|
Register | Sign In |
|
QuickSearch
EvC Forum active members: 64 (9164 total) |
| |
ChatGPT | |
Total: 916,767 Year: 4,024/9,624 Month: 895/974 Week: 222/286 Day: 29/109 Hour: 2/3 |
Thread ▼ Details |
|
Thread Info
|
|
|
Author | Topic: The Twins Paradox and the speed of light | |||||||||||||||||||||||
cavediver Member (Idle past 3669 days) Posts: 4129 From: UK Joined: |
were it possible to observe two clocks separated by half a universe would they keep the same time? Not bloody likely The problem that both you and Iblis are having is that you are not being specific enough, and therefore it is equally easy to agree or disagree with anything you both say. If one of your clocks is next to me, and the other is "half a universe" away then, definitely, there will be a difference. But if I observe them from a point equidistant between the two clocks, then they will appear to run at the same rate. Because they are at different points in space, you cannot make any direct comparisons. Absolute (non-observer dependent) comparisons can only be made at a point. Just because the two clocks' time directions are different, does not mean that the two clocks are running at different rates, any more than it means that they are running at the same rate. The terminology used is far too loose to be able to make sensible, well-defined statements about what is actually happening. Edited by cavediver, : No reason given.
|
|||||||||||||||||||||||
cavediver Member (Idle past 3669 days) Posts: 4129 From: UK Joined: |
Would we have worlds in our own galaxy which would experience time very much faster than we ourselves experience? Do we ever get scenarios along the line of a day is equal to a year or more? Only by way of gravitational-well time-dilation, and that would take some g to generate the day/year ratio.
|
|||||||||||||||||||||||
tis---strange Junior Member (Idle past 5270 days) Posts: 14 From: Oslo, Norway Joined: |
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
|
|||||||||||||||||||||||
lyx2no Member (Idle past 4742 days) Posts: 1277 From: A vast, undifferentiated plane. Joined: |
The birdbrain bows to the egghead.
It's not the man that knows the most that has the most to say. Anon
|
|||||||||||||||||||||||
cavediver Member (Idle past 3669 days) Posts: 4129 From: UK Joined: |
I hope I have not offended you ;-) No, not at all I'm sorry I speak harshly but it is for a purpose: the internet is full of incorrect information regarding relativity, and I want to ensure that any readers will not be confused or persuaded that your view is correct.
all I wrote can be found in the standard textbook 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? So, you cannot possibly change the metric of space-time by simply observing it from an accelerating point of view!! The Universe likes you but it doesn't care about you that much Your coordinate system may look weird from your point of view, but the space-time of Special Relativity starts off flat, and stays flat. The Einstein Field Equations are trivially satisfied (guv=diag[1,-1,-1,-1], guv,w=[0], Ruvwx=0, Ruv=0, R=0) and a change of coordinates to an accelerating frame is not going to change this.
|
|||||||||||||||||||||||
tis---strange Junior Member (Idle past 5270 days) Posts: 14 From: Oslo, Norway Joined: |
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
|
|||||||||||||||||||||||
Trae Member (Idle past 4332 days) Posts: 442 From: Fremont, CA, USA Joined: |
Thanks Cavediver,
If we cannot get a large 'to us' time shift within our own galaxy under the conditions I set forth, what about between planets each in galaxies somewhat far apart? Would say galaxies 100,000 light years be sufficiently far apart from each other to be accelerating fast enough away from each other to be effectively rocket ships. What amount of time dilation might occur then (keeping the idea that each planet is 1-G).
|
|||||||||||||||||||||||
cavediver Member (Idle past 3669 days) Posts: 4129 From: UK Joined: |
I know that he uses GR in his discription of the problem in the book. There's nothing wrong with using GR to solve the problem - and this is by using a global homogeneous gravitational field as an equivalence to the acceleration - it's claiming that you can't use SR that I object to. We can simply calculate the proper length of two different length paths from Event A to Event B, and voila, we have the twins paradox. ABE: he may be objecting to using SR in a real world example, which I can appreciate. As ever, you have to be very specific in relativity. Edited by cavediver, : No reason given.
|
|||||||||||||||||||||||
lyx2no Member (Idle past 4742 days) Posts: 1277 From: A vast, undifferentiated plane. Joined: |
Would say galaxies 100,000 light years be sufficiently far apart from each other You'd want to multiply that by a thousand for it to even start getting interesting, yet alone relativistic. With Hubble's constant being 71 km/sec/Mps, these galaxies are experiencing only 2.2 km/sec of expansion. 20% of Earth's escape velocity. AbE: ‘yvind Grn is, himself, a twin so may have a special interest in this problem. Edited by lyx2no, : AbE. It's not the man that knows the most that has the most to say. Anon
|
|||||||||||||||||||||||
Iblis Member (Idle past 3921 days) Posts: 663 Joined: |
the two clocks' time directions are different What do you mean when you say that? I'm not begging the question, I'm trying to understand what statements like that one are describing. Maybe I need another picture? Don't get me wrong, T+S=C, Pythagoras is right even when one of your legs is drawn in time, that's a wonderful picture! But I persist in thinking of time as one-dimensional, and therefore having only two directions, forward and backward. Space is 3-dimensional, and therefore it's obvious that which dimensions we pick are totally arbitrary, as long as they are at right angles to one another. If spacetime is 4-dimensional, then it seems to follow that all 4 dimensions would be arbitrary? So just as I can swap length from North to West just by rotating, I ought to able to swap South for Past somehow. No? But in that case, what would a different "direction" in time actually mean? A different speed? a different route? Feel free to whack me about the ears, I'm trying but I'm pretty sure I'm still missing important stuff.
|
|||||||||||||||||||||||
tis---strange Junior Member (Idle past 5270 days) Posts: 14 From: Oslo, Norway Joined: |
I will tell you when they resolve the problem by flying themselves
|
|||||||||||||||||||||||
tis---strange Junior Member (Idle past 5270 days) Posts: 14 From: Oslo, Norway Joined: |
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
(my emphasis)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.(...) 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
|
|||||||||||||||||||||||
cavediver Member (Idle past 3669 days) Posts: 4129 From: UK Joined: |
quote: No, it doesn't. There is no one frame of reference in which Eva remains at rest.
quote: And this is where this goes wrong. There's nothing wrong with using GR, but SR is just as good, and there are even papers demonstrating the equivalence of the two.
I just would like to know how you calculate the age of the homestaying twin as seen by the travelling twin only using SR. 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.
|
|||||||||||||||||||||||
tis---strange Junior Member (Idle past 5270 days) Posts: 14 From: Oslo, Norway Joined: |
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.
|
|||||||||||||||||||||||
lyx2no Member (Idle past 4742 days) Posts: 1277 From: A vast, undifferentiated plane. Joined: |
Well, I think it follows from the wrong assumption that you can ignore the acceleration in the problem One can ignore the acceleration in the problem in the same way one can ignore the acceleration in the problem: I travel from Boston to Denver, a trip of 1770 miles, on a train at 85 miles per hour. How long did it take? It's not a assumption; it's a simplification. It's not the man that knows the most that has the most to say. Anon
|
|
|
Do Nothing Button
Copyright 2001-2023 by EvC Forum, All Rights Reserved
Version 4.2
Innovative software from Qwixotic © 2024