Register | Sign In


Understanding through Discussion


EvC Forum active members: 65 (9162 total)
5 online now:
Newest Member: popoi
Post Volume: Total: 915,817 Year: 3,074/9,624 Month: 919/1,588 Week: 102/223 Day: 13/17 Hour: 1/1


Thread  Details

Email This Thread
Newer Topic | Older Topic
  
Author Topic:   The Twins Paradox and the speed of light
tis---strange
Junior Member (Idle past 5244 days)
Posts: 14
From: Oslo, Norway
Joined: 11-11-2009


Message 103 of 230 (534801)
11-11-2009 5:56 AM


Why it is called the Twin PARADOX
Hi,
The thing is, it is not possible to explain the twin paradox using only special relativity. In special relativity (which is valid in inertial systems only, or in english, non-accelerated systems only) the space-time intervall is constant (as stated before). The whole point of special relativity is that we cannot find out who is moving and who is not, meaning that we could look at the problem from the position of the traveling twin(clock) and find out that the other twin (on earth) is younger, thous we can not use SR to find our solution without ending in a paradoxal state.
However, the one twin is in a spaceship which is accelerated to get too the speed it needs, and decelerated to turn when it gets to the moon. As we all know, when we are low in a gravitational field, the clocks run slower than clocks higher in a grav-field.
(a simple example: the light coming in from the roof of your room has a slightly lower frequency than the light on the floor. We do not want the light to gain energy on the way to the floor, this would be in conflict with energy conservation. We can however solve this dilemma by using that the time goes slower on your floor, therfore the frequencys are the same.)
In the system of the traveler, he experiences a strong gravitational field/acceleration (the main assumption of GR is that these are physical equal) when he turns while arriving on the moon. This field has upwards direction towards earth, in the travelers system, earth is very high in the gravitational field the traveler experiences, therfore time goes slower for him then for earth. The two other accelerations (landing and starting) are negelicable, because the two twin are virtually in the same place (meaning exactly as high in the gravitational field)
THAT is the reason that the earthbound twin is older than the flying twin.

"Hey, hvor hen du er i verden...
Det er deilig slve i skyggen!"
-Dumdum Boys

Replies to this message:
 Message 105 by cavediver, posted 11-11-2009 7:29 AM tis---strange has replied

  
tis---strange
Junior Member (Idle past 5244 days)
Posts: 14
From: Oslo, Norway
Joined: 11-11-2009


Message 108 of 230 (534848)
11-11-2009 11:02 AM
Reply to: Message 105 by cavediver
11-11-2009 7:29 AM


Re: Why it is called the Twin PARADOX
Hi there :-)
I might very good be completely wrong. Just to be clear: I do not, as a practice, get my information from the internet , all I wrote can be found in the standard textbook on tensor mathematics and general relativity i am studying at the time being, but I have shure as hell not a complete grasp of the theory...
The thing is (as far as I have understood): using Minowski metric to calculate the time diltation is no problem , as long as the frame of reference you are calculating from is an inertial system. If it isn't, you must use another metric calculating it (governed by the einstein field equations).
But to find out what happens in the reference frame of the twin traveling on the turning point, you would have to use a different metric (depending on the exact form of acceleration). You have to rotate the minovski metric, such that you can view earth as coming towards you after the turn. How would you calculate the turn using special relativity? (as information: I have seen calculations using a rotated minovski diagram and only special relativity a couple of years ago in my introduction to astrophysics class, but my professor in the GR subject is quite clear about the subject and is saying that this is an incomplete explanation by the reasons given above...)
I hope I have not offended you ;-)
Tis--- the little german
Edited by tis---strange, : small language fix...
Edited by tis---strange, : even more text-retardations

This message is a reply to:
 Message 105 by cavediver, posted 11-11-2009 7:29 AM cavediver has replied

Replies to this message:
 Message 110 by cavediver, posted 11-11-2009 3:59 PM tis---strange has replied

  
tis---strange
Junior Member (Idle past 5244 days)
Posts: 14
From: Oslo, Norway
Joined: 11-11-2009


Message 111 of 230 (534897)
11-11-2009 4:20 PM
Reply to: Message 110 by cavediver
11-11-2009 3:59 PM


Re: Why it is called the Twin PARADOX
Hmmm, are you sure? Which textbook is it? It is an old mistake to think that accelerating frames require GR, and some old textbooks may have this error. If it is a new book, then I may write to the author If your GR professsor says this, then there is a problem. What is his name? Is he a professor of relativity, or some other area of physics such as particle physics, astrophysics, etc?
His name is ‘yvind Grn and he is the man that has written the book we use. But as I said: I am a beginner to this subject, and I might have misunderstood him.
I know however that he stopped me rather harshly talking about the SR way of resolving the problem, saying I needed GR to properly resolve the twin paradox (in a popular sience lecture he gave about a year ago) and I know that he uses GR in his discription of the problem in the book.
I will read a bit of that and ask him again.
You have of course a point about the metric... I think I blended principles here. I will make my mind up about that, and propably accept what you say. And then I will write you again .
I am sorry, I almost forgot: He is a professor in Relativity as far as I know.
Edited by tis---strange, : not all questions answered

This message is a reply to:
 Message 110 by cavediver, posted 11-11-2009 3:59 PM cavediver has replied

Replies to this message:
 Message 113 by cavediver, posted 11-11-2009 5:40 PM tis---strange has replied

  
tis---strange
Junior Member (Idle past 5244 days)
Posts: 14
From: Oslo, Norway
Joined: 11-11-2009


Message 116 of 230 (534958)
11-12-2009 2:56 AM
Reply to: Message 114 by lyx2no
11-11-2009 6:00 PM


Re: Back to the basics
I will tell you when they resolve the problem by flying themselves

This message is a reply to:
 Message 114 by lyx2no, posted 11-11-2009 6:00 PM lyx2no has not replied

  
tis---strange
Junior Member (Idle past 5244 days)
Posts: 14
From: Oslo, Norway
Joined: 11-11-2009


Message 117 of 230 (534959)
11-12-2009 3:22 AM
Reply to: Message 113 by cavediver
11-11-2009 5:40 PM


Re: Why it is called the Twin PARADOX
Ok, just to give you the reason for my confusion:
from Einstein's General Theory of Relativity by ‘yvind Grn and Sigbjrn Hervik, ISBN 978-0-387-69199-2
(I have cut some paragraphs where I write (...), and I had of course to cut the illustrations)
page 34-35
2.9 The twin paradox
Rather then discussing the life-time of elementary particles, we may as well apply Eq.(2.46) to a person. (...) Assume that Eva is rapidly accelerating from rest at the point of time t=0 at origin to a velocity v along the x-axis of a (ct,x) coordinate system in an inertial reference frame S. (...) At a point in time tp she has come to a position xp. She then rapidly deceletates until reaching a velocity v in the negative x-direction. At a point of time TQ, as measured on clocks at rest in S, she has returned to her starting location. If we neglect the brief periods of aceleration, Eva's travelingntime as measured on a clock which she carries with her is:
teva = (1-v^2/c^2)^(1/2)tQ) (2.49)
Now assume that Eva has a twin-sister named Elizabeth who remains at rest at the origin of S.
Elizabeth has become older by tau-elizabeth = tQ during Eva's travel, so that:
teva = (1-v^2/c^2)^(1/2)tau-elizabeth (2.50)
For example, if Eva travelled to Proxima Centauri(the Sun's nearest neighbour at four light years) with a velocity v=0.8c, she would be gone for ten years as measured by Elizabeth. Therefore Elizabeth has aged 10 years during Eva's travel. According to Eq.(2.50), Eva has only aged 6 years. According to Elizabeth, Eva has aged less then herself during her travels.
The principle of relativity, however, tells that Eva can consider herself at rest and Elizabeth as the traveller. According to Eva, it is Elizabeth who has only aged by 6 years, while Eva has aged by 10 years during the time they are apart.
(...)In order to arrive at a clear answer (...) we shall have to use a result from the general theory of relativity.(...)
(my emphasis)
I suspect you will tell me that things go wrong when we assume that the acceleration effect is negliable (I know that is where things go wrong).
In a later chapter he does the same calculation and arrives at the same result for both twins perspective by using a global homogenous gravitational field, as you say, for calculating the age of the homestaying twin in the system of the travelling twin.
It might have sounded like this, but I have no problem calculating tne difference with SR (of course not, if you calculate from the perspective of an inertial system), I just would like to know how you calculate the age of the homestaying twin as seen by the travelling twin only using SR.
With my best regards
tis--- the german in norway

This message is a reply to:
 Message 113 by cavediver, posted 11-11-2009 5:40 PM cavediver has replied

Replies to this message:
 Message 118 by cavediver, posted 11-12-2009 3:32 AM tis---strange has replied

  
tis---strange
Junior Member (Idle past 5244 days)
Posts: 14
From: Oslo, Norway
Joined: 11-11-2009


Message 119 of 230 (535059)
11-12-2009 5:16 PM
Reply to: Message 118 by cavediver
11-12-2009 3:32 AM


Re: Why it is called the Twin PARADOX
No, it doesn't. There is no one frame of reference in which Eva remains at rest.
Well, I think it follows from the wrong assumption that you can ignore the acceleration in the problem (you can not find one reference frame in which Eva is at rest, true, but you can find two, one for the journey up to the turning point, and one back). But that is not the point: I see that this explanation is wrong, as I did before, I just didn't know any way of explaining it without GR before. Thank you for clearing that up for me
Actually, the very next part of the chapter is about hyperbolic motion, so I would probably have seen it had I read on after the example. Something just shorted in my brain. But I don't mind being wrong, how else would I learn
What do you mean, as "seen" by the twin? Do you mean his observations of the home bound twin as he travels? This can be done by basic hyperbolic geometry - just trace the null rays from the home path to the travelling path. In terms of the actual age experienced - this is just the proper time along the respective paths.
Yes, that was exactly what I mean. But after reading the part about hyperbolic motion I understand much better. I only do not see why my professor was so agressivly against using SR to discribe the system, but I will ask him that myself.

This message is a reply to:
 Message 118 by cavediver, posted 11-12-2009 3:32 AM cavediver has not replied

Replies to this message:
 Message 120 by lyx2no, posted 11-12-2009 5:55 PM tis---strange has replied

  
tis---strange
Junior Member (Idle past 5244 days)
Posts: 14
From: Oslo, Norway
Joined: 11-11-2009


Message 121 of 230 (535129)
11-13-2009 2:35 AM
Reply to: Message 120 by lyx2no
11-12-2009 5:55 PM


Re: Why it is called the Twin PARADOX
But the problem does not make any physical sense if you make this simplification. If you just assume that you can solve the problem for the time dilitation on the one way and add the time dilitation on the other way (as the book does), the two twins come to different conclusions about the age of the other twin. Therefore, we can not ignore the acceleration.

This message is a reply to:
 Message 120 by lyx2no, posted 11-12-2009 5:55 PM lyx2no has replied

Replies to this message:
 Message 122 by lyx2no, posted 11-13-2009 8:23 PM tis---strange has replied

  
tis---strange
Junior Member (Idle past 5244 days)
Posts: 14
From: Oslo, Norway
Joined: 11-11-2009


Message 126 of 230 (535281)
11-14-2009 11:08 AM
Reply to: Message 122 by lyx2no
11-13-2009 8:23 PM


Re: School Me
I am certanly not qualified to school you, but what I have of recollections of the two ways of computing the time age of the twin on earth from the perspective of the twin on the spaceship, most of the aging of the twin on earth happens while the second twin is accelerating.
We can try calculating that using GR and assuming an infinately short(in the referance frame of the spaceship) acceleration with an infinite acceleration g (as cavediver said, it is not wrong to do it that way, and I am not used to using hyperbolic motion... so therefore I use GR ;-)
We will split the problem in three parts. Part one is moving from earth in direction proxima centauri (4.1ly away at rest distance) and part three is moving back with a constant speed 0.8c.
Part two is the acceleration part, I will come to that.
We consider sitting in the spaceship, thus "being at rest". At start, earth (and proxima centauri) is moving with a speed 0.8c. Since the bodys are moving, we observe a lorentz contracted distance between the two bodies s=s0sqrt(1-v^2/c^2) = 2.4ly. When proxima centauri arrives at our ship we will have aged 2.4ly/0.8c = 3 years. Earth however has moved, so we calculate that the time elapsed on earth is given by 3years * sqrt(1-v^2/c^2) = 1.8 years.
Then we experience a gravitational field (we accelerate) with a constant gravitational-acceleration g. Earth is falling freely in this field. At this point we observe earth at a distance x1=2.4ly. We will then observe earth slowing down until it comes to rest at a distance x2=4.1ly. Earth will then accelerate back against us until it reaches v=0.8c again at the distance x1=2.4ly.
We will now need to use the proper time intervall deltatau (the time experienced in the free falling frame/earth while the acceleration lasts). It is not trivial to find the expression for this (the way I was taught to do this, you need to know a bit about lagrange functions), and I hope you can believe me when I say that in this case it is given as:
delta tau = c/g *sqrt(((1+gx2)/c^2)^2 - ((1+gx1)/c^2)^2)
Now, we want the acceleration to last a very short amount of time (this makes calculating the age of the traveler from the earth frame easyer), meaning that we want g to approach infinity giving (by using L'hopitales rule):
delta tau = 1/c sqrt(x2^2 - x1^2)
for x1 = 2.4ly and x2 = 4.1ly, this gives:
delta tau = 6.4 years
on the way back, we calculate the time elapsed on our planet in the same way as in part one, giving:
Time elapsed for us: 3 years + 3 years + nearly nothing = 6 years
Time elapsed for earth: 1.8 years + 1.8 years + 6.4 years = 10 years
Which is the same result as expected by calculating the age of the traveler from earths frame.
So if we eliminate the acceleration by making it very short and then saying it doesn't have any influence, we are 6.4 years off in the end result.
I realize that this is not exactly what you had in mind, you wantet to integrate over an infinite amount of frames moving with constant velocity to get the result. I think the result would be the same (if you take the amount of time you accelerate to be very short)...

This message is a reply to:
 Message 122 by lyx2no, posted 11-13-2009 8:23 PM lyx2no has replied

Replies to this message:
 Message 128 by lyx2no, posted 11-17-2009 7:59 PM tis---strange has replied

  
tis---strange
Junior Member (Idle past 5244 days)
Posts: 14
From: Oslo, Norway
Joined: 11-11-2009


Message 129 of 230 (535794)
11-18-2009 1:36 AM
Reply to: Message 128 by lyx2no
11-17-2009 7:59 PM


Re: School Me More
I don't know why we consider a gravitational field at all. Are you just saying that an acceleration is akin to a gravitational field? Isn't "g" Earth's surface gravity? We'd not feel that from 2.4 ly.
Yes! In GR we can associate an acceleration with a gravitational field. In fact: We consider ourselves standing still in this gravitational field (kind of like standing on the surface of earth) while earth is falling freely in this field. It seems strange, but it is physical identical with what you experience in the space ship: You feel a "force" (this picture on wikipedia makes a point of this) and earth is accelerating towards you at the right rate. I am shure cavediver has a better explanation up his sleeve ;-)
g is only a variable I defined. Call it for a if you think that is easier. But I did not consider Earth's surface gravity at all (we neglect the suns gravitational field and proxima centauris grav field, earth grav field is much less), and remember g goes to infinity (to simulate an infinitly short acceleration).
Of course I can believe ya', buddy What's a lagrange function? (I have been reading up I don't expect you to do my work but my maths aren't there yet.)
A lagrange function describes a system in classical mechanics (in this case we had to generalize it to apply to GR, but the principal is the same).
In the case of uniform acceleration it is given by:
L = -1/2(1+ gx/c^2)^2c^2dt/dtau + 1/2dx/dtau
Where tau is the proper time/time expirienced in the accelerated system.
The Lagrange function is a way to discribe a physical system using conservation of energy and conservation of momentum. The point is: When you have the Lagrange function, you can use the Euler-Lagrange functions of motion to discribe the system.

This message is a reply to:
 Message 128 by lyx2no, posted 11-17-2009 7:59 PM lyx2no has not replied

  
Newer Topic | Older Topic
Jump to:


Copyright 2001-2023 by EvC Forum, All Rights Reserved

™ Version 4.2
Innovative software from Qwixotic © 2024