Exploring Twins Paradox & Length Contraction

[Ibix]

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.

No - we'll know when the event's light cone intersects our worldline. In other words, when the light from the supernova reaches us.

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...
$\sqrt{100^2-\sqrt{120^2-100^2}^2} = \sqrt{20000-14400} = \frac{\sqrt{140}}{20}$

But if you're happy with "the route is shorter because I took the hypotenuse", why aren't you happy with "one twin's time is longer because he took the direct route"?

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.

But an observer near the supernova (if he can accelerate it to 0.1c, he can avoid being destroyed by it), would know it had started moving immediately, while he wouldn't know we had started moving for 1000 years. The situation is completely symmetrical.

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.

No one ever experiences time dilation or length contraction. It's always something that one measures in someone else. So, to you, the supernova's clocks will be ticking slowly and it will appear length contracted (in the direction pointing towards us, so careful measurement will be needed). To an observer on the supernova, they are at rest and you are in motion, so it is your clocks that are running slow and you who will be length contracted in the direction of motion.

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.

No one ever experiences time dilation or length contraction. It's always something that one measures in someone else. So, to you, the supernova's clocks will be ticking slowly and it will appear length contracted (in the direction pointing towards us, so careful measurement will be needed). To an observer on the supernova, they are at rest and you are in motion, so it is your clocks that are running slow and you who will be length contracted in the direction of motion.

There's a reason those two paragraphs are identical: your two scenarios are identical except for who accelerates. You can tell if it was you who accelerated because you'll have felt the force. If you were somehow incapacitated during the acceleration phase, you have no way to know if it was you who accelerated just now, or the supernova that accelerated 1000 years ago.

None of the above is relevant to the twin paradox.

The space-time diagrams you've been drawing with Mentz are effectively maps of space-time, showing where and when things happen and tracing out how they move. The Lorentz transforms relate the map drawn by one person to the map drawn by another in relative motion. All the stuff about length contraction and time dilation and the relativity of simultaneity and differential aging is there.

All the stuff about light travel time is an additional layer of complexity on top of that. I strongly advise you not to worry about it until you've got the maps sorted out. Make sure you understand the terrain on Google Maps before you switch to Street View, or you'll just get lost.

Yes, we cannot know the supernova has started moving until it's light reaches us. That does not stop us from, after the fact, drawing a map of what must have happened. Here is a pair of space-time diagrams showing us in blue and the supernova in red. I have added dots every 500 years on each worldline.

In the left-hand diagram, the supernova is in motion and we are at rest. Note that the red dots are slightly wider spaced than the blue ones because of time dilation - there are 20 red dots and 21 blue dots. In the right-hand diagram, we are in motion and the supernova is at rest. In this frame our clocks are time-dilated, but we started them early - so there are still 20 red dots and 21 blue dots.

I haven't drawn on any light pulses because they aren't necessary to understand what's going on with the clocks. I repeat that I would advise not worrying about them - I think you are confusing yourself by trying to sort out the details of what we would see without having a grasp of the big picture.

 

[Stephanus]

Thanks Ibix for your answer.
Mentz114 has really helped me understanding Space time diagram, at least I know about that.

No - we'll know when the event's light cone intersects our worldline. In other words, when the light from the supernova reaches us.

Okay, okay I understand now about what light cone is, and what worldline is. Thanks for introducing me light cone and worldline. I thought they are the same. But I realize now, that they aren't. I'll contemplate your answer.

But if you're happy with "the route is shorter because I took the hypotenuse", why aren't you happy with "one twin's time is longer because he took the direct route"?

"Happy" is not the word . It's just that I need a time longer to really grasp the theory.

Thanks.

 

[Stephanus]

Yes, we cannot know the supernova has started moving until it's light reaches us. That does not stop us from, after the fact, drawing a map of what must have happened. Here is a pair of space-time diagrams showing us in blue and the supernova in red. I have added dots every 500 years on each worldline.
View attachment 84886
In the left-hand diagram, the supernova is in motion and we are at rest. Note that the red dots are slightly wider spaced than the blue ones because of time dilation - there are 20 red dots and 21 blue dots. In the right-hand diagram, we are in motion and the supernova is at rest. In this frame our clocks are time-dilated, but we started them early - so there are still 20 red dots and 21 blue dots.

I haven't drawn on any light pulses because they aren't necessary to understand what's going on with the clocks. I repeat that I would advise not worrying about them - I think you are confusing yourself by trying to sort out the details of what we would see without having a grasp of the big picture.

Thanks for your trouble for answering me. It gives me enlightment. I'll contemplate your answer.
[EDIT] and graph

 

[Mentz114]

"Happy" is not the word . It's just that I need a time longer to really grasp the theory.

I made graph of the two objects approaching each other on the x-axis with images of their clocks as they progress. The dots on the worldlines are clock ticks.

1. Note that all the spatial movement is in the x-direction ( or -x direction )
2. The worldlines show the progress through spacetime.
3. The time and place ( t,x) where they meet is an example of an event.

 

[Stephanus]

I made graph of the two objects approaching each other on the x-axis with images of their clocks as they progress. The dots on the worldlines are clock ticks.

1. Note that all the spatial movement is in the x-direction ( or -x direction )
2. The worldlines show the progress through spacetime.
3. The time and place ( t,x) where they meet is an example of an event.

Let me try to solve the problem.

At rest frame:
A speed is 11/23
C speed is 15/23
A meet C at time 23

Is this the space time diagram if Blue/A is rest? The Black line is the moving rest line in previous graph.
Green is green line from previous graph.
Blue is at (0,17) because $\sqrt{23^2 - 11^2} = \sqrt{408} = 16.94$

 

[Mentz114]

Let me try to solve the problem.
..
..
At rest frame:
A speed is 11/23
C speed is 15/23
A meet C at time 23
..
..
Is this the space time diagram if Blue/A is rest? The Black line is the moving rest line in previous graph.
Green is green line from previous graph.
Blue is at (0,17) because $\sqrt{23^2 - 11^2} = \sqrt{408} = 16.94$

The 'shape' is right. Blue stays at x=0, green approaches and the line that was at 0,0 moves to the right.

The proper time on Blues clock is about 20.2 (counting the dots) but on your diagram the time on Blues clock is about 17. That is not right. They should be the same.

If you want to transform the diagram to Blues rest frame, you should transform the points A,B,C with a Lorentz transformation with velocity -v where v is Blues velocity.

I transformed B ( t=23, x=11) with $\beta=-0.478, \gamma=1.138$

$x'=(11)*1.138-(0.478)*(1.138)*23=0.005$
$t'=(23)*1.138-(0.478)*(1.138)*11=20.2$

There are rounding errors but you can see that $x'$ goes to zero.

I'm encouraged that you seem to be understanding worldlines. If you complete the transformation for B and C then on the new diagram you will see

1. Invariance of proper times
2. Relativity of simultaneity.

It is worth doing.

 

[Stephanus]

The 'shape' is right. Blue stays at x=0, green approaches and the line that was at 0,0 moves to the right.

The proper time on Blues clock is about 20.2 (counting the dots) but on your diagram the time on Blues clock is about 17. That is not right. They should be the same.

If you want to transform the diagram to Blues rest frame, you should transform the points A,B,C with a Lorentz transformation with velocity -v where v is Blues velocity.

I transformed B ( t=23, x=11) with $\beta=-0.478, \gamma=1.138$

$x'=(11)*1.138-(0.478)*(1.138)*23=0.005$
$t'=(23)*1.138-(0.478)*(1.138)*11=20.2$

There are rounding errors but you can see that $x'$ goes to zero.

I'm encouraged that you seem to be understanding worldlines. If you complete the transformation for B and C then on the new diagram you will see

1. Invariance of proper times
2. Relativity of simultaneity.

It is worth doing.

Hi Mentz114, thanks for your answer. I'm using someone computer now. I don't have my "spreadsheet formula" and graph with me. I'll try to make the correction later. I'm out of town now.

 

[Stephanus]

Blue is at (0,17) because $\sqrt{23^2 - 11^2} = \sqrt{408} = 16.94$

What? $\sqrt{408}=16.94$? Not even 20?? I must have dozed off. Argghhh.

 

[Stephanus]

$x'=(11)*1.138-(0.478)*(1.138)*23=0.005$
$t'=(23)*1.138-(0.478)*(1.138)*11=20.2$

There are rounding errors but you can see that $x'$ goes to zero.

Yes, yes. You don't have to convince me. Rounding errors. But as I said, when I get back to my own computer, I'll do the calculation again. I think the green line is not right. My green line.
$\sqrt{408} = 16.94$ Arghh

 

[Mentz114]

Yes, yes. You don't have to convince me. Rounding errors. But as I said, when I get back to my own computer, I'll do the calculation again. I think the green line is not right. My green line.
$\sqrt{408} = 16.94$ Arghh

'Arghh' indeed.

These diagrams are fairly accurate but not the same numbers. I won't be available for a while now.