YT2095 Posted March 1, 2006 Share Posted March 1, 2006 this has been bugging me for ages, this whole twins and one in a rocket at near c thing. if I`m stationary in space (I`ll do it there as opposed to a train and station for reasons that`ll become clear later), and a spaceship with my twin passes me at near c BUT at a constant speed, time will go slower for him than it will for me, but WHY? I`m moving just as fast relative to him in the other direction, so Who is getting "Younger"? Gravity is playing little part as is mass (that would occur on a train and station on a large Earth). who is to say WHO is actualy moving, what dictates the one whos time goes slower:confused: Link to comment Share on other sites More sharing options...
insane_alien Posted March 1, 2006 Share Posted March 1, 2006 Oh great i thought i had the whole relativity thingymujig sussed until you said that. Link to comment Share on other sites More sharing options...
timo Posted March 1, 2006 Share Posted March 1, 2006 if I`m stationary in space (I`ll do it there as opposed to a train and station for reasons that`ll become clear later), and a spaceship with my twin passes me at near c BUT at a constant speed, time will go slower for him than it will for me, but WHY? Because the frame of reference you have chosen is one in which he moves faster than you (zero <-> near c). I`m moving just as fast relative to him in the other direction, so Who is getting "Younger"? Depends on whose frame you use. If you use the frame in which the rocket is at rest then your time passes slower. Strictly speaking, you can only compare aging when you manage to meet your twin a 2nd time because otherwise you are comparing your age to an arbitrary age of your twin. I´ve tried to visualize the problem in the attached image with two lines in R² which have a relative angle of 45°. On the left-hand side, the coordinate system is chosen such that the red line lies on the y-axis. If you go one unit along the red line, the length of the green one will increase by 1.4. So the length of the green line increases faster than that of the red line (that statement doesn´t make much sense which is what I actually want to show with the example). On the right-hand side, there is exactly the same scenario except that this time I´ve chosen the coordinate system such that the green line lies on the y-axis now. One step along the green line will increase the length of the red one by 1.4. This apparent paradox is the one you have with your problem above (with the exceptoin that the metric of spacetime is dy²-dx² instead of dy²+dx² so that the length of the line not on the axis will increase slower). Effectively, what I did in above R²-example was comparing the length from the origin to two points which have nothing in common other than having the same y-coordinate in the currently chosen (arbitrary) coordinate system. who is to say WHO is actualy moving, what dictates the one whos time goes slower:confused: You do. Oh, and "time passed for someone" is the length of his/her/its path through spacetime, in case you didn´t know that and wondered why I´m talking about the length on a line all the time. Link to comment Share on other sites More sharing options...
Connor Posted March 2, 2006 Share Posted March 2, 2006 and the twin paradox, where your twin goes out and comes back younger than you: you'd think there would be problem there because YOU should seem younger than HIM, but it works out because he changes his frame of reference when he turns around... Link to comment Share on other sites More sharing options...
pcs Posted March 2, 2006 Share Posted March 2, 2006 if I`m stationary in space (I`ll do it there as opposed to a train and station for reasons that`ll become clear later), and a spaceship with my twin passes me at near c BUT at a constant speed, time will go slower for him than it will for me, but WHY? I`m moving just as fast relative to him in the other direction, so Who is getting "Younger"? This is why it's important to remember that we transform across inertial frames of reference. Your stationary friend is not the one experiencing an applied force (in this case, the thrust of your rocket). Link to comment Share on other sites More sharing options...
YT2095 Posted March 2, 2006 Author Share Posted March 2, 2006 so if there are only these 2 fellas and one is moving very fast for a while and returns to the other afterwards, OR they both travel off in diferent directions in a loop (like a figure 8 path) at both the same speed there should be no change in each when they eventualy meet up (other than the time taken to travel the loop). it`s the when they meet up again part I find confusing, I can perhaps understand the initial acceleration having something to do with it, but when you get to max speed and maintain it, you`re not really aware that you`re moving anyway (not pinned back in your seat). the velocity remains constant for these examples as I want to rule out as much non relavent factors as possible to boil it down to bare bones basics edit: here`s what started the thought, when that plane took off with an atomic clock and flew around the world and then returned back to base and the clocks were a little out of synch, the one in the plane "aged less" I want to know why it happened That way around, if everythings relative then the plane was stationary and it was the earth that was moving fast! Link to comment Share on other sites More sharing options...
m4rc Posted March 2, 2006 Share Posted March 2, 2006 In order to use special relativity to calculate the time spent in another reference frame, you must stay in the same reference frame. So in the case of a plane that takes off, accelerates and then returns, you can apply special relativity from the earth's frame of reference to calculate the time elapsed by a clock on the plane. You can't start from the frame of reference of the plane and use special relativity to determine the time spent on earth because the plane accelerated. You will need to apply general relativity which is much more complicated. In the case of twins that take off in separate space ships, I believe (although not certain of this) that you can solve for the difference in the age of the two twins if you know their acceleration profile. You will need to assume an observer in a un-accelerated frame of reference and make all calculations from that frame of reference. It should not matter which un-accelerated frame of reference you chose. Link to comment Share on other sites More sharing options...
YT2095 Posted March 2, 2006 Author Share Posted March 2, 2006 so if 2 planes took off around the world but in opposite directions and at the same speed, both their clocks should be the same on landing back at base but the earth based clock would seem to have been running a little slow? Link to comment Share on other sites More sharing options...
m4rc Posted March 2, 2006 Share Posted March 2, 2006 so if 2 planes took off around the world but in opposite directions and at the same speed, both their clocks should be the same on landing back at base I agree with the previous comment but not the following comment: but the earth based clock would seem to have been running a little slow It is the moving clocks that run slow. Link to comment Share on other sites More sharing options...
YT2095 Posted March 2, 2006 Author Share Posted March 2, 2006 ah ok, sorry about that, I knew it was One way or the other but the concensus is that there Would be a difference with the earth clock when compared to each plane, and that each plane would have the same time as the other (if all barring the dirrection of travel was equal), would this sound about right? Link to comment Share on other sites More sharing options...
Severian Posted March 2, 2006 Share Posted March 2, 2006 You might find these threads useful: http://www.scienceforums.net/forums/showthread.php?t=12538 http://www.scienceforums.net/forums/showthread.php?t=8190 http://www.scienceforums.net/forums/showthread.php?t=6986 Link to comment Share on other sites More sharing options...
m4rc Posted March 2, 2006 Share Posted March 2, 2006 Sounds right. Link to comment Share on other sites More sharing options...
YT2095 Posted March 2, 2006 Author Share Posted March 2, 2006 well the problem I have with it, is the whole "out of Phase" thing, if upon return to the same rest frame time doesn`t "Snap back" to normal again, wouildn`t the pilots also be out of phase by a few mili-seconds or what ever it was upon return also, seeing things either before or later after they occur etc... and wouldn`t be able to interact with us properly. this clearly isn`t the case, so is it something perculiar to People that we can and that clocks don`t share? Link to comment Share on other sites More sharing options...
m4rc Posted March 2, 2006 Share Posted March 2, 2006 The returning pilot would be able to interact with others without any problem. He would simply be a few miliseconds younger than expected. Lets look at a situation where the difference is larger. If an astronaught leave earth. If he travels near the speed of light his clock might show that his return trip lasted 1 year while a clock left on earth shows that he has been gone for 10 years. As before it is the moving clock that shows a shorter time interval. The astronaught will have aged one year (the same amount of time indicated by the clock that travelled with him) while everybody on earth will have aged 10 years. Once he is back, he will be able to hold conversations with the people that stayed. All of his friends that he grew up with however will be 10 years older while he will be only 1 year older than when he left. All that has happened is that the astronaught has slowed down his aging while travelling. Link to comment Share on other sites More sharing options...
YT2095 Posted March 2, 2006 Author Share Posted March 2, 2006 so where does that missing 9 years go? it would seem that only a Ceratin part of time "snaps back" like an elastic when both are in the same frame again, so there`s no 9 years realTime phase difference, so where DOES that 9 years go or happen? Link to comment Share on other sites More sharing options...
Severian Posted March 2, 2006 Share Posted March 2, 2006 It is not a case of 'snapping back', it is a case of time not being the same in different reference frames. So after you accelerate/decelerate, the measurement of time you are making is a different quantity to what you were measuring before. It is no different (in principle) to measuring a relative velocity. Driving in your car, you may measure the relative velocity of an overpass as 50mph. If you stop the car, the relative velocity become 0mph. It doesn't snap-back - it is a different measurement. So in your plane example, the pilot's would actually see things on the Earth as running slow, becaue he is comparing their clocks with his clocks. The clocks are measuring different quantities. When he decellerates to the Earth's rest frame, his clock will now be in the same frame as an Earth clock, so it will be measuring the same quantity again. Link to comment Share on other sites More sharing options...
YT2095 Posted March 2, 2006 Author Share Posted March 2, 2006 but the missing time, where does that go? Link to comment Share on other sites More sharing options...
swansont Posted March 2, 2006 Share Posted March 2, 2006 so if 2 planes took off around the world but in opposite directions and at the same speed, both their clocks should be the same on landing back at base but the earth based clock would seem to have been running a little slow? No. The earth is not an inertial frame of reference, since it is rotating, so you cannot use it as a rest frame. What you have described is basically the Hafele-Keating experiment: the reference frame you can use would be one where the observer is at rest, travelling with the earth (assume a straight path, i.e. ignore for the moment the fact that we are orbiting the sun). The plane going east is going faster that the earth, so its clock slows more. The one going west is moving slower, so the dilation is reduced, and its clock speeds up, relative to the one on the earth. This is what was observed. (When you're ready we can discuss the Sagnac effect, which is present in rotating coordinate systems) Link to comment Share on other sites More sharing options...
swansont Posted March 2, 2006 Share Posted March 2, 2006 but the missing time, where does that go? It's not missing, as such. Any person in each frame will have experienced time passing normally. The clocks from the two frames will disagree, but it's not valid to use the other person's clock; from your standpoint the other clock wasn't running correctly. Link to comment Share on other sites More sharing options...
YT2095 Posted March 2, 2006 Author Share Posted March 2, 2006 the other clock wasn't running correctly. why? Link to comment Share on other sites More sharing options...
m4rc Posted March 2, 2006 Share Posted March 2, 2006 When electrons(or anything else) travel at velocities, their interactions between each other and other particles slow down. Chemical reaction using these electrons will also slow down. Any biochemical reaction will slow down as well so while travelling near the speed of light we will move slower, think slower and age slower. However we will not notice unless we observe the outside world aound us because the rate of everything travelling with us is slowed down by the same amount. Even our clock will run slower (both our internal body clock and any mechanical clock we bring with us). When we return to earth, and meet with people who didn't have their internal body clocks slowed down they will have aged more than we did. Link to comment Share on other sites More sharing options...
YT2095 Posted March 2, 2006 Author Share Posted March 2, 2006 ok then, going on that, were different orientations of the clock inside the plane tried? maybe it would speed up if the atomic beam was traveling WITH the direction of the plane instead of against it? don`t get me wrong, I accept that it Does happen, I just want to know Why, and far from trying to pick holes in the idea, I`m actualy trying to fill in gaps! My gaps in knowledge. Link to comment Share on other sites More sharing options...
Severian Posted March 2, 2006 Share Posted March 2, 2006 No. The earth is not an inertial frame of reference' date=' since it is rotating, so you cannot use it as a rest frame. What you have described is basically the Hafele-Keating experiment: the reference frame you can use would be one where the observer is at rest, travelling with the earth (assume a straight path, i.e. ignore for the moment the fact that we are orbiting the sun). The plane going east is going faster that the earth, so its clock slows more. The one going west is moving slower, so the dilation is reduced, and its clock speeds up, relative to the one on the earth. This is what was observed. (When you're ready we can discuss the Sagnac effect, which is present in rotating coordinate systems)[/quote'] That is a very 'practical' response. One could imagine, for the purposes of a thought experiment, that the Earth is not rotating (or only rotating very slowly compared to the planes velocity) and that the 'Earth' is so large that the acceleration (from the circular motion of the planes) is small enough to be neglected. Then your thought experiment can be handled purely with SR. Link to comment Share on other sites More sharing options...
YT2095 Posted March 2, 2006 Author Share Posted March 2, 2006 well actualy, the earth rotation and stuff is what I wanted to factor out in the 1`st place (my not using the Train/Station stuff in the OP). just 2 bodies in space with an accurate watch each. and don`t get me wrong, I did consider the idea that maybe gravity was an issue at 1`st, hence using Earth (massive) and a Plane (tiny) was unfair. Link to comment Share on other sites More sharing options...
Severian Posted March 2, 2006 Share Posted March 2, 2006 why? It was running correctly. It just looks to the (unaware) observer that it was running slow, because it does not match his own clock. Let me put it a little differently. Imagine you want to measure the distance between two points. You take a very long tape measure and stretch it between the points and have your answer. But your friend, whose tape measure is too short, has to measure it in lots of little steps instead of one big one. It is difficult to align his tape measure with the endpoints, so sometimes his tape measure is pointing in a slightly wrong direction and he measures a longer distance than you did. He has measured a distance just as you did, but you measured different distances so got different answers. The clocks are similar to the short tapemeasure. They measure very short timesteps (with their tick-tock), and we want to measure the time between the plane leaving and coming back in the Earth's rest frame.The guy on the ground's clock is automatically 'pointing' in the right 'direction' so he meausures the correct time displacement (in the Earth's rest frame). The guy in the plane's clock is pointing in the wrong direction so its steps are slightly out. He still measures a time displacement, but it is not the same displacement as the guy on the ground. they have measured different things, so we should not be surprised that they get different answers. Link to comment Share on other sites More sharing options...
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