ralfcis:

Thanks for your replies to post #71 and #72. I'll comment on those next. But this post is a reply to post #127.

Here's the Loedel diagram of two atomic locks leaving Earth at 1/3c in opposite directions.

Following your replies to #71/72, I think I now understand what you mean when you refer to Loedel diagrams and Loedel lines.

To me, your Loedel diagrams are essentially the same as Minkowski diagrams. One addition you make is to label the proper times of the "moving" frames on the worldlines of the objects/observers who are moving. The other main addition you make is to add "Loedel lines", like the green ones on your graph in #127, which appear to be just straight lines that connect equal "proper times" of selected "moving" observers.

I see that you have noted that, on your diagram here, the proper time of the Earth or lab frame is different from the proper time of either of the moving clocks. Your green "Loedel line" at proper time t=2 does not pass through the t=2 position on the coordinate axis (i.e. for the Earth frame). In other words, that "Loedel line" does not connect the proper times of all three observers (the two moving clocks and the stationary Earth), but only two out of the three.

You have also drawn the hyperbola that corresponds to the equation

$s^2=(ct')^2=(ct)-x^2$.

In relativity, this would be a curve of constant $s$ (call it 'the spacetime interval', maybe).

In fact, for your two moving clocks there, there are

*two* such hyperbolas, but due to the symmetry of this scenario, with the two moving clocks have velocities $\pm c/3$ in the Earth frame, they happen to coincide on this diagram.

Those hyperbolas are, I think you would say (?), curves of "equal proper time".

I don't really understand what use your straight green "Loedel lines" are, since only the endpoints of those lines have "equal proper time", as I'm sure you'll agree. Intermediate points have no special significance, as far as I can see. Is that correct, or do intermediate points on those straight green lines mean something in particular to you?

The light signals you have drawn are each emitted at proper time t=2 on the "moving clocks", and are received at proper time t=4. As we see from the diagram, the lines are symmetrical, which is again due to the $\pm c/3$ velocities of the two moving clocks (i.e. they both have the same speed, just in opposite directions relative to Earth).

What does not appear on your diagram are any lines of simultaneity that would show us which events occur simultaneously in either of the moving clock frames. Note: I am not talking about "proper time simultaneity" - those would be the curves of "equal proper time" discussed above - but the times which all the (synchronised) clocks in the relevant reference frame would agree were simultaneous (i.e. clocks everywhere display the same time reading, in that frame). Obviously, any horizontal line drawn on the graph is a line of simultaneity for the Earth frame, but lines of simultaneity in either of the two "moving clock" frames would not be horizontal. They would also not be parallel for the two moving clocks because those two clocks are moving in opposite directions, away from one another.

We could ask a question like this, for example: for an observer in the frame moving at $v=+c/3$, at what time did the clock moving at $v=-c/3$ emit its light signal? The answer to that question would not be "At $t'=2$", where t' is the time coordinate in the frame of the $v=+c/3$ clock; it would be a time coordinate with $t'>2$ in that frame. In the same frame, the answer to the question "At what time was the signal received?", the answer would be $t'=4$, of course, because that event occurs at $x'=0$, whereas the sending of the signal happened at some $x'<0$ coordinate -i.e. at a distance from the $v=+c/3$ moving clock.

The elapsed

*coordinate time* in the frame of the $v=+c/3$ moving clock, between the emission of the signal from the $v=-c/3$ clock and the receipt of that signal, is

*less than* 2 seconds. Note: this would be what an observer in the frame of the $v=+c/3$ clock would

*actually measure.* But it is

*also* the case that that observer would measure the distance the light had to travel from the other clock to be less than 2 light-seconds (or light-years, or whatever distance units we're using). Thus, when that observer calculates the speed of the light, the calculated value would be $c$, as usual.

With all of this in mind, there's not a lot more to say about the this "Loedel diagram". It might be worth mentioning that, from the $v=+c/3$ clock's perspective, the light signal from the other clock was emitted

*after *$t'=2$, while in the $v=-c/3$ clock's perspective, the light signal was emitted

*at* $t''=2$, where the double-prime indicates the $v=-c/3$ frame's clocks, which of course are not synchronised with the $v=+c/3$ frame's clocks.

To summarise all of this: there's nothing "wrong" with the spacetime diagram (call it a "Loedel diagram" if you like; it's just a label). It doesn't tell us everything we might like to know about the situation, especially from the (different) perspectives of the two moving clocks, but that's okay.

The Loedel stationary frame allows a glimpse of proper time simultaneity of the two separating clocks. This means the green lines of Loedel simultaneity share the same proper times of the moving clocks whereas the perspective lines of simultaneity would connect two different perspective times on both ends.

The straight green "Loedel" lines only connect the "proper times" at the two ends. SR lines of simultaneity (I don't know why you want to use the term "perspective lines") have the property that every point on the line is an event that happens simultaneously in whichever frame the line of simultaneity applies to.

Each ship would have star charts of its distance from earth and their on-board clocks would be programmed to send out light signals at t=2 proper time within their own frames. They would receive the light signals at t=4 proper time within their own frames.

Not a problem. You'd have to bear in mind that the star charts would have been drawn up using Earth rulers, of course. What you say about proper times here is 100% correct.

Since it's proper time they don't need the other ship's perspective time.

Correct, if all they are interested in is when they sent their own signal and when they received the other one, since both of those events obviously happen at

*this* ship, not at the

*other* one.

In a lab, of course, the two clocks would register that they indeed sent out and received the signals proper time simultaneously...

More accurately, an observer in the lab frame would agree that the two "moving" clocks sent out their signals at the same (Earth) time, and, at that time, both displayed the same "proper time" readings, according to the Earth observer.

The reference frame itself does not share the same proper time as the other two clocks but its time is calculable using relativity.

Note that

*all three *clocks have different "proper times". We can see that all three time axes head in different directions on the diagram. It is incorrect to say that the two "moving" clocks share the same "proper time". They do not. What is true is that they both appear to tick at the same rate

*according to the Earth observer* and always display the same times

*according to the Earth observer*.

If we consider, for example, the clock moving at $v=+c/3$, any observer in the rest frame of that clock would see the $v=-c/3$ clock as ticking

*slower* than the $v=+c/3$ clock, which is the reason why, in the $v=+c/3$ observer's frame, the $v=-c/3$ clock emits its light signal

*later than* $t'=2$, rather than at $t'=2$.