[fontwell]

There is, as I see it, no 'set distances'. You need to use the concepts of clocks and rulers to make them 'connect' for two different frames. And there I'm still wondering what they really are:) Distance is such an 'normal' everyday concept to us, we expect them to be the same from day to day, and they are too at least inside our own frame.

This made me smile.


I had a look for links and all I can find (that look helpful) are these two...

http://wapedia.mobi/en/Minkowski_diagram [nofollow]

http://www.einsteins-theory-of-relativity-4engineers.com/loedel-diagrams.html [nofollow]

Both of them dive in quite quickly.

The way I think about the diagrams is like this (not why they work, only how to interpret)...

A normal space vs time diagram, which you know from school looks like this, except I have swapped x axis (distance) and time axis (t) over, because this is what Loedel diagrams do. 

[BTW, I've put events on all these diagrams but they are not related, they are just examples of how to read coordinates from the axis.]

[diagram=575_0]

Note how there is a grid.

The horizontal lines represent things which all happen at the same time (from the origin). The vertical lines represent things which all happen at the same distance (from the origin). In relativity an observer makes all measurements relative to themselves i.e. they never move. So their 'world line' is the vertical 't' axis at x=0.

The units on the axis are defined in terms of C. Any unit can be used for distance, but the time unit used will then be "the time it takes light to travel one distance unit (in a vacuum)". Thus a light beam is always at 45 degrees on the diagram (to move one distance unit along, it takes one time unit up).

At low relative speeds two observers can use the same diagram. But at high relative speeds the effects of relativity can be modelled by 'compressing' the observers grid. It is just as if you had a trellis (like for plants) and squeeze it.

So for one observer we get a grid like this.

[diagram=576_0]

[Note how due to our choice of units for the time and distance axis, C is still at 45 degrees. The grey lines represent light signals which pass through the origin.]

And the other observer gets this.
[For some reason the events dots don't show on this diagram, just imagine ]

[diagram=577_0]

Note that (apart from drawing accuracy) the two grids are mirror images of each other, both up-down and left-right. The only thing that doesn't reflect is the +/- on the axis.

So the Loedel diagram works by making a reference event where both observers were present and synchronised clocks, and puts this event at the origin. Then the two grids can then be overlaid. Note, I am not explaining why this works, just how to do it.

The resulting grids are not easy to draw on one diagram and it would just look a mess anyway. What usually happens is that only the axis are drawn, the events marked, and the lines from events to the axis to read off times and distances.

Note that for events remote to an observer, they see its time as being along their time grid line, and the distance as being along their distance grid lines.

[diagram=578_0]

So using this diagram as an example, there are two events marked and the construction lines to show what times and distances the observers measure.

Mr Green, from along his time line measures that there was some time before an event occurred, and an even longer time before the next one. However, Mr Red measures that he waited for some time in between what Mr Green thinks, and then both events happened at almost the same time, he even thinks they happened in reverse order.

Now, neither observer can define absolute positions where any event occurs, only the distance from themselves. But by measuring how far away from them each event is, they can determine the distance between events. By looking at the total distance between events on each observer's 'x' axis it can be seen that Mr Red think there was less distance between events than Mr Green, but not a huge difference compared to the time discrepancy.

There are good reasons why these diagrams work [to do with s^2 = (Ct)^2 - x^2] but I can't remember the full explanation any more. Suffice to say that if you try out any of the common 2 party scenarios, such as the OP, or why observers each think the other has a slow clock and a short ruler, or how causality is preserved even though observers can disagree on the order of events, then they always seem to give the correct explanation.

As a true exercise in relativity, you should do the following, which is actually quite tricky but worth the effort.

Place a straight edge (like a piece of paper, or a ruler) along one observers 'x' axis.

Mark the position of the 't' axis (x=0) on the straight edge. This mark represents the observer.

Slide the edge up the diagram, always keeping it parallel to that observers 'x' axis and always keeping the mark on that observers 't' axis (x=0). This takes a bit of practice.

As you move it up the page you can see what the observer (the mark) thinks is happening.

Events hit the straight edge in the time order he measures and at the distance he measures. It is his world view.

You can now do the same thing for the other observer using their axis and see their version of events.

Once you have done this enough times you can kind of do it in your head.



Note that for either observer, light beams either move toward them or away from them at velocity C. For example, the grey diagonals shown could represent light beams that coincided at the origin, along with our observers. Both observers measure that these beams move away from themselves in opposite directions, at velocity C.

It goes without saying that these diagrams only show one space dimension but for the armchair physicist that seems to be enough.

Have fun!

[yor_on (OP)]

Wow, Fontwell, I will, in fact i will
Kind'a love this one.. But it will take me some time to 'melt' this. And I'm expecting to find good use of your explanation here while assimilating it.

It was very well done, and thought through. I'm sure I will have use for it as I look at your links. It's a pity when it comes to things like this that we don't have an area where we can put explanations like yours for others that are interested, at least I don't think we have? Do we? In an area of their own, sort of, search able for example on 'Minkowski diagram' 'Loedel diagram' etc. It couldn't hurt and would simplify for those not needing to make the same explanation twice.

As for why it works Oh yes, I'm wondering about that too, but I'm expecting to learn as I read. Otherwise I will have to pester you again I guess And thanks again Mr Fontwell, it's been a pleasure reading you.

[fontwell]

You are most welcome yor_on, I'm glad you are interested enough to read my posts. Yes, it will take a while for this to 'melt' but once it does, it can be surprisingly easy to use.

The test of if you have understood it all will be if you can use my original diagram to help with the OP

[Farsight]

Farsight, thanks for your response, although it isn't the 'mainstream idea' that confounds me here, just some implications that I'm not even sure that I'm getting right, it may just as easily be that I need to think it through again.

The real problem here is what time is. People rather think it's something that "flows", but there's no scientific evidence at all for this. Every measure of time relies on some form of motion, be it in a mechanical clock, an atomic clock, or a a light clock. If you're moving fast through the universe, the rate of local motion is of necessity reduced as per the parallel mirrors example. This affects all processes, so your thoughts and bodily functions slow down too. Hence you can't measure your time dilation locally.     

As I said, I might need to rethink how the question should look. But if you feel you have two specific answers to give me, one for each sceniario I will read them with interest. F.ex, I'm more or less stating that the answers you'll get will be different, depending on time of acceleration, don't I? Is that right? Why?

I'm not quite clear what you mean, but the acceleration is only necessary to achieve a different velocity. The total time dilation experienced by say the travelling twin in the Twins Paradox is more if he coasts more with the same acceleration.

We spend the exact same amount of energy (Let's assume so for the Q.) So why would the result differ, and how will it differ? I will get two 'distances' but the same 'time displacement' relative the rest of the universe? Or, I will get the same 'distance' and 'Time displacement'? Or, I will the same 'distance but not 'time displacement'?

There is no reality to time displacement. In the Twins Paradox, the twins depart at the same "time", and at a later "time", meet up at the same time. They don't miss one another by six months! The interval is measuring how far the light has moved between their parallel mirrors, and it's the same distance irrespective of any travel. The time-dilated twin has merely experienced less local motion because of his motion through the universe. Hence he's time-dilated, and has to turn around and come back so that both twins agree on who was doing the moving.   

Also. When I look at that signal it will according to relativity come at me at 'c' relative my own frame, no matter how you observe it, and me, from an 'inertial frame' like Earth f.ex if we assume that this were the origin for us both. As I'm moving away relative Earth very near the speed of light (99.99999999~) what does that state about this light? If we now, as SoulSurfer pointed out, could have observed it. In reality the observation won't matter for my statement though, assuming that the concept of relativity is correct it have to be true.

Relativity is correct, but people don't quite understand why. It's just because the local motion of light defines your time. If the rate of propogation of all electromagnetic phenomena in the universe was halved, your thinking and your bodily processes and your clocks etc would all run at half the previous rate, so you wouldn't measure a reduced c. Look at the definition of the second and the metre again.     

And it do make me wonder about the concept of 'distance'?

That's defined by the motion of light too. But it's more fundamental than time. Hold your hands up a metre apart and you can see the distance between them. Waggle your hands and you can see motion. We use regular motion to mark out time, but you can't see time.

So if you have ideas about the scenarios feel free to give them Farsight, and if you want to state them from another format than relativity, you're welcome, just make sure that I know you do So we're speaking the same language here, sort of.

The things I talk about are relativity, because it's based on what Einstein said. Sadly people don't know about everything Einstein said. Things like the forgotten legacy of Godel and Einstein. However that means the correct fundamentals are space and motion, not space and time, and this is counter to Minkowski. Hence issues arise. 

[fontwell]

In the Twins Paradox...both twins agree on who was doing the moving.  

"both twins agree on who was doing the moving"

There is something here which I'm not sure if people know and take for granted or if it is an area of confusion, so here goes...

If the two twins start in constant relative motion with respect to each other they might pass each other at one point and agree to synchronise clocks. Then they would naturally move apart. In this case their situation is exactly symmetrical. They each think the other has a slow clock and they are 'both correct'. Of course, since they are no longer both in the same place any more, this is because they both disagree about times and distances between any events.

However, when one of them turns round he really is changing direction and not the other. It could appear that everything is still symmetrical but it isn't. One of the twins will undergo acceleration as he changes direction. He will know this by, say, dropping a stone next to him which will continue to follow him as long as he moves at constant velocity. For the twins to meet again there is no avoiding that one twin must give up his inertial frame and he would see the dropped stone move away from him.

This unsymmetrical action is what results in the returning twin having experienced less time, rather than the situation being equal.

In fact as I understand it, if the twin always changes velocity instantaneously (including at the start) the part of the journey which causes a final difference in clocks is really the mid-point acceleration. During the constant motion part of the two journeys, the situation is identical for each twin.

For instance, (I believe this is true) when the twins have reached their furthest point apart, one twin could decide to change his motion so that he drifts at a constant distance to the first twin for a while i.e in the same inertial frame. Before he does this he will consider that he has moved a certain distance away from the first twin and that the first twin's clock is reading a certain (slow) time. As he changes his motion to match the other twin, he will observe the clock of his twin to suddenly race ahead (overtaking his own) and also the distance between them to suddenly grow.

After that, they would both agree about times and distances between subsequent events for as long as they both shared the same motion.

So the point is that even for two constant velocity journeys one twin will have to do the accelerating, and it is during the accelerating that the situation unbalances.

[Farsight]

I endorse that. If two twins A and B are moving apart in empty space with no external references, they can't say A is moving away from B, or vice versa. Their situation is symmetrical. It's only when twin B turns round and comes back that they can agree that it was twin B doing the moving. That turnaround requires a change in velocity - an acceleration. As you suggest, removing initial acceleration and final deceleration simplifies matters - start with the twin B passing twin A, as they pass they "tag" or touch each other to establish a synchronised event, and after B turns round they pass each other again and do the same to establish another synchronised event.

[PhysBang]

Relativity is correct, but people don't quite understand why. It's just because the local motion of light defines your time.

It is important to note at this point that this position is essentially the religious position of this poster. Farsight has no scientific evidence for this theory other than the literary criticism he performs on select quotations and he has no plans to actually demonstrate that his particular theories match what scientists investigate. He admits as much:

But actually, I don't want to cast this as a coherent mathematical model myself. That might sound odd, but think about it. If I locked myself away and came up with something that really flew, every theoretical physicist in the world would then be redundant. It's too late for them to get involved once it's finished. Moreover they'd look like crystal-sphere fools, and the public would feel betrayed. There would be a backlash, and the upshot would be a disaster. I'm trying to help physics, not destroy it.

Note the contempt for contemporary physicists.

[Farsight]

There's nothing religious or contemptuous about what I'm saying here. Here's the official definition of the second:

Since 1967, the second has been defined to be the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium 133 atom.

An atomic clock employs microwave radiation, which is essentially light. We count 9,192,631,770 microwave peaks going past and call it a second. Hence we're defining the second using the motion of light. We then use the second to measure the motion of light. Hence we always measure a constant value. Physicists are saying this too. See Comments on "Note on varying speed of light theories" by Magueijo and Moffat at http://arxiv.org/abs/0705.4507, which includes:

"Following Ellis [1], let us first consider c as the speed of the photon. Can c vary? Could such a variation be
measured? As correctly pointed out by Ellis, within the current protocol for measuring time and space the answer
is no. The unit of time is defined by an oscillating system or the frequency of an atomic transition, and the unit of
space is defined in terms of the distance travelled by light in the unit of time. We therefore have a situation akin to
saying that the speed of light is “one light-year per year”, i.e. its constancy has become a tautology or a definition."

[BenV]

Physbang, two of your three posts in this thread have been nothing but personal attacks on Farsight.

If his science is wrong, kindly explain so without resulting to personal attacks.  If you have nothing to contribute but attacks, please don't contribute at all.

I think you've been warned for this before on a different thread.

[Geezer]

An atomic clock employs microwave radiation, which is essentially light. We count 9,192,631,770 microwave peaks going past and call it a second. Hence we're defining the second using the motion of light. We then use the second to measure the motion of light. Hence we always measure a constant value.

(I put the third sentence in bold)

This is quite erroneous. We might as well say something like:

"A guitar string employs acoustic radiation, which is really sound. We count n acoustic events and call it a second. Hence, we are defining the second using the motion of air. We then use the second to measure the motion of light. Hence we always measure a constant value."

Obviously, that makes no sense, but it is little different from Farsight's statement.

It might be possible to stretch things a bit and say that microwaves are a form of light, but that's not commonly accepted. It is however possible to say that they are both forms of electromagnetic radiation. However, the real problem lies in the third sentence.

We are absolutely not defining the second using the motion of light. What we are doing is defining the second in terms of atomic events. We are merely using the microwave energy released to detect and count the atomic events.

[yor_on (OP)]

There are some things here I need you to define Mr Fontwell.

By saying "constant relative motion with respect to each other they might pass each other at one point" are you thinking of them uniformly accelerating, or uniform moving? Also, do you mean that they are traveling in opposite directions meeting each other? To be able to pass each other it seems to me that this is what you are talking about?

If we assume that they are uniformly moving past each other in opposite directions and then say that 'A' do that 'turn around' and catches up to 'B' he will have to accelerate, right? And as you say an acceleration equals a 'compressed' timeframe for 'A' relative 'B'. But are you then saying that the uniform motion he will use after it to 'drift' up to 'B' and overtake him have nothing to do with the time dilation created? That's one of the things I'm really wondering about. If uniform motion also creates a time dilation. To me it seems that it should. That is if we assume that he does this turnaround after one years travel and then caches up to 'B' will give him a lesser time dilation relative 'B' than if he did it first after have traveled three years away from 'B' before doing the turn around.

And there is also the complication with it that he in both my examples will have to pace himself at a faster rate than 'B' to catch up with him, which sort of destroy the equivalence there, no matter which scenario you choose, at least I think so? If now not all 'uniform motion', no matter their velocity relative something else is equivalent of course. But thats the scenario that confuses me. To get it into perspective I actually have asked a physicist about it, but he wasn't sure himself of how to see it, there seems to be some heavy math involved in it.

As I said, to me it comes back to how to look at 'uniform motion' and whether all 'free falling' then could be said to be equivalent, no matter velocity? If it would be so that it is only the acceleration that creates a 'time dilation' and your uniform motion, for however long after the fact, won't have a bit to do with it then all free falling frames have to be equivalent (uniform moving). But if uniform moving do have something to do with the 'time dilation' observed then i can't say that any free fall will be equivalent to another 'free fall' as they will have to be absolutely the same velocity, and if so one starts to wonder about if their invariant mass also will play a role for it?

And if they (all free falling frames) on the other hand are equivalent then acceleration really becomes weird, at least to me. But probably I missed what you meant right

And I am interested in all views here. PhysBang, Farsight, Geezer, BEnV and all you  others that have a view on it. But first of all naturally you Mr Fontwell As it was your scenario that I, ah, manhandled here )

Hey, don't blame me. I'm just confused

[Geezer]

Fine, but let's try make it clear when we are within the bounds of supportable scientific theory rather than flying in the face of it.

[fontwell]

There are some things here I need you to define Mr Fontwell.

By saying "constant relative motion with respect to each other they might pass each other at one point" are you thinking of them uniformly accelerating, or uniform moving?

Neither party is accelerating. The effect is that the parties move at a constant velocity with respect to each other (really their frame of reference). It is understood in SR that if two parties are both in uniform motion (i.e. they don't experience personal acceleration) then they will move at a constant velocity with respect to each other.

Also, do you mean that they are traveling in opposite directions meeting each other? To be able to pass each other it seems to me that this is what you are talking about?

For uniform motion any direction is possible. If it is convenient to have them in the same place at t=0, then before that time they would have been moving together and after that time they move apart. Like trains on two parallel tracks which pass each other. 

If we assume that they are uniformly moving past each other in opposite directions and then say that 'A' do that 'turn around' and catches up to 'B' he will have to accelerate, right? And as you say an acceleration equals a 'compressed' timeframe for 'A' relative 'B'.

Yes

But are you then saying that the uniform motion he will use after it to 'drift' up to 'B' and overtake him have nothing to do with the time dilation created? That's one of the things I'm really wondering about. If uniform motion also creates a time dilation. To me it seems that it should. That is if we assume that he does this turnaround after one years travel and then caches up to 'B' will give him a lesser time dilation relative 'B' than if he did it first after have traveled three years away from 'B' before doing the turn around.

The dilation due to uniform motion is symmetrical. This is an absolutely fundamental point. Two parties in constant motion both think the other has a slow clock and a short ruler. This is because they disagree about times and distances between events. This time dilation is what they each infer about the other through observation at a distance. One point to note is that in uniform motion they can only ever meet once, thus they can never get to see 'who is really slow'

Time dilation due to acceleration causes a difference between the two parties that is permanent and not symmetrical. this is because one party absolutely does the accelerating and the other doesn't.

When 'A' turns round and accelerates back to 'B' the total time 'A' has been drifting does affect the dilation between them, this is true. But up to that point it is still symmetrical, no one could say 'A' is actually slower than 'B'. But by accelerating back to 'B', 'A' fixes the dilation between them in favour of 'B' being static.

Also, the acceleration determines the new velocity between 'A' and 'B' and thus when they meet up again. So actually, by determining the return journey it already defines the next symmetrical time dilation of the next drift journey. They will each have a measurement of how long this journey takes. But at the end they meet at the same event. This means that the difference in their observations of the journey time has to be 'absorbed' by 'A' during the the turn around at the start of this portion.

Actually, assuming the return velocity is the same as the outgoing velocity, the entire journey is symmetrical in time. The start and end both has 'A' and 'B' at the same events. The drift away/toward from that event is seen the same way by 'A' and 'B'. Thus all the unbalanced dilation occurs during the turn around.

So the time dilation which causes the difference in their clocks when they finally meet again is all caused (or at least unbalanced) by the turn around acceleration. The duration of the drift does alter the amount of time by which the final dilation occurs, but it is the acceleration which swings this dilation round to put one party older than the other.

And there is also the complication with it that he in both my examples will have to pace himself at a faster rate than 'B' to catch up with him, which sort of destroy the equivalence there, no matter which scenario you choose, at least I think so? If now not all 'uniform motion', no matter their velocity relative something else is equivalent of course. But thats the scenario that confuses me. To get it into perspective I actually have asked a physicist about it, but he wasn't sure himself of how to see it, there seems to be some heavy math involved in it.

Are you talking about a light signal between them? I'm not sure what you are asking here.

As I said, to me it comes back to how to look at 'uniform motion' and whether all 'free falling' then could be said to be equivalent, no matter velocity? If it would be so that it is only the acceleration that creates a 'time dilation' and your uniform motion, for however long after the fact, won't have a bit to do with it then all free falling frames have to be equivalent (uniform moving). But if uniform moving do have something to do with the 'time dilation' observed then i can't say that any free fall will be equivalent to another 'free fall' as they will have to be absolutely the same velocity, and if so one starts to wonder about if their invariant mass also will play a role for it?

For parties in uniform motion the time dilation between them (symmetrical) is bigger as their relative velocity increases. Two parties will each observe the other's clock to be running slow and the faster they move with respect to each other, the slower that clock will seem to run. There can be many inertial observers and when they look at each other the time dilation they observe is a function of relative velocities.

Also, to me, 'free fall' is a phrase which implies falling toward a mass. That would be GR and something else. We have only been talking about SR here. The equivalent situation here is uniform motion = constant velocity = an inertial frame.

[Farsight]

An atomic clock employs microwave radiation, which is essentially light. We count 9,192,631,770 microwave peaks going past and call it a second. Hence we're defining the second using the motion of light. We then use the second to measure the motion of light. Hence we always measure a constant value.

(I put the third sentence in bold) This is quite erroneous.

It isn't, Geezer. This is what we do. We define our time using motion. In the old days we used the motion of the earth, later it was the motion of a pendulum, later still it was the motion within a mechanical clock. Nowadays our best clocks are atomic clocks, and they use the motion of microwave radiation.

We might as well say something like:

"A guitar string employs acoustic radiation, which is really sound. We count n acoustic events and call it a second. Hence, we are defining the second using the motion of air. We then use the second to measure the motion of light. Hence we always measure a constant value."

Obviously, that makes no sense, but it is little different from Farsight's statement.

That doesn't work because light and air are not the same thing.

It might be possible to stretch things a bit and say that microwaves are a form of light, but that's not commonly accepted. It is however possible to say that they are both forms of electromagnetic radiation.

Accepted. I was using the word "light" in the widest sense. Electromagnetic radiation is a better term.

However, the real problem lies in the third sentence. We are absolutely not defining the second using the motion of light. What we are doing is defining the second in terms of atomic events. We are merely using the microwave energy released to detect and count the atomic events.

We absolutely are. The atomic event concerned is the hyperfine transition. We aren't counting hyperfine transitions. That's like counting the number of plucks of the guitar string. We are counting the resultant EM wavepeaks going past. See http://en.wikipedia.org/wiki/Hyperfine_structure#Use_in_defining_the_SI_second_and_meter for more information.

[Farsight]

Yor-on, everything fontwell has said is IMHO bang on, but you seem to be thinking it's more complicated than it is. See the wikipedia article on time dilation, and look at this section:

http://en.wikipedia.org/wiki/Time_dilation#Simple_inference_of_time_dilation_due_to_relative_velocity

Don't worry about the arithmetic, just think about the parallel-mirror light clock. Now imagine you're twin A and you've got one of these light clocks in front of you. The light is moving up and down, so you can draw a light path like this ║, multiplied a zillion times whilst twin B is on his trip out and back. Now think of B's light clock, and draw the light path as you'd see it, something like this:

→ /\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\ →
                                                                                ↓
← /\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\/\ ←

The total light path length between your parallel mirrors is the same as his. He comes back time-dilated, and because of his motion through space, his light has bounced back and forth by less than a zillion times. When you get back together you can compare counters to prove it. But if that turnaround didn't happen, you can't, and twin B would assert that your light path was the one that was zigzag, not his.

[yor_on (OP)]

Okay Fontwell. If I got you right time dilation takes place both at uniform motion and acceleration, right? so assuming this, knowing that there are no 'preferred frames' defining a 'gold standard' in the universe, we can't really define the time dilation for any uniformly moving frame, can we?

It's all about comparison and definition via one 'inertial frame' relative the frame we observe (think Earth and a rocket). And our definition of that 'inertial frame' is arbitrarily made as we have no 'gold standard' for that either, right?

So, how would you define the circumstances describing two 'uniformly moving' frames as equivalent to each other?

[yor_on (OP)]

Farsight please, look at the thread before you answer

I do understand what you are saying, as you would see if you read it through. That all frames internally have the same time (more or less:) relative those/that inside that frame is 'self consistent' to me. That's not my discussion. I'm looking at how to define uniform motion relative acceleration, and time dilation.

It may be all self clear to you, but it's not to me.

And. .

[Farsight]

I've looked at the thread. I've read it all. Motion is relative, uniform motion is motion that is measurably occuring at a constant rate, relative acceleration is a change in relative motion, and time dilation is what you call it when your cumulative measure of local motion is occurring at a reduced rate compared to somebody else's. It always comes back to motion. Once you appreciate this, it's clear. 

[yor_on (OP)]

Yes Farsight, that's your take on it and it might be mine too, but not yet. First I will ask my questions and then we will see if I understands it

[fontwell]

Okay Fontwell. If I got you right time dilation takes place both at uniform motion and acceleration, right?

yes...

so assuming this, knowing that there are no 'preferred frames' defining a 'gold standard' in the universe, we can't really define the time dilation for any uniformly moving frame, can we?

It depends on what you mean by gold standard, I'll come back to this...

I don't mean to sound rude but you don't seem to have understood what I mean't about symmetrical and unbalanced (non-symmetrical) time dilation.

For two observers in uniform motion, both think the other has a slow clock. There is no preferred position. The time dilation is purely what they infer about the other party by looking at their clock as it moves. They look toward the moving clock (also allowing for the time it takes light from the clock to arrive) and what they observe is that the moving clock runs too slow. This is one form of time dilation. It is symmetrical because for each party they both deduce the other to have a slow clock. This can work because they can only ever meet once, so they have no way to meet again to see who is correct (as long as they remain in uniform motion).

If you like, there is a gold standard of time dilation. For any particular velocity between to inertial parties, they both observe the same gold standard time dilation (slowness) in the other.

But this isn't an absolute gold standard because a third party in uniform motion will observe both of them and have a different time dilation with each one. And the simple sum of each dilation will not sum to the total between the first two parties.

It's all about comparison and definition via one 'inertial frame' relative the frame we observe (think Earth and a rocket). And our definition of that 'inertial frame' is arbitrarily made as we have no 'gold standard' for that either, right?

If I understand you correctly, not right. As I've said before, there is an easy test to see if you are in an inertial frame. You drop a stone. If it moves away from you then you are accelerating. On Earth we are not in an inertial frame because stones move away from us. Actually, on Earth we are feeling gravity and SR doesn't like to mention gravity.

In deep space our two parties can move away from each other in two inertial frames. They each drop a stone and it doesn't move away from either party. The stones move along side each party.

But for the parties to meet again one must turn round. It could be either but it has to be one. Whoever turns round will see his stone move away from him. It will track the path he would have taken.

This is the un-symmetrical time dilation which makes one clock absolutely ahead of the other when they meet again.

The gold standard is that any two parties with the same relative paths and acceleration would have the same time dilation experience.

But I could watch these two parties as I move at 0.99C past one of them and the whole situation would look different to me. So it is not a gold standard.

So, how would you define the circumstances describing two 'uniformly moving' frames as equivalent to each other?

Hopefully this is clear now. Each one experiences no acceleration (using the stone test). Each sees the other as having a slow clock. They can only meet once, so there is no way to get the two clocks together again to see 'who is right'. They can only meet twice and compare clocks twice if one of them ceases uniform motion.