I can't fully understand why a person who makes a journey into space in a high-speed rocket will return home to find his age less than an identical twin who stayed on Earth. It makes since for the twin who stayed on earth, but for the twin who travelled into space, he sees himself at rest and sees earth moving at constant velocity. so he should feel that time on earth is moving slower than his time, this way he will find his twin younger than him not older. Please tell me what I am missing to help me understand this thought experiment. Thank you.
The trick is that one twin (the one in the rocket ship) undergoes increased acceleration (much more than 9.8 m/s), while the other does not. http://en.wikipedia.org/wiki/Twin_paradox
That only proves that the two twins aren't in identical situations. It implies that something can be different, but it doesn't explain why something is different. The part that's hard to get is that there's no way for the astronaut twin to turn his ship around without having his brother's age "jump ahead" by a large amount. You need to understand simultaneity in SR before you can understand that. Check out this spacetime diagram. I'm calling the twin on Earth "A" and the twin in the rocket "B". Blue lines: Events that are simultaneous in the rocket's frame when it's moving away from Earth. Red lines: Events that are simultaneous in the rocket's frame when it's moving back towards Earth. Cyan (light blue) lines: Events that are simultaneous in Earth's frame. Dotted lines: World lines of light rays. Vertical line in the upper half: The world line of the position (in Earth's frame) where the rocket turns around. Green curves in the lower half: Curves of constant -t^2+x^2. Points on the two world lines that touch the same green curve have experienced the same time since the rocket left Earth. Green curves in the upper half: Curves of constant -(t-20)^2+(x-16)^2. Points on the two world lines that touch the same green curve have experienced the same time since the rocket turned around.
I wrote a longer post, but it got lost. Fundamentally, I don't buy the discontinuity at the turn around point. "A" appears to age 7 or so years on "B"'s outward leg, and around 33 years on the return leg. You can think about it with "A" sending out constant updates by radio signal, "I have aged 1 day" - once a day, every day and working out when and where these messages and "B" will be collocated. You should be able to see that "B" will receive 7 years worth of messages on the way out and 33 years worth on the way home. There will be a few messages during the turn around period, but not 25.6 years worth. (How many depends entirely on the rates of acceleration and thus how quickly "B" turns around.) cheers, neopolitan
You are correct, what you are describing is essentially the relativistic Doppler effect for a signal frequency of 12 uHz, but presumably you would want to correct for the light travel time. So you would get an "I have aged 1 day" signal, note how far you are from Earth at that time (in your frame), and calculate when the signal was sent to determine when he had aged. You wouldn't assume that he had aged when you recieved the signal, but rather when he sent the signal.
Yep, that is why I only talk about receiving the signals. But the signals are just reification of the _information_ about the aging of "A" according to "B". It is the only valid way of thinking about it, of putting labels on a chart like the one that Fredrik put together, imho. cheers, neopolitan
its absolutely true that there is a jump at the turn around point. in fact if he accelerates decelerates reaccelerates again and again then his calculation of the twins age will jump back and forth again and again (the twin will appear to move backward in time). now I dont buy that this is a 'real' effect but it certainly happens. you forget that to calculate the twins age he must take into account the speed of light. but he always measures the speed of light to be c relative to himself in spite of the fact that he just changed velocity. obviously that much have a drastic effect on his calculation. whether such relativistic effects are 'real' or not is another discussion entirely
It's 7.2 during the outbound trip, 7.2 during the return trip, and 25.6 during an instantaneous turnaround. Nothing important changes if you make the turnaround phase last longer. You might as well say that the definition of simultaneity in SR that's accepted by everyone isn't valid, because that's what I'm using.
But are you using precisely what is "accepted by everyone"? And if you are using that, is that definition of simultaneity supposed to be a helpful device or a proclamation on how things actually are? Ponder this ... in your example, "B" travels for 12 years. "A" ages 25.6 years in the turnaround period. How does the universe know to age "A" 25.6 years, and not 51.2 years or 102.4 years (which apply if "B" plans to travel for 24 or 48 years? Suppose, just suppose, that "B" travels to the turnaround point at a velocity of 0.8c, the heads off at 0.8c on the way back, but half-way home suffers a breakdown and has to limp the rest of the way at 0.001c. We'd have to assume that A and B are long lived, of course, but your figures will no longer work out. The instantaneous aging you posit at the turnaround point is an implication that the universe somehow sees the future and applies the right instantaneous aging to "A" (based on "B"'s future speeds). I don't think that is part of the standard understanding of simultaneity. cheers, neopolitan
Try thinking about it with "B" calculating "A"'s age purely on the basis of the number of reports of one day's aging. You will see that the discontinuities just don't happen. I will recant of course, if you can show how 25.6 years worth of daily reports will be received by "B" instantaneously, or over a relatively short period, if "B" chooses to decelerate and accelerate back homewards. (I would then also like to see how the other discontinuities work, for example if "B" decelerates on the way home and "A" moves backward in time - are the daily reports rescinded and reissued? and how?) cheers, neopolitan
That definition of simultaneity tells you how to associate an inertial frame with a physical observer moving with constant velocity. There's nothing controversial about it. Any coordinate system is a "helpful device" because you can use it to describe "how things actually are". The problem is of course that if you use a coordinate system to describe how things actually are, the description may sound very different from a description made using another coordinate system. (Take the pole-and-barn paradox for example: In the pole's frame, there's a time when the whole pole is inside the barn. In the barn's frame, there isn't). Because of these apparent contradictions, I can't encourage the use of expressions like "how things actually are". It's better to talk about e.g. "how things are in this particular frame". The universe isn't doing anything, so it doesn't have to know anything. The reason it's 25.6 and not any other number is that B is 16 light-years from Earth and that the slope of a simultaneity line is always v when the slope of the world line is 1/v. You can verify for yourself that with v=0.8c, that works out to 25.6 years. If B slows down to 0.001c on the way back, he will be in a frame where his "now" is simultaneous with an earlier event on Earth, so A will be younger than "before". That doesn't mean that the universe has "seen" or "done" something. It only means that B is using a different coordinate system. I don't see the problem.
Maybe you have the right idea about how simultaneity in SR is applied, but I really doubt it after long discussions elsewhere. To bring in simultaneity, if you must have it, consider "A" as a simple transmitter, pulsing out "A has aged x days" repeatedly (where x(m) = x(m-1)+1). Consider "B" as a similar transmitter with a similar message, "B has aged z days" (similar values of z). Using the relevant equations, "A" may consider the sending of one message to "B" and "B"'s sending of another message to be simultaneous, whereas "B" would not agree. Once "B" reaches the turn around point, there will a number of signals from "A" still on the way. Since the speed was 0.8c, for a period of 12 years, then in _classical_ terms there would be about 9.6 years worth of signals inbound (and last signal from "A" should be "A has aged 876 days"). In SR terms there will be 12.8 years worth of signals still on the way and the last message will be "A has aged 2629 days". (I am using your figure of 7.2 years btw, I haven't checked its validity. In any event, the actual figures are not important, it is the whole concept that matters). "B" then turns around and goes back through the incoming barrage of signals. Over a period of 12 years ship time, in _classical_ terms, "B" would experience 12 years of signals, plus the 9.6 years worth which were inbound at the turnaround point. In SR terms, "B" will travel through 32.8 years worth of signals before getting home and asking what the hell is wrong with "A"'s transmission schedule. What won't happen is the sudden delivery of 25.6 years worth of signals at the turn around point or the double delivery of signals (which would be the implication of "B" ending up in 'a frame where his "now" is simultaneous with an earlier event on Earth'). The only way you could organise double delivery of signals is by travelling faster than the speed of light (and it would involved some doubling back too, unless "A" started sending out pulses before "B" left Earth). Your mathematics might be spot on, Fredrik, but the application of them is questionable. cheers, neopolitan
A stream of signals arriving does not define simultaneity. You also have to calculate how far they travelled and hence when they originated. It is that calculation which has an abrupt discontinuity at the turn-around point.
But you agree that "B" will receive a rational stream of signals from "A"? Additionally, why exactly is there an abrupt discontinuity at the turn around? When pointed in one direction and getting "A has aged x days", B will calculate that A is a certain distance away (with the assumption that the message travelled c). Once turned around and getting "A has aged x+1 days" and being in approximately the same position, why does B make a calculation that A must have aged 25.6 years over the past day? My thinking is that B should make a calculation like this: No discontinuity. Is there any other sensible way that B should calculate how great the separation between the two of them is? cheers, neopolitan PS Thinking about it some more ... "A" will appear to have been sending out signals at a greater rate than "B" might have otherwise suspected, so that by the turnaround point "B" will already have received 2.4/0.6 years of signals. If "B" is clever, then an assumption will be made to the effect that "A" has and will continue to transmit signals at the same rate, then "B" should expect that there are 9.6/0.6 years worth of signals on the way at the turnaround point. 9.6/0.6 + 2.4/0.6 = 16 + 4 = 20 years On the way back "B" will expect to intercept a total of 36 years worth of messages. Still no discontinuity :)
The change of B's frame of reference at the turn around makes all the difference to the time and space separation. You can illustrate the ordinary 2D rotation equivalent by using "forwards" and "sideways" instead of "time" and "space" when moving in a plane. If two objects start from the same spot and move off in different directions at an acute angle relative to one another, then they are both moving less in each others' "forward" direction than the other. However, if one then turns a corner towards where the other one is going, the other one is suddenly much further "forward".
I already told you the reason why in post 8. he has to take the speed of light into account. the speed of light remains the same even when he changes velocity. cant you see that thet wall change his calculation?
Neopolitan, you need to make an effort to understand simultaneity. Suppose that a light signal is emitted in the positive x direction at (t_{1},0), gets reflected at some point along the x axis and returns at (t_{2},0), then the reflection event must be simultaneous with (t_{1}+(t_{2}-t_{1})/2,0). That's how simultaneity works. Pick t_{1}=-T and t_{2}=T for simplicity. The emission event is (-T,0) and the return event is (T,0). The reflection event must be simultaneous with (0,0), and since the speed of light is 1, the spatial coordinate must be T, so the reflection event is (0,T). Now draw the world line of someone who's moving with speed v in the positive x direction, and imagine that this person is doing the same thing we just did to find out which events are simultaneous. Note that the slope ([itex]\Delta t/\Delta x[/itex]) of the world line is 1/v. You will find that to this person, two events are simultaneous if and only if they are on a line with slope v. That's what I used to draw the diagram. Before the rocket turns around, B considers events on a blue line simultaneous. After the turnaround, he considers events on a red line simultaneous.
This isn't the way to "bring in simultaneity". See my previous post. You need to specify a frame if you're going to refer to a certain time. Since you're saying "12 years" instead of "20 years", I assume that we're considering B's frame, and since you're talking about things that happened before the turnaround, I assume that we're talking about the first of B's frames, the one that's moving away from Earth. The 9.6 and 876 figures are wrong. When (in B's frame) the rocket turns around, only 7.2 years have passed on Earth. The last message that arrived was sent by A when he had aged 4.0 years. (The turnaround event is (20,16) in A's frame, so a light signal that arrives at that event must have been emitted at (4,0)). I don't think it matters. It doesn't have anything to do with simultaneity. He's going to experience 36 years of signals, since he only received 4 years of signals before he turned around. (I see now that you realized that yourself). Huh. You lost me here. First of all, he's only ending up in 'a frame where his "now" is simultaneous with an earlier event on Earth' if he slows down on the way back (which is the scenario you were considering at the time). The turnaround puts him in a frame where his "now" is simultaneous with a much later time on Earth. 25.6 years later to be exact. I have no idea why you're talking about double delivery of signals and stuff like that. Simultaneity. (There's obviously no discontinuity in the incoming messages, but the simultaneity lines get tilted the other way when the rocket turns around).
"B" may have to a new basis of determining, in his new frame, which distant events about which he receives information in the future are simultaneous with which local events (which may be in his past), but he won't reassess events which were simultaneous with events in his previous frame. All such events are in the absolute past and cannot be resurrected. I do not think that a range of spatially colocated events at "A" (not colocated in time) can be simultaneous for "B", ever. That would violate causality. (For example, "B" cannot receive two events which one which is dependant on the other with a delay. Imagine an egg dropping to the floor under the influence of gravity, "B" cannot receive the event "egg whole" and "egg broken" together since there is a time component to the fall between.) Your "turning the corner" simile only works if "B" changes speed, which he doesn't really ... except during the actual turnaround. "B" changes velocity, speed is the same, only the direction changes. cheers, neopolitan