What Causes the Twins Paradox and Length Contraction in Relativity?

[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?

 

[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.