I How do you resolve seeming contradictions in SR?

[curiousburke]

This seems like a contradiction: if I have two spatially separated observers in one frame (A and B), and one observer in another frame (C) at the same point in space as A. As I understand it, B will have the same definition of "now" as A, and A will have the same "now" as B and C, but B will have a different "now" from C. Expressed as age, B might say they are the same age as A, and also say that A is the same age as C, but still say that they (B) are a different age from C. How does this get resolved?

 

[Hill]

and A will have the same "now" as [ ] C

No, they will not.

 

[curiousburke]

why not if A and C are at the same space and time?

 

[Hill]

why not if A and C are at the same space and time?

Because "now" is not an event but a spacelike hypersurface of all events which are simultaneous in a given frame. These hypersurfaces differ between A and C although they might intersect, for example in that space-time event you refer to.

 

[curiousburke]

So, you're saying that my misunderstanding is that participants in space-time events don't have to agree on "now"?

 

[Hill]

So, you're saying that my misunderstanding is that participants in space-time events don't have to agree on "now"?

Yes, they don't.

 

[PeterDonis]

As I understand it, B will have the same definition of "now" as A

Yes.

and A will have the same "now" as B

Yes.

and C

No. A and C have the same "here and now" at the event you describe, where A and C meet; but they do not have the same "now" because "now" is not just a single event but a set of events, and the set of events "now" for C is not the same as the set of events "now" for A.

Expressed as age

"Age" is a different thing from "now".

B might say they are the same age as A, and also say that A is the same age as C, but still say that they (B) are a different age from C.

No, this is not possible. Think of A, B, and C as all carrying clocks that read their age. At the event where A and C meet, if A and C's clocks both read the same age, that is an invariant; it doesn't matter what frame you choose. And if B, spatially separated from A and C, is at an event which is in the same set of "now" events for A and B as the event where A and C meet, says that his (B's) age--the reading on B's clock--is the same as A's age--the reading on A's clock--at the event where A and C meet, then B's age must also be the same as C's age--the reading on C's clock--at that event, because A's and C's clock readings don't depend on what frame you choose.

 

[curiousburke]

I don't understand what that means that two observers at the the same point in space-time don't agree on "now"; do there agree on "here"?

 

[PeterDonis]

I don't understand what that means that two observers at the the same point in space-time don't agree on "now"; do there agree on "here"?

As I said, they agree on "here and now": they are both at the same point in spacetime.

If they are moving relative to each other, then they do not agree on either "here" or "now" separately: both "here" and "now" are sets of events in spacetime, not single events, and they are both different sets of events for A and C. They only agree on "here and now" at the single event at which they meet.

 

[Hill]

do there agree on "here"?

They agree on "here" at that moment but generally they do not. Need an example?

 

[curiousburke]

Yes.


Yes.


No. A and C have the same "here and now" at the event you describe, where A and C meet; but they do not have the same "now" because "now" is not just a single event but a set of events, and the set of events "now" for C is not the same as the set of events "now" for A.


"Age" is a different thing from "now".


No, this is not possible. Think of A, B, and C as all carrying clocks that read their age. At the event where A and C meet, if A and C's clocks both read the same age, that is an invariant; it doesn't matter what frame you choose. And if B, spatially separated from A and C, is at an event which is in the same set of "now" events for A and B as the event where A and C meet, says that his (B's) age--the reading on B's clock--is the same as A's age--the reading on A's clock--at the event where A and C meet, then B's age must also be the same as C's age--the reading on C's clock--at that event, because A's and C's clock readings don't depend on what frame you choose.

This is exactly how I was thinking, but wouldn't the velocity difference and spatial separation mean that C and B would interpret 'now' differently?

 

[Ibix]

So, you're saying that my misunderstanding is that participants in space-time events don't have to agree on "now"?

They will agree on "now at the same place". So if you swing your left hand and your right hand and they meet and make a noise, everyone will agree that your hands were at the same place at the same time. Thus no-one hears a noise with no cause, nor sees a cause with no noise. But they will, in general, disagree that your hands started their swinging motions at the same time - because they are not at the same place there's no immediate physical consequence of whether they start to move simultaneously or not. Only whether they meet - and everyone will agree on that.

 

[curiousburke]

Okay, so I should have been explicit and said "here and now", which A and C will agree on.

I started thinking about this because of the triplet paradox posed by FloatHeadPhysics on youtube. He makes the claim that C would not see A and B as the same age. I just turned this around and asked the question how B would see their three ages. So, is FloatHead wrong, and A B and C will all see their ages being the same? In which case his resolution of the paradox using simultaneity is wrong.

 

[PeroK]

How does this get resolved?

One resolution is to note that global simultaneity (global "now", in your terms) is not a thing. Locally, it looks like simultaneity is a thing. In a race, all the athletes ought to hear the gun at the same time. But, that's a local scenario, where any lack of absolute simultaneity is hardly measurable. If we try to define simultaneity between Earth and Mars, say, then we have the significant (20 minutes plus) time lag of light signals each way. This creates some ambiguity in when precisely "now" on Mars is. There's a 20-minute variation each way.

If we try to define now on Earth and now somewhere in the Andromeda galaxy, then there is a 2 million years ambiguity each way. So, global now is just not a thing.

Note that the clocks on the GPS satellite systems have to take this ambiguity in "now" into account. It's only a fraction of a second, but enough to through the GPS system off if you try to impose a global now on them.

 

[Ibix]

I started thinking about this because of the triplet paradox posed by FloatHeadPhysics on youtube.

Perhaps you could state the triplet paradox that you are talking about, rather than expecting us to hunt through YouTube and watch videos on subjects we already understand.

All (almost all?) "paradoxes" in relativity boil down to failing to apply the relativity of simultaneity correctly.

 

[PeterDonis]

wouldn't the velocity difference and spatial separation mean that C and B would interpret 'now' differently?

The relative velocity is what makes C and B interpret "now" differently. The spatial separation does not affect that, as should be obvious from the fact that A, who is not spatially separated from C at the event where they meet, also interprets "now" differently from C (but the same as B, who is spatially separated from A).

 

[curiousburke]

Perhaps you could state the triplet paradox that you are talking about, rather than expecting us to hunt through YouTube and watch videos on subjects we already understand.

All (almost all?) "paradoxes" in relativity boil down to failing to apply the relativity of simultaneity correctly.

Sorry, you're right. It's basically the scenario I posted with the PeterAdonis' addition of synchronizing the clocks when A and C are at the same here and now. The question is how old are B and C when C gets to B. C is traveling parallel to the line between A and B.

 

[PeterDonis]

C would not see A and B as the same age

More precisely, consider the two different sets of "now" events for A and C that both contain the event where A and C meet. The event on B's worldline that is in A's "now" set shows B's "age clock" reading the same as A's and C's "age clock" at the event where A and C meet. But the event on B's worldline that is in C's "now" set shows B's "age clock" with a different reading from A's and C's "age clock" at the event where A and C meet. This is just a simple consequence of A's and C's "now" sets being different.

Notice how being precise about exactly what is being said helps to resolve any apparent "paradox"?

 

[Dale]

So, you're saying that my misunderstanding is that participants in space-time events don't have to agree on "now"?

That is correct. They agree on “here and now” at the event where they meet, but not “now” even at that event.

 

[curiousburke]

More precisely, consider the two different sets of "now" events for A and C that both contain the event where A and C meet. The event on B's worldline that is in A's "now" set shows B's "age clock" reading the same as A's and C's "age clock" at the event where A and C meet. But the event on B's worldline that is in C's "now" set shows B's "age clock" with a different reading from A's and C's "age clock" at the event where A and C meet. This is just a simple consequence of A's and C's "now" sets being different.

Notice how being precise about exactly what is being said helps to resolve any apparent "paradox"?

I think I do. Like, I ... kinda get it. It's about looking at two different world lines for each question.

 

[curiousburke]

That is correct. They agree on “here and now” at the event where they meet, but not “now” even at that event.

"now" being the entire line, not just this event.

 

[PeterDonis]

It's about looking at two different world lines for each question.

A's and C's worldlines are different, yes. (And B's is different from both.) Each observer's worldline defines "here" for that observer.

 

[PeterDonis]

"now" being the entire line, not just this event.

Yes.

 

[curiousburke]

Thank you all for your insights. It's a difficult concept for me, and really hard to eliminate all vestiges of the global "now" assumption.

 

[Ibix]

Sorry, you're right. It's basically the scenario I posted with the PeterAdonis' addition of synchronizing the clocks when A and C are at the same here and now. The question is how old are B and C when C gets to B. C is traveling parallel to the line between A and B.

So, we have two observers, A and B, at rest relative to each other and a third travels from one to the other. The issue is, presumably, that B sees the traveller's clock tick slow and the traveller sees B's clock tick slow. Which one is older when they meet?

The relevant application of the relativity of simultaneity is to note that the traveller's frame does not see A and B's clocks as synchronised - he says B's clock is ahead of A's. Thus he is unsurprised that B's clock shows a larger value than his own - it was initially ahead, and although it is ticking slowly there has not been enough time for it to fall behind.

 

[curiousburke]

So, we have two observers, A and B, at rest relative to each other and a third travels from one to the other. The issue is, presumably, that B sees the traveller's clock tick slow and the traveller sees B's clock tick slow. Which one is older when they meet?

The relevant application of the relativity of simultaneity is to note that the traveller's frame does not see A and B's clocks as synchronised - he says B's clock is ahead of A's. Thus he is unsurprised that B's clock shows a larger value than his own - it was initially ahead, and although it is ticking slowly there has not been enough time for it to fall behind.

Exactly. So, it made me think: if C sees B's clock as ahead, why doesn't B see C's clock as starting ahead or maybe behind?

 

[Ibix]

Exactly. So, it made me think: if C sees B's clock as ahead, why doesn't B see C's clock as starting ahead or maybe behind?

C sets his clock to match A and everyone will agree that they read the same as they pass. But different frames will have different views on what "at the same time as" means in the question "what time does B's clock show at the same time as C and A meet?". B says his clock shows the same time as A because it's a tautology: they synchronised their clocks by their own definition of synchronisation and therefore they show the same time. C uses a different synchronisation convention, by which A and B's clocks aren't synchronised.

Perhaps you mean to ask why different frames have different definitions of "simultaneous"?

 

[curiousburke]

C sets his clock to match A and everyone will agree that they read the same as they pass. But different frames will have different views on what "at the same time as" means in the question "what time does B's clock show at the same time as C and A meet?". B says his clock shows the same time as A because it's a tautology: they synchronised their clocks by their own definition of synchronisation and therefore they show the same time. C uses a different synchronisation convention, by which A and B's clocks aren't synchronised.

Perhaps you mean to ask why different frames have different definitions of "simultaneous"?

No, I think I understand why simultaneity is frame dependent. I just would expect symmetry between C's perspective on B's time and B's perspective on C's time.

 

[curiousburke]

No, I think I understand why simultaneity is frame dependent. I just would expect symmetry between C's perspective on B's time and B's perspective on C's time.

I'm sorry I'm rehashing: does B view them all (A,B,C) as simultaneous at synchronization but C does not? That's where I am tripped up.

 

[Dale]

Thank you all for your insights. It's a difficult concept for me, and really hard to eliminate all vestiges of the global "now" assumption.

This is indeed the single most challenging concept in SR. You are by far not alone in this challenge. Once you learn this then you will be most of the way to learning relativity.

 

[Ibix]

No, I think I understand why simultaneity is frame dependent. I just would expect symmetry between C's perspective on B's time and B's perspective on C's time.

There is symmetry in the perspectives, but this scenario is asymmetric. You've got one observer using one frame and two observers using the other. So in one frame you haven't defined a second fixed point to make measurements with.

If you add an observer, D, at rest with respect to C and as far away in their mutual rest frame as A and B are in their frame, you can ask how old A or B is when they get to C (or D, whichever they meet later). That'll give you the same answer, which is the symmetry you expect.

 

[curiousburke]

There is symmetry in the perspectives, but this scenario is asymmetric. You've got one observer using one frame and two observers using the other. So in one frame you haven't defined a second fixed point to make measurements with.

If you add an observer, D, at rest with respect to C and as far away in their mutual rest frame as A and B are in their frame, you can ask how old A or B is when they get to C (or D, whichever they meet later). That'll give you the same answer, which is the symmetry you expect.

Ironically, that's exactly what I tried to think about when I got really confused. If frame 1 is contracted wrt frame 2 and frame 2 is contracted wrt frame 1, how can A be next to C and B be next to D? That's a rhetorical question, not meant to be answered, which would take this tread down a different path.

 

[Ibix]

Ironically, that's exactly what I tried to think about when I got really confused. If frame 1 is contracted wrt frame 2 and frame 2 is contracted wrt frame 1, how can A be next to C and B be next to D? That's a rhetorical question, not meant to be answered, which would take this tread down a different path.

Look up Minkowski diagrams. I can draw one for this scenario later. I find them very helpful - essentially the answer is that when two frames say they are measuring "distance" they are measuring different lines in spacetime, and there end up being four different things being measured when you say "they measure the distance between them". There would only be a contradiction if they were measuring one or two things.

 

[Dale]

Ironically, that's exactly what I tried to think about when I got really confused. If frame 1 is contracted wrt frame 2 and frame 2 is contracted wrt frame 1, how can A be next to C and B be next to D? That's a rhetorical question, not meant to be answered, which would take this tread down a different path.

Because of things like that, a spacetime diagram can be a very helpful tool in both organizing your thoughts and communicating them to others.

 

[PeroK]

Ironically, that's exactly what I tried to think about when I got really confused. If frame 1 is contracted wrt frame 2 and frame 2 is contracted wrt frame 1, how can A be next to C and B be next to D? That's a rhetorical question, not meant to be answered, which would take this tread down a different path.

You have to be careful with such specious arguments. That means seemingly plausible, but actually wrong. (It's a good word to know!) Consider this: two ambulances are racing in opposite directions towards each other. The first ambulance hears the second's siren at a higher pitch. And, the second ambulance hears the first's at a higher pitch. If someone suggests that is a contradiction, you can examine the basis of that argument. In this case, however, you are probably convinced from the outset that there is no contradiction - even if you can't immediately see the flaw.

 

[curiousburke]

Yes, I certainly didn't mean to suggest this problem I was having in thinking about two frames contracting was a contradiction as much as it was frying my brain.

 

[robphy]

On a spacetime diagram,
simultaneity is described using the Minkowski version of the "tangent-line to a circle".
More specifically, when one intersects a circle by a radial line,
the tangent-line at that intersection is orthogonal to the radial line...
in the geometry determined by the "circle".

Try it at:
robphy SIMULTANEITY spacetime diagrammer for relativity (simplified from v8c) - 2023
https://www.desmos.com/calculator/ajktes8bp5
Euclidean (E= -1), Minkowski/Special Relativity (E = +1), Galilean Relativity (E = 0)

 

[Ibix]

Here are a few spacetime diagrams. If you haven't come across them, they're simply plots of the position of an object over time, the custom being that time goes up the page. So a vertical line represents an object that isn't moving, and a line slanted to the left indicates one moving to the left. You may have come across these (usually with time horizontally) as "displacement-time graphs" in high school physics. We take them a little more seriously in relativity, since spacetime is a thing - these are maps of spacetime, and the lines are the 4d objects inhabiting it. You see one 3d slice at a time.

So here's your scenario, with my extra "D" observer. A and B are marked in red and are stationary, C and D are marked in blue and moving to the right.

Note that, in this frame, C and D are closer together than A and B - the horizontal distance between the lines ("the space between them") is shorter.

We could mark on the diagram the start and end of the experiment - when C passes A and then B. Let's do that with fine red lines:

Now comes the interesting part. C and D don't use the same definition of simultaneity. The lines in spacetime they call "now" are not horizontal lines on this diagram - they're sloped. In fact they are the fine blue lines on this graph:

Note that these lines still go through the A/C and B/C meetings. But if you look at the lower fine line, you can see it passes through B much later than the fine red line did. So according to C, at the start of the experiment B's clock was ahead. And the part of B's line that is "during the experiment" according to C (the part between the fine blue lines) is much shorter than the part that is "during the experiment" according to A and B - that is, it starts ahead and ticks slower.

This is also why length contraction doesn't lead to a contradiction. Length is measured at one time, and the frames have different definitions of "at one time". A and B measure length horizontally; C and D measure it along the fine blue lines. So there are four different measures: the A-B and C-D distances along a horizontal line, and the A-B and C-D distances along a fine blue line (warning: "distance" here is a slight lie - you have to measure "interval", which is $\sqrt{\Delta x^2-c^2\Delta t^2}$ rather than the Euclidean distance you see on the graph, $\sqrt{\Delta x^2+c^2\Delta t^2}$ - this has the consequence that some lines that are longer on the graph are shorter in reality).

You asked about symmetry. That's why I've drawn D, even though I haven't used that letter yet. I can draw the equivalent graph using the frame where C and D are at rest. It looks like this:

This is a mirror of the first graph - now the blue lines are vertical and the red lines are sloped and closer together. The "same experiment" is A moving from C to D, and the same graphs with horizontal and slanted fine lines (just left-right reversed) can be drawn:

That's the symmetry you were missing. But you need to add D to get it, or in this frame there's nothing for A to move to and you can't do the same experiment as you did with the first frame. If you ignore D for a moment, the only thing you can measure is the time between A and B reaching C - and here you can see that the time between those events as measured by C is shorter than the time measured by A and B, by comparing the vertical distance between the crossings.

I hope that helps. You may need a more thorough intro to Minkowski diagrams if you haven't seen them before, but they really are the easiest way to visualise special relativity.

 

[curiousburke]

I still have to digest this more to fully get it, but thank you for doing this!

 

[curiousburke]

Okay, I think I get most of it. The only thing I'm not fully grasping is the later part of the last, actually second last, paragraph in which you say how to read time from the diagram and from whose perspective that time is. What do I measure on the diagram to say find A and C's proper elapsed times during the experiment?

 

[Ibix]

What do I measure on the diagram to say find A and C's proper elapsed times during the experiment?

For a straight line, measure the vertical distance $\Delta t$ and horizontal distance $\Delta x$ between the endpoints. The "length" squared (properly called the interval) of the line is $\Delta s^2=c^2\Delta t^2-\Delta x^2$. When this is positive the proper time along the line is the square root of it over $c$. When it's zero the line represents something travelling at $c$, and proper time is not defined. When it's negative it represents a spatial distance with a length equal to the square root of the absolute value.

Note that some sources define $\Delta s^2$ with the opposite sign. As long as you don't use both conventions in one calculation, it doesn't matter which you use - just reverse the positive and negative in my paragraph above. Note also that the graphs above use years for time and light years for distance, and using these units $c=1\mathrm{ly/y}$, which simplifies the algebra. It also lets you see the sign of $\Delta s^2$ immediately by seeing if the line has a slope greater than, less than, or equal to 1.

 

[curiousburke]

so, without getting a real number, the longer the line the shorter the interval because of the negative sign?

So, vertical lines are easy, like in the last diagram if I'm interested in C between A and B. Using the same diagram, if I want to know the elapsed time for A or B, when C goes from A to B, is this a line we have on the diagram? Also, while the length of a line will be related to the elapsed time, do we need to include differences in the initial clock time to calculate which clock will have more time between events?

2nd edit: I think I see, in the first diagram, B's elapsed time is the vertical line from t=0 to the crossing point with C. Shorter line than C from A to B, so longer proper time

 

[Ibix]

so, without getting a real number, the longer the line the shorter the interval because of the negative sign?

It's not that simple - lines of equal interval starting at the origin all end on a hyperbola (the upward curving line in @robphy's second diagram above is a unit hyperbola, and all lines from the origin to it have interval 1). A line that goes past a hyperbola will have a larger interval than one that ends on it, but its representation on the diagram will be longer than some lines that end on the hyperbola and shorter than others.

With lines that start at the same place you can probably visualise the hyperbolae and estimate which is longer. In general, you probably have to do some maths.

I think I see, in the first diagram, B's elapsed time is the vertical line from t=0 to the crossing point with C. Shorter line than C from A to B, so longer proper time

Yes - but as above, careful with "longer is shorter" as a general rule. Since the lines have the same $\Delta t$ it works out here, though.

 

[curiousburke]

View attachment 341323



View attachment 341329

Checking if I understand this. I seems you can't read A, B and C's proper times directly off a single diagram, but we can we get the elapsed proper times as C goes from A to B for all three from the vertical lines in two diagrams. So, A and B from the first diagram is ~1.7 and C from the second diagram is ~1.4?

 

[Ibix]

Checking if I understand this. I seems you can't read A, B and C's proper times directly off a single diagram

You can get A and B's proper times directly from the first diagram because $\Delta x=0$, so in this case $\Delta \tau=\Delta t$. Similarly you can get C's from the fourth diagram because $\Delta x'=0$ so C's $\Delta\tau=\Delta t'$. You can't get all three off one diagram without doing a fairly simple bit of maths, no. (Although note that @robphy has a clever graphical method of measuring arbitrary proper times).

So, A and B from the first diagram is ~1.7 and C from the second diagram is ~1.4?

The diagram is draw with the speed of C and D being 0.6c relative to A and B, and with A and B separated by 1ly. So the time A and B measure should be 1.66y, and C should measure 1.33y. So you are pretty close.

 

[curiousburke]

So the time A and B measure should be 1.66y, and C should measure 1.33y. So you are pretty close.

I almost said 1.35, but the added precision seemed unjustified :)

I see this method of getting to the answer shows an asymmetry: a single observer/clock in frame A that is assessed at two points in frame B will have less elapsed time then clocks in frame B. Maybe that interpretation is wrong, it's what I guessed when thinking through the triple paradox, which is the same setup, so I could just be holding on to it.

Unfortunately, now I'm taking a big step backwards (sorry). I think I'm understanding the space-time diagrams, but I'm still not really getting how the asymmetrical setup leads to different elapsed times. If B travels from some initial position to C, the space between them contracts and it takes less time. If C travels to B, the same. How does adding A change anything?

In thinking about the youtube triplets, I convinced myself that B was being required to travel the distance to A in uncontracted space because A and B are in the same frame. This would of course lead to C being younger from either perspective since C travels a contracted distance. However, I don't think that is the case in your spacetime diagram analysis (maybe it wasn't on youtube either), because when A travels from C to D, A experiences less elapsed time (there is symmetry). So what is it about comparing a clock in one frame with two in another that causes an asymmetry in the elapsed time?

Is the synchronization somehow involved, or is that just an arbitrary choice of t=0?

 

[robphy]

Here's a Desmos visualization that reconstructs @Ibix 's spacetime diagram:

robphy-PFreply-resolveSeemingContradictions-CausalDiamonds
https://www.desmos.com/calculator/y5aiiixdvx

You can reposition the two events to find the interval between them,
which is calculated as the square-root of the
area of the causal diamond enclosed by two causally-related events.

With the given graph paper, one small diamond has diagonal with interval (1/3).
The causal diamond of OP has 16 diamond areas.
So, the (diagonal's interval) = (sqrt(16))*(1/3) =4/3.
With O at the origin, one can consider events Z only on the hyperbola of radius (4/3).

 

[PeroK]

Unfortunately, now I'm taking a big step backwards (sorry). I think I'm understanding the space-time diagrams, but I'm still not really getting how the asymmetrical setup leads to different elapsed times. If B travels from some initial position to C, the space between them contracts and it takes less time. If C travels to B, the same. How does adding A change anything?

In thinking about the youtube triplets, I convinced myself that B was being required to travel the distance to A in uncontracted space because A and B are in the same frame. This would of course lead to C being younger from either perspective since C travels a contracted distance. However, I don't think that is the case in your spacetime diagram analysis (maybe it wasn't on youtube either), because when A travels from C to D, A experiences less elapsed time (there is symmetry). So what is it about comparing a clock in one frame with two in another that causes an asymmetry in the elapsed time?

Is the synchronization somehow involved, or is that just an arbitrary choice of t=0?

If you are serious about learning SR, you could try Morin's book. The first chapter is free online:

https://scholar.harvard.edu/david-morin/special-relativity

The amount of time you've spent on this random video could have been spent building a solid foundation for the whole subject.

 

[PeroK]

PS I had a look at that video. It's very insightful, but I suspect the ideas might be too advanced until you've grasped the basics of SR yourself. IMO, Mahesh is spot on with everything he says. Having the radio transmitter in the middle is a great idea. But, there's a lot to digest in that 20-minute video.

So, rather than trying to learn SR from that video, I would knuckle down with Morin's book and use Mahesh's video as a test piece. When you reach the stage that his video is obvious, then you know you've understood SR!

 

[Ibix]

Unfortunately, now I'm taking a big step backwards (sorry). I think I'm understanding the space-time diagrams, but I'm still not really getting how the asymmetrical setup leads to different elapsed times. If B travels from some initial position to C, the space between them contracts and it takes less time. If C travels to B, the same. How does adding A change anything?

Forget about Minkowski spacetime for a minute and just look at the diagrams as pictures. Look at the second one, with the fine red lines marked. The two thick and two thin red lines form a rectangle. One blue line is the diagonal. The length of blue line between the fine red lines is longer than the length of thick red line between the fine lines, because one's a diagonal and the other's an edge.

Now go back to thinking of it as a Minkowski diagram. The elapsed time along the blue line between the fine red lines is shorter than the elapsed time along the thick red line between the fine lines, because one's a diagonal and the other's an edge.

That's why the times are different. It's because you picked the red lines to form the rectangle, so the blue is inevitably the diagonal. The third diagram (with slanted fine blue lines) shows what blue did wrong with intuitive reasoning: there is less elapsed time for the destination red clock during the experiment by blue's definition of "during". But the red clocks are not synchronised like that, and blue forgot to allow for the extra time.