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# The Twins Paradox and the speed of light

Author Topic:   The Twins Paradox and the speed of light
New Cat's Eye
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 Message 16 of 230 (473643) 07-01-2008 2:15 PM Reply to: Message 11 by cavediver07-01-2008 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?

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Son Goku
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 Message 17 of 230 (473654) 07-01-2008 4:19 PM Reply to: Message 16 by New Cat's Eye07-01-2008 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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New Cat's Eye
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 Message 18 of 230 (473663) 07-01-2008 5:01 PM Reply to: Message 17 by Son Goku07-01-2008 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
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 Message 19 of 230 (473671) 07-01-2008 6:33 PM Reply to: Message 18 by New Cat's Eye07-01-2008 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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lyx2no
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 Message 20 of 230 (473678) 07-01-2008 7:50 PM Reply to: Message 19 by NosyNed07-01-2008 6:33 PM

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

Kindly

âˆžâˆžâˆžâˆžâˆžâˆžâˆž

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

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NosyNed
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 Message 21 of 230 (473681) 07-01-2008 8:38 PM Reply to: Message 20 by lyx2no07-01-2008 7:50 PM

 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.

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lyx2no
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 Message 22 of 230 (473685) 07-01-2008 9:34 PM Reply to: Message 21 by NosyNed07-01-2008 8:38 PM

x=âˆš(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

âˆžâˆžâˆžâˆžâˆžâˆžâˆž

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

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NosyNed
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 Message 23 of 230 (473687) 07-01-2008 9:37 PM Reply to: Message 22 by lyx2no07-01-2008 9:34 PM

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
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 Message 24 of 230 (473707) 07-02-2008 6:10 AM Reply to: Message 19 by NosyNed07-01-2008 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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 Message 25 of 230 (473708) 07-02-2008 6:15 AM Reply to: Message 1 by Jester4kicks06-30-2008 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
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 Message 26 of 230 (473718) 07-02-2008 10:34 AM Reply to: Message 24 by Son Goku07-02-2008 6:10 AM

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

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onifre
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 Message 27 of 230 (473723) 07-02-2008 11:52 AM Reply to: Message 14 by NosyNed07-01-2008 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
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 Message 28 of 230 (473759) 07-02-2008 6:41 PM Reply to: Message 16 by New Cat's Eye07-01-2008 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
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 Message 29 of 230 (473764) 07-02-2008 7:08 PM Reply to: Message 24 by Son Goku07-02-2008 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| ?

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fallacycop
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 Message 30 of 230 (473789) 07-02-2008 11:05 PM Reply to: Message 29 by PaulK07-02-2008 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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