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Author Topic:   The Twins Paradox and the speed of light
New Cat's Eye
Inactive Member


Message 16 of 230 (473643)
07-01-2008 2:15 PM
Reply to: Message 11 by cavediver
07-01-2008 12:43 PM


in the topsy turvy geometry of space-time, a wavy path must be SHORTER than the straight line.
Got a simple explanation for why this is the case?

This message is a reply to:
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Replies to this message:
 Message 17 by Son Goku, posted 07-01-2008 4:19 PM New Cat's Eye has replied
 Message 28 by randman, posted 07-02-2008 6:41 PM New Cat's Eye has replied

  
Son Goku
Inactive Member


Message 17 of 230 (473654)
07-01-2008 4:19 PM
Reply to: Message 16 by New Cat's Eye
07-01-2008 2:15 PM


The spacetime of special relativity, formally called Minkowski space, has a different way of calculating the distance between points.
In a bit more detail:
Take a space with 4 space dimensions, which I'll label with x, y, z, w.
Also take Minkowski spacetime with 3 space and 1 time dimension and coordintes x, y, z, t.
If I pick two points in either space, then ds denotes the distance between them. dx, dy, dz, dt or dw denote the difference in the values of that coordinate between the two points.
Then distance in the purely spatial space is calculated by:
ds^2 = dw^2 + dx^2 + dy^2 + dz^2
And distance in Minkowski space is calculated by:
ds^2 = -dt^2 + dx^2 + dy^2 + dz^2
The "weird" geometry comes down to that minus sign.

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 Message 16 by New Cat's Eye, posted 07-01-2008 2:15 PM New Cat's Eye has replied

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New Cat's Eye
Inactive Member


Message 18 of 230 (473663)
07-01-2008 5:01 PM
Reply to: Message 17 by Son Goku
07-01-2008 4:19 PM


I guess I'll have to take your word for it.
But I still don't see how traveling in the spatial deminsions can make the spacetime distance between two points SHORTER.

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NosyNed
Member
Posts: 8996
From: Canada
Joined: 04-04-2003


Message 19 of 230 (473671)
07-01-2008 6:33 PM
Reply to: Message 18 by New Cat's Eye
07-01-2008 5:01 PM


spatial
But I still don't see how traveling in the spatial deminsions can make the spacetime distance between two points SHORTER.
I'm not sure I get the question but in SG's example the "all spatial" universe is not ours.
We move through a 4 dimensional spacetime and distance is calculated as shown by the x, y, z and T example. That is the correct method of calculating distance through our spacetime and produces the odd results. At least that is my understanding (again, waiting for correction).
But I'd love to see an actual calculation because I can't show it in mathematical detail. (and I tried )

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Replies to this message:
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 Message 24 by Son Goku, posted 07-02-2008 6:10 AM NosyNed has replied

  
lyx2no
Member (Idle past 4716 days)
Posts: 1277
From: A vast, undifferentiated plane.
Joined: 02-28-2008


Message 20 of 230 (473678)
07-01-2008 7:50 PM
Reply to: Message 19 by NosyNed
07-01-2008 6:33 PM


Re: spatial
Set y and z to 0 then solve for x.

Kindly
Everyone deserves a neatly dug grave. It is the timing that's in dispute.

This message is a reply to:
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Replies to this message:
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NosyNed
Member
Posts: 8996
From: Canada
Joined: 04-04-2003


Message 21 of 230 (473681)
07-01-2008 8:38 PM
Reply to: Message 20 by lyx2no
07-01-2008 7:50 PM


More help please
Set y and z to 0 then solve for x.
So I get dx2 = ds2 + dt2
Reading that in something like English I get.
The spatial distance through Minkowski spacetime is related to the sum of the total distance PLUS the timelike distance.
Is that right? But that doesn't help me see why the "wavy" trip is shorter.

This message is a reply to:
 Message 20 by lyx2no, posted 07-01-2008 7:50 PM lyx2no has replied

Replies to this message:
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lyx2no
Member (Idle past 4716 days)
Posts: 1277
From: A vast, undifferentiated plane.
Joined: 02-28-2008


Message 22 of 230 (473685)
07-01-2008 9:34 PM
Reply to: Message 21 by NosyNed
07-01-2008 8:38 PM


Re: More help please
x=(s2 + t2).
It's the Pythagorean theorem. x, the spatial separation is the hypotenuse of a right triangle. s, the spacelike spacetime interval; and, t, the time interval make up the adjacents. The hypotenuse is always the longest side.
Edited by lyx2no, : Meant time interval, not timelike spacetime interval. Got carried a way with parallel sentence structure.
Edited by lyx2no, : Typos, confusing ones.

Kindly
Everyone deserves a neatly dug grave. It is the timing that's in dispute.

This message is a reply to:
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NosyNed
Member
Posts: 8996
From: Canada
Joined: 04-04-2003


Message 23 of 230 (473687)
07-01-2008 9:37 PM
Reply to: Message 22 by lyx2no
07-01-2008 9:34 PM


Re: More help please
That doesn't do anything for me .
I understand that much but can't make the leap from there to the twin paradox.

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Son Goku
Inactive Member


Message 24 of 230 (473707)
07-02-2008 6:10 AM
Reply to: Message 19 by NosyNed
07-01-2008 6:33 PM


Calculation
If nobody minds I'll do the calculation for a twin that stays on Earth and one that goes to the moon and back. All calculations are done in the twin who remains on Earth's frame.
I'm also going to use ds^2 = dt^2 - dx^2 - dy^2 - dz^2, which is equivalent to ds^2 = -dt^2 + dx^2 + dy^2 + dz^2, but easier to use in this case.
Basic set up:
First of all I have to get the problem of units out of the way.
Since I'm computing a spacetime distance I'm going to have to use the same units for all quantities. Distances in space are measured using meters. So let's take a meter as our measurement. However humans use seconds for measuring distances in time, in order to compute the spacetime distance I'll have to convert a second into meters.
Well basically a second is (roughly) 300,000,000 meters in the temporal direction.
Now I'll ignore y and z. So I'll be using ds^2 = dt^2 - dx^2.
First of all the starting location of both twins will be labelled as:
(0 ; 0). That is t=0, x=0.
Stationary twin:
If one twin sits where they are for four seconds they end up with coordinates:
(1,200,000,000 ; 0) or t=1,200,000,000 and x=0.
Now I'll compute the spacetime distance. Since dx=0 (no difference or change in spatial coordinate) we just have ds^2 = dt^2.
dt = 1,200,000,000 - 0 = 1,200,000,000
dt^2 = 1,440,000,000,000,000,000
Hence ds^2 = 1,440,000,000,000,000,000 and taking the square root:
ds = 1,200,000,000 meters.
Moving twin:
The moon is roughly 384,000,000 meters from Earth. The second twin starts at Earth and travels to the moon in two seconds.
So they start at (0 ; 0) and end up at (600,000,000 ; 384,000,000).
The spatial difference is dx = 384,000,000 - 0 = 384,000,000
Similarly, dt = 600,000,000.
dx^2 = 147,456,000,000,000,000
dt^2 = 360,000,000,000,000,000
ds^2 = 360,000,000,000,000,000 - 147,456,000,000,000,000 = 212,544,000,000,000,000.
Taking the square root, ds = 461,024,945 meters.
Assuming the twin takes an exactly similar journey back to Earth, that is they return in two seconds, then the distance for the return journey is again ds = 461,024,945 meters.
Hence the total spacetime distance of the moving twin is
ds = 2 x 461,024,945 meters = 922,049,890 meters.
Which is significantly less than the 1,200,000,000 meters of the stationary twin. Hence spacetime distance is reduced by moving through space.

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Replies to this message:
 Message 26 by NosyNed, posted 07-02-2008 10:34 AM Son Goku has not replied
 Message 29 by PaulK, posted 07-02-2008 7:08 PM Son Goku has replied
 Message 36 by Jester4kicks, posted 07-03-2008 9:38 AM Son Goku has not replied

  
Dr Adequate
Member (Idle past 284 days)
Posts: 16113
Joined: 07-20-2006


Message 25 of 230 (473708)
07-02-2008 6:15 AM
Reply to: Message 1 by Jester4kicks
06-30-2008 4:57 PM


So far as I can understand it, the travelling twin is not affected by his velocity, but by his acceleration. One of the twins is accelerated, the other isn't, and so far as I understand it, that's the difference between them.
Now I'd like to make a couple of points.
First, Einstein's ideas have been borne out by meticulous experiments: his ideas may sound crazy, but they're right.
Second, I'm a mathematician. Of the two of us, I am much, much closer to being able to understand the math of general relativity. Nonetheless, I can't be bothered. There is so much more interesting stuff to learn and do that I shall live and die without bothering to learn what it was Einstein was trying to tell us, and on my deathbed I won't be saying "Damn, I should have spent more time learning about General Relativity", I'll be saying "Damn, I should have spent more time having sex".
This is kind of a serious point. The people who study physics know that Einstein was right, and they know why Einstein was right. Unless we're going to try to become physicists, and learn the math, we might as well shrug our shoulders and say "yeah, it's weird, but apparently it's true".
Edited by Dr Adequate, : No reason given.

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NosyNed
Member
Posts: 8996
From: Canada
Joined: 04-04-2003


Message 26 of 230 (473718)
07-02-2008 10:34 AM
Reply to: Message 24 by Son Goku
07-02-2008 6:10 AM


Re: Calculation
Thank you SG. That is simple and clear. I should have managed it myself.

This message is a reply to:
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onifre
Member (Idle past 2950 days)
Posts: 4854
From: Dark Side of the Moon
Joined: 02-20-2008


Message 27 of 230 (473723)
07-02-2008 11:52 AM
Reply to: Message 14 by NosyNed
07-01-2008 1:28 PM


Re: An alternate view??? (cavediver to check)
However, relativity theory has to transform between the different frames of reference and that produces different numerical values for a "tick" as you transform space and time variables. Neither clock (on earth or in the GPS satellites) are changed. But the calculations to compare them to one another (in whichever reference frame you pick) changes the numbers attached.
Thanks for the explanation, it made perfect sense.

All great truths begin as blasphemies
I smoke pot. If this bothers anyone, I suggest you look around at the world in which we live and shut your mouth.

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randman 
Suspended Member (Idle past 4899 days)
Posts: 6367
Joined: 05-26-2005


Message 28 of 230 (473759)
07-02-2008 6:41 PM
Reply to: Message 16 by New Cat's Eye
07-01-2008 2:15 PM


Here is the simple explanation....maybe
Point A in spacetime to Point B in space time is calculated by the distance in time (hours, days, etc,...) plus the distance moved during that time within the 3 dimensions of space.
So the spacetime distance from A to B is calculated by t + l + w + h (time plus length plus width plus height). Let's call A to B....A/B
A/B = t + l + w +h
If l and t and h are large, t is less.
So if you move around a lot, but arrive at Point B in space-time, then you had to have had spent less time doing it. Keep in mind we are not talking about Point B in space, but Point B in space-time.
If you don't move at all, then you only spent time getting there, and so had to have spent more time getting there. The longest way to get there as far as time is sitting still.
At least that's my idea of a simple explanation......hmmmm...maybe not right though.

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PaulK
Member
Posts: 17822
Joined: 01-10-2003
Member Rating: 2.2


Message 29 of 230 (473764)
07-02-2008 7:08 PM
Reply to: Message 24 by Son Goku
07-02-2008 6:10 AM


Re: Calculation
Maybe I'm being a bit pedantic, but shouldn't it be ds^2 = |-dt^2 + dx^2 + dy^2 + dz^2| ?

This message is a reply to:
 Message 24 by Son Goku, posted 07-02-2008 6:10 AM Son Goku has replied

Replies to this message:
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fallacycop
Member (Idle past 5520 days)
Posts: 692
From: Fortaleza-CE Brazil
Joined: 02-18-2006


Message 30 of 230 (473789)
07-02-2008 11:05 PM
Reply to: Message 29 by PaulK
07-02-2008 7:08 PM


Re: Calculation
No. ds^2 can be either positive or negative, depending on which of dt and dx is bigger. But note that wheather ds^2 is defined as
ds^2 = dt^2 - dx^2
or as
ds^2 = -dt^2 + dx^2
is a matter of convention

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