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Author  Topic: The Twins Paradox and the speed of light  
New Cat's Eye Inactive Member 
Got a simple explanation for why this is the case?


Son Goku Member Posts: 1153 From: Ireland Joined: 
The spacetime of special relativity, formally called Minkowski space, has a different way of calculating the distance between points.
In a bit more detail: Then distance in the purely spatial space is calculated by: And distance in Minkowski space is calculated by: The "weird" geometry comes down to that minus sign.


New Cat's Eye Inactive Member 
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.


NosyNed Member Posts: 8868 From: Canada Joined: Member Rating: 7.2 
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 ) :(


lyx2no Member (Idle past 3056 days) Posts: 1277 From: A vast, undifferentiated plane. Joined: 
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.


NosyNed Member Posts: 8868 From: Canada Joined: Member Rating: 7.2 
So I get dx 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.


lyx2no Member (Idle past 3056 days) Posts: 1277 From: A vast, undifferentiated plane. Joined: 
x=âˆš(s^{2} + t^{2}).
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.


NosyNed Member Posts: 8868 From: Canada Joined: Member Rating: 7.2 
That doesn't do anything for me :(.
I understand that much but can't make the leap from there to the twin paradox.


Son Goku Member Posts: 1153 From: Ireland Joined: 
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: 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: Stationary twin: 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 Moving twin: 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 Which is significantly less than the 1,200,000,000 meters of the stationary twin. Hence spacetime distance is reduced by moving through space.


Dr Adequate Member Posts: 16107 Joined: Member Rating: 8.3 
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.


NosyNed Member Posts: 8868 From: Canada Joined: Member Rating: 7.2 
Thank you SG. That is simple and clear. I should have managed it myself. :o


onifre Member (Idle past 1291 days) Posts: 4854 From: Dark Side of the Moon Joined: 
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.


randman Suspended Member (Idle past 3239 days) Posts: 6367 Joined: 
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 spacetime, 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 spacetime. 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.


PaulK Member Posts: 15644 Joined: Member Rating: 2.8 
Maybe I'm being a bit pedantic, but shouldn't it be ds^2 = dt^2 + dx^2 + dy^2 + dz^2 ?


fallacycop Member (Idle past 3860 days) Posts: 692 From: FortalezaCE Brazil Joined: 
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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