Quote by spikenigma reading through this thread has been interesting instead of creating a new thread, I'd like to offer a little variation if I may: A Rocket is an arbitrary distance from Earth, it accelerates up to a constant 0.7c As the rocket hurtles past Earth at that constant speed, two twins are born, one on Earth and one on the rocket. Rocket-twin and Earth-twin know that they were both born at the same time. They then communicate with each other with (VERY powerful lasers). Rocket-twin asserts that he is stationary and that the Earth is moving away from him at 0.7c, Earth-twin asserts that the Earth is stationary and Rocket-twin is moving. If Rocket-twin is older than Earth-twin, doesn't that make Earth the preferred reference frame?
Why should the rocket twin be older. After the acceleration, which ends before the birth, they are both in inertial frames in relative motion and so any calculation of time elapsed is reciprocal.They will each "see" each others clock running slower than their own.

Matheinste.

 Quote by matheinste Why should the rocket twin be older. After the acceleration, which ends before the birth, they are both in inertial frames in relative motion and so any calculation of time elapsed is reciprocal.They will each "see" each others clock running slower than their own. Matheinste.
ok then, two further things for clarification:

1)

let's say that Rocket-twin travels back to Earth (very) slowly as to minimise any time dilation and meets up with his twin. Which twin will be older/younger when he lands?

2)

during travel, why will each twin view the others clock as running more slowly than their own?

Mentor
 Quote by spikenigma using this, they can calculate [x] years have passed on Earth and [y] years have passed on the rocket, and be in agreement at any given time
No, they will always be in disagreement unless they agree on a reference frame in which to do the calculations.

 Quote by spikenigma ok then, two further things for clarification: 1) let's say that Rocket-twin travels back to Earth (very) slowly as to minimise any time dilation and meets up with his twin. Which twin will be older/younger when he lands? 2) during travel, why will each twin view the others clock as running more slowly than their own?
1) The earth twin will be older because he has remained inertial (for the purpose of this discussion) and so has traversed a longer spacetime interval and so accumulated more time on his clock than the spaceship twin.

2) Because that is what relativity says will happen.

Matheinste.

 Quote by Al68 Why not just look at a one way trip where the ship doesn't return, but just comes to rest with earth and stays there indefinitely? The answer is the same (divided by two) and the reason for it is clearer. Then just double that answer.
Quite right - you and I have tried previously to get this across when the twin trip analysis creeps into the forum - and as always no one seems to appreciate how simple it is to do the one way trip and double the answer

 Quote by DaleSpam No, they will always be in disagreement unless they agree on a reference frame in which to do the calculations.
Earth

 Quote by Matheinste 1) The earth twin will be older because he has remained inertial (for the purpose of this discussion) and so has traversed a longer spacetime interval and so accumulated more time on his clock than the spaceship twin. 2) Because that is what relativity says will happen.
doesn't this then imply a prefered reference frame?, which was my original point

both twins know that an object that has accelerated will experience time dilation with reference to one that has not.

When they both meet up, they can conclude that it is in fact the rocket that has accelerated and not Earth, even though rocket-twin never underwent any acceleration during his lifetime

 Quote by spikenigma Earth doesn't this then imply a prefered reference frame?, which was my original point both twins know that an object that has accelerated will experience time dilation with reference to one that has not. When they both meet up, they can conclude that it is in fact the rocket that has accelerated and not Earth, even though rocket-twin never underwent any acceleration during his lifetime
Only preferred in the sense that one is inertial and the other is not. As far as physics is concerned there is nothing that makes this inertial frame stand out from the infinite number of other inertial frames.

Acceleration is not the cause of time dilation. Each twin will consider the other to have experienced time dilation because they are in relative motion with respect to each other.

For the twins to meet up again one of them must have undergone acceleration. As the earth twin has remained inertial (for the purposes of this discussion) then the rocket twin must have undergone acceleration.The difference in ages is not directly due to acceleration but to the differences in spacetime paths due to the acceleration of the rocket twin.

I don't really want to get into a deep discussion of the twin "paradox" as it always causes a lot of grief.

Matheinste.

Mentor
 Quote by spikenigma Earth
Then their answers will only apply to the Earth's frame. This reference frame is "prefered" only in the sense that they agreed to use it, it is not prefered in any physical sense. They could have picked any other inertial frame and the laws of physics would look the same (which is the physics meaning of a "prefered" frame).

 Quote by DaleSpam Then their answers will only apply to the Earth's frame. This reference frame is "prefered" only in the sense that they agreed to use it, it is not prefered in any physical sense. They could have picked any other inertial frame and the laws of physics would look the same (which is the physics meaning of a "prefered" frame).
perhaps I'll clarify clearly what I mean

relativity (as far as I understand it) means that there is no prefered reference frame, i.e. if two bodies are moving at a constant velocity, no one body can say that it is the other is the one that is moving, or has moved and visa versa. It is supposed to be impossible to tell.

However, in the scenario, both of the twins can tell which one has accelerated - because one is older. Even though neither of them have ever undergone any acceleration

Recognitions:
 Quote by spikenigma perhaps I'll clarify clearly what I mean relativity (as far as I understand it) means that there is no prefered reference frame, i.e. if two bodies are moving at a constant velocity, no one body can say that it is the other is the one that is moving, or has moved and visa versa. It is supposed to be impossible to tell. However, in the scenario, both of the twins can tell which one has accelerated - because one is older. Even though neither of them have ever undergone any acceleration
Which scenario are you talking about? The one where the rocket travels past Earth at constant velocity and the two twins are born at the moment the rocket is next to Earth, then the rocket continues onward at constant velocity forever without turning around? In this case there is no objective truth about which twin is older, in the frame where the Earth is at rest the rocket-twin ages more slowly, in the frame where the rocket is at rest the Earth-twin ages more slowly. Are you familiar with the relativity of simultaneity? In the Earth frame it might be true that the event of the Earth-twin's 40th birthday is simultaneous with the event of the rocket-twin's 32nd birthday, while in the rocket frame it would then be true that the event of the Earth twin's 40th birthday is simultaneous with the event of the rocket-twin's 50th birthday, so in each frame the moving twin is only aging at 0.8 the rate of the at-rest twin. Only if you bring the twins back together to a single location in space will both frames have to agree on their respective ages at a single moment.

On the other hand, if you're talking about the scenario in post #53 where you said "let's say that Rocket-twin travels back to Earth (very) slowly as to minimise any time dilation and meets up with his twin", in this case the rocket must have turned around at some point to travel back to Earth, so the rocket did accelerate in the rocket-twin's lifetime.
 I cant understand why many of you, including "PF mentors", talk about acceleration. According to http://www.physicsforums.com/library...tem&itemid=166 "Time dilation does not depend on the acceleration of the clock."

 Quote by alviros I cant understand why many of you, including "PF mentors", talk about acceleration. According to http://www.physicsforums.com/library...tem&itemid=166 "Time dilation does not depend on the acceleration of the clock."

I think it is often mentioned by questioners because they think tme dilation is related to acceleration and is mentioned by responders to explain that this is not the case.

It is alway mentioned in the twin "paradox" because it is an integral part of it in so much that for the twins to reunite acceleration must be involved. In the twins "paradox" it is thought neccessary to point out that the difference in spacetime path length, which is ultimately the cause of the dfferential ageing, involves acceeration.

Matheinste

Recognitions:
 Quote by alviros I cant understand why many of you, including "PF mentors", talk about acceleration. According to http://www.physicsforums.com/library...tem&itemid=166 "Time dilation does not depend on the acceleration of the clock."
See my post #44 for an explanation of what is meant by this. The time dilation at any given instant depends solely on the the velocity in whatever frame you're using, the factor by which a moving clock is slowed down is always $$\sqrt{1 - v^2/c^2}$$ where v is that clock's instantaneous velocity. However, if you have two worldlines that cross paths at two times t0 and t1, and you know the velocity as a function of time v(t) on each worldline, then you can do the integral $$\int_{t_0}^{t_1} \sqrt{1 - v(t)^2/c^2} \, dt$$ for both of them to find the total time elapsed on each worldline between the two points where they cross. If one worldline is inertial (constant value for v(t)) and the other involves some acceleration (the value of v(t) changes with t), it will always work out that when you do the integral above, you'll find that the total time elapsed is greater on the inertial worldline than the worldline that involved an acceleration. That's just a property of the way the integral works, and it's totally compatible with the idea that the time dilation at each moment depends solely on the velocity at that moment, not the acceleration.

If it helps, there's a direct analogy for this in ordinary Euclidean geometry. Suppose we have two paths on a 2D plane which cross at two points, and one is a straight-line path while the other involves some bending. Since we know a straight line on a 2D plane is the shortest distance between points, we know the straight-line path will have a shorter total length. But suppose we want to measure the length of each path by driving cars along them with odometers running to measure how far the cars have travelled. Suppose we also have an x-y coordinate system on this 2D plane, so we can talk about "the rate a car is accumulating distance as a function of its x-coordinate"--if you think about it, it's not hard to show that this is solely a function of the slope of the path at that point in the coordinate system you're using. If you know the function for the path in this coordinate system y(x), then the slope at x is defined by looking at a small interval from x to (x + dx), and seeing the amount dy that the y-coordinate of the path changes between those points, with the slope defined as dy/dx. Since dx and dy are assumed to be arbitrarily small, the path can be assumed to be arbitrarily close to a straight line between the points (x,y) and (x+dx,y+dy), so the distance accumulated on the car's odometer as it travels between those points is just given by the pythagorean theorem, it'll be $$\sqrt{dx^2 + dy^2}$$, which is equal to $$dx*\sqrt{1 + dy^2/dx^2}$$, and since the "slope" at a given coordinate S(x) is defined to be dy/dx, this means the distance accumulated on the car's odometer as it travels between these points can be written as $$dx * \sqrt{1 + S(x)^2}$$.

So, the ratio of (increment odometer increases)/(increment x-coordinate increases), i.e. "the rate the car is accumulating distance as a function of its x-coordinate", will just be $$\frac{dx*\sqrt{1 + S(x)^2}}{dx}$$ which is just $$\sqrt{1 + S(x)^2}$$, purely a function of the slope. On the other hand, if you want to know how much distance accumulates on the odometer over a non-incremental change in the x-coordinate, say from some value $$x_0$$ to $$x_1$$, then we have to integrate the amount the odometer increases over each increment over the entire range from $$x_0$$ to $$x_1$$, giving the integral $$\int_{x_0}^{x_1} \sqrt{1 + S(x)^2} \, dx$$. Since we know a straight path is the shortest distance between two points, and we know straight implies constant slope, this means that if we have two different paths which cross once at $$x_0$$ and then again at $$x_1$$, and one has a constant S(x) while the other has a varying S(x), that means if we do the above integral for both paths the answer for the constant-slope path is guaranteed to be smaller.

Obviously all this is very closely analogous to the situation in relativity, where the rate a clock accumulates time as a function of the t-coordinate is just $$\sqrt{1 - v^2}$$ (in units where c=1, like seconds and light-seconds), while the total time accumulated on a path with a specific v(t) is $$\int_{t_0}^{t_1} \sqrt{1 - v(t)^2} \, dt$$, and a path with constant v is guaranteed to have a longer total time than a path with a v that changes (the reason a straight path in SR is guaranteed to have the largest time while a straight path in geometry is guaranteed to have the shortest distance has to do with the fact that there's a plus sign in front of the geometric slope but a minus sign in front of the velocity in the two square roots).
 To Whom... I hoped for some substantive response to my posts before being sent off to read more. I have read more SR books and papers than I can remember, and my comments come from careful reading of them. At least Einstein, French, Taylor and Wheeler, Feynman, and Bondi. Nothing in these sources or in the sources mentioned in this thread addresses my concerns. I stand by my previous posts. I hope that you will also apply 'good faith' effort in response. Here is a little more detail. Einsteins 1905 paper is based on the kinematics of rigid bodies. He says so in his introduction, and applies the principle in part 4, where he uses the equations for inertial motion in a straight line to determine the behavior of a clock moving in a closed curve consisting of a series of connected straight lines. Refering to Wikipedia-Kinematics, the definition of Kinematecs is 'A branch of classsical mechanics which describes the motion of objects without consideration of the causes leading to the motion.' It doesn't deny there are forces, it just defers their consideration to a later, Dynamic, analysis. This eliminates rockets etc. from possible explanation of Einsteins Twin/clock paradox, don't you think? Einstein asserts that all inertial frames are equal, or in his terms from Relativity,1952, '... every motion must be considered only as a relative motion', and '...two forms,both of which are equally justifiable: (a) The carriage is in motion relative to the embankment.(b) The embankment is in motion relative to the carriage.' Notice that he has no trouble with the massive earth being in motion. Thus for every result obtained with A at 'rest' and B in 'motion' there is an equal result with B at 'rest' and A in 'motion'. Thus my early post, when they reunite each twin thinks the other one is younger. I believe that this is what the 1905 paper says. How it's explained has not yet known, is it?

Recognitions:
 Quote by JM Here is a little more detail. Einsteins 1905 paper is based on the kinematics of rigid bodies. He says so in his introduction, and applies the principle in part 4, where he uses the equations for inertial motion in a straight line to determine the behavior of a clock moving in a closed curve consisting of a series of connected straight lines. Refering to Wikipedia-Kinematics, the definition of Kinematecs is 'A branch of classsical mechanics which describes the motion of objects without consideration of the causes leading to the motion.' It doesn't deny there are forces, it just defers their consideration to a later, Dynamic, analysis. This eliminates rockets etc. from possible explanation of Einsteins Twin/clock paradox, don't you think?
All that matters is whether the path through spacetime is a "straight" inertial (constant velocity path, or one involving different velocities at different times (like 'a closed curve consisting of a series of connected straight lines). Knowing the path through spacetime is sufficient to determine the time elapsed by a clock that takes that path, and a "straight" path will always have a greater time elapsed than a non-straight one.
 Quote by JM Einstein asserts that all inertial frames are equal, or in his terms from Relativity,1952, '... every motion must be considered only as a relative motion', and '...two forms,both of which are equally justifiable: (a) The carriage is in motion relative to the embankment.(b) The embankment is in motion relative to the carriage.' Notice that he has no trouble with the massive earth being in motion. Thus for every result obtained with A at 'rest' and B in 'motion' there is an equal result with B at 'rest' and A in 'motion'.
All inertial frames are equal, but the situation is not symmetrical because one twin's path is not straight while the other is--all inertial frames will agree that one twin moved at constant velocity while the other changed velocities. And again, it's always true that a straight path has the greatest proper time, in much the same way that a straight line in 2D Euclidean geometry always has the shortest distance (see my previous post #64 for a discussion of this analogy).
 Quote by JM Thus my early post, when they reunite each twin thinks the other one is younger.
Not if they calculate things from the perspective of an inertial frame, they won't--no matter which inertial frame you use, you'll always end up predicting that the inertial twin aged more than the twin that changed velocities midway through the trip.

Here's a simple example. Suppose twin A remains at rest on Earth (which for the sake of the problem we can assume is moving inertially rather than orbiting the Sun), while in the Earth's rest frame, twin B travels away from Earth at 0.6c for 25 years, then instantaneously turns around and travels back towards Earth at 0.6c for another 25 years. Thus twin A will be 50 years older when twin A returns, but because twin B experiences a time dilation factor of $$\sqrt{1 - 0.6^2}$$ = 0.8 on both the outbound leg and the inbound leg, twin B only ages 25*0.8 = 20 years on the outbound leg and another 20 years on the inbound leg, for a total of 40 years.

Now consider things from the perspective of a different inertial frame, the one where twin B is the one who's at rest during the outbound leg, while twin A moves away at 0.6c. In this frame twin B remains at rest for 20 years, but then instantaneously accelerates to catch up to A while A continues to move away at 0.6c--using the formula for addition of relativistic velocities, in this frame B must have a velocity of (0.6c + 0.6c)/(1 + 0.6*0.6) = 1.2c/1.36 = 0.88235c during the inbound leg. At the moment B turns around in this frame, A has been traveling away from B at 0.6c for 20 years, so A is 0.6*20 = 12 light-years away at this moment. With A continuing to move away at 0.6c and B moving at 0.88235c to catch up, the distance between B and A will be shrinking at a rate of (0.88235c - 0.6c) = 0.28235c in this frame, so it'll take another 12/0.28235 = 42.5 years for B to catch up with A. During this leg B is experiencing a time dilation factor of $$\sqrt{1 - 0.88235^2}$$ = 0.4706, so B only ages 42.5*0.4706 = 20 years during the second leg of the trip. Meanwhile A was moving at 0.6c during both the first leg which lasted 20 years and the second leg which lasted 42.5 years, a total of 62.5 years in this frame, and A's time dilation factor was $$\sqrt{1 - 0.6^2}$$ = 0.8, so A aged 0.8*62.5 = 50 years in this frame. So you see that even when we calculate things in a totally different frame, we still find the same conclusion as before: A aged a total of 50 years between B leaving and B returning, while B aged 20 years during the first leg of the trip and 20 years during the second leg, for a total of 40 years.

Quote by yogi
 Quote by Al68 Why not just look at a one way trip where the ship doesn't return, but just comes to rest with earth and stays there indefinitely? The answer is the same (divided by two) and the reason for it is clearer. Then just double that answer.
Quite right - you and I have tried previously to get this across when the twin trip analysis creeps into the forum - and as always no one seems to appreciate how simple it is to do the one way trip and double the answer
I guess it's just more interesting to do it the hard way. The thing is, showing two separate one way trips (correctly) would completely eliminate every objection I've seen to the standard resolutions.

Mentor
 Quote by JM Einstein asserts that all inertial frames are equal, ... Thus my early post, when they reunite each twin thinks the other one is younger.
One of the twins was inertial and the other was non-inertial, they both agree which one is younger.