Looking to understand time dilation

[Grimble]

Are observer A and the clock co-moving?

Yes the observer is permanently adjacent to the clock, he may even be holding it.

grimble said:
"So, if I say that I have a clock in space and specify no relationship to anything else; let there be an observe A with the clock and as far as he knows he and the clock are alone in space.

Then, as there is nothing to relate to he is neither stationary nor moving but he is inertial.
The clock will be on its world line and will be keeping proper time.
For A and the clock proper time and coordinate time will be identical."

Isn't he saying that the person and clock are moving together at some indeterminant speed?

One could certainly say that, but would it not be true to say that they were stationary as they cannot be moving if there is nothing to relate the movement to. i.e. movement has to be relative to something and if there is nothing to move relative to they cannot be moving?

 

[ghwellsjr]

...would it not be true to say that they were stationary as they cannot be moving if there is nothing to relate the movement to. i.e. movement has to be relative to something and if there is nothing to move relative to they cannot be moving?

In Special Relativity, things move or are stationary in a defined frame of reference. So you could say that this observer and his clock are stationary in a frame at the co-ordinates of x=1000 m, y=-2345 m, z-0 m at time 345 s and he is facing in the x direction with y to his right and z above his head OR you could define a different scenario in which he is moving in a frame with the same conditions as before but that he is traveling at vx=345 m/s, vy=0 m/s and vz=-4235 m/s, but most people just like to say he is stationary in an inertial frame or he's traveling a .6c in the x direction or not even specify a direction. We usually know what they mean.

But even if there is nothing else in the universe, you can still define him to be stationary or have any kind of motion and/or acceleration you desire (short of c) within a frame.

 

[Grimble]

In Special Relativity, things move or are stationary in a defined frame of reference. So you could say that this observer and his clock are stationary in a frame at the co-ordinates of x=1000 m, y=-2345 m, z-0 m at time 345 s and he is facing in the x direction with y to his right and z above his head OR you could define a different scenario in which he is moving in a frame with the same conditions as before but that he is traveling at vx=345 m/s, vy=0 m/s and vz=-4235 m/s, but most people just like to say he is stationary in an inertial frame or he's traveling a .6c in the x direction or not even specify a direction. We usually know what they mean.

But even if there is nothing else in the universe, you can still define him to be stationary or have any kind of motion and/or acceleration you desire (short of c) within a frame.

OK, then let us define him as stationary. Then can I say:
The FoR, that he is permanently at the origin of, is an Inertial Frame of Reference.
The clock will be on its world line and will be keeping proper time.
For A and the clock proper time and coordinate time will be identical.?

 

[Dale]

Yes the observer is permanently adjacent to the clock, he may even be holding it.

Then the proper time measured by the clock is indeed equal to the coordinate time in A's frame.

 

[phyti]

Yes the observer is permanently adjacent to the clock, he may even be holding it.



One could certainly say that, but would it not be true to say that they were stationary as they cannot be moving if there is nothing to relate the movement to. i.e. movement has to be relative to something and if there is nothing to move relative to they cannot be moving?

I don't accept Newtons definitions about motion and rest. Rest is not a state of motion, but a relation about the difference in motion of two objects. For me, when two objects have the same velocity (vector), one is at rest relative to the other, i.e., rest is a special case of motion but not a 'lack of motion' as Newton defines it.
Since as you say, there is no outside reference object (excepting dark matter), there is no motion of the frame, which by the previous definition eliminates 'rest'.
I agree with your 2nd statement in red.

But...if we look deeper, there is motion at the molecular and atomic levels, therefore the inertial condition is only an approximation, although a practical one.

Now if you could just get light-clock 101...

 

[Grimble]

Let us now add a second clock, 'held' by observer B, at rest at the origin of a second Inertial Frame of Reference.

The clock will be on its world line and will be keeping proper time.
For B and the clock proper time and coordinate time will be identical.?

As A and B are each at rest at the origin of their own Inertial Frames of Reference; we can say that their identical clocks will be keeping identical time. Identical in that their units of time will be of the same duration as judged by an independent observer.

This is necessary as they are both keeping proper time and as that is a requirement of Einstein's first postulate.

 

[Dale]

Let us now add a second clock, 'held' by observer B, at restat the origin of a second Inertial Frame of Reference.

The clock will be on its world line and will be keeping proper time.
For B and the clock proper time and coordinate time will be identical.?

Yes. Assuming that B is also inertial then the proper time measured by the clock held by B will be equal to the coordinate time in B's frame.

As A and B are each at rest at the origin of their own Inertial Frames of Reference; we can say that their identical clocks will be keeping identical time. Identical in that their units of time will be of the same duration as judged by an independent observer.

How does the independent observer judge the duration of their units of time? This is a non-standard phrase so you need to specify what you mean by this.

 

[Grimble]

Yes. Assuming that B is also inertial then the proper time measured by the clock held by B will be equal to the coordinate time in B's frame.

How does the independent observer judge the duration of their units of time? This is a non-standard phrase so you need to specify what you mean by this.

Well you are right of course.

So let us specify an independent inertial observer, who can measure the time passing in A and B's own frames and that, by applying the LT transforms, he can calculate how time passes in each of those frames according to an observer in that frame.

Secondly, as each clock is keeping proper time each tick of those clocks must be an event in space-time. And as JesseM pointed out

If you are looking at some specific event on the worldline of the clock, then all frames agree on what reading it shows at that event.

So if each clock's one second ticks are events in space-time they must be ticking at identical rates.

 

[Dale]

Well you are right of course.

So let us specify an independent inertial observer, who can measure the time passing in A and B's own frames and that, by applying the LT transforms, he can calculate how time passes in each of those frames according to an observer in that frame.

So the independent observer calculates the ratio of (proper time A)/(coordinate time A) in A's frame and (proper time B)/(coordinate time B) in B's frame? If that is your intention then what is the purpose of the independent observer? Since he is transforming into each frame he doesn't add anything to the question just makes the problem more confusing. And we have already specified that both of those ratios are 1.

 

[Grimble]

So the independent observer calculates the ratio of (proper time A)/(coordinate time A) in A's frame and (proper time B)/(coordinate time B) in B's frame? If that is your intention then what is the purpose of the independent observer? Since he is transforming into each frame he doesn't add anything to the question just makes the problem more confusing. And we have already specified that both of those ratios are 1.

Yes, you are quite right of course that is all unnecessary!

 

[ghwellsjr]

Let us now add a second clock, 'held' by observer B, at rest at the origin of a second Inertial Frame of Reference.

The clock will be on its world line and will be keeping proper time.
For B and the clock proper time and coordinate time will be identical.?

As A and B are each at rest at the origin of their own Inertial Frames of Reference; we can say that their identical clocks will be keeping identical time. Identical in that their units of time will be of the same duration as judged by an independent observer.

This is necessary as they are both keeping proper time and as that is a requirement of Einstein's first postulate.

Why do you want to introduce a duplicate of your first scenario with no realtionship between them? If you want to introduce as second observer/clock, then you should define their position/motion/whatever in your first frame OR you could define your second frame in relation to your first frame and then define the second observer/clock in that second frame and then you could transform the observer/clock from the second frame to see how they appear in the first frame. Eventually you need to "put" all observers/clocks into a single frame, that is, if you want to discuss this in the context of Special Relativity.

 

[Grimble]

Why do you want to introduce a duplicate of your first scenario with no realtionship between them? If you want to introduce as second observer/clock, then you should define their position/motion/whatever in your first frame OR you could define your second frame in relation to your first frame and then define the second observer/clock in that second frame and then you could transform the observer/clock from the second frame to see how they appear in the first frame. Eventually you need to "put" all observers/clocks into a single frame, that is, if you want to discuss this in the context of Special Relativity.

Because now we have established that as two individual inertial FoR A and B are, as far as can be determined in identical situations. The one second 'ticks' of one clock keeping proper time, and therefore being Time-space Events, have to be equal in duration and frequency to the one second ticks of the the second clock also keeping proper time, and its 'ticks' therefore, also being Time-space Events.
There is nothing to say, so far concerning the simultaneity of the said cloaks' 'ticks'.

Let us therefore have these two clock's, and as they are at rest at the origin of their respective frames, the frames themselves moving at 0.6c relative to one another.

 

[ghwellsjr]

Because now we have established that as two individual inertial FoR A and B are, as far as can be determined in identical situations. The one second 'ticks' of one clock keeping proper time, and therefore being Time-space Events, have to be equal in duration and frequency to the one second ticks of the the second clock also keeping proper time, and its 'ticks' therefore, also being Time-space Events.

They are identical simply and only because you have defined them to be identical. That's why I asked why you wanted to make a duplicate of your first scenario with no relationship to your original one. I think your understanding of Special Relativity would be advanced more directly if you simply follow Einstein's example from your referenced book in which he looks at an object first at rest in one frame and then from another frame in motion relative to the first one. But if you want to take this indirect approach, then let's see where you go with it.

There is nothing to say, so far concerning the simultaneity of the said cloaks' 'ticks'.

Let us therefore have these two clock's, and as they are at rest at the origin of their respective frames, the frames themselves moving at 0.6c relative to one another.

Are you implying that now there is something to say concerning the simultaneity of the said clocks' 'ticks'?

 

[Grimble]

They are identical simply and only because you have defined them to be identical. That's why I asked why you wanted to make a duplicate of your first scenario with no relationship to your original one. I think your understanding of Special Relativity would be advanced more directly if you simply follow Einstein's example from your referenced book in which he looks at an object first at rest in one frame and then from another frame in motion relative to the first one. But if you want to take this indirect approach, then let's see where you go with it.

Are you implying that now there is something to say concerning the simultaneity of the said clocks' 'ticks'?

Not at all I'm trying to be clear and accepted that this is the case. Or someone will say I am not being specific enough.

Next point let us say that these two clocks pass one another and at the point of their passing they synchronise their clocks.
And let us say that there is another observer, C, who is at rest at the point of their intersection who observes A and B each pass him at speeds of 0.3c in opposite directions
Giving us a single space-time event involving all 3 observers.

 

[Dale]

Not at all I'm trying to be clear and accepted that this is the case. Or someone will say I am not being specific enough.

Next point let us say that these two clocks pass one another and at the point of their passing they synchronise their clocks.
And let us say that there is another observer, C, who is at rest at the point of their intersection who observes A and B each pass him at speeds of 0.3c in opposite directions
Giving us a single space-time event involving all 3 observers.

Actually, relative to observer C the other two observers will be going at 0.33333333 c.

 

[ghwellsjr]

Not at all I'm trying to be clear and accepted that this is the case. Or someone will say I am not being specific enough.

Next point let us say that these two clocks pass one another and at the point of their passing they synchronise their clocks.
And let us say that there is another observer, C, who is at rest at the point of their intersection who observes A and B each pass him at speeds of 0.3c in opposite directions
Giving us a single space-time event involving all 3 observers.

You haven't specified which frame you want to analyze these three observers in and it sounds like you might even be thinking in terms of a third frame where the third observer is at rest. You need to be focusing on a single frame before you start transforming between frames.

 

[Mike_Fontenot]

Einstein, in his little book "Relativity" by Crown publishers, derives the time dilation result in a very clear way. If you want to understand why any inertial observer concludes that a clock which is stationary in some other inertial frame is ticking slower than his own stationary clocks, THAT'S where I'd recommend that you spend some quality time.

Mike Fontenot

 

[ghwellsjr]

"some other inertial frame"

Other than what?

Every inertial observer concludes that every moving clock is ticking slower than his own stationary clocks, no matter what frame we use to define, describe, or analyze the entire scenario in. All observers and clocks "reside" equally and simultaneously in all frames we want to consider. There is no reason or need to introduce a separate frame for each and every observer and clock in a scenario.

 

[Grimble]

You haven't specified which frame you want to analyze these three observers in and it sounds like you might even be thinking in terms of a third frame where the third observer is at rest. You need to be focusing on a single frame before you start transforming between frames.

No, you are right I haven't specified any frame and I am not going to until I have agreement as to the situation/circumstance that I am concerned with.

I have tried before to ask a question or state where I have concerns and it just becomes an endless series of questions, suggestions and alternatives that I should use and what I am asking, trying to understand, or, in my previous attempt, trying to show what my logic tells me in order to find where I am going wrong.

So this time I am trying to build a picture, just as I see it, so we can all agree what I am talking about.

 

[Grimble]

Einstein, in his little book "Relativity" by Crown publishers, derives the time dilation result in a very clear way. If you want to understand why any inertial observer concludes that a clock which is stationary in some other inertial frame is ticking slower than his own stationary clocks, THAT'S where I'd recommend that you spend some quality time.

Mike Fontenot

Thank you Mike, but that is not the problem I understand the relationship and the derivation of time dilation and length contraction. Einstein derives them very clearly.

All I am doing is attempting to describe a relationship between certain bodies: clocks, rulers, and observers. When I can be sure that that arrangement is accepted I can raise the points I find trouble with logically.
Otherwise it all descends into endless arguments over minutiae.

 

[Grimble]

Actually, relative to observer C the other two observers will be going at 0.33333333 c.

Yes, of course, thank you

 

[Grimble]

Now as the one second 'ticks' of the clocks 'held' by A and B are events and they were synchronised at their passing C, then if they emitted a pulse of light at the start of each tick the light from those pulses would arrive at C simultaneously?

And if we allow for the time for the passage of light between frames A and B would each observe the pulses of light from the other to be synchronous with their own pulses of light.

(Each pulse of light from either source is being emitted once every proper second and as those emissions are events their timing will be agreed upon by any inertial observer.)

Am I right so far?

 

[ghwellsjr]

If by "synchronizing" the one-second ticks of the clocks at the moment they are all co-located, then, yes, C will from then on see the respective "one-second" ticks arriving from A and B simultaneously, but they will arrive at less than at one-second intervals as measured by C. In fact, you can use the Relativistic Doppler factor to determine that C will measure the time intervals at 1.414 seconds.

And, in a symmetrical way, A and B will "see" and measure C's one-second ticks coming in also at 1.414 second intervals.

But there is no sense in which we can say that any of the three clocks continue to be synchronized after the one monent when they are co-located. Only clocks that are at rest with respect to each other can be synchronized and remain synchronized.

Furthermore, since A and B are traveling apart from each other at .6c, they will "see" and measure the one-second ticks coming from each other at 2-second intervals.

And we can calculate the time-dilation factors that each observer will determine that each of the other clocks are running at. For the clock/observers involving C (C-A, A-C, B-C, C-B) where the relative speed is 1/3 c, the time dilation is 1.125 even though they "see" each other's clocks ticking at 1.414-second intervals, they "know" that the clocks are actually ticking at 1.125-second intervals. And since A and B have determined that their relative velocity is .6c, they "know" that the time dilation of the other one's clock is 1.25 even though the "see" each other's clocks ticking at 2-second intervals.

Note that this analysis is based on what each observer sees and measures of the other clock's ticks and what they can calculate. No frame has been assumed or used in this analysis which means it has nothing to do with Special Relativity. It is simply a description of what is actually going on for each observer. Now if you want, you can specify a frame of reference, any frame of reference, and you can further anayze the relative simultaneity of the clocks, but all the previous analysis will be true in any frame of reference.

 

[Dale]

For some concise notation, let Exy be the emission event from clock X as expressed in frame Y coordinates and Dxyz be the detection event of the pulse from clock X detected by observer Y as expressed in frame Z coordinates.

Now as the one second 'ticks' of the clocks 'held' by A and B are events and they were synchronised at their passing C, then if they emitted a pulse of light at the start of each tick the light from those pulses would arrive at C simultaneously?

Yes.

Dacc = (1.41 cn, 0,0,0)
Dbcc = (1.41 cn, 0,0,0)
Therefore the pulse from A is detected by C at the same time (in C's frame) as the pulse from B is detected by C.

And if we allow for the time for the passage of light between frames A and B would each observe the pulses of light from the other to be synchronous with their own pulses of light.

No.

Ebb = (cn,0,0,0)
Dabb = (2cn,0,0,0)
Therefore in B's frame the pulses emitted by B are twice as fast as the pulses detected by B from A.

Eaa = (cn,0,0,0)
Dbaa = (2cn,0,0,0)
Therefore in A's frame the pulses emitted by A are twice as fast as the pulses detected by A from B.

(Each pulse of light from either source is being emitted once every proper second and as those emissions are events their timing will be agreed upon by any inertial observer.)

The emissions are indeed events, but the time coordinate of an event is not frame invariant.

Eaa = (cn,0,0,0)
Eac = (1.06 cn, -0.35 cn, 0,0)
Therefore the time coordinate of the emission from A are not the same in A's frame and C's frame.

Ebb = (cn,0,0,0)
Ebc = (1.06 cn, 0.35 cn, 0,0)
Therefore the time coordinate of the emission from B are not the same in B's frame and C's frame.

 

[Grimble]

If by "synchronizing" the one-second ticks of the clocks at the moment they are all co-located, then, yes, C will from then on see the respective "one-second" ticks arriving from A and B simultaneously, but they will arrive at less than at one-second intervals as measured by C. In fact, you can use the Relativistic Doppler factor to determine that C will measure the time intervals at 1.414 seconds.

And, in a symmetrical way, A and B will "see" and measure C's one-second ticks coming in also at 1.414 second intervals.

But there is no sense in which we can say that any of the three clocks continue to be synchronized after the one monent when they are co-located. Only clocks that are at rest with respect to each other can be synchronized and remain synchronized.

Furthermore, since A and B are traveling apart from each other at .6c, they will "see" and measure the one-second ticks coming from each other at 2-second intervals.

And we can calculate the time-dilation factors that each observer will determine that each of the other clocks are running at. For the clock/observers involving C (C-A, A-C, B-C, C-B) where the relative speed is 1/3 c, the time dilation is 1.125 even though they "see" each other's clocks ticking at 1.414-second intervals, they "know" that the clocks are actually ticking at 1.125-second intervals. And since A and B have determined that their relative velocity is .6c, they "know" that the time dilation of the other one's clock is 1.25 even though the "see" each other's clocks ticking at 2-second intervals.

Note that this analysis is based on what each observer sees and measures of the other clock's ticks and what they can calculate. No frame has been assumed or used in this analysis which means it has nothing to do with Special Relativity. It is simply a description of what is actually going on for each observer. Now if you want, you can specify a frame of reference, any frame of reference, and you can further anayze the relative simultaneity of the clocks, but all the previous analysis will be true in any frame of reference.

You see here we go straight away with a whole lot of extra information.

C has no clock as seen by A or B.

I have already stated that I am not concerned with the time a signal takes. We may assume that all observers allow for the transmission times which only add another layer of complication.

 

[ghwellsjr]

You see here we go straight away with a whole lot of extra information.

C has no clock as seen by A or B.

I have already stated that I am not concerned with the time a signal takes. We may assume that all observers allow for the transmission times which only add another layer of complication.

I'm sorry if you think I'm adding "a whole lot of extra information" because it is very important information and I'm trying to help you. Please pay attention to what I'm about to say. You need to understand this because your questions belie a fundamental misunderstanding of the problem you're trying to understand.

You have repeated your caveat that you are "not concerned with the time a signal takes. We may assume that all observers allow for the transmission times which only add another layer of complication."

Here are the facts: no observer, no person, nobody, knows "the time a signal takes". We cannot "assume that all observers allow for the transmission times" since they don't know the transmission times. Nobody does. If we knew them or if we could figure them out, we wouldn't have Special Relativity. You need to understand this or the entire rest of this learning exercise will be a waste of time because if you do learn something, it will be the wrong thing.

What I described for you in my previous post was all that the observers can know and as I said, it has nothing to do with SR and nothing to do with transmission times.

As soon as you apply SR to a scenario, you are arbitrarily assigning transmission times and these times vary depending on the frame of reference you select. You may think that there is some reality to using each observer's rest frame to determine the transmission times, but there is not. SR arbitrarily defines the transmission times even within a single frame of reference. So your statement that "all observers allow for the transmission times" does not comport with reality, it's the same as saying that you (and your observers) believe in an absolute aether rest frame and that you know where it is.

Am I getting through to you?

 

[ghwellsjr]

Ebb = (cn,0,0,0)
Dabb = (2cn,0,0,0)
Therefore in B's frame the pulses emitted by B are twice as fast as the pulses detected by B from A.

Ebb = (cn,0,0,0)
Dbaa = (2cn,0,0,0)
Therefore in A's frame the pulses emitted by A are twice as fast as the pulses detected by A from B.

Dalespam, I believe your second pair of equations should start with Eaa rather than Ebb, correct?

And just for the fun of it, could you also calculate those two ratios in all three frames, please?

 

[bobc2]

Sometimes it can help to visualize the problem in the context of a 4-D spatial universe. Depicting the motion of a 4-D structure with respect to some system at rest (supressing X2 and X3), the velocity is manifest directly as the slope of the world line that constitutes the 4th dimension for that structure. The peculiar and mysterious aspect of Special Relativity is that if the X4 axis is rotated with respect to a rest system, then the X1 axis is rotated as well---rotated such that the photon 4-D world line always bisects the angle between the X1 and X4 axis (this goes for all observers). A little reflection on this circumstance leads to the realization that when a photon is observed (for any observer, no matter his velocity), the distance traveled by the photon along the observer's X4 axis (ct) is always the same as the distance traveled by the photon along the X1 axis. The ratio of dX1 to dX4 is always 1. Thus, the speed of light is the same for all observers. Each observer moves along his own X4 axis at the speed of light.

The first sketch above (upper left) depicts an observer moving at some velocity with respect to a rest system (X1, X4). The instantaneous 3-D universe he lives in is represented by the slanted X1' axis. Observers having different velocities will have X1' axes with different slants--they are living in different cross-sections of the same 4-D universe. Coordinate systems for observers having different velocities are shown in a sequence across the bottom sketches (they are all living in different 3-D cross-sections of the same 4-D universe). Above Right: An observer is changing velocity as time passes--his coordinate system, thus his cross-section view of the universe changes accordingly.

 

[Dale]

Dalespam, I believe your second pair of equations should start with Eaa rather than Ebb, correct?

And just for the fun of it, could you also calculate those two ratios in all three frames, please?

Oops, thanks for catching the typo, I will go edit it.

Which two ratios are you interested in? I certainly can calculate them in any frame quite easily.

 

[ghwellsjr]

Which two ratios are you interested in? I certainly can calculate them in any frame quite easily.

Dabx to Ebx and Dbax to Eax, where x is a, b, and c. You did two of the six possible calculations and got ratios of 2 for each of them, I'd like you to do the other four, please.

 

[Dale]

Dabx to Ebx and Dbax to Eax, where x is a, b, and c. You did two of the six possible calculations and got ratios of 2 for each of them, I'd like you to do the other four, please.

Sure. However, since the Lorentz transform is linear you automatically know that the ratio will always be 2. But here are the results:

Daba = (2.5cn,1.5cn,0,0)
Eba = (1.25cn,.75cn,0,0)
Dbaa = (2cn,0,0,0)
Eaa = (cn,0,0,0)


Dabb = (2cn,0,0,0)
Ebb = (cn,0,0,0)
Dbab = (2.5cn,-1.5cn,0,0)
Eab = (1.25cn,-.75cn,0,0)


Dabc = (2.12cn,.71cn,0,0)
Ebc = (1.06cn,.35cn,0,0)
Dbac = (2.12cn,-.71cn,0,0)
Eac = (1.06cn,-.35cn,0,0)

 

[ghwellsjr]

Sure. However, since the Lorentz transform is linear you automatically know that the ratio will always be 2.

Of course the Lorentz Transform has to get the correct answer of 2 no matter what frame you do the calculation from because this ratio is 2 and has nothing to do with Special Relativity. I just didn't want Grimble to think that since you originally only calculated the ratio that A sees from the frame in which A is at rest and the ratio that B sees from the frame in which B is at rest, that a conclusion could be drawn that these ratios were somehow associated with rest frames.

This ratio, as I pointed out in post #323 is the Relative Doppler factor (or its reciprocal depending on whether the clocks are advancing or retreating) and has nothing to do with Special Relativity. It is a real physical measurement made by observer/clocks in constant relative motion (with prior knowledge that they have identical clocks, of course). One of the points of Special Relativity is that you can use any inertial frame of reference and it will calculate that observer/clocks make the same measurements. It better, or it wouldn't match reality.

 

[Grimble]

But there is no sense in which we can say that any of the three clocks continue to be synchronized after the one monent when they are co-located. Only clocks that are at rest with respect to each other can be synchronized and remain synchronized.

Let us assign the space-time coordinates (0,0,0,0) to the event A and B pass C.

The interval between that event and A emitting the first flash of light, first 'tick' of the clock will be given by S² = (x²+y²+z²-c²t²)

Where S is the space-time interval
x is the displacement in proper length units = 0.333 light seconds
y and z = 0, c=1, t =1 second proper time.

which gives us S² = (0.333² + 0 + O - 1) = -0.889 and
s = 0.943

For B the displacement will be -0.333 giving the same interval of 0.943 and in which the displacement and time difference are the same as for A.

So by my reckoning the emission of light from A and B are synchronous, everything is symmetrical.

And if that is true for the first pulse of light from each clock it will also be true for the second pulses and all subsequent pairs of pulses.

 

[Dale]

The interval between that event and A emitting the first flash of light, first 'tick' of the clock will be given by S² = (x²+y²+z²-c²t²)

Where S is the space-time interval
x is the displacement in proper length units = 0.333 light seconds
y and z = 0, c=1, t =1 second proper time.

Careful here. x is the coordinate displacement and t is the coordinate time. Then s^2 is the square of the interval so that sqrt(-s^2/c^2) is the proper time for timelike intervals and sqrt(s^2) is the proper length for spacelike intervals.

That said, you are correct that the time coordinate in frame C is the same for Eac and Ebc. But they are not synchronized with coordinate time in frame C.

 

[Grimble]

Careful here. x is the coordinate displacement and t is the coordinate time. Then s^2 is the square of the interval so that sqrt(-s^2/c^2) is the proper time for timelike intervals and sqrt(s^2) is the proper length for spacelike intervals.

That said, you are correct that the time coordinate in frame C is the same for Eac and Ebc. But they are not synchronized with coordinate time in frame C.

So are you agreeing that the emission event for A is synchronous with the emission event at B? I didn't say it would be synchronised with coordinate time in C, yet is it not the coordinate time in frame C for the emission events at A and B that is equal?

I do agree that the emission from A as observed by B and that of B observed by A that will be subject to Time dilation - (and length contraction).

Yet the light emitted from those events will not be affected, will it? As light travels at c and this is not affected by the speed of the source. So the two pulses of light will travel the same distance at c.

Consider too that after 100 seconds each clock will have emitted 100 pulses of light. It is the arrival of these pulses of light that will be observed by the other observers who by subtracting the transmission time will count 100 pulses emitted in 100seconds. At the same rate by each clock.
There is not and cannot logically be a two for one ratio between the emission and receipt of the pulses.
The only way that we could have a two second coordinate time between the emission of the said pulses is if two seconds coordinate time have the same duration as one second proper time, which is what the two clocks are keeping.

 

[Dale]

So are you agreeing that the emission event for A is synchronous with the emission event at B?

In frame C, yes, in any other frame, no. That is what the relativity of simultaneity means.

Yet the light emitted from those events will not be affected, will it? As light travels at c and this is not affected by the speed of the source. So the two pulses of light will travel the same distance at c.

Correct. Dacx=Dbcx (where x is anyone of a,b,c). The distances traveled by the light pulses is only equal in frame c. In other frames the distances traveled are different, and because of the relativity of simultaneity the times of emission are also different, such that they arrive at C at the same event.

Consider too that after 100 seconds each clock will have emitted 100 pulses of light.

No, after 100 seconds in any given frame only the clock at rest in that frame will have emitted 100 pulses of light. For example, in A's frame:
Eaa = (cn,0,0,0)
Eba = (1.25cn,.75cn,0,0)
so after 100 seconds there will be 100 pulses emitted from clock A but only 80 emitted from clock B.

It is the arrival of these pulses of light that will be observed by the other observers who by subtracting the transmission time will count 100 pulses emitted in 100seconds. At the same rate by each clock.

No, even after accounting for the delay of light you will still find that fewer than 100 pulses were emitted.

There is not and cannot logically be a two for one ratio between the emission and receipt of the pulses.

Sure there can.

The only way that we could have a two second coordinate time between the emission of the said pulses is if two seconds coordinate time have the same duration as one second proper time, which is what the two clocks are keeping.

The coordinate time between the emissions is 1.25 seconds (for Eba), not 2 seconds. The 2 seconds is the coordinate time between the detections (for Dbaa)

 

[Grimble]

l3sm506608gc

No, after 100 seconds in any given frame only the clock at rest in that frame will have emitted 100 pulses of light. For example, in A's frame:
Eaa = (cn,0,0,0)
Eba = (1.25cn,.75cn,0,0)
so after 100 seconds there will be 100 pulses emitted from clock A but only 80 emitted from clock B.

I am still finding this way of expressing things a little confusing - what does the 'cn' mean?

No, even after accounting for the delay of light you will still find that fewer than 100 pulses were emitted.

And yes, of course, from one frame the time in another will be dilated. BUT In each frame, as observed from within that frame, 100 pulses will be emitted in 100 seconds. And as they frames that are not moving, according to that observer they are proper seconds (I am not sure whether this is significant or not)

And light from the pulse from each frame will arrive at C simultaneously. (And this has nothing to do with relative simultaneity as they are simultaneous at the same place, i.e. together they comprise a single event.)
And if a mirror was in place at C then the light from one frame would pass in the same time, the same distance and arrive at the other frame simultaneously with that frame's light reflected back from C, so each has to observe the other frame emitting 100 pulses in 100 seconds.

If one frame only observed 80 flashes in 100 seconds, what happens to the other 20 flashes that we know it emitted in 100 seconds in its own frame?

And so, regardless of how you calculate time passing what happens doesn't change, it is how it is observed that changes.

The coordinate time between the emissions is 1.25 seconds (for Eba), not 2 seconds. The 2 seconds is the coordinate time between the detections (for Dbaa)

I'm sorry but can you explain that? How do we have a greater time between detections than between emissions. Or are you adding in the transmission times again?

 

[yuiop]

I'm sorry but can you explain that? How do we have a greater time between detections than between emissions. Or are you adding in the transmission times again?

In the above diagram the green line represents a person firing projectiles (blue lines) at rate of one per second at a fleeing person represented by the red line. The red guy receives the projectiles at a rate of one every two seconds because he is going away from the emitter. The transmission time for the projectiles between being emitted and being received is steadily increasing from 1 to 5 seconds and yet the victim continues to receive them at a constant rate of one every 2 seconds. Therefore the rate of reception is independent of the transmission time and can differ from the emitter rate. This is a simple geometrical property that is the basis of classic Doppler shift and in this example we are not even considering relativistic velocities or time dilation / length contraction etc.

 

[Dale]

I am still finding this way of expressing things a little confusing - what does the 'cn' mean?

The c is the speed of light and n is simply an arbitrary integer indicating which pulse. For example, for the 5th pulse you would set n=5.

And yes, of course, from one frame the time in another will be dilated. BUT In each frame, as observed from within that frame, 100 pulses will be emitted in 100 seconds.

Only by an emitter which is at rest in that frame.

If one frame only observed 80 flashes in 100 seconds, what happens to the other 20 flashes that we know it emitted in 100 seconds in its own frame?

Nothing has happened to them. They simply haven't been emitted yet in the other frames. That is what the relativity of simultaneity implies.

I'm sorry but can you explain that? How do we have a greater time between detections than between emissions. Or are you adding in the transmission times again?

Yes, of course, by definition the time of detection includes the time of emission plus the time of transmission. If you negelct the transmission time then by definition you have the emission, not the detection.

 

[Grimble]

The c is the speed of light and n is simply an arbitrary integer indicating which pulse. For example, for the 5th pulse you would set n=5.

Thank you, that makes sense.

Only by an emitter which is at rest in that frame.

But those two Frames are in every way the same (apart from their relative velocity), their clocks are synchronised and identical operating in identical conditions. viz.The Principle of Relativity – The laws by which the states of physical systems undergo change are not affected, whether these changes of state be referred to the one or the other of two systems in uniform translatory motion relative to each other.

So if that 1st postulate is adhered to they must be emitting pulses simultaneously.

* But you will no doubt say simultaneous to which observer?

And I would answer to Space time itself.

The emission of each 'pair' of simultaneous pulses are equal space time intervals from the origin event where/when A and B pass C.

Not only do their Space time intervals have to be equal but those space time intervals comprise a space element and a time element and in each case the space intervals are equal, as are the time elements.
So if the spacetime intervals are equal and the time elements of them are equal how can they not be simultaneous.

These events observed by A or B however will not be silultaneous, distances and time swill differ BECAUSE of where they are observed from.

 

[Dale]

But those two Frames are in every way the same (apart from their relative velocity), their clocks are synchronised and identical operating in identical conditions. viz.The Principle of Relativity – The laws by which the states of physical systems undergo change are not affected, whether these changes of state be referred to the one or the other of two systems in uniform translatory motion relative to each other.

So if that 1st postulate is adhered to they must be emitting pulses simultaneously.

No, simultaneity is not a law of physics. It is, in fact, merely an arbitrary human-made convention.

Do you believe that if two events occur at the same x coordinate in one frame that they must therefore occur at the same x coordinate in all other frames? If not, then what would make you believe that is the case with the t coordinate?

The emission of each 'pair' of simultaneous pulses are equal space time intervals from the origin event where/when A and B pass C.

Yes, in all frames.

those space time intervals comprise a space element and a time element and in each case the space intervals are equal, as are the time elements.

Yes in frame C, but not in any other frame.

So if the spacetime intervals are equal and the time elements of them are equal how can they not be simultaneous.

They are simultaneous in frame C, but not in any other frame.

These events observed by A or B however will not be silultaneous, distances and time swill differ BECAUSE of where they are observed from.

Yes. And not only are the detection events not simultaneous, but the emission events are also not simultaneous in A and B, as shown above.

 

[Grimble]

No, simultaneity is not a law of physics. It is, in fact, merely an arbitrary human-made convention.

I have to agree with you there
It depends on how one defines the convention and how it is measured.
So let us say that if we take the difference in proper time from a single event A in space time and a second event B and compare that with the difference in proper time from A and a third event C.
If the proper time difference AB is equal to AC then, in space-time, is B not simultaneous with C?
Surely as A,B and C are all individual events in Space time there relationship is fixed, wherever it is measured from, inertially in spacetime.

Do you believe that if two events occur at the same x coordinate in one frame that they must therefore occur at the same x coordinate in all other frames? If not, then what would make you believe that is the case with the t coordinate?

No, and I do not.

Yes, in all frames.

Yes in frame C, but not in any other frame.[/quote]
Because from other frames the measurements will be coordinate measurements? And by the application of the LT equations they can be converted to proper measurements? And as proper measurements they will then be equal? (I am asking here, not telling)

Yes. And not only are the detection events not simultaneous, but the emission events are also not simultaneous in A and B, as shown above.

Yes that is just what I am saying they are simultaneous when measured from an impartial frame of reference yet from nowhere else.

Spacetime is a way of measuring the relationship of events.

If two intervals from a common event in space time are not only equal but identical then not only are their Space-time intervals equal but their space elements are equal as are their time elements.

As I understand it Space-time is described by a four dimensional coordinate system. With three space-like coordinates and one time-like coordinate.

Everything in time and space may be plotted with reference to those coordinates. By any inertial observer. The only things that can be different are the designated origin (0,0,0,0) that is used and which inertial frames of reference are at rest and which are moving relative to that designated origin.

All measurements made by an observer at rest within said inertial FoR are proper measurements: made by clocks at rest (proper times) and between the world lines of such clocks at rest (proper lengths/distances).

All measurements of moving objects, made by taking the measurements made within the moving object's own FoR (in which it is at rest) and transforming them using LT, will be coordinate measurements.

(I have a mind that sees things in pictures/ diagrams/ images/ in multiple dimensions and so I am trying to find a starting point from which I can show where I find things that don't seem to fit)

Thank you one and all for your efforts.

 

[Dale]

let us say that if we take the difference in proper time from a single event A in space time and a second event B and compare that with the difference in proper time from A and a third event C.
If the proper time difference AB is equal to AC then, in space-time, is B not simultaneous with C?

Not in general, no. B is simultaneous with C in a given frame iff the coordinate time difference AB is equal to the coordinate time difference AC.

Surely as A,B and C are all individual events in Space time there relationship is fixed, wherever it is measured from, inertially in spacetime.

I am not sure what you mean here by "relationship". The spacetime intervals AB, AC, and BC are indeed fixed, but of course the coordinate differences depend on the coordinate system.

And by the application of the LT equations they can be converted to proper measurements? And as proper measurements they will then be equal? (I am asking here, not telling)

You don't need the LT in order to transform the measurements into proper measurements. All you need is the formula for the spacetime interval:

ds² = -c²dt² + dx² + dy² + dz²

The spacetime interval squared, ds², is invariant, meaning that it is the same in all reference frames. The Lorentz transform is the transform which preserves the spacetime interval.

If two intervals from a common event in space time are not only equal but identical then not only are their Space-time intervals equal but their space elements are equal as are their time elements.

I don't know what you mean by "not only equal but identical".

As I understand it Space-time is described by a four dimensional coordinate system. With three space-like coordinates and one time-like coordinate.

Yes.

Everything in time and space may be plotted with reference to those coordinates. By any inertial observer. The only things that can be different are the designated origin (0,0,0,0) that is used and which inertial frames of reference are at rest and which are moving relative to that designated origin.

In spacetime a point particle is represented by a line, called a worldline, which indicates the particle's position at each moment of time. If the particle is inertial then the worldline is straight. If two particles are at rest wrt each other then their worldlines are parallel to each other.

The origin is an event, which is a point rather than a line in spacetime. It makes no sense to say that something is moving relative to the origin any more than it does to say that a given line is parallel to a point.

(I have a mind that sees things in pictures/ diagrams/ images/ in multiple dimensions and so I am trying to find a starting point from which I can show where I find things that don't seem to fit)

Me too. I highly recommend going through the exercise of making a few spacetime diagrams where you carefully plot the coordinate lines of a moving frame on the same diagram so that you can see how the coordinates differ. I also recommend learning about four-vectors, they are much more suited to a geometrically-oriented mind.

 

[ghwellsjr]

If two intervals from a common event in space time are not only equal but identical then not only are their Space-time intervals equal but their space elements are equal as are their time elements.

I don't know what you mean by "not only equal but identical".

Dalespam, it looks to me like Grimble is thinking that since the spacetime interval between two events is the same for two observers at rest in two different frames of reference, then this means that the space "distances" are equal in those two FoRs and the time "intervals" are equal in those two FoRs. He doesn't understand that it is only the "spacetime intervals" that are equal which are a combination of both the space and time "intervals". I know you have been repeating this over and over again, but he's still stuck on this point, I believe.

 

[Grimble]

Not in general, no. B is simultaneous with C in a given frame if the coordinate time difference AB is equal to the coordinate time difference AC.

But that was not what I was specifying was it? I am trying to relate to the concept of Space-Time which is a way of describing what is. Coordinate times and distances are measurements taken from different Reference Frames: different reference frames are, surely, how that same single reality is measured by different observers?[/quote]

I am not sure what you mean here by "relationship". The spacetime intervals AB, AC, and BC are indeed fixed, but of course the coordinate differences depend on the coordinate system.

I mean the relationship between events as defined by the space-time intervals between them

You don't need the LT in order to transform the measurements into proper measurements. All you need is the formula for the spacetime interval:

ds² = -c²dt² + dx² + dy² + dz²

The spacetime interval squared, ds², is invariant, meaning that it is the same in all reference frames. The Lorentz transform is the transform which preserves the spacetime interval.

I'm sorry but how does one convert between coordinate and proper measurements using the space time interval?
I understood (or thought that I did) that;

proper measure/coordinate measure= gamma

I don't know what you mean by "not only equal but identical".

Well to me 'equal' compares the totals, whereas identical compares the individual terms; i.e. the time vales are equal and the distance values are equal.

In spacetime a point particle is represented by a line, called a worldline, which indicates the particle's position at each moment of time. If the particle is inertial then the worldline is straight. If two particles are at rest wrt each other then their worldlines are parallel to each other.

Yes, as I said "Everything in time and space may be plotted with reference to those coordinates "
But this next bit has me confused:

The origin is an event, which is a point rather than a line in spacetime. It makes no sense to say that something is moving relative to the origin any more than it does to say that a given line is parallel to a point.

But if the origin (0,0,0,0) defines an event, a point in space at a specific time, then the world line of that point in space is a straight line – agreed.

Yet any other reference frame must also have an origin that will be in the form (ct, x, y, z) and which will either have a constant Space-time interval or a changing one. If it is constant then that Reference frame is at rest relative to the space-time origin and if the space-time interval is changing then that reference frame is moving with respect to the defined space-time origin.

Now it seems to me that within our 4 dimensional coordinate system any inertial frame of reference may be considered to have a clocks at rest with respect to it. Those clocks will have straight worldlines that are parallel one to the other. The world line of the one at the origin of that frame of reference will describe the time coordinate of our 4D set of coordinates. The world lines of all the clocks will measure proper time and the distances between their parallel world lines will therefore measure proper distances.

Me too. I highly recommend going through the exercise of making a few spacetime diagrams where you carefully plot the coordinate lines of a moving frame on the same diagram so that you can see how the coordinates differ. I also recommend learning about four-vectors, they are much more suited to a geometrically-oriented mind.

Space diagrams I am happy with but I will look into 'four-vectors'.

 

[Dale]

But that was not what I was specifying was it? I am trying to relate to the concept of Space-Time which is a way of describing what is. Coordinate times and distances are measurements taken from different Reference Frames: different reference frames are, surely, how that same single reality is measured by different observers?

Yes, although usually you would say something more along the lines of "different observer's perspective" rather than "measured by different observers". This is because what is actually measured by an observer is affected by the finite speed of light and the reference frame is what the observer calculates after accounting for the finite speed of light.

I'm sorry but how does one convert between coordinate and proper measurements using the space time interval?
I understood (or thought that I did) that;

proper measure/coordinate measure= gamma

The formula you just posted is only correct in certain particular circumstances. The one I posted is more general and also it easily generalizes further to non-inertial coordinates and curved spacetimes.

In the formula
ds² = -c²dt² + dx² + dy² + dz²

the dt, dx, dy, and dz are all coordinate times and distances in some inertial frame while the ds is the frame-invariant spacetime interval. If ds² > 0 then the interval is called "spacelike" and ds is the proper distance. If ds² < 0 then the interval is called "timelike" and dτ = sqrt(-ds²/c²) is the proper time. If ds² = 0 then the interval is called "lightlike" or "null" and represents the path of a pulse of light.

The value of this formula is that all reference frames can use their own coordinate values and come up with the same value for ds². This formula also let's you easily see where the formula you posted comes from:

dτ² = -ds²/c² = dt² - dx²/c² - dy²/c² - dz²/c²
dτ²/dt² = 1 - (dx²/dt² - dy²/dt² - dz²/dt²)/c²
dτ²/dt² = 1 - v²/c²
dτ/dt = 1/γ

identical compares the individual terms; i.e. the time vales are equal and the distance values are equal.

Then yes, clearly "if two intervals from a common event in space time are not only equal but 'the time vales are equal and the distance values are equal' then not only are their Space-time intervals equal but their space elements are equal as are their time elements".

Yes, as I said "Everything in time and space may be plotted with reference to those coordinates "
But this next bit has me confused:
But if the origin (0,0,0,0) defines an event, a point in space at a specific time, then the world line of that point in space is a straight line – agreed.

Yet any other reference frame must also have an origin that will be in the form (ct, x, y, z) and which will either have a constant Space-time interval or a changing one. If it is constant then that Reference frame is at rest relative to the space-time origin and if the space-time interval is changing then that reference frame is moving with respect to the defined space-time origin.

I think you are confusing the origin of spacetime, the event (0,0,0,0), with the coordinate time axis, the worldline (ct,0,0,0). Objects may be resting or moving wrt the time axis (parallel worldlines), but not wrt the origin (lines cannot be parallel or not parallel to a point).

Now it seems to me that within our 4 dimensional coordinate system any inertial frame of reference may be considered to have a clocks at rest with respect to it. Those clocks will have straight worldlines that are parallel one to the other. The world line of the one at the origin of that frame of reference will describe the time coordinate of our 4D set of coordinates.

Yes, exactly. That clock defines the time axis of the coordinate system, (ct,0,0,0).

 

[grav-universe]

It depends on how one defines the convention and how it is measured.
So let us say that if we take the difference in proper time from a single event A in space time and a second event B and compare that with the difference in proper time from A and a third event C.
If the proper time difference AB is equal to AC then, in space-time, is B not simultaneous with C?
Surely as A,B and C are all individual events in Space time there relationship is fixed, wherever it is measured from, inertially in spacetime.

Hi, Grimble. It looks like you are still getting stuck on this simultaneity thing. The difference in times between the events that occur at A and B, and at A and C, depend upon which frame is measuring the time of those events and the simultaneity convention by which their clocks are set. As Dalespam mentioned, simultaneity conventions are man-made, meaning there is no absolute concept of simultaneity.

For instance, let's say that events A and B are both measured from within the same frame and it is determined that they occurred simultaneously. What does that mean? It simply means that the clocks that an observer carries at A and another observer carries at B read the same when the events occurred, right? Now what if we were to set observer B's clock forward 5 minutes? Then according to the new simultaneity convention, event B now occurs five minutes after event A, not simultaneously at all anymore. That's what simultaneity is, it is man-made and depends upon the convention we have declared for how each frame should set their clocks.

The Einstein simultaneity convention gives a convenient way to set clocks by using light since all frames measure the same two way time in any direction for light to travel away and back over the same distance, between points 1 and 2 say, so a clock at the other end of that distance at point 2 can simply be set such that the time for the light to travel from point 1 to point 2 is the same as that from point 2 to point 1. After the simultaneity has been set accordingly for each frame, Relativity applies as measurements are then made between frames for events. This is what applies to space-time, not just space and not just time, since a simple re-adjustments in the clocks of a frame will cause that frame's observers to measure lengths in other frames differently as well, for example.

You can establish an altogether different simultaneity convention if you wish, but then you must first determine what it should be and by what procedure it would be physically established, then work out the details for the simultaneity issues between frames while still acknowledging the results of experiments that confirm Relativity.

 

[Grimble]

Yes, although usually you would say something more along the lines of "different observer's perspective" rather than "measured by different observers". This is because what is actually measured by an observer is affected by the finite speed of light and the reference frame is what the observer calculates after accounting for the finite speed of light.

Thank you, that makes sense, I will try and remember that in the future.

The formula you just posted is only correct in certain particular circumstances. The one I posted is more general and also it easily generalizes further to non-inertial coordinates and curved spacetimes.

In the formula
ds² = -c²dt² + dx² + dy² + dz²

the dt, dx, dy, and dz are all coordinate times and distances in some inertial frame while the ds is the frame-invariant spacetime interval. If ds² > 0 then the interval is called "spacelike" and ds is the proper distance. If ds² < 0 then the interval is called "timelike" and dτ = sqrt(-ds²/c²) is the proper time. If ds² = 0 then the interval is called "lightlike" or "null" and represents the path of a pulse of light.

The value of this formula is that all reference frames can use their own coordinate values and come up with the same value for ds². This formula also let's you easily see where the formula you posted comes from:

dτ² = -ds²/c² = dt² - dx²/c² - dy²/c² - dz²/c²
dτ²/dt² = 1 - (dx²/dt² - dy²/dt² - dz²/dt²)/c²
dτ²/dt² = 1 - v²/c²
dτ/dt = 1/γ

Thank you

I think you are confusing the origin of spacetime, the event (0,0,0,0), with the coordinate time axis, the worldline (ct,0,0,0). Objects may be resting or moving wrt the time axis (parallel worldlines), but not wrt the origin (lines cannot be parallel or not parallel to a point).

Oh yes, I see that now, I was becoming a little muddled there; talking about the path of a point in space at one particular point in time!
What I should have said was that the world line of the origin of any other inertial frame of reference would be parallel if that FoR was at rest with the first one.

Yes, exactly. That clock defines the time axis of the coordinate system, (ct,0,0,0).

And that time shown by that clock will be proper time? and the distance between any two clocks at rest will be a proper distance?
Which leads me to the conclusion that Proper time and Proper lengths are the units of an inertial FoR's coordinates? Of any inertial FoR?

 

[Grimble]

Hi, Grimble. It looks like you are still getting stuck on this simultaneity thing.

Yes I suppose that is true. It seems to me that something, that I see as very simple, is very hard to refer to without everyone becoming confused.

Are not all conventions, by definition, man-made?

To me simultaneity is is a straight forward concept – that two things happen at the same time.

The difficulty that is immediately seized upon is 'how does one define “at the same time”'.

The obvious way to do that, is to see how those two events are plotted onto a coordinate system that has time as an axis. Such as Space-time...

Then if two events have the same time coordinate, in one FoR, they must be simultaneous? For that is what having the same time coordinate means? Surely.

Now we know that the three space coordinates may be aligned any way we like in Space time, but what about the time axis? How is that aligned? Is it aligned as such?

Coordinate time depends on the observer's perspective, but what of proper-time, i.e. the time of space-time itself? Every observer can translate their coordinate time into proper time but is that into a common 'proper-time' that all agree upon or does it still vary according to the observer's perspective?

After all Space-time is not fixed is it? It is only a set of coordinates and the Space-time coordinates that are at rest with respect to One FoR will be moving with respect to another FoR.

 

[grav-universe]

Most of the statements you made are pretty much as you have said, but I will elaborate on these.

To me simultaneity is is a straight forward concept – that two things happen at the same time.

Just to be clear, if two events occur simultaneously according to a frame, then they happen at the same time according to that frame's coordinate system, yes, but simultaneity issues in general actually refer to differences in simultaneity, where events that occur simultaneously according to the coordinate system of one frame do not necessarily occur simultaneously according to the coordinate system of another frame, depending upon the simultaneity convention that is set for each frame.

For instance, if we have two observers in a frame that are separated by some distance, and they both then instantly and simultaneously accelerate to the same speed in the same direction and they leave their clocks alone, then if the original frame measures events simultaneously, the two observers in their new frame will still measure them as simultaneous also, but the two observers will also measure the speed of light anisotropically. But if the two observers re-synchronize their clocks to measure the same speed of light in every direction in accordance with the Einstein simultaneity convention, then they will no longer measure the events as simultaneous as the original frame does, because in order to synchronize in this way to measure an isotropic speed of light, the front observer's clock must be set back somewhat or the rear observer's clock set forward.

Now we know that the three space coordinates may be aligned any way we like in Space time, but what about the time axis? How is that aligned? Is it aligned as such?

Yes. We start by placing clocks at all points within a frame, all stationary to each other within the frame. Now let's say we want to re-synchronize the clocks along the x axis. We can begin at the origin and add, say, one second on the clocks per meter along the positive x direction, so add one second at one meter, two seconds at two meters, and so on, and subtract one second per meter in the negative x direction. Observers within the frame will consider the clocks to be synchronized regardless of how they are set since the readings upon clocks for events can only be measured directly by clocks that coincide in the same place, just as long as the synchronization method is linear along any direction so that objects that travel inertially will still be measured as such.

Coordinate time depends on the observer's perspective, but what of proper-time, i.e. the time of space-time itself? Every observer can translate their coordinate time into proper time but is that into a common 'proper-time' that all agree upon or does it still vary according to the observer's perspective?

Only one observer in a particular frame can measure the proper time between two events, since that observer's clock must coincide with both events to measure the time of the events directly. In other words, the proper time can only be measured directly by a single clock. However, since all of the observer's clocks in the same frame as that observer, no matter how they have actually been synchronized, are considered by observers within that frame to be in perfect synch with each other, then all observers within the frame with the observer that measures the proper time of the two events will agree upon the time of both events also, each occurring in the same place as that particular observer, so with zero spatial difference as measured by that frame. All other frames, however, moving at some relative speed to the frame in which the proper time is measured, will measure some distance between the events due to the relative speed. They will also measure a different difference in time between the events, depending upon how their clocks are synchronized. If both frame's clocks have been synchronized according to the Einstein simultaneity convention, then the difference in times and distances measured by each frame will be related by the Minkowski metric.