Exploring Twins Paradox & Length Contraction

[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?

 

[Mentz114]

I don't know. I can't follow what you've written and I'm not going to finish your calculations.

The answers are

red worldline $\tau = \sqrt{(14.54)^2 - (3.64)^2} = 14.08$
other worldline $\tau = \sqrt{(14.54)^2 - (8-3.64)^2} = 13.87$

stationary worldline $\tau = 14.54$

I'm hoping you might grasp some of the princples of relativity from this exercise.

You have made spacetime diagram, and calculated some proper times, so it could be progress.

Can you work out the time on the blue worldline when a light beam from the origin hits it ?

 

[Stephanus]

I don't know. I can't follow what you've written and I'm not going to finish your calculations.
The answers are
red worldline $\tau = \sqrt{(14.54)^2 - (3.64)^2} = 14.08$
other worldline $\tau = \sqrt{(14.54)^2 - (8-3.64)^2} = 13.87$
stationary worldline $\tau = 14.54$

Yes, the our calculations matches. Now I'm using calculator.

Can you work out the time on the blue worldline when a light beam from the origin hits it ?

Wait...

 

[Stephanus]

Can you work out the time on the blue worldline when a light beam from the origin hits it ?

Hits it.
Hits who? Blue worldline?
$\sqrt{14.54^{2} - (8-3.64)^2}$?
13.87 you mean?
The lights hits Blue when his clocks is 13.87?

 

[Stephanus]

No, it's not that way.
The light hits blue world line I think is an intersection
Between blue and the vertical line.
A perpendicular line from blue to vertical line. Is that so?

 

[Stephanus]

No, not perpendicular. But 45⁰ from vertical lines. That's how x = t
Tell me. Which one?

 

[Mentz114]

No, not perpendicular. But 45⁰ from vertical lines. That's how x = t
Tell me. Which one?

This one. The WL of the light is as you say $x=t$ and the thing is on $x=8-vt$ so you can get the intersection on the appraoching WL.

Then make a triangle and work out a $t^2-x^2$ and you got it.

 

[Stephanus]

I think the latter, let me calculate.
for x = t from the origin, so...
$t = x - \frac{2}{0.55}$ this is 45⁰ from origin
$0.3t = 8-x$ this is b velocity
They will meet at...
$1.3x - \frac{0.6}{0.55} - 8 = 0$
$x ≈ 7$
$t = \frac{1.85}{0.55} = 3.36$
The signal from the origin hits B at (7,3.36)

 

[Stephanus]

This one. The WL of the light is as you say $x=t$ and the thing is on $x=8-vt$ so you can get the intersection on the appraoching WL.

Then make a triangle and work out a $t^2-x^2$ and you got it.

Thanks. But I already calculated before your alert popped up. Is it time dilation in PF forum?
Thanks a lot Mentz, I have to go to sleep. It's 02:00 AM in the morning here.
Why time runs faster when I chat with you? Hmmhh... It's the relativity I think. Don't realize it's already 2:00AM.
Shall we continue tomorrow? Or today noon
Thanks, bye.

 

[Mentz114]

Solving $x=t$ with $x=8-vt$ gets $t=8-vt$ so $t=8/(1+v)=6.15$
and $x=6.15$.

I'm sorry to keep you awake.

 

[Stephanus]

Solving $x=t$ with $x=8-vt$ gets $t=8-vt$ so $t=8/(1+v)=6.15$
and $x=6.15$.

I'm sorry to keep you awake.

No, no, no. Don't be sorry. I liked that.
It helps me much.

 

[Stephanus]

Solving $x=t$ with $x=8-vt$ gets $t=8-vt$ so $t=8/(1+v)=6.15$
and $x=6.15$.

I'm sorry to keep you awake.

If by origin you mean (0,0), then $8/1.3 = 6.15$
I tought it's from D (3.36,0) Point D in your graph.

 

[Mentz114]

If by origin you mean (0,0), then $8/1.3 = 6.15$
I tought it's from D (3.36,0) Point D in your graph.

I drew the light ray on the graph. It is thin yellow line.

 

[Stephanus]

I drew the light ray on the graph. It is thin yellow line.

The light that comes from the origin.

If this is a Spacetime diagram. And C is the centre.
Is it true, that C were/will never be in magenta area? Not even in 45⁰ lines?
Is it true, that C could be in the cyan area below?

 

[Mentz114]

Events can exist at every point on the diagram. An event is a pair of numbers (t,x).

 

[Stephanus]

But, for events to happen at the magenta region, that events should travel x>t?
I mean travel faster than light from c point of view?
But I think, I know why there is Twins Paradox.
It is simultaneity of events which responsible, is it so?

 

[Ibix]

Events can exist anywhere, and they don't "travel". Things that are traveling appear as sloped lines in your graph. Events are points on your graph. More physically an event is something like turning on your rockets. That happens at a point in space at an instant in time, which can't be said to be moving.

That said, you are correct that events in the magenta region cannot be reached from the origin without exceeding the speed of light. An example would be a car crash in London and a simultaneous (in some frame) car crash in New York. If one was at the origin the other was in the magenta region.

 

[Stephanus]

Events can exist anywhere, and they don't "travel". Things that are traveling appear as sloped lines in your graph. Events are points on your graph. More physically an event is something like turning on your rockets. That happens at a point in space at an instant in time, which can't be said to be moving.

That said, you are correct that events in the magenta region cannot be reached from the origin without exceeding the speed of light. An example would be a car crash in London and a simultaneous (in some frame) car crash in New York. If one was at the origin the other was in the magenta region.

Back again to Twins Paradox.
Supposed there's a Type 1a Supernova, 1000 lys away. We know that it's 1000 lys because of its brightness. And suddenly it moves toward us at 0.1 c.
What would we see? Nothing changes, is this right?
Suddenly, 1000 years later, we'll see that the star is blue shifted toward us or at least it changes. Is this right? And keeps getting bigger and bigger. And also 1000 years later we'll experience that time in the supernova runs faster
I know, I know, supernova only lasts a few weeks or years.

Now, what if WE move toward the supernova. We'll see it blue shifted right away. Is this right? We'll experience doppler effect right away and we'll experience that time in the supernova runs faster.

In both cases, we'll experience that times slows for us, but at different date. Time dilation and length contraction work both ways.
But if we move, we'll experience those effect RIGHT AWAY.
But if we at the 'rest frame' and the other object is moving toward us. We'll notice that time/length,etc from that object also dilated/contracted WHEN the object's WORLDLINE reaches us?
Is that so?
Is that why there's Twins Paradox? Simultaneity of events?

 

[Ibix]

Back again to Twins Paradox.
Supposed there's a Type 1a Supernova, 1000 lys away. We know that it's 1000 lys because of its brightness. And suddenly it moves toward us at 0.1 c.
What would we see? Nothing changes, is this right?

Correct. Although I'm not sure how you intend to accelerate a supernova to 0.1c.

Suddenly, 1000 years later, we'll see that the star is blue shifted toward us or at least it changes. Is this right? And keeps getting bigger and bigger.

The supernova would be blue-shifted, yes. You don't need to say "towards us" in this context.

And also 1000 years later we'll experience that time in the supernova runs faster
I know, I know, supernova only lasts a few weeks or years.

Not really. We will see the supernova evolving rapidly due to the decreasing lightspeed delay. But if we correct for the changing delay we will find that the supernova is evolving more slowly than a non-moving supernova, and this is due to time dilation.

Now, what if WE move toward the supernova. We'll see it blue shifted right away. Is this right? We'll experience doppler effect right away and we'll experience that time in the supernova runs faster.

Yes, although again there is an effect from changing light travel time and an effect from time dilation. So we would, again, see that the supernova was running fast, but could subtract out the light travel time and deduce that time was running slow at the supernova.

In both cases, we'll experience that times slows for us, but at different date. Time dilation and length contraction work both ways.
But if we move, we'll experience those effect RIGHT AWAY.

We will always experience time for us at one second per second. That's more or less what time is.

But if we at the 'rest frame' and the other object is moving toward us. We'll notice that time/length,etc from that object also dilated/contracted WHEN the object's WORLDLINE reaches us?
Is that so?

No. When an object's worldline reaches you, it has crashed into you. When an event's light cone reaches you, it is possible to see that event. You can't see it before because the light didn't have time to reach you yet.

Is that why there's Twins Paradox? Simultaneity of events?

You aren't really talking about simultaneity here. You are talking about when people perceive events, which is a separate topic.

The reason for differential aging in the twin paradox is that the twins take different "length" routes through space-time. The point (or a point, anyway) of the twin paradox is to show that you must always take relativity of simultaneity into account, and if you don't you will get silly answers.

Let me ask you a question. I drive from A to B in a straight line, and my odometer reads 100 miles. You take a triangular route from A to B via C, and your odometer reads 120 miles. But we started in the same place and finished in the same place. How would you explain the difference in our odometer readings to me?

 

[Stephanus]

Correct. Although I'm not sure how you intend to accelerate a supernova to 0.1c.

Come on..., I was just trying to say that even though an object moves toward us, we'll never know if it HAS MOVED toward us until it worldline "crashes" us. Or I should have said, if it's a probe with laser signal emitted to us, we'll never know that the probe is moving toward us, until the last signal emitted when the probe still at rest reaches us. But if we move toward the supernova/probe, we'll realize at once that that thing is getting closer. Especially a type 1a supernova, as you certainly know it's a standard candle.

The supernova would be blue-shifted, yes. You don't need to say "towards us" in this context.

Yes, yes. Thanks. It greatly improves my logic. Thank you. We could have flown away from Andromeda 100 km/s, still its light is blueshifted.

Let me ask you a question. I drive from A to B in a straight line, and my odometer reads 100 miles. You take a triangular route from A to B via C, and your odometer reads 120 miles. But we started in the same place and finished in the same place. How would you explain the difference in our odometer readings to me?

You're taking the hypotenuse route.
To calculate the angle you take, we should have used $asin(\frac{100}{120})$.
And of course in space time. The hypotenuse should be...
$\sqrt{100^2-\sqrt{120^2-100^2}^2} = \sqrt{20000-14400} = \frac{\sqrt{140}}{20}$

 

[Stephanus]

And even though MOTION IS RELATIVE and length contraction and time dilation is MUTUAL, in Supernova case, if we move toward the supernova, we'll know AT ONCE that the distance between the supernova and us is receding. But if the supernova moves toward us, we'll have to WAIT till its wordline reaches us, so we know that the distance is receding.
I think there are two cases here.
A. We move toward the supernova.
Time dilation and length contraction effect only experienced by us. The supernova doesn't "experience" time dilation and length contraction.
Time dilation here is, if somehow there's clock or signal from the supernova traveling toward us, we'll see that the clock in the supernova is faster and somehow the supernova will looks oblate not sphere, in short we'll see that the supernova is length contracted.
B. The supernova moves toward us.
We'll never know that it has alread moved toward us. And once its worldline reaches us, then we'll see that its clock is faster and its length is contracted. Of course an observer in the neutron star/black hole/supernova ignoring the strong gravity, will see that our clock is mutually faster and our length is contracted.