How Ebeb's Diagram Reveals Time Dilation

[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 traveled 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 traveled 0.6 light seconds, the light in B will have traveled 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 traveling 1 light second measured within clock B, yet, relative to clock A, it has traveled 1.25 light seconds taking 1.25 seconds to arrive at the mirror.
• Relative to clock A, the light has traveled 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 traveled 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.

 

[Dale]

For when I shave it with Ockham's Razor I see some significant differences

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.

 

[Ibix]

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.

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.

 

[Dale]

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.

 

[Vanadium 50]

Grimble, you have previously indicated displeasure

That would be a Grimble Grumble?

 

[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.

 

[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 traveled 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 traveled 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)

 

[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 realize 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.

For me this shews exactly where time dilation comes from:
measured from A, after 1 second the light in B will have traveled 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)

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.

 

[Dale]

a simple 2 dimensional slice through time

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.

When clock B is 0.6 light seconds from clock A the lights in the clocks will have reached the mirrors;

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 passed, (time physically measured by each clock)

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

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

Yes

For me this shews exactly where time dilation comes from:

Yes

 

[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 traveled 1 light second to its mirror. From the perspective of clock A, the light in clock B has traveled 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.

 

[Dale]

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.

Yes

 

[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.

 

[Grimble]

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

Yes, of course, thank you.

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

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 traveled 0.6ls after 1 second - that is measured by A who knows that B is traveling 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

 

[Ibix]

A's light reaches his mirror after 1 second.
Clock B has traveled 0.6ls after 1 second - that is measured by A who knows that B is traveling 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.

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 traveled 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.

 

[Dale]

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.

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

No clock actually measures 1.25 seconds

A synchronized clock in A's frame measures 1.25

Clock B is identical to clock A and will also read 1 second as measured by observer B who is 'holding' clock B.

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

 

[Ebeb]

...
Here's a simplified version of Ebeb's diagram.
...

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

 

[Grimble]

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.

Yes, that is what I am describing.

So they do not function identically as described in the rest frame of one or other.

Yes, as I described

but the time A calculates for B is the dilated time of a moving clock...

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.

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.

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.

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 traveled 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.
https://www.physicsforums.com/posts/5799548/report

 

[Dale]

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.

Yes

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.

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.

it is measured along the rotated time axis of clock B, as viewed from clock A.

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.

 

[Nugatory]

In clock A's frameUsing coordinates in which clock A is at rest, after 1 second has passed, the light has traveled 1 light second and reached the mirror of clock A but hasn't yet reached the mirror of clock. 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 secondssomething less than one second - the light hasn't yet reached the mirror in clock B, and the distance it has traveled using coordinates in which B is at rest is less than one light-second. So no, neither clock is reading 1.25 seconds at the same time that the light reaches A's mirror and using A's definition of "at the same time"; but a bit later, at the same time (still using A's definition) that the light does reach B's mirror, A's clock will read 1.25 seconds, both clock are read 1 second when event B occurs;(but you have to be careful when you say "when event B occurs", as it occurs at different times depending on the frame) it is the time elapsed for moving clock B, calculated by A, that is 1.25 seconds. Dilated (increased) Ξtime.

I've made the necessary corrections above.

 

[Grimble]

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.

Presumably that of the observer permanently mid way between the clocks?

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.

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

Any clock measures proper time. It doesn't matter if they are at rest or moving, if they are inertial or non inertial, in curved spacetime or flat. They always measure proper time along their worldline.

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.

 

[Grimble]

I've made the necessary corrections above.

Thank you,
Everything everyone tells me makes a lot of sense and all fits with what we know about special relativity.
The problem is that it won't all fit at the same time.

Let us look at clock B for example.
At time 0, clock A and clock B are side by side. The light is emitted from the base of clock B simultaneously with the light being emitted in clock A.
After 1 second the light in clock B has traveled 1 light second to mirror B.
A will be 0.6 light seconds from clock B, and the mirror in clock B will be 1.25 seconds from the base of clock A.
That is the light in B will have traveled 1 light second to its mirror at c, and must therefore have taken 1 second. that is the definition of a second after all.
Clock B traveling at 0.6c will be 0.6 light seconds from clock A.

So in B's frame after 1 second, proper time, measured by the perfect clock in frame B, the light in the clock has reached the mirror. Event B.
After 1 second, measured by Clock B, clock A will be 0.6 light seconds from clock B and the lamp in A will be 1.25 light seconds from mirror B.
So what is happening? And the answer is found by nothing more than straightforward mechanics.
In frame A
the proper time = τ, the distance to the mirror is cτ
the coordinate time = t the distance to B's mirror is ct
the distance between A and B is vt
and (cτ)² = (ct)² - (vt)²
Yes the proper time squared (which is another way of saying the Spacetime interval) is equal to the coordinate time squared minus the distance squared.

Yes, the time for the light to reach the mirror measured from A is grater than the time measured in B by the factor x, which is the additional distance from A

The time displayed on a moving clock must be exactly the same as shewn on the same clock when it is stationary.
Stationary of moving it is the same clock in its own reference frame.
The dilated time, the clock slowing, must be an effect due to the clock moving relative to the observer. It is, after all, the observer's measurement that measures time dilation. It has to be, for if it were the clock's time that physically changed as that clock displays a single time, which time would it display if the clock moved t different speeds relative to 3, or 10, or an infinite number, of observers at different speeds?

For what property of a moving clock could affect the time the clock displayed? In its own frame it is at rest, so how could its speed relative to an observer affect the time display? The clock measure its proper time, as all clocks must do, while the observer measures the dilated coordinate time.

That is how I see it and everything fits and it all works - just as Einstein described...

So can you explain where I am mistaken?

 

[Dale]

Presumably that of the observer permanently mid way between the clocks?

For the special case here, yes. In more complicated cases there is still such a frame, but it may not be so easy to calculate.

I understood that clocks synchronised in frame A would be synchronised to the clock at A's null point.

Yes

Which, as I see it, means they would be synchronised to proper time in A's frame.

Since you are considering a partial tick there is no clock at rest in either A's frame or B's frame whose proper time is relevant in this analysis. All you can do is consider coordinate time.

If you wish to consider proper time then you need to do a full cycle and talk about the proper time on the emitter/receiver end.

Also, the phrase "synchronized to proper time" doesn't make any sense. You synchronize to coordinate time, not proper time.

 

[Ibix]

For what property of a moving clock could affect the time the clock displayed?

The time the clock displayed when?

The dilated time, the clock slowing, must be an effect due to the clock moving relative to the observer.

Yes. Observers in relative motion don't agree what direction in spacetime to call coordinate time. So, they can agree that a given clock is measuring its own definition of time while noting that it is not measuring their own definition of time.

 

[Grimble]

Since you are considering a partial tick there is no clock at rest in either A's frame or B's frame whose proper time is relevant in this analysis. All you can do is consider coordinate time.

So let us extend my thought experiment and say there are clocks measuring nanoseconds at every point in Spacetime, all synchronised with clock A at Event 0. So we could measure the time at any point on the photon's path.

Also, the phrase "synchronized to proper time" doesn't make any sense. You synchronize to coordinate time, not proper time.

I am sorry but I don't follow that. I thought one synchronised to the frame's clock. The one at the null point that measures 'the frame's time. Which I thought was proper time...

Any clock measures proper time. It doesn't matter if they are at rest or moving, if they are inertial or non inertial, in curved spacetime or flat. They always measure proper time along their worldline.

There seems to be a bit of a sticking point here about A's measurement of the time for the light to travel from event A to event B, being 1.25 seconds - which it has to be because, measured from frame A the light travels 1.25 light seconds; and the fact that in frame B it only takes one second because it travels a shorter distance. (also that the light in frame A only travels 1 light second in one second to the mirror)
That these are two different measures of time between two events.
Yet I do not see a problem with that, for we (or rather A) is measuring different paths for the light.
A measures light travel to the mirror in their clock 1 light second...
B measures light travel to their mirror 1 light second...
A measures light travel to B's mirror 1.25 light seconds...
The light travels different paths between the same two events in different frames because, relative to A the mirror in B has moved away while the light is traveling to B's mirror.
Different times between the same events depending on the frame of reference, different observers measuring different times - isn't that what relativity is all about?
Different distances, different times relative to different observers... Isn't that why absolute time was abandoned?

 

[Ibix]

So let us extend my thought experiment and say there are clocks measuring nanoseconds at every point in Spacetime, all synchronised with clock A at Event 0. So we could measure the time at any point on the photon's path.

You can't synchronise at an event - that's like talking about the direction of a point. Clocks are either synchronised (implying a choice of synchronisation convention if they are not in the same place) or not. So you could measure the time at any point on the light pulse's path in the frame of your choice.

I am sorry but I don't follow that. I thought one synchronised to the frame's clock. The one at the null point that measures 'the frame's time. Which I thought was proper time..

Two clocks in relative motion both measure their own proper time but will not, in general, agree that the other is ticking at the same rate. So "synchronised to proper time" is an incomplete sentence in the same way that "this house belongs to" is an incomplete sentence. You need to specify whose proper time you are talking about, and you need to specify your synchronisation convention. A reference frame is, essentially, a choice of clock and a choice of synchronisation convention.

 

[sdkfz]

There seems to be a bit of a sticking point here about A's measurement of the time for the light to travel from event A to event B, being 1.25 seconds - which it has to be because, measured from frame A the light travels 1.25 light seconds; and the fact that in frame B it only takes one second because it travels a shorter distance. (also that the light in frame A only travels 1 light second in one second to the mirror)

It's not clear to me exactly where you currently think you and everyone else are diverging.

To fill out the scenario above (to my understanding) ...

At 1 second, according to A's clock, the light hits A's mirror.
At 1 second, according to A's clock, the light has not yet hit B's mirror.
At 1.25 seconds, according to A's clock, the light hits B's mirror.
A agrees with B that B's clock reads 1 second at that time (light hitting B's mirror).
According to B, their clock is fine. For them, 1 second now feels like any second before it.
According to A, B's clock is slow, that's why it reads 1 second (for light hitting B's mirror) when A has the time for that as 1.25 seconds.

At 1 second, according to B's clock, the light hits B's mirror.
At 1 second, according to B's clock, the light has not yet hit A's mirror.
At 1.25 seconds, according to B's clock, the light hits A's mirror.
B agrees with A that A's clock reads 1 second at that time (light hitting A's mirror).
According to A, their clock is fine. For them, 1 second now feels like any second before it.
According to B, A's clock is slow, that's why it reads 1 second (for light hitting A's mirror) when B has the time for that as 1.25 seconds.

i.e. they both agree with each other about what their own clock reads (no contradiction); but proper time is not absolute time ... because they are in relative motion their clocks may have been synchronised at some time (e.g. both set to zero by some convention) but will not tick at the same rate.

Which bit of the above, Grimble, does not match how you see things?

 

[Dale]

So let us extend my thought experiment and say there are clocks measuring nanoseconds at every point in Spacetime, all synchronised with clock A at Event 0. So we could measure the time at any point on the photon's path.

That is coordinate time, not proper time. I have told you several times, but you don't seem to pay attention and you never respond to such comments, but I will try again:

Coordinate time is defined over a 4D region of spacetime and includes a concept of simultaneity. Proper time is defined only along the 1D worldline of a specified clock and has no concept of simultaneous.

I hope you pay attention this time.

I am sorry but I don't follow that. I thought one synchronised to the frame's clock. The one at the null point that measures 'the frame's time. Which I thought was proper time...

No. Feel free to go back and count how many times I have mentioned this. All I know is it feels like I am making a sincere effort to pay attention to your questions, explain, and teach, but I see no progress on your end. You seem to prefer to go back to what you know rather than confront points that you do not understand like the difference between coordinate time and proper time.

There seems to be a bit of a sticking point here about A's measurement of the time for the light to travel from event A to event B, being 1.25 seconds - which it has to be because, measured from frame A the light travels 1.25 light seconds

Note that this is 1.25 s of coordinate time in frame A.

and the fact that in frame B it only takes one second because it travels a shorter distance.

Again, this is coordinate time in frame B

also that the light in frame A only travels 1 light second in one second to the mirror

Sloppy wording here. Which mirror? I assume A's mirror.

That these are two different measures of time between two events.

Now I am confused. You speak of two measures of time between two events, but with the parenthetical comment introducing A's mirror you now have three events of interest: the event where the light was emitted, the event where the light reaches B's mirror, and the event where the light reaches A's mirror.

I know that it bothers you when we harp on the semantics, but it is likely that the confused semantics is a symptom of your confusion over the concepts.

Also, do you understand that the measures of time discussed here are measures of coordinate time?

I do not see a problem with that, for we (or rather A) is measuring different paths for the light.

Yes. Although the path of the light which goes from the emission event to B's mirror is considered to be the same path in A's frame and B's frame. It is just labeled using different coordinates, but the path itself is a geometric object which is frame-independent.

Different times between the same events depending on the frame of reference, different observers measuring different times - isn't that what relativity is all about?

Note again, this is different coordinate times, not different proper times. Proper times are frame invariant.

 

[Grimble]

That is coordinate time, not proper time. I have told you several times, but you don't seem to pay attention and you never respond to such comments, but I will try again:

I am sorry that it appears that way, but there are so many comments that it would take forever (or so it seems) to answer every point made.
But let me assure you that I appreciate the effort you are making.
It has seemed that each time I have thought I was understanding the difference between proper and coordinate time the definitions were defined differently - I obviously have been missing the essential points.

Proper time then is the time 'counted' by a clock. It is the same whoever reads that clock. It measures the passage of time for that clock? The elapsed time measured along that clocks worldline?

Coordinate time is time measured relative to a frame of reference?

Does that mean that there can be more than one coordinate time measured in a single frame?
I am thinking here of the coordinate time which is the (I don't know what the correct label is here but I would lean towards 'local' time or 'home' time as descriptors for the time measured by the frame's clock. which I have been thinking of as the proper time of the frame's clock.
The [virtual] clock at rest at the frame's null point, like all clocks measures proper time(1D); but when it is being used in relation to the frame (4D) we refer to it as coordinate time?

Yes. Although the path of the light which goes from the emission event to B's mirror is considered to be the same path in A's frame and B's frame. It is just labeled using different coordinates, but the path itself is a geometric object which is frame-independent.

OK. Then that one single path, measured relative to frame B, is a direct passage 1 ls long between two points (events?)
Relative to frame A one point is fixed (null point) while the other (B's mirror) is moving away and is fixed by the time coordinate, 1.25 seconds. So the length measured for the same path in A's frame is longer because the coordinates of B's mirror are changing... because A's coordinate system, it frame of reference, is moving relative to that single "geometric object which is frame-independent"?

 

[Dale]

Proper time then is the time 'counted' by a clock. It is the same whoever reads that clock. It measures the passage of time for that clock? The elapsed time measured along that clocks worldline

Yes, and it is only defined on that given clock's world line.

Coordinate time is time measured relative to a frame of reference?

Yes, it is the time coordinate of a given reference frame. It is defined everywhere the reference frame is defined (usually all of 4D spacetime)

Does that mean that there can be more than one coordinate time measured in a single frame?

No, there are three spatial coordinates and only one time coordinate for a given frame. Of course, different frames will have different coordinate times from each other.

I am thinking here of the coordinate time which is the (I don't know what the correct label is here but I would lean towards 'local' time or 'home' time as descriptors for the time measured by the frame's clock. which I have been thinking of as the proper time of the frame's clock

And once again I repeat myself again over and over some more yet again. The coordinate time of a given frame is defined over the whole 4D spacetime region covered by the frame. The proper time of a given clock is defined only on the 1D worldline of the clock.

The [virtual] clock at rest at the frame's null point, like all clocks measures proper time(1D);

Yes

but when it is being used in relation to the frame (4D) we refer to it as coordinate time?

They are different things, not just the same thing referred to differently in different contexts.

Do you remember from grade school the definition of a function and its domain and range? Proper time and coordinate time are functions with different domains. The domain of coordinate time is the set of all events in spacetime. The domain of proper time is the set of all events on the worldline of the clock. Even if those functions have the same value on the worldline, they are still different functions because they have different domains.

 

[Grimble]

Thank you Dale, I think I get it now!
I mistakenly got the wrong end of the stick when I first read of proper time and coordinate time - I formed the wrong idea that proper time was the time scale for the whole frame and that coordinate time was the time taken from another frame. But now I see how wrong that was.

In the special case of an inertial observer in special relativity, by convention the coordinate time at an event is the same as the proper time measured by a clock that is at the same location as the event, that is stationary relative to the observer and that has been synchronised to the observer's clock using the Einstein synchronisation convention.

 

[Grimble]

because they are in relative motion their clocks may have been synchronised at some time (e.g. both set to zero by some convention) but will not tick at the same rate.

Hmmm.
You see, this is the sort of thing I am referring to; a completely sensible and rational summary that fits perfectly with what we know. It is fine taken on its own; but I have a problem making it fit with Einstein's first Postulate:

The laws of physics are invariant (i.e. identical) in all inertial systems (non-accelerating frames of reference).

Take any number of identical, inertial, light clocks; each and every one will read 1 second after the light in that clock has traveled 1 second to its mirror 1 light second away. These are identical light clocks in identical frames that happen to be traveling at different speeds relative to one another, but each is at rest in its own frame, subject to identical scientific laws. The proper time for each clock on its worldline is invariant being the same wherever it is measured from, as it is the time for light to travel 1 light second. That, the proper time of one second is what that clock will read after 1 second has passed.

i.e. each and every clock in this scenario will read the same time = 1second (How could they not?)

Note: this is not absolute time, as each observer will measure the other clocks to be time dilated.

That interval measured by the clock in one frame for a clock(A) in another, moving frame(B), can be found by applying the Lorentz Transformation Equation in order to transform that measurement of 1 second coordinate time; to calculate what that interval would be relative to the observer's frame. At 0.6c it would be 1.25 seconds.

Taking A as the stationary system K, and B as the moving system K',
then

As judged from K, the clock is moving with the velocity v; as judged from this reference-body, the time which elapses between two strokes of the clock is not one second, but

seconds, i.e. a somewhat larger time.

t' = γt = 1.25seconds.

In the special case of an inertial observer in special relativity, by convention the coordinate time at an event is the same as the proper time measured by a clock that is at the same location as the event, that is stationary relative to the observer and that has been synchronised to the observer's clock using the Einstein synchronisation convention.

Therefore if we place a clock adjacent to B's mirror at the moment B's light hits the mirror (at the same location as the event), in frame A(that is stationary relative to the observer) synchronised to A's clock, then the coordinate time of that clock must be the same as the proper time (i.e. 1 second).
the coordinate time is 1 second, the clock calculated to read 1.25 seconds is the moving clock in frame B.
The time 1.25 is the time for the light in B to travel the distance to the clock plus the distance traveled by the clock. (by vectorial addition)

 

[Dale]

Take any number of identical, inertial, light clocks; each and every one will read 1 second after the light in that clock has traveled 1 second to its mirror 1 light second away

I am not sure that what you are saying is correct. Again, the semantics are off. To avoid getting bogged down in semantics, can you write mathematically what you mean?

Be sure to use different symbols for coordinate time and proper time. Make sure that all coordinate times are associated with a specific reference frame and that all proper times are associated with a specific clock. Label any important events or times that you are comparing, particularly whatever you think is in conflict with SR.

For instance I might use $A$, $B$, and $C$ to label reference frames, $t_A$ to label the coordinate time in frame $A$, and $\tau_a$ to label the proper time along the worldline of the clock which is at rest at the spatial origin of $A$. I might use labels like $a_0$ to designate the event along the worldline of clock $a$ where the clock reads 0, and $a_1$ to designate the event on the worldline of $a$ where it reads 1. And I would then use labels like $t_{Ba1}$ to indicate the coordinate time in frame B at which clock $a$ reads 1. I would use units where c=1 and I would write coordinates as (t,x,y,z) and I would use the convention where spacelike intervals $ds^2>0$ so $d\tau^2=-ds^2$. You can do it differently as long as you are clear.

 

[PeterDonis]

That interval measured by the clock in one frame for a clock(A) in another, moving frame(B), can be found by applying the Lorentz Transformation Equation

No, it can't. The interval--the actual elapsed time on the clock--is an invariant, it doesn't change when you change frames.

The coordinate time difference between the two events that define the interval--i.e., the two events on the clock's worldline that define the "elapsed time"--will change when you change frames; but the coordinate time difference, by itself, has no physical meaning.

 

[sdkfz]

The point of post #26 was to try to boil down to where you and everyone else differs, simply, and away from the minutia. It's disappointing that you ignored it and instead tried again to add more sloppy complexity.

Hmmm.
You see, this is the sort of thing I am referring to; a completely sensible and rational summary that fits perfectly with what we know. It is fine taken on its own; but I have a problem making it fit with Einstein's first Postulate:

That's because you read that postulate (whether you realize it or not) as meaning "absolute time". In fact that postulate is well covered by the results in post #26: both A and B consider their own mirror to be 1 light second away from themselves, and their own clock ticks 1 second when the light strikes it. Yes, the laws of physics were invariant for A and B.

Take any number of identical, inertial, light clocks; each and every one will read 1 second after the light in that clock has traveled 1 second to its mirror 1 light second away. These are identical light clocks in identical frames that happen to be traveling at different speeds relative to one another, but each is at rest in its own frame, subject to identical scientific laws. The proper time for each clock on its worldline is invariant being the same wherever it is measured from, as it is the time for light to travel 1 light second. That, the proper time of one second is what that clock will read after 1 second has passed.

i.e. each and every clock in this scenario will read the same time = 1second (How could they not?)

Yes. As noted in post #26.

Note: this is not absolute time, as each observer will measure the other clocks to be time dilated.

Yes. As noted in post #26.

That interval measured by the clock in one frame for a clock(A) in another, moving frame(B), can be found by applying the Lorentz Transformation Equation in order to transform that measurement of 1 second coordinate time; to calculate what that interval would be relative to the observer's frame. At 0.6c it would be 1.25 seconds.

Taking A as the stationary system K, and B as the moving system K',
then
t' = γt = 1.25seconds.

Therefore if we place a clock adjacent to B's mirror at the moment B's light hits the mirror (at the same location as the event), in frame A(that is stationary relative to the observer) synchronised to A's clock, then the coordinate time of that clock must be the same as the proper time (i.e. 1 second).
the coordinate time is 1 second, the clock calculated to read 1.25 seconds is the moving clock in frame B.
The time 1.25 is the time for the light in B to travel the distance to the clock plus the distance traveled by the clock. (by vectorial addition)

This is where your thinking gets sloppy (I took great pains to be very clear about who was measuring what in post #26) and you end up off the rails. Simply placing a clock at some location doesn't make it read the same as some other clock at the location. If that clock is "in frame A", i.e. stationary with respect to A, then it's in a frame where the B light moves more than 1 light second to hit the B mirror. The 1.25 seconds is B's light-hit-mirror event, as measured by A. You can't claim it's the proper time for B, as according to B the light only moved 1 light second to hit the mirror.
Just above, you wrote "as each observer will measure the other clocks to be time dilated", now you seem to argue the opposite. Why is your scenario so self-contradictory?
You seem to be trying hard to again make absolute time, and demote relativity to mere illusion.

 

[Grimble]

No, it can't. The interval--the actual elapsed time on the clock--is an invariant, it doesn't change when you change frames.

The coordinate time difference between the two events that define the interval--i.e., the two events on the clock's worldline that define the "elapsed time"--will change when you change frames; but the coordinate time difference, by itself, has no physical meaning.

Sorry... I was using the word 'interval' literally rather than scientifically...

The proper time displayed by clock B, (that is the coordinate time in frame B?), transformed by LTE gives the measurement relative to the observer A.