me: as it has further to travel it will take longer but its arrival at the mirror is the same event as the light taking 1 second from B measured within clock B.

11. Jul 8, 2017

### Staff: Mentor

Yes

12. Jul 8, 2017

### sweet springs

Hi. Go one way is a plain discussion because every twins are on equal foot. Return to the selected one is a spice. Everybody can say I'm Home where other guys should gather. If brother on the Earth is not chosen as Home, he must make great effort to accelerate the Earth as I watched in a Japanese SF movie.

13. Jul 9, 2017

### Grimble

Yes, of course, thank you.

Looking at this again, I can see where you are coming from, but as I see it, when observer A calculates 1.25 seconds to have passed for the moving clock it is his perception of time passing in the moving frame - dilated time - A's own clock will still read only 1 second..

A's light reaches his mirror after 1 second.
Clock B has travelled 0.6ls after 1 second - that is measured by A who knows that B is travelling at 0.6c.
Clock B is identical to clock A and will also read 1 second as measured by observer B who is 'holding' clock B.

(I know it is unconventional to draw B's frame, in red, in the same diagram but that is what happens in the Loedel diagram that Ebeb posted, where the two time lines have a common origin - but so do mine! - only shewn a different way.)

Surely the time displayed on clock A, is 1 second (let us say my light clock displays half ticks too, before anyone complains about that) ;
the time displayed on clock B is 1 second,
but the time A calculates for B is the dilated time of a moving clock...
No clock actually measures 1.25 seconds that is the proper time on clock A plus the travel time of clock B added by vector addition -
τ2 + (vt)2 = (ct)2
or
s2 = (ct)2 - (x)2

14. Jul 9, 2017

### Ibix

Clock A and clock B are designed identically, but are not being operated identically, at least in the rest frame of either clock - the other one is moving. So they do not function identically as described in the rest frame of one or other. So, no. The light in clock B has not reached the end of clock when the light has reached the end of clock A, not in the rest frame of A.

Your diagram has to be in the green frame because the red clock is moving at 0.6c. But for some reason you've drawn the green clock at t=1s in this frame and the red clock at t=1.25s. These two things aren't simultaneous, so you're mashing together two different times in one frame, not two different frames. That's why it's nonsensical.

Here's a simplified version of Ebeb's diagram. It shows the paths of the red clock and green clock in the frame where they are moving with equal and opposite speeds of c/3. I've marked the time (t=$3/\sqrt 8$=1.06s) when both clocks read 1s.

There are three "slices" that you can sensibly take through these. One, marked as a black dashed line, is what the frame of the Loedel diagram regards as simultaneous. I've draw what that looks like on a black grid. The second, marked as a green dashed line, is what the rest frame of the green clock regards as simultaneous. I've drawn what that looks like on the green grid. The third, marked as a red dashed line, is what the rest frame of the red clock regards as simultaneous. I've drawn what that looks like on the red grid.

Start with the black grid. In this frame, both clocks are doing c/3 in opposite directions. Both light pulses (orange dots) have reached the top mirror having travelled 1ls "up" and 0.35ls "across". The clocks are operating identically in this frame.

Now look at the red grid. In this frame, the pulse in the red clock, which is at rest, has reached the top mirror, so 1s has passed. But the pulse in the green clock has only moved up 0.8ls because it has also moved across 0.6ls, for a total distance of 1ls. So the green clock does not yet read 1s - it reads 0.8s. This is consistent with the red dashed line on the Loedel diagram - it crosses the red solid line when the red clock reads 1s, but it crosses the green solid line before the green clock reads 1s.

I shan't bother discussing the green grid - it's a mirror image of the red grid.

That's how you need to draw the kind of diagram you are trying to draw. Mashing together two times is just confusing.

15. Jul 9, 2017

### Staff: Mentor

You are neglecting the second postulate. When 1.25 s of coordinate time have passed in A's frame then both pulses of light will have traveled 1.25 light-seconds in A's frame. That is required by the second postulate.

The pulse of light in A's clock will have hit the mirror and returned 1/4 of the way. So the proper time* will be 1.25.

The pulse of light in B's clock will have just hit the mirror. So the proper time will be 1.00.

*technically for proper time we should only talk about complete cycles from the emitter to the mirror and back again, but that is not what you drew

A synchronized clock in A's frame measures 1.25

Why would you say this? What is the condition for B's clock to read 1 s? Has that condition been met?

16. Jul 10, 2017

### Ebeb

Thanks for clarification of diagram, Ibix. I'm short of time to post in this thread...

17. Jul 10, 2017

### Grimble

Yes, that is what I am describing.
Yes, as I described
No, you are quite right, not in the rest frame of A, Not in the rest frame of B.

But the light reaching the mirror in B is a single event, Let us call he light hitting the mirrors Events A and B; they each happen at one location at one moment in time. Single events.
In B's frame it takes one second for the light to travel 1 light second from the base to the mirror, measured by B, to the event B.
In A's frame it takes one second for the light to travel 1 light second from the base to the mirror, measured by A, to the event A.
Both lights start at event 0, the null point of both frames.
Both lights take one second measured within that frame to arrive at their mirrors (events A and B).
Although Events A and B both happen 1 second from event 0, because those measurements are made with respect to different observers; they are cannot be measured to be simultaneous in any frame.

With respect, I have drawn my diagram to shew the Event B. Which can only be the one event in Spacetime when the light in clock B hits the mirror in clock B.

In clock A's frame, after 1 second has passed, the light has travelled 1 light second and reached the mirror. that is an unassailable fact because of the 2nd Postulate. That is the time measure measured by a perfect clock. It must tell the correct time.
The time axis for clock A would be drawn along the z axis.
The measurement along the path to B's mirror is not measured along the time axis of frame A, it is measured along the rotated time axis of clock B, as viewed from clock A. That is why after 1 second has passed in frame A, the time passed in clock B, moving at 0.6c will be γct = 1.25t = 1.25 seconds. So no, neither clock is reading 1.25 seconds, both clock are read 1 second when event B occurs; it is the time elapsed for moving clock B, calculated by A, that is 1.25 seconds. Dilated (increased) Ξtime.
Report

#### Attached Files:

• ###### Light clocks for forum.2.png
File size:
34.6 KB
Views:
24
18. Jul 10, 2017

### Staff: Mentor

Yes

No, for any two spacelike separated events there is always a reference frame where they are simultaneous. Neither A's frame nor B's frame is such a frame, but such a frame does exist.

That sounds like A's frame. In which case your other comments are simply wrong. At event B the coordinate time in A's frame is 1.25 and all A synchronized clocks read 1.25.

19. Jul 10, 2017

### Staff: Mentor

I've made the necessary corrections above.

20. Jul 11, 2017

### Grimble

Presumably that of the observer permanently mid way between the clocks?
Pardon me, but I understood(I know, that would be a rash claim from me! hehehe!) but I understood that clocks synchronised in frame A would be synchronised to the clock at A's null point...
And
Which, as I see it, means they would be synchronised to proper time in A's frame.

Imagine that there was another identical clock, C, that was also synchronised at event 0 and was moving away from A at 0.866c. A's coordinate time for clock C would be 2 seconds. those synchronised clocks would not be synchronised to the coordinate times for there could be as many coordinate times as there were clocks moving at different speeds.