What Causes the Twins Paradox and Length Contraction in Relativity?

[Stephanus]

Dear PF Forum
I want to know about these questions that are still bothering me,
Does the universe have preferred frame of refference?
Why there's twins paradox?
Motion is relative, why 1 clock experiences time dilation while the other doesn't?

V = \sqrt\frac{3}{4} ≈ 86.6\%
If V define ratio of speed of light, then Lorentz factor is 2.

There are 4 probes.
Clocks are synchronized and reset. Time = T0
Their distance is 100 lys according to Pic 01

It's been 100 years since the clocks are synchronized and reset.
They have been sending digital signals.
The signals contain respective time and status, the time when they fire their rocket.

Then BB and B, preprogrammed, fire their rockets for 1 second, catapult them for V and stop.

A would have said. "Wow, B is traveling 86.6%c, but BB still stays". Is that right?
Because there's no way that A will know that BB has fired its rocket.

And after 10 years in A clock, will A still consider BB still at rest?
Because A only receives BB time and status, which is the rocket has never been fired, yet.
Does A still see that BB hasn't been contracted yet, and sees that B is already contracted?
Does A read BB clock showing T0+10 years?
And one more thing, tell me if I calculate correctly.
After 10 years in A clock, B keep sending signals
A will read B clock... at
L = V * Tb,
L = Distance traveled by B to send signal so that the signal will be received by A in 10 years.
Ta = L, Time taken by Light to reach A from L
Ta + Tb = 10
L + L/V = 10
L=10V/(1+V)
L = 4.641 Ly
Tb = 5.359 Ly
Lorentz factor is 2, so the clock in B reads 2.6795 years.
Is that right, that in 10 years, A will read the signal that B is T0+102.6795 years?

Back to T100
Would A see Pic03 the instant if rocket B and BB are fired and shot at V?

B is contracted while BB isn't, is this the right picture?

What if rocket A (and AA) is fired?
Would A see Pic04 the instant A (and AA) is shot at V?

Both BB, B and the distance is contracted by 50%

Does this why twins paradox occur.
Because in both cases the distance between A-AA and B-BB decreases but there is asymmetry?
But I don't ask when B reaches AA or A reaches BB, it's much more complicated I think with length contraction, where BB can reach A before B reach AA. I'll ask once I get this picture.

 

[pervect]

Dear PF Forum
I want to know about these questions that are still bothering me,
Does the universe have preferred frame of refference?
Why there's twins paradox?
Motion is relative, why 1 clock experiences time dilation while the other doesn't?

I'm not sure why these questions are in a strike-out font, or why you're asking them again. I guess what we said earlier didn't register :(. This does not make me hopeful that what we say will register this time, either, but I'll try.

To recap:

1) No, the universe doesn't have a preferred frame of reference. To clarify : experimental efforts to detect such a preferred frame of reference have to date all failed, and special relativity (conceived of in part beccaue of these failures of experiment to detect any such preferred frame) does not allow one.

2) The twin paradox is not an actual paradox. Rather, it's supposed to be an educational tool. Unfortunately, some people never figure it out :(

3) Motion is relative, and both clocks experience time dilation. I would recommend the PF thread https://www.physicsforums.com/threa...on-implies-relativity-of-simultaneity.805210/ "symmetrical time dilation implies the relativity of simultaneity" which describes how this is possible. I am tempted to repeat the arguments here, but I don't think it's asking too much to click on a link that answers your questions.

 

[PeterDonis]

A would have said. "Wow, B is traveling 86.6%c, but BB still stays".

A cannot see (in the literal sense of receiving light signals) BB moving until light signals emitted by BB after he starts moving reach A. Does that answer your question?

after 10 years in A clock, will A still consider BB still at rest?

Again, A cannot see (in the literal sense of receiving light signals) BB moving until light signals emitted by BB after he starts moving reach A. Does that answer your question?

Does A still see that BB hasn't been contracted yet, and sees that B is already contracted?

A does not see (in the literal sense of receiving light signals) either B or BB as contracted. If B is large enough that light signals from different parts of B take significantly different times to reach A, then A will see (in the literal sense of receiving light signals) B as rotated (google "Penrose-Terrell rotation"). Similarly for BB, when light signals from BB emitted after he starts moving reach A.

Does A read BB clock showing T0+10 years?

A sees (in the literal sense of receiving light signals) BB's clock showing whatever reading was on BB's clock when the light signals were emitted.

Is that right, that in 10 years, A will read the signal that B is T0+102.6795 years?

As above, A sees (in the literal sense of receiving light signals) B's clock showing whatever reading was on B's clock when the light signals were emitted. I haven't verified your calculation, but if that was the intent of your calculation, then you have the right method.

Would A see Pic03 the instant if rocket B and BB are fired and shot at V?

A does not see (in the literal sense of receiving light signals) the distance from BB to B. A only sees (in the literal sense of receiving light signals) light signals coming from B and BB, with B's and BB's clocks showing certain readings. A can deduce the distance from BB to B, at some instant of A's time, only by using additional information: what A knows of BB's and B's state of motion, and whatever simultaneity convention A is using.

B is contracted while BB isn't, is this the right picture?

See above on "seeing" length contraction.

What if rocket A (and AA) is fired?
Would A see Pic04 the instant A (and AA) is shot at V?

See above on "seeing" distances.

Does this why twins paradox occur.
Because in both cases the distance between A-AA and B-BB decreases but there is asymmetry?

No.

 

[Stephanus]

I'm not sure why these questions are in a strike-out font, or why you're asking them again. I guess what we said earlier didn't register :(. This does not make me hopeful that what we say will register this time, either, but I'll try.

Thank you very much Pervect for your effort.
The reason why those lines were struck is: Actually that's what I want to know, why there's twins paradox.
But to know about twins paradox, I have to know what is inertial frame of reference, why the universe has no frame of reference.
And to understand twins paradox, I have to learn what is Lorentz transformation/contraction
All those questions are too much for me.
So I just settled with motion in space. That's why I struck those lines.
Thanks for your effort.

 

[Stephanus]

Thank you very much PeterDonis for your help.
I'll contemplate your answer first

 

[Stephanus]

Again A cannot see (in the literal sense of receiving light signals) BB moving until light signals emitted by BB after he starts moving reach A. Does that answer your question?

Yes and yes and yes

A does not see (in the literal sense of receiving light signals) either B or BB as contracted. If B is large enough that light signals from different parts of B take significantly different times to reach A, then A will see (in the literal sense of receiving light signals) B as rotated (google "Penrose-Terrell rotation")

I'll contemplate.

A sees (in the literal sense of receiving light signals) BB's clock showing whatever reading was on BB's clock when the light signals were emitted

Ok.

A does not see (in the literal sense of receiving light signals) the distance from BB to B. A only sees (in the literal sense of receiving light signals) light signals coming from B and BB, with B's and BB's clocks showing certain readings. A can deduce the distance from BB to B, at some instant of A's time, only by using additional information: what A knows of BB's and B's state of motion, and whatever simultaneity convention A is using.

So every thing A knows about BB (and about everything else in the universe) depends ONLY after light signals emitted by BB reach A.
I've seen you type this phrase very often.

As above, A sees (in the literal sense of receiving light signals) B's clock showing whatever reading was on B's clock when the light signals were emitted. I haven't verified your calculation, but if that was the intent of your calculation, then you have the right method.

Thanks for your confirmation. Okay..., now even if I haven't reached my target yet, but at least I'm no the right track.

Okay guys, thanks for you invaluable helps.

 

[Mentz114]

Dear PF Forum
I want to know about these questions that are still bothering me,
..
Why there's twins paradox?
..
.

The reason for the twins 'paradox' is because every clock measures its own path through spacetime. No clock or thing experiences universal time.

That is the way things appear to work.

 

[Stephanus]

Aww, please don't answer the struck lines now.
That's the questions that I would like to know in the end. But now, I have to understand motion in space

 

[Mentz114]

Aww, please don't answer the struck lines now.
That's the questions that I would like to know in the end. But now, I have to understand motion in space

Non-accelerated motion in space is easy - it does not exist.

If someone is in a closed box and feels no weight - how can they tell if they and the box are moving ?

 

[harrylin]

Non-accelerated motion in space is easy - it does not exist. [..]

I have the impression that your philosophy is at odds with the ones I know of, and possibly it's incompatible with SR. Can you explain how clocks in space can "record" time differently according to the philosophy that their motions do not exist?

 

[Mentz114]

We may be using different definitions of 'motion in space' since I'm not sure what the OP means.

Using the word 'motion' without the 'relative' qualifier has no meaning except if there is absolute motion. I wanted to discourage that idea.

I think your philosophy is unimpugned.

 

[Stephanus]

I would recommend the PF thread https://www.physicsforums.com/threa...on-implies-relativity-of-simultaneity.805210/ "symmetrical time dilation implies the relativity of simultaneity" which describes how this is possible. I am tempted to repeat the arguments here, but I don't think it's asking too much to click on a link that answers your questions.

What a great thread. Yours? Thanks

 

[pervect]

Thank you very much Pervect for your effort.
The reason why those lines were struck is: Actually that's what I want to know, why there's twins paradox.
But to know about twins paradox, I have to know what is inertial frame of reference, why the universe has no frame of reference.
And to understand twins paradox, I have to learn what is Lorentz transformation/contraction
All those questions are too much for me.
So I just settled with motion in space. That's why I struck those lines.
Thanks for your effort.

I'm interpreting this as the sticking point for you is the Lorentz transform. Why is this so difficult?

The Lorentz transform is based on two simpler notions, the notion of addition and the notion of proportionality.

Addition is simple enough - it's just addition. I'd feel funny trying to explain how to add. And I really think that the concept is well understood by most (I could be fooling myself, I suppose).

Proportionality is a bit trickier but not a lot trickier. It's just multiplication, it says that you multiply one number by a constant (or maybe a parameter) in order to get another number. I really think it's likely that most PF posters understand proportionality too, I've seen many lay posters successfully use the concept, and use arguments based on it.

It seems to me that a fair number of posters just give up (I'm guessing you're one of them) when confronted with combining the two concept - one has an equation that involves the sum of two terms (addition), each term f which is a multiplication of a variable by a paramter (a proportionality). And I don't understand why this is perceived as such an obstacle.

In math, if we choose units such that c=1, we write:

$x' = \gamma x - \beta \gamma t \quad t' = \gamma t - \beta \gamma x$

What's hard to understand about this?

 

[Stephanus]

In math, if we choose units such that c=1, we write:

$x' = \gamma x - \beta \gamma t \quad t' = \gamma t - \beta \gamma x$

What's hard to understand about this?

Okay,,, if C = 1, than we can get rid of C from the equation as long as it's a multiplication not addition such as in $\beta$
Is it the time dilation and length contraction formula?
But.., whose x, whose t are we talking here?
One participant will see the other participant is contracting, the other will see the same.
A will see B is contracting, B will see A is contracting.
But when they meet, the clocks are not the same.
Okay..., I'll try again next time. I'll contemplate your answer.

 

[Mentz114]

Okay,,, if C = 1, than we can get rid of C from the equation as long as it's a multiplication not addition such as in $\beta$
Is it the time dilation and length contraction formula?
But.., whose x, whose t are we talking here?
One participant will see the other participant is contracting, the other will see the same.
A will see B is contracting, B will see A is contracting.

No one can se a Lorentz transformation. It is an abstract formula that connects the rulers and clocks of 2 coordinate systems in inertial motion wrt each other.

What you will see if another something approaches you is blue shift in the light they send towards you. From this you can say that their clocks are running faster.
If they are receding from you then you get the opposite - their clocks are running slower. That is the physical facts.

Have another look at the diagrams ghwells made earlier.

But when they meet, the clocks are not the same.

That is because all clocks show their own spacetime interval. Forget time dilation/length contraction. Clocks and Doppler is where the physics is.

 

[Stephanus]

That is because all clocks show their own spacetime interval. Forget time dilation/length contraction. Clocks and Doppler is where the physics is.

Ahh, Doppler.
Nugatory and PeterDonis have said Doppler in my previous thread and I ignore it.
Okayy... I'll contemplate your answer. Thanks Mentz114

 

[harrylin]

We may be using different definitions of 'motion in space' since I'm not sure what the OP means.

Using the word 'motion' without the 'relative' qualifier has no meaning except if there is absolute motion. I wanted to discourage that idea.
[..].

But what is your idea, by which you want to replace an idea that you dislike? Once more: according to you, motion relative to what causes a difference in recorded "time"? Do you suggest that motion relative to a construct of our mind (an inertial "reference frame") can cause one clock to get behind on another clock?

 

[A.T.]

Do you suggest that motion relative to a construct of our mind (an inertial "reference frame") can cause one clock to get behind on another clock?

Physics is about predicting what the clocks will show, and the concept of inertial reference frames is useful for that. Philosophizing about "causation" on the other hand, is often futile.

 

[harrylin]

Physics is about predicting what the clocks will show, and the concept of inertial reference frames is useful for that. Philosophizing about "causation" on the other hand, is often futile.

Sure; I merely ask Mentz114 to clarify the philosophy that he is promoting here to Stephanus.

 

[Mentz114]

But what is your idea, by which you want to replace an idea that you dislike? Once more: according to you, motion relative to what causes a difference in recorded "time"? Do you suggest that motion relative to a construct of our mind (an inertial "reference frame") can cause one clock to get behind on another clock?

I have no idea what you are talking about.

 

[1977ub]

Okay,,, if C = 1, than we can get rid of C from the equation as long as it's a multiplication not addition such as in $\beta$
Is it the time dilation and length contraction formula?
But.., whose x, whose t are we talking here?
One participant will see the other participant is contracting, the other will see the same.
A will see B is contracting, B will see A is contracting.
But when they meet, the clocks are not the same.
Okay..., I'll try again next time. I'll contemplate your answer.

The clocks are not the same because one traveler has moved in multiple inertial frames and one has not. The apparent slowness of one person's clock as measured by another during a particular leg of the journey turns out not to be important.

Two people A & B are moving roughly toward one another and pass close to each other. When they do, they synchronize their clocks, and then they are coasting apart. What could ever cause their clocks to come together again? Either A or B will have to turn on the jets and change to a different inertial frame to catch up with the other again in the future. Whichever person does that will feel the acceleration as they change frames, and it is this person's clock which will read less time when the clocks meet up again.

 

[Mentz114]

The clocks are not the same because one traveler has moved in multiple inertial frames and one has not. The apparent slowness of one person's clock as measured by another during a particular leg of the journey turns out not to be important.

Two people A & B are moving roughly toward one another and pass close to each other. When they do, they synchronize their clocks, and then they are coasting apart. What could ever cause their clocks to come together again? Either A or B will have to turn on the jets and change to a different inertial frame to catch up with the other again in the future. Whichever person does that will feel the acceleration as they change frames, and it is this person's clock which will read less time when the clocks meet up again.

That may be true for the case where one twin remains inertial. But how do you tell the ages of the twins if the both go on a trip in opposite directions and then meet up ?

 

[1977ub]

If they "go in opposite directions" they are both inertial during that initial period. There is no way to say which is the "stay at home" twin. Whichever one changes direction, stopping the separation, and starts moving toward the other will have aged less when they meet again. That person's trip takes place in 2 inertial frames.

 

[Mentz114]

If they "go in opposite directions" they are both inertial during that initial period. There is no way to say which is the "stay at home" twin. Whichever one changes direction, stopping the separation, and starts moving toward the other will have aged less when they meet again. That person's trip takes place in 2 inertial frames.

But in my setup they both stop and turn round and meet.

The answer is - whichever twin has the shortest path through spacetime will have aged less.

 

[Nugatory]

The clocks are not the same because one traveler has moved in multiple inertial frames and one has not.

This is not exactly wrong, but it is terribly misleading. Clocks moving on different paths through spacetime generally register different amounts of time because the lengths of the paths are generally different - it's like finding that cars driven on different routes through space generally end up recording different mileages on their odometers. Inertial frames are pretty much a red herring here, as the amount of time elapsed on the clock on its path through spacetime is the same in all frames, whether inertial or not.

 

[1977ub]

In the classic Twin Paradox, people often get hung up on the fact that each sees the other's clock moving slowly, when in fact it is the traveling twin's clock which has moved in more than one inertial frame, and it is this fact which tells us that the traveling twin will have aged less. A clock which doesn't change inertial frames is the one which has aged more than any other path through spacetime.

 

[PeterDonis]

A clock which doesn't change inertial frames is the one which has aged more than any other path through spacetime.

This is true in flat spacetime, but it does not generalize to curved spacetime (where gravity is present).

 

[1977ub]

It strikes me as the simplest way to look at the Twin Paradox.

 

[PeterDonis]

It strikes me as the simplest way to look at the Twin Paradox.

Sure, if you're just looking at the standard twin paradox in flat spacetime. But if you search through all the umpteen threads here on PF on the twin paradox (never mind all the other places on the web where it's discussed), you'll find that nobody does that. Everybody always wants to know about more general scenarios, many of which involve gravity, and so they soon bump up against the fact that your rule does not generalize. Rather than have to continually get people to unlearn that rule, IMO it's better to not adopt it in the first place; just tell people, right up front, that the general rule is that you look at the length of each person's path through spacetime and compare them.

 

[1977ub]

To know the path through spacetime, you have to be able to say which frames are inertial. In SR, the way to do that is to specify who experienced forces to change their trajectory. Many people come to the TP preoccupied with coordinates and don't understand that inertia is empirically determined and is not simply a matter of whose coordinates are being used.

 

[PeterDonis]

To know the path through spacetime, you have to be able to say which frames are inertial.

More generally, you need to know the geometry of spacetime and the specification of the particular worldlines you're interested in. Also, "which frames are inertial" is in general a local question, not a global one; in a curved spacetime there are no global inertial frames. So just knowing which states of motion are inertial at one event does not, in general, help you know which states of motion are inertial at another event.

In SR, the way to do that is to specify who experienced forces to change their trajectory.

This isn't just in SR; the general definition of "inertial" motion, applicable in any spacetime, is that it is motion with zero experienced force.

The problem is that, in a curved spacetime, there is no guarantee that a particular inertial worldline between two given events is the longest in length. So just knowing which worldlines are inertial is not sufficient to know who experiences the most proper time.

 

[1977ub]

A clock which doesn't change inertial frames is the one which has aged more than any other path through spacetime.

This is true in flat spacetime, but it does not generalize to curved spacetime (where gravity is present).

I'm curious here - can you describe a scenario using GR / gravity wherein the "clock which doesn't change inertial frames" is not "the one which has aged more than any other path through spacetime" ?

 

[A.T.]

I'm curious here - can you describe a scenario using GR / gravity wherein the "clock which doesn't change inertial frames" is not "the one which has aged more than any other path through spacetime" ?

Very fast objects on circular orbits are inertial all the time, but age less than a non-inertial object hovering at the same distance from the mass

 

[PeterDonis]

can you describe a scenario using GR / gravity wherein the "clock which doesn't change inertial frames" is not "the one which has aged more than any other path through spacetime" ?

In curved spacetime, there is no such thing as "a clock which doesn't change inertial frames". There are no global inertial frames in curved spacetime. Every clock "changes inertial frames" along its worldline, regardless of whether it is in free fall or accelerated, because inertial frames in curved spacetime are local. The usual way in GR to specify the state of motion you are describing is "freely falling", or "zero proper acceleration", or "geodesic motion".

As A.T. said, an example of a freely falling worldline that does not have the longest elapsed proper time between two events is the worldline of an object in a circular orbit about a central mass. This worldline is freely falling, but it has less elapsed proper time than an accelerated object hovering at the same altitude has between successive events where the two pass each other.

 

[PeterDonis]

Very fast objects on circular orbits

The object's orbital velocity doesn't have to be "very fast"; it will have less elapsed proper time than a hovering object at the same altitude regardless of the orbital velocity. The magnitude of the difference does depend on the orbital velocity, but even very slow orbital velocities, relativistically speaking, are fast enough to see the difference with modern atomic clocks. A body in a circular orbit around Earth at low altitude has an orbital velocity of about 8 km/s, i.e., about $3 \times 10^{-5}$ of the speed of light, which leads to about a one part in a billion difference in elapsed proper time between an object in a circular orbit and a hovering object at the same altitude, which is easily measurable with current technology. (This exact experiment has not been done, but enough similar ones have been done for us to have very high confidence in the predictions of GR in this regime.)

 

[harrylin]

I have no idea what you are talking about.

I would have appreciated it if you clarified some of your claims. But never mind, Stephanus can enquire more if he likes.

 

[Stephanus]

Can we go on?
I have a new question here.

There are four probes, each has a clock.
A1, A2, B1, and B2.
The clocks are synchronized.
And then B1-B2 moves to A2 direction.
$v = 0.6c \implies \gamma = 1.25$
B travels for 20 years wrt B's clock then stops.
So B is in the same frame again as A
B1 will meet A2, B2 will meet A1.
Q1: Is this right?
B is preprogrammed to stop when their clock read 20 years. Because for B2, there's no way B2 knows that B1 has stopped.

The instant B accelerates, B1 will see this.

From B1 point of view
But, when B1 reaches A2, B2 still haven't me A1, yet. There's still 3 lys away.or...
The instant B accelerates, this is how B2 will see

Pic 02 and Pic 03 don't agree each other. I try to understand SR. I think we just can't simply draw the picture.

Q2: Is this caused by simultaneity of events?

Dear mentors,...
Before I ask further, I want to know if I'm on the right track, so I'll ask this first.
If you explained to me in details, perhaps I can't understand that. Perhaps a yes/no question would suffice this time.
Q1: Will B1 meet A2 and B2 meet A1 in 20 years wrt B's clock?
Q2: Is Pic02 and Pic03 don't match caused by simultaneity of events?
Thanks for the answer.

 

[Mentz114]

Can we go on?
I have a new question here.
View attachment 84791
There are four probes, each has a clock.
A1, A2, B1, and B2.
The clocks are synchronized.
And then B1-B2 moves to A2 direction.
$v = 0.6c \implies \gamma = 1.25$
B travels for 20 years wrt B's clock then stops.
So B is in the same frame again as A
B1 will meet A2, B2 will meet A1.
Q1: Is this right?
B is preprogrammed to stop when their clock read 20 years. Because for B2, there's no way B2 knows that B1 has stopped.

The instant B accelerates, B1 will see this.
View attachment 84792

If you could only show these set ups as spacetime diagrams, the answers would be clearly seen

I'll try to explain how to make an ST plot.

If something moves along the x direction starting at x=0 with velocity v=0.25 we can work out its position at different times

t=0, x=0
t=1, x=v
t=2, x=2v
t=3, x=3v

and so on. If the(x,t) pairs are plotted on a graph with the x-axis horizontal and t-axis vertical, they lie on a straight line inclined to the right.

Repeat this for something coming the other way starting at x=8 with v=-0.3

t=0, c=8
t=1, x=8-0.3
t=2, x=8-(2*0.3)
t=3, x=8-(3*0.3)

and so on.

This is a straight line inclined to the left. The point where the lines cross is where they meet

From this diagram we can deduce the answers to many questions - eg what will their clocks show when they meet ?

I urge you strongly to plot this graph - and show it here so I can tell if it is correct.

If you can understand this concept you will make a great step forward.

 

[Stephanus]

If you could only show these set ups as spacetime diagrams, the answers would be clearly seen

I'll try to explain how to make an ST plot.

If something moves along the x direction starting at x=0 with velocity v=0.25 we can work out its position at different times

t=0, x=0
t=1, x=v
t=2, x=2v
t=3, x=3v

and so on. If the(x,t) pairs are plotted on a graph with the x-axis horizontal and t-axis vertical, they lie on a straight line inclined to the right.

Repeat this for something coming the other way starting at x=8 with v=-0.3

t=0, c=8
t=1, x=8-0.3
t=2, x=8-(2*0.3)
t=3, x=8-(3*0.3)

and so on.

This is a straight line inclined to the left. The point where the lines cross is where they meet

From this diagram we can deduce the answers to many questions - eg what will their clocks show when they meet ?

I urge you strongly to plot this graph - and show it here so I can tell if it is correct.

If you can understand this concept you will make a great step forward.

Thank you.
Okay...

 

[Stephanus]

The point where the lines cross is where they meet
From this diagram we can deduce the answers to many questions - eg what will their clocks show when they meet ?
I urge you strongly to plot this graph - and show it here so I can tell if it is correct.
If you can understand this concept you will make a great step forward.

This looks like a basic math?
The first one, moving at speed 2.5 from 0 is
$2.5t - x = 0$
and the second one.
$0.3t+x = 8$
They will meet at
$t = 2\frac{6}{7}$
$x = 7\frac{1}{7}$
Sorry, drawing the graphs by no mean an easy task.
Okay...
So they meet at $x = 7\frac{1}{7}$ and their clocks show, say $2\frac{6}{7}$
I don't know if this is how we translate the equation in English.
And tell me something from this graph.
Does the direction always to the top. They can go left and right, but always to the top?

 

[Mentz114]

Sorry, drawing the graphs by no mean an easy task.
Okay...
So they meet at $x = 7\frac{1}{7}$ and their clocks show, say $2\frac{6}{7}$
I don't know if this is how we translate the equation in English.
And tell me something from this graph.
Does the direction always to the top. They can go left and right, but always to the top?
View attachment 84811

No. The red line is wrong. The endpoint is more like (t=14, x=6)

 

[Stephanus]

No. The red line is wrong. The endpoint is more like (t=14, x=6)

Ahhh, the other way around.
Okay...
$0.25t - x = 0 \text{, velocity is } 0.25$
$0.3t + x = 8 \text{, velocity is } - 0.3$
Okay,...
They will meet at
$t = \frac{8}{0.55} = 14.54$, calculator here
$x = \frac{-2}{-0.55} = 3.63$

I hope, I do it right.

 

[Mentz114]

Excellent. That is correct.
Remember that the things are moving only along the x-axis. The vertical axis is 'motion in time'.

Now imagine someone who is not moving in this diagram. Their worldline is a vertical line. Can you see that is logical because they have no motion in the x-direction but their clock is always ticking. The times and distances on this chart are in the frame of that observer.

Now to work out the clock times for those worldlines ( yes, you have drawn worldlines) we use $\tau^2 = dt^2-dx^2$ to work out the hypotenuses of the two triangles ( see the pic). Could you do that ? It is just arithmetic.

 

[Stephanus]

Excellent. That is correct.
Remember that the things are moving only along the x-axis. The vertical axis is 'motion in time'.

Now imagine someone who is not moving in this diagram. Their worldline is a vertical line. Can you see that is logical because they have no motion in the x-direction but their clock is always ticking. The times and distances on this chart are in the frame of that observer.

Now to work out the clock times for those worldlines ( yes, you have drawn worldlines) we use $\tau^2 = dt^2-dx^2$ to work out the hypotenuses of the two triangles ( see the pic). Could you do that ? It is just arithmetic.

Wait...

 

[Stephanus]

we use $\tau^2 = dt^2-dx^2$ to work out the hypotenuses of the two triangles ( see the pic).

What do you mean by dt?
Limit t = 0 or $\Delta t$?
And if it's $\Delta t$, where should I began? Right in the intersection?
If it is so...
$\Delta \tau^2 = \Delta t^2 - \Delta x^2 = \frac{8^2}{0.55^2} - \frac{-2^2}{0.55^2} = \frac{60}{0.3025}$ well...
I have to use calculator, but $\Delta \tau = \sqrt{\frac{60}{0.3025}}$
Is that so?

 

[Stephanus]

[..]to work out the hypotenuses of the two triangles ( see the pic).

It just hit me. Are the red and blue line angles limited between 45⁰ and 135⁰?
45⁰>angles<135⁰, not
45⁰≥angles≤135⁰?

 

[Mentz114]

Sorry my notation is not clear.

Please look at the picture herewith. Can you see the lines AC, AB and so on ?

The proper time for the red worldline is $\sqrt{AD^2-CD^2}$

The worldline angle must be more greater than 45^o and less that -45^o ( I think that is what you said)

 

[Stephanus]

Sorry my notation is not clear.

Please look at the picture herewith. Can you see the lines AC, AB and so on ?

The proper time for the red worldline is $\sqrt{AD^2-CD^2}$

$\sqrt{AD^2 - CD^2}$, are you sure? that will put $\sqrt{<0}$

 

[Mentz114]

$\sqrt{AD^2 - CD^2}$, are you sure? that will put $\sqrt{<0}$

Come on ! You are replying too quickly ! Think first !

It is $\sqrt{(14.54)^2 - (3.64)^2}.$

 

[Stephanus]

Come on ! You are replying too quickly ! Think first !

It is $\sqrt{(14.54)^2 - (3.64)^2}.$

Mea culpa, I tought AD - AC, sorry.
Okay then, so...
$\Delta \tau^2 = \frac{8^2}{0.55^2} - (8-\frac{2}{0.55})^2 = \frac{64}{0.55^2} - (64-\frac{32}{0.55}+\frac{4}{0.55^2}) = \frac{60}{0.55^2}-64+\frac{32}{0.55}$
$\Delta \tau = \sqrt{\frac{77.6}{0.55^2} - 64}$ Do you want it in decimal?
But 8-3.64 is not DC it's BD

And for CD
$\Delta \tau^2 = \frac{8^2}{0.55^2} - \frac{-2^2}{-0.55^2}$

$\Delta \tau = \frac{\sqrt{60}}{0.55}$
Is this so?