I Proper (and coordinate) times re the Twin paradox

[Grimble]

The proper time between two events is measured along the wordline of a clock that is present at both events. If the clock does not experience any acceleration this worldline will be a straight line. (If the clock is accelerated its worldline will not be straight, and the proper time along the non-straight worldline will be less than the proper time along the straight worldline of an unaccelerated clock). There is no reason why that straight line has to be vertical and parallel to the t axis.

But here you say:

If we assign coordinates using a frame in which the object is at rest we'll label the points that the worldline passes through (t,0,0,0) and if we use a frame in which the object is moving these points might be labeled (t',vt',0,0), but they're the same points and the same worldline either way. When we change frames we're changing the axes of the coordinate system we're using to assign coordinates to points, but this doesn't change the points themselves.

In the clock's frame it is at rest and the worldline and the proper time will be on a vertical line parallel to the t axis.
This measurement is invariant.
The proper time is measured between two events measured by the same clock at which the clock is present (t,0,0,0) which are necessarily at the same location.
In a frame in which the clock is moving the coordinates, as you say, are given by (t',vt',0,0). t' is greater than t (t' = γt) and the clock moves further - vt'

So although the clock does not move vertically in a frame in which it is moving, its invariant measurement of proper time is the vertical distance to E2 from the x axis.

 

[Dale]

The proper time is: $$\tau=\int \sqrt{-g_{\mu\nu}dx^{\mu}dx^{\nu}}$$ Where in an inertial frame $g_{\mu\nu}dx^{\mu}dx^{\nu}=-c^2 dt^2+dx^2+dy^2+dz^2$

This quantity is the same in all frames. If there is some frame where $dx=dy=dz=0$ then the worldline “is measured vertically parallel to the t-axis”, but even if the worldline is not parallel to the t axis the proper time is still the same. The whole point is that proper time does not need to be parallel to the t axis.

If you draw a line both you and someone looking at an angle will agree on its length. It is not necessary for that line to be parallel to any coordinate line.

 

[Grimble]

The proper time is: $$\tau=\int \sqrt{-g_{\mu\nu}dx^{\mu}dx^{\nu}}$$ Where in an inertial frame $g_{\mu\nu}dx^{\mu}dx^{\nu}=-c^2 dt^2+dx^2+dy^2+dz^2$

I am afraid that is right over my head!

This quantity is the same in all frames. If there is some frame where $dx=dy=dz=0$ then the worldline “is measured vertically parallel to the t-axis”, but even if the worldline is not parallel to the t axis the proper time is still the same. The whole point is that proper time does not need to be parallel to the t axis.

If you draw a line both you and someone looking at an angle will agree on its length. It is not necessary for that line to be parallel to any coordinate line.

Yes, I do understand exactly what you are saying. Indeed you said the same earlier:

In any reference frame. It is invariant.

The worldline is the geometric figure itself, irrespective of the coordinate system that you might use for describing it. One worldline will have different coordinates in different coordinate systems but it is the same worldline.

But can you clear this up for me? In what way does the Worldline have the same length?
At rest its worldline is parallel to the time axis - it has to be because it does not physically move.
In motion it also travels between locations so it is moving in time and space; which seems to me to lead to a longer worldline.

And this post from Mister T seems to say the opposite to your post:

If the two events occur in the same place (in some frame) then the time that elapses in that frame is a proper time.

It is a proper time if the events are at the same location - i.e. parallel to the t axis.

In the rest frames of observers who move relative to those events

that is frames in which the worldline is sloping...

the time that elapses will not be a proper time, for them those two events occur in different places. They will therefore need two clocks, one located (local!) at each of the two events. And the time that elapses in those frames will always be larger than the proper time.

 

[jbriggs444]

In motion it also travels between locations so it is moving in time and space; which seems to me to lead to a longer worldline.

This is Minkowski geometry, not Euclidean geometry. Increasing the spatial displacement (the difference between starting and ending x coordinates) while holding the temporal displacement (the difference between starting and ending t coordinates) constant reduces the length of the (timelike) interval between the end points.

 

[Dale]

And this post from Mister T seems to say the opposite to your post

You will have to ask @Mister T to clarify his own post, not me. I would not phrase it that way because proper time is only defined on a worldline and coordinate time is defined throughout spacetime.

I am afraid that is right over my head!

If you want to understand this stuff then you cannot simply give up on this part. You must pursue it until it is not over your head.

Look at the right hand side of the formula in the inertial frame. Do you understand that? Do you see the similarity with the Pythagorean theorem?

But can you clear this up for me? In what way does the Worldline have the same length?

If you take the Pythagorean theorem and calculate the distance between two points and then you rotate your coordinates and recalculate you get the same result. In spacetime the proper time is the equivalent of the Pythagorean theorem and a Lorentz transform (boost) is the equivalent of a rotation. A rotation/boost does not change the length/proper-time. It is not that hard to prove, I would recommend doing it to convince yourself.

At rest its worldline is parallel to the time axis - it has to be because it does not physically move.
In motion it also travels between locations so it is moving in time and space; which seems to me to lead to a longer worldline.

Note that in a rotation both the x and y coordinates are changed in such a way that the distance is preserved. Similarly here both the t and x coordinates are changed in such a way that the proper time is preserved. In your analysis above you neglected the change in t.

 

[Grimble]

If you want to understand this stuff then you cannot simply give up on this part. You must pursue it until it is not over your head.

Look at the right hand side of the formula in the inertial frame. Do you understand that? Do you see the similarity with the Pythagorean theorem?

Yes, τ in the first line and the expression on the right in the second line are what I understand as a way of expressing the proper time formula – only with the signs reversed – but that is just the convention used ( I believe).

If you take the Pythagorean theorem and calculate the distance between two points and then you rotate your coordinates and recalculate you get the same result.

Yes, because the hypotenuse is the same length and the sum of the squares of the catheti will always equal the square of the hypotenuse.

In spacetime the proper time is the equivalent of the Pythagorean theorem and a Lorentz transform (boost) is the equivalent of a rotation.

I'm feeling a bit stupid here...
but how is part of the Pythagorean theorem is equivalent to proper time?

A rotation/boost does not change the length/proper-time. It is not that hard to prove, I would recommend doing it to convince yourself.

When we say a rotation, what exactly are we rotating and how? I think that if we take two events with different time coordinates and increase the x coordinate of one of them then the connecting line is rotated, and the slope of the connecting line changes as the relative speed increases, from 0 when they are vertical to horizontal at c (because at c time stops?)

Note that in a rotation both the x and y coordinates are changed in such a way that the distance is preserved. Similarly here both the t and x coordinates are changed in such a way that the proper time is preserved. In your analysis above you neglected the change in t.

Which part of my analysis are you referring to?
I thought I had used τ for proper time and t for coordinate time - should I have used t and t' instead?

 

[jbriggs444]

When we say a rotation, what exactly are we rotating and how? I think that if we take two events with different time coordinates and increase the x coordinate of one of them then the connecting line is rotated, and the slope of the connecting line changes as the relative speed increases, from 0 when they are vertical to horizontal at c (because at c time stops?)

We are rotating the coordinate system. This is not the same thing as moving the events being considered.

As an analogy, consider two points drawn on a blank sheet of paper. Call them point A and point B. Now take a sheet of transparent film on which some grid lines have been drawn and lay it on top of the sheet of paper. This amounts to a coordinate system. You can number the grid lines and identify the points by the numbered lines that intersect nearby.

So, for instance, you might line up the film so that the point A is at (0,0). And you might then rotate the film so that the point B is at (0,5). Or you could rotate the film so that the point B is at (3,4).

You will find that there is an invariant. No matter how you rotate the film you will find that $$(x_b-x_a)^2 + (y_b-y_a)^2=5^2$$
Note well that this works not just for rotations, but also for translations. No matter how you slide or rotate the film, the above equation still holds. Its truth is an invariant property of Euclidean geometry.

Now then, this was an analogy. In Special Relativity, the geometry is not Euclidean. It is hyperbolic. One of the directions is singled out as "time-like". The remaining directions are "space-like". The grid is difficult to visualize because of the geometry. (Space-time diagrams are an attempt to twist the grid lines so that they can be presented on a Euclidean piece of paper. Reading and understanding them takes a bit of effort).

The invariant equation for this geometry is:
$$(t_b-t_a)^2-(x_b-x_a)^2=(\Delta \tau)^2$$
The $\Delta \tau$ here is the invariant length of the interval between events A and B.

 

[Grimble]

This is Minkowski geometry, not Euclidean geometry. Increasing the spatial displacement (the difference between starting and ending x coordinates) while holding the temporal displacement (the difference between starting and ending t coordinates) constant reduces the length of the (timelike) interval between the end points.

I understand that Minkowski geometry is considered the ideal way to consider time, space and movement in Spacetime, but what is it? How does it differ from Euclidean geometry and what makes it wrong to use Euclidean geometry?

I looked at the article for Minkowsi Geometry in Wkipedia and did not understand a word of it!
The Maths was way over my head, I had no idea of the meaning of any of the symbols used.

Why cannot simple mechanics and geometry be used? They may not be as appropriate for mathematicians but it seems to me that the real difference is that time has different properties from space in Spacetime. Principally the second postulate. The constancy of c.

Can you explain to me what other factors require the use of Minkowski geometry, rather than it being the preferred tool to use.

It seems to me that time dilation and length contraction and the relativity of simultaneity are all easy to comprehend in classical mechanics once the 2nd postulate is included.

Obviously I am getting this wrong or being confused somewhere, but I am trying to resolve these difficulties that I have.
Thank you for being patient with me.

 

[jbriggs444]

I understand that Minkowski geometry is considered the ideal way to consider time, space and movement in Spacetime, but what is it? How does it differ from Euclidean geometry and what makes it wrong to use Euclidean geometry?

The difference is in the topology. That is, in the notion of distance between points in a geometric space. The notion of distance for a geometry is known as the "metric".

[I may stumble a bit with this exposition. I've never taken a course in linear algebra or topology]

The idea of a metric is that you have a function that takes two points/events as inputs and produces a scalar distance as an output.

In Euclidean geometry we would call the end points "points". In Minkowski geometry we would usually call them events. In Euclidean geometry, we would refer to the universe within which these points live as "space". In Minkowski geometry we might call it "space-time".

It is hard to write down a formula for such a function without having coordinates for the the end points. So we lay down a coordinate system on the space. Think of a transparent film with grid lines being laid over a blank sheet of paper where the end points live.

With the coordinate representations in hand, we can write down the Euclidean metric:
$$s(a,b)=\sqrt{(x_a-x_b)^2+(y_a-y_b)^2}$$

Or for Minkowski geometry, we could write down the Minkowski metric:
$$s(a,b)=\sqrt{(t_a-t_b)^2-(x_a-x_b)^2}$$

There is a tricky part to doing this. How do we lay down a transparent film on space-time so that we have coordinates to use?

The answer is that we pick a standard of rest. A "frame of reference". Within that frame of reference, we can measure spatial distances as usual -- with rulers and such. We can measure time with clocks. But if we pay attention to details, we need to synchronize those clocks carefully. Einstein showed that the synchronization depends on a standard of rest. That frame of reference we chose is important.

If we change the standard of rest, we end up with a different coordinate system. But the important part is that the metric remains unchanged. We have re-labelled the same events with different coordinates, but the distance between them (reflected by the metric) is the same.

Why cannot simple mechanics and geometry be used?

The laws of the universe are what they are. We do not get to choose them. It turns out that our real universe follows the Minkowski metric.

The metric distance between two [timelike separated] events is given by the elapsed time on a clock that is present at both events. That is a physical measurable quantity. That is how we can tell whether our universe matches the geometry. Or fails to do so.

 

[Grimble]

@Dale said:

You will have to ask @Mister T to clarify his own post, not me.

So let me ask @Mister T Can you help resolve this apparent conflict between what you say and what Dale is saying?

You need two events to measure an elapsed time. If the two events occur in the same place (in some frame) then the time that elapses in that frame is a proper time. In the rest frames of observers who move relative to those events the time that elapses will not be a proper time, for them those two events occur in different places. They will therefore need two clocks, one located (local!) at each of the two events. And the time that elapses in those frames will always be larger than the proper time.

and

This quantity is the same in all frames. If there is some frame where dx=dy=dz=0dx=dy=dz=0 then the worldline “is measured vertically parallel to the t-axis”, but even if the worldline is not parallel to the t axis the proper time is still the same. The whole point is that proper time does not need to be parallel to the t axis.

If you draw a line both you and someone looking at an angle will agree on its length. It is not necessary for that line to be parallel to any coordinate line.

I am not trying to be difficult but I am not in a position to resolve differences in what I am being told by different Experts...

 

[Orodruin]

The difference is in the topology.

There is no difference in the topology of Minkowski space and Euclidean space, they are homeomorphic. The difference is geometrical, not topological. I agree with the rest. The main difference between Minkowski space and Euclidean space is the inner product. Among other things, the inner product on Minkowski space results in a Pythagorean theorem that is similar to, but yet very different from, the one we are used to from Euclidean space, namely
$$
ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2.
$$
With $s = c\tau$ and $\tau$ being proper time along a world-line, computing the proper time along a curve is completely analogous to computing the length of a curve in Euclidean space - the only difference being in the different Pythagorean theorem.

As to why Nature is better described by Minkowski geometry than by Euclidean geometry, it is just how Nature seems to work.

 

[SiennaTheGr8]

@Grimble

As for the "contradiction" between what @Mister T and @Dale said, it comes down to a semantic issue that I think I can help resolve.

1. "Proper time" is what a traveler's wristwatch logs.
2. Sometimes we loosely speak of the "proper time between events"—this is shorthand for "the proper time that would elapse during an inertial journey between a pair of timelike-separated events" (equivalent to the "spacetime interval" between them).
3. For every inertial journey there's an inertial frame for which the traveler is at rest (all events on the traveler's world line occur at the same spatial location in this frame). The "coordinate-time clocks" in this frame tick at the same rate as the traveler's wristwatch, so in practice we don't always make a distinction between an inertial traveler's proper time and the coordinate time of the traveler's rest frame (but strictly speaking they are different concepts).

When @Mister T says "If the two events occur in the same place (in some frame) then the time that elapses in that frame is a proper time," he means that there's an equality between the coordinate-time that elapses in that frame and the proper time that would elapse for a traveler during an inertial journey between the events in question.

 

[Herman Trivilino]

Can you help resolve this apparent conflict between what you say and what Dale is saying?

But I don't see an apparent conflict. Perhaps if you could explain what you find conflicting?

 

[Dale]

I'm feeling a bit stupid here...
but how is part of the Pythagorean theorem is equivalent to proper time?

Look at the formulas:
$ds^2=dx^2+dy^2+dz^2$
$ds^2=-dt^2+dx^2+dy^2+dz^2=-d\tau^2$
They both represent an arc length, one in a 3D space with signature (+++) and the other in a 4D space with signature (-+++) which is equivalent to (+---).

When we say a rotation, what exactly are we rotating and how?

We are rotating our coordinate system. How is with a Lorentz transform, that is a rotation (boost) in spacetime.

I think that if we take two events with different time coordinates and increase the x coordinate of one of them then the connecting line is rotated,

No. That is not a rotation, that is a different line between different points. We leave the events the same and rotate the coordinate system. It is important to make a distinction between the geometric objects and their coordinates.

Think of, for example, the lightning strikes on the train thought experiment. The two flashes of light are two separate events. Those events have coordinates in the train frame, and those same two events have different coordinates in the embankment frame. The coordinate system is rotated (boosted), but the events stay the same.

Which part of my analysis are you referring to?

The part where you said “In motion it also travels between locations”. You are neglecting the change in time coordinates.

Do this exercise. Use units where c=1. Start with the origin as your first point since it just trivially transforms to the origin. Now, for your second point start with $(t,x)=(1,0)$. Calculate the proper time. Then do a Lorentz transform to a frame with $v=0.6$ relative to the first. Then calculate the proper time again.

How does it differ from Euclidean geometry and what makes it wrong to use Euclidean geometry?

Euclidean geometry can’t work because it only has one type of length. Spacetime needs two, one for the kind of “distance” we measure with rulers and the other for the kind of “distances” we measure with clocks. If $ds^2>0$ we call the interval spacelike and we measure it with a ruler. If $ds^2<0$ we call the interval timelike and we measure it with a clock.

 

[Orodruin]

Look at the formulas:
$ds^2=dx^2+dy^2+dz^2$
$ds^2=-dt^2+dx^2+dy^2+dz^2=-d\tau^2$
They both represent an arc length, one in a 3D space with signature (+++) and the other in a 4D space with signature (-+++) which is equivalent to (+---).
We are rotating our coordinate system. How is with a Lorentz transform, that is a rotation (boost) in spacetime.

No. That is not a rotation, that is a different line between different points. We leave the events the same and rotate the coordinate system. It is important to make a distinction between the geometric objects and their coordinates.

Think of, for example, the lightning strikes on the train thought experiment. The two flashes of light are two separate events. Those events have coordinates in the train frame, and those same two events have different coordinates in the embankment frame. The coordinate system is rotated (boosted), but the events stay the same.

The part where you said “In motion it also travels between locations”. You are neglecting the change in time coordinates.

Do this exercise. Use units where c=1. Start with the origin as your first point since it just trivially transforms to the origin. Now, for your second point start with $(t,x)=(1,0)$. Calculate the proper time. Then do a Lorentz transform to a frame with $v=0.6$ relative to the first. Then calculate the proper time again.

Euclidean geometry can’t work because it only has one type of length. Spacetime needs two, one for the kind of “distance” we measure with rulers and the other for the kind of “distances” we measure with clocks. If $ds^2>0$ we call the interval spacelike and we measure it with a ruler. If $ds^2<0$ we call the interval timelike and we measure it with a clock.

Arguably the most important type is the third, ie, null intervals with $ds^2=0$ although the points are distinct.

 

[Grimble]

Well, now I begin to see...
I seem to be a little hazy with exactly what some terms mean. (duh!)
rotation is not a line being rotated and stretched, as for example, if the later event is moved in space but keeps the same time value; but it is the line rotating and keeping the same length.
I am going to review and check where else I am getting confused; particularly:
worldlines, proper time, coordinate time, intervals and how I see them reflected in Minkowski diagrams.

 

[vanhees71]

Have a look here:

https://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

 

[Grimble]

Have a look here:

https://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf

Thank you for the reference but I'm sorry that I do not understand that level of maths, not even the symbols or techniques used.

 

[Grimble]

I am going to surprise you now, Dale, by doing just what you say:

Do this exercise. Use units where c=1. Start with the origin as your first point since it just trivially transforms to the origin. Now, for your second point start with (t,x)=(1,0). Calculate the proper time.

The proper time is the time difference in an inertial frame. In this case it will be 1.

Then do a Lorentz transform to a frame with v=0.6v=0.6 relative to the first. Then calculate the proper time again.

With a Lorentz factor γ = 1.25, the transformed coordinates of the second point of the world line will be: x = 0.6, t = 0.8. So the rotated length of the worldline remains at 1. The distance, measured along the x-axis is 0.6 and the time difference is 0.8.

In relativity, proper time along a timelike worldline is defined as the time as measured by a clock following that line.

Now in this case it seems to me that the time measured by a clock would be the time difference = 0.8.
So that should be the proper time, τ = 0.8

 

[jbriggs444]

I am going to surprise you now, Dale, by doing just what you say:
The proper time is the time difference in an inertial frame. In this case it will be 1.

The proper time is only going to match the coordinate time difference when using an inertial frame within which the object is motionless.

In any other frame, you can calculate the proper time using the formula $\sqrt{(\Delta t)^2 - (\Delta x)^2}$ = $\sqrt{1^2-0^2}$ = 1.

With a Lorentz factor γ = 1.25, the transformed coordinates of the second point of the world line will be: x = 0.6, t = 0.8.

No.

Rather than just play pretend Lorentz transformation by throwing time dilation and length contraction factors willy nilly at numbers and pretending that they stick, you need to be doing real Lorentz transformations.

As it stands, you have performed a Euclidean rotation, taking (1,0) to (0.8,0.6). The Lorentz transformation is a hyperbolic rotation. So the result you obtained cannot be correct.

You also need to show the work, not just the results.

So the rotated length of the worldline remains at 1. The distance, measured along the x-axis is 0.6 and the time difference is 0.8.

No, no, no. Wrong length formula.

Now in this case it seems to me that the time measured by a clock would be the time difference = 0.8.
So that should be the proper time, τ = 0.8

The [squared] invariant length of the interval is given by $(\Delta t)^2-(\Delta x)^2$ = $0.8^2 - 0.6^2$ = 0.64 - 0.36 = 0.18. It should have come out to be 1.0. So it is clear that you have messed up somewhere.

Let us Google up the Lorentz transformation.

$$t' = \gamma (t-\frac{vx}{c^2})$$ $$x' = \gamma (x-vt)$$

As @Dale has suggested, we will use v=0.6c which results in a gamma of 1.25

For the starting point (0,0) we have t=0, x=0
The transform yields: [YOU DO THE MATH]

For the ending point (1,0) we have t=1, x=0
The transform yields: [YOU DO THE MATH]

The length formula for the interval between the transformed starting point and the transformed ending point yields: [YOU DO THE MATH]

 

[Dale]

The proper time is the time difference in an inertial frame. In this case it will be 1.

Sure, but use the correct formula to calculate it.

With a Lorentz factor γ = 1.25, the transformed coordinates of the second point of the world line will be: x = 0.6, t = 0.8.

This is incorrect. Use the Lorentz transform formula.

Now in this case it seems to me that the time measured by a clock would be the time difference = 0.8.
So that should be the proper time, τ = 0.8

Please use the actual formula, not just guessing from Wikipedia. We have covered the formula for proper time several times in this thread. It is the formula that looks like the Pythagorean theorem almost.

 

[PeterDonis]

The proper time is the time difference in an inertial frame.

No, it isn't. It's the spacetime interval along the worldline.

in this case it seems to me that the time measured by a clock would be the time difference

No. Proper time is the spacetime interval along the worldline. It is not the coordinate time difference.

 

[Grimble]

The proper time is only going to match the coordinate time difference when using an inertial frame within which the object is motionless.

In any other frame, you can calculate the proper time using the formula $\sqrt{(\Delta t)^2 - (\Delta x)^2}$ = $\sqrt{1^2-0^2}$ = 1.

Rather than just play pretend Lorentz transformation by throwing time dilation and length contraction factors willy nilly at numbers and pretending that they stick, you need to be doing real Lorentz transformations.

Ah, yes! I can see I am getting confused by two frames in one diagram. I have been seeing the Primed frame in terms of the inertial frame, rather than using its own coordinates.

As it stands, you have performed a Euclidean rotation, taking (1,0) to (0.8,0.6). The Lorentz transformation is a hyperbolic rotation. So the result you obtained cannot be correct.

I am trying to grasp this terminology, but what is the difference? What exactly is a hyperbolic rotation?

You also need to show the work, not just the results.

Yes, I see that.

The [squared] invariant length of the interval is given by $(\Delta t)^2-(\Delta x)^2$ = $0.8^2 - 0.6^2$ = 0.64 - 0.36 = 0.18. It should have come out to be 1.0. So it is clear that you have messed up somewhere.

Let us Google up the Lorentz transformation.

$$t' = \gamma (t-\frac{vx}{c^2})$$ $$x' = \gamma (x-vt)$$

As @Dale has suggested, we will use v=0.6c which results in a gamma of 1.25

For the starting point (0,0) we have t=0, x=0
The transform yields: [YOU DO THE MATH]

t' = γ (t - ^vx/_c²), x' = γ (x - vt) using c = 1
t' = 1.25 (0 -0) = 0 x' = 1.25 (0 - 0) = 0

For the ending point (1,0) we have t=1, x=0
The transform yields: [YOU DO THE MATH]

t' = γ (t - ^vx/_c²), x' = γ (x - vt) using c = 1
t' = 1.25 (1 - (0.6 x 0)), x' = 1.25 (0 - (0.6))
t' = 1.25, x' = - 0.75

The length formula for the interval between the transformed starting point and the transformed ending point yields: [YOU DO THE MATH]

(Δs)² = ((Δt)² - (Δx)²)
(Δs)² = ((1.25)² - (-0.75)²)
(Δs)² = ((1.5625 - 0.5625) = 1

 

[Herman Trivilino]

I have been seeing the Primed frame in terms of the inertial frame, rather than using its own coordinates.

This seems to be a point of confusion. Both frames are inertial, that is, both the primed frame and the unprimed frame are inertial frames.

Moreover, proper time is the time that elapses, not in just any inertial frame, but in the inertial frame where the two events happen in the same place. In this case, it's the unprimed frame since the value of $x$ is the same for both events.

 

[Ebeb]

Grimble, let me visualize this for you.
(I use a Loedel diagram. A Loedel diagram uses same unit lengths for both reference frames (Minkowski diagram doesn't), which makes it easier to understand that the 'time dilation' is actually not due to a stretching of reference units... or worse, a stretching of proper time units... )

---------------------------For clarity I show how green ref frame gives coordinates to the event[red clock diplays 1.25] :

 

[Dale]

t' = γ (t - ^vx/_c²), x' = γ (x - vt) using c = 1
t' = 1.25 (1 - (0.6 x 0)), x' = 1.25 (0 - (0.6))
t' = 1.25, x' = - 0.75(Δs)² = ((Δt)² - (Δx)²)
(Δs)² = ((1.25)² - (-0.75)²)
(Δs)² = ((1.5625 - 0.5625) = 1

Notice how both the x’ and the t’ coordinates differ from the x and t coordinates of the same event. And yet the spacetime interval was the same.

 

[Orodruin]

Notice how both the x’ and the t’ coordinates differ from the x and t coordinates of the same event. And yet the spacetime interval was the same.

Also, not only the spacetime interval - but the events themselves are the same! Even if they are described by different numbers.

 

[Grimble]

This is Minkowski geometry, not Euclidean geometry. Increasing the spatial displacement (the difference between starting and ending x coordinates) while holding the temporal displacement (the difference between starting and ending t coordinates) constant reduces the length of the (timelike) interval between the end points.

Yes, I can see that, but just what do those measurements represent?
It seems to me that:
the spatial displacement is the distance along the x axis;
the temporal displacement is the distance along the ct axis;
then what is the timelike interval between the endpoints? If both events have the same x coordinate, it is in a resting frame, and the timelike interval = the temporal displacement.
If they have different x coordinates then the frame is moving, the temporal displacement could be labelled ct', and the timelike interval must be the vertical distance to the x axis...
So which of them is the proper time?
if Δτ² = (Δct² - (Δx²)
then surely τ must be the vertical cathetus, x the other catheus and ct the hypotenuse

Yet this is confusing if τ is the proper time, it is the temporal displacement along the rotated worldline which must be ct - the temporal displacement...

Please help me I am confused again...

 

[jbriggs444]

vertical distance to the x axis

This is not Euclidean geometry. Getting out a ruler and measuring the distance on a piece of ruled paper with perpendicular grid lines will not work. I would suggest either learning to read a Minkowski diagram or dropping the visualization entirely and working with the abstract math. You have a distance metric and a formula for transforming coordinates.

 

[Nugatory]

If they have different x coordinates then the frame is moving,

No. The frame is not moving, and indeed the words "The frame is moving" are just some words strung together without meaning, making no more sense than something like "The quadratic formula is moving". A frame is a mathematical convention for assigning coordinates to events, and mathematical conventions aren't thing sthat can move... And until you are absolutely clear on what a frame is you will be wasting your time trying to understand anything more.

If two events have different x coordinates that alone tells us nothing about the two events.

If two events have different x coordinates and the same t coordinates, then the spacetime interval between them is the difference between the x coordinates and is the spatial distance between them in whatever frame we are using to assign the coordinates. There exists no frame in which that distance is zero - that is, we are necessarily working with events that happen at different points in space so there is no frame in which the two events happen at the same place (have the same x coordinate).

If two events have different t coordinates and the same x coordinates, then the spacetime interval is the difference between the the t coordinates and is the amount of time that a freefalling clock present at both events will measure between the two events. There will be no frame in which the two events happen at the same time (have the same t coordinate).

If both the x and the t coordinates are different, then the spacetime interval tells us whether there exists a frame in which both events happen at the same place or both events happen at the same place - at most only one of these is possible, but one or the other will exist as long as the spacetime interval is non-zero.