Twins Paradox: Why One Twin is Older When Reunited

Moris526
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Hi All. Layman question.

In the twin's paradox, why one of the brothers is older when they meet again?. if movement is relative. What determines which of them ends up older?

Hope i explained myself.

Thanks.
 
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Moris526 said:
Hi All. Layman question.

In the twin's paradox, why one of the brothers is older when they meet again?. if movement is relative. What determines which of them ends up older?

Hope i explained myself.

Thanks.
You explained your question very well. The answer is the Principle of Relativity and the Principle that the speed of light is independent of the speed of the source of the light, both of which have been experimentally verified.

This means that as the brothers are separating at a constant speed, they will each see the other one's clock as running slower than their own by exactly the same ratio. We'll call this ratio R. Then when one of the brothers turns around and approaches his brother at the same speed going the opposite direction, he will see his brothers clock running faster than his own by exactly the inverse of the first ratio, which is 1/R. Since he spends the same amount of time traveling away as traveling back, we can average these two ratios to get the final ratio of his brother's accumulated age to his own accumulated age. The average will be (R+1/R)/2. If this average is greater than one, then the twin who turned around sees his brother is older, correct?

Why don't you pick any value of R less than 1 and see what you get?

Does that answer your question?
 
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I should also explain why the ratios for going and coming at the same speed are the inverse of each other.

Let's say that you are looking at two observers with clocks. One of them is a fixed distance away from you so you will see his clock ticking at the same rate as your own. They may have different times on them but we don't care about that, we only care about tick rates. The second observer is traveling at a constant speed from the first one directly toward you. Therefore, you will see his clock ticking faster than yours. You also know that you are seeing time progress on the traveler's clock compared to the first observer's clock just as the traveler is seeing them but delayed in time, provided that the light from both clocks travels towards you at the same speed. Therefore the ratio of the tick rates that you see of the distant fixed clock to the traveler's clock is some ratio that we called R. But the ratio of the tick rates that you see of the traveler's clock compared to your own (which is the same as the distant fixed clock) is the inverse of R.

Knowing that the speed that the traveler is going away from the distant fixed clock is the same as the speed that the traveler is approaching you wraps up the proof that we are looking for.

Does that make sense?
 
Moris526 said:
Hi All. Layman question.

In the twin's paradox, why one of the brothers is older when they meet again?. if movement is relative. What determines which of them ends up older?

Hope i explained myself.

Thanks.

The twin paradox refers to one twin staying at home while the other twin travels to a planet, turns around an comes back to where it started. Suppose that one twin is on eath and at t = 0 the other twin is moving past the stay at home twin at a significant fraction of the speed of light. This means that their motion is not symmetric since each twin has different experiences. The twin that goes away and comes back must accelerate during his journey and its this acceleation which breaks the symmetry which existed between the two twins. The twin who is accelerating can think of himself as being in a gravitational field and clocks higher up in a gravitational field run at different rates than ones lower in the gravitational field. The accelerating twin can think of himself as being at the place where the potential energy of any object is zero. The clock which is with his twin back on Earth (perhaps his wrist watch or his biological clock) will run at a different rate than his is. When all is said and done the calculations will show that the stay at home twin aged less.

Essentially it all boils down to this: The worldline of the stay at home twin is a straigt line while the twin on the journey isn't straight. You can idealize the accelerating twins worldline as two straight line segments, one of the ends of one segement is at the event "traveling twin is at same location a stay at home twin" while the other end of the segent is at the event "traveling twin turns around and heads for home." The second segment of his worldline starts at this last event and eneds at the event "traveling twin comes back home to meet his twin." Draw these worldlines on a spacetime diagrame and you'll see that one is straight and the other is a bent line where the bend is sharp bend, the sharper the bend the larger the accelertion of the traveling twin.
 
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Thank you both for your replies.

But i still don't get something.

If the twin on the ship accelerates away, turns back and accelerates again, doesn't that mean that from his/her perspective his/her brother at Earth does the same thing?. I don't understand what differentiates one of the frames of references. We could as well say that one brother stays on planet Ship and the other flies away on ship earth.

Or that's just a mathematical consequences and won't make common sense?

Thanks again.

Ive always liked physics, never could study. So now I am trying to learn something on my own.
 
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Moris526 said:
I don't understand what differentiates one of the frames of references.

One twin, the one we call the "traveling" twin, has to fire rockets to turn around. He feels acceleration when he does that. The other twin, the one we call the "stay-at-home" twin, doesn't do anything; he just floats freely the whole time, and feels no acceleration. That's the difference between the two.
 
Moris526 said:
Thank you both for your replies.

But i still don't get something.

If the twin on the ship accelerates away, turns back and accelerates again, doesn't that mean that from his/her perspective his/her brother at Earth does the same thing?. I don't understand what differentiates one of the frames of references. We could as well say that one brother stays on planet Ship and the other flies away on ship earth.

Or that's just a mathematical consequences and won't make common sense?

Thanks again.

. I've always liked physics, never could study. So now I am trying to learn something on my own.
Did you do the calculation I asked you to do to figure out which one aged more? I don't see that you have gotten an answer to your question. What is the correct answer?

And please keep in mind that my answer to your question does not require you to identify any frame of reference but even if you want to do that, it won't matter which frame of reference you pick, they all will give the same answer so your statement that you don't understand what differentiates one of the frames of reference doesn't have any bearing on the answer to your question.
 
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PeterDonis said:
One twin, the one we call the "traveling" twin, has to fire rockets to turn around. He feels acceleration when he does that. The other twin, the one we call the "stay-at-home" twin, doesn't do anything; he just floats freely the whole time, and feels no acceleration. That's the difference between the two.

Thanks. I get it now.
 
Moris526 said:
Thanks. I get it now.
OK, so you understand that there is a difference between the twins, but how does that lead to an answer to your question? Have you determined which twin ends up older? What is your answer?
 
  • #10
ghwellsjr said:
I should also explain why the ratios for going and coming at the same speed are the inverse of each other. Let's say that you are looking at two observers with clocks. One of them is a fixed distance away from you so you will see his clock ticking at the same rate as your own. They may have different times on them but we don't care about that, we only care about tick rates. The second observer is traveling at a constant speed from the first one directly toward you. Therefore, you will see his clock ticking faster than yours. You also know that you are seeing time progress on the traveler's clock compared to the first observer's clock just as the traveler is seeing them but delayed in time, provided that the light from both clocks travels towards you at the same speed. Therefore the ratio of the tick rates that you see of the distant fixed clock to the traveler's clock is some ratio that we called R. But the ratio of the tick rates that you see of the traveler's clock compared to your own (which is the same as the distant fixed clock) is the inverse of R.

That explanation doesn't seem very clear to me. I think a clear explanation would have to convey why the frequency ratios are inverses according to special relativity, even though they would not be inverses according to classical Doppler. This difference is the key factor that distinguishes special relativity from pre-relativistic physics, and of course it's also key to understanding the twins paradox, but I don't see an explanation of the difference here. Also it seems confusing and unnecessary to introduce a third observer/clock that is stationary relative to one of the other two observer/clocks... but perhaps it just seems that way because I don't follow your explanation.

By the way, in order to explain why "the ratios for going and coming at the same speed" are reciprocals, I think you will need to specify how you define "speed". You can't base it on the Doppler ratios, because that would be circular. You need some independent means of defining the "speed" of the twin. Normally this is done by defining a suitable system of space and time coordinates. If you think you can dispense with this, and simply (for example) define equal outgoing and incoming speeds as those for which the Doppler ratios are reciprocal, then the proposition you're trying to prove is actually tautological... but then you face the task of showing that this definition of "speed" agrees with the normal definition of speed - which of course is based on a suitable system of space and time coordinates, the very thing you are trying to do without.
 
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  • #11
ghwellsjr said:
OK, so you understand that there is a difference between the twins, but how does that lead to an answer to your question? Have you determined which twin ends up older? What is your answer?

I didn't understand why should any be older than the other if movement is relative and there were no difference between the frames of reference. Now, the acceleration of the ship makes a difference between them. Thou i have to admit i can't fully visualize why the Earth man ends up older.

I have the notion that time slows down as you accelerate but i can't make that clear in my head. I don't kown if it can be done or is just a logical conclusion of the math.

You should also know that i have just basic knowledge of mathematics.

By the way, thanks for your time.
 
  • #12
Samshorn said:
That explanation doesn't seem very clear to me. I think a clear explanation would have to convey why the frequency ratios are inverses according to special relativity, even though they would not be inverses according to classical Doppler.
You must have understood my explanation enough to determine that there is no difference between Relativistic Doppler and Classical Doppler for what I described and I agree.

Samshorn said:
This difference is the key factor that distinguishes special relativity from pre-relativistic physics, and of course it's also key to understanding the twins paradox, but I don't see an explanation of the difference here. Also it seems confusing and unnecessary to introduce a third observer/clock that is stationary relative to one of the other two observer/clocks... but perhaps it just seems that way because I don't follow your explanation.
I should have carried the explanation two steps further and applied the Principle of Relativity to point out that whatever Doppler ratio you see of the traveling observer's clock coming toward you compared to your clock, he sees the exact same Doppler ratio of your clock to his. And whatever Doppler ratio the traveling observer sees of the remote stationary clock to his, is the same that the remote observer sees of the traveling observer's clock to his own. In other words, there are only two ratios to consider among the three observers (not counting the unity ones between you and the remote observer). In Classical Doppler, there can be four.

By the way, this little exercise is not part of the Twin Paradox scenario--it is just meant to be an independent proof that the Doppler ratios are inverses for approaching and retreating at the same speed.

Samshorn said:
By the way, in order to explain why "the ratios for going and coming at the same speed" are reciprocals, I think you will need to specify how you define "speed". You can't base it on the Doppler ratios, because that would be circular. You need some independent means of defining the "speed" of the twin. Normally this is done by defining a suitable system of space and time coordinates. If you think you can dispense with this, and simply (for example) define equal outgoing and incoming speeds as those for which the Doppler ratios are reciprocal, then the proposition you're trying to prove is actually tautological... but then you face the task of showing that this definition of "speed" agrees with the normal definition of speed - which of course is based on a suitable system of space and time coordinates, the very thing you are trying to do without.
I'm not defining the speed of the traveling twin nor am I determining what the relationship is between speed and Doppler ratio--only that they are inverses for approaching and retreating at the same speed. I didn't even think it was necessary to state how the traveling twin determines that he is traveling in both directions at the same speed--that could be a mere assumption. But in any case, the traveling twin could legitimately use the inverse Doppler ratio to determine that his speed was the same on the return trip as on the outgoing (without identifying what that speed is) and it is important that the speeds be equal in order for the travel times to be equal so that the calculation can be legitimate.

Keep in mind, I'm only trying to answer the OP's question: which twin ages more?
 
  • #13
Moris526 said:
I didn't understand why should any be older than the other if movement is relative and there were no difference between the frames of reference. Now, the acceleration of the ship makes a difference between them.
Do you understand that looking at any activity while traveling away from it will make it appear to be in slow motion and while approaching, it will make it appear in fast motion? And do you know that this is called Doppler?

Moris526 said:
Thou i have to admit i can't fully visualize why the Earth man ends up older.
Are you sure the Earth man ends up older? Popper said the opposite:
Popper said:
When all is said and done the calculations will show that the stay at home twin aged less.
Moris526 said:
I have the notion that time slows down as you accelerate but i can't make that clear in my head.
You don't have to know about time slowing down or anything about acceleration to know how to get the answer to your question.

Moris526 said:
I don't kown if it can be done or is just a logical conclusion of the math.

You should also know that i have just basic knowledge of mathematics.

By the way, thanks for your time.
Was the math of post #2 within your grasp?
 
  • #14
ghwellsjr said:
I should have carried the explanation two steps further and applied the Principle of Relativity to point out that whatever Doppler ratio you see of the traveling observer's clock coming toward you compared to your clock, he sees the exact same Doppler ratio of your clock to his. And whatever Doppler ratio the traveling observer sees of the remote stationary clock to his, is the same that the remote observer sees of the traveling observer's clock to his own.

Yes, by invoking the principle of relativity along with the independence of light speed from the speed of the source, I agree that you can infer that the Doppler shift observed by the traveling twin on his outbound journey is the reciprocal of the shift on his return journey (with symmetrical speeds). And from this it follows that the total elapsed time for the traveling twin is less than for the stay-at-home twin. If I was to paraphrase your argument, I'd say you are considering three clocks a,b,c, with a and c mutually at rest, with clock b directly in between a and c, and moving away from a and toward c. Letting Da/Db denote the ratio of clock ticks from a reaching b to clock ticks of b, and so on, we have (Da/Db)(Db/Dc) = Da/Dc = 1, and by relativity we have Dc/Db = Db/Dc. Therefore, Da/Db is the reciprical of Dc/Db, and these represent the Doppler shift ratios for approaching and receding at the speed of b. Then, letting R denote the receding Doppler ratio, and 1/R the approaching Doppler ratio, you point out that for a symmetrical trip the average Doppler ratio for the whole trip is (R + 1/R)/2, which is strictly greater than 1 as R differs from 1. Hence the traveling twin ages less.

ghwellsjr said:
You must have understood my explanation enough to determine that there is no difference between Relativistic Doppler and Classical Doppler for what I described and I agree.

I don't follow that. Classical Doppler (for a theory in which the speed of the wave is independent of the speed of the source, like sound waves) would not lead to the same result, because classical Doppler gives different frequency ratio for the cases when the source or the receiver is moving.

Also I think this line of reasoning doesn't enable us to quantify the magnitude of the relativistic time dilation for any specific trip velocity, since it doesn't relate the speed to the Doppler ratios.

ghwellsjr said:
Keep in mind, I'm only trying to answer the OP's question: which twin ages more?

Yes, although I think it's debatable how much this explanation really clarifies why the twin who turns around ages less. It shows that if we impose the two principles, the turn-around twin must age less, but for a beginning student those two principles seem to be irreconcilable, and everyone knows we can derive infinitely many false conclusions from self-contradictory premises. So, for this reasoning to be persuasive, we need to explain why those two principles are not mutually contradictory (as they would be classically). And of course the light-speed principle applies only to inertial coordinates, so it comes back to the question of what makes one path inertial and the other not (if we imagine a purely relational world, for example, although the OP doesn't seem to be concerned about that - yet). This, combined with the fact that this approach doesn't enable us to quantify the effect, may explain why this approach is not often followed.
 
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  • #15
ghwellsjr said:
Are you sure the Earth man ends up older? Popper said the opposite:
When all is said and done the calculations will show that the stay at home twin aged less.

Well,according to Wikipedia It is opposite of what popper said!

http://en.wikipedia.org/wiki/Twin_paradox

"""In physics, the twin paradox is a thought experiment in special relativity involving identical twins, one of whom makes a journey into space in a high-speed rocket and returns home to find that the twin who remained on Earth has aged more."""
 
  • #16
ash64449 said:
Well,according to Wikipedia It is opposite of what popper said!

http://en.wikipedia.org/wiki/Twin_paradox

"""In physics, the twin paradox is a thought experiment in special relativity involving identical twins, one of whom makes a journey into space in a high-speed rocket and returns home to find that the twin who remained on Earth has aged more."""
The inertial twin is always the one with the maximum proper time (i.e. stay at home twin ages more). I suspect Popper knew that and just wrote it wrong.
 
  • #17
DaleSpam said:
The inertial twin is always the one with the maximum proper time (i.e. stay at home twin ages more). I suspect Popper knew that and just wrote it wrong.

How is maximum proper time indicate that time dilation is less?(just curious)
 
  • #18
ash64449 said:
How is maximum proper time indicate that time dilation is less?(just curious)
For me, the best way to keep it straight is to think of it geometrically. Think about Euclidean geometry. Given two points you can connect them with an infinite number of paths, each of which will have a certain length. The shortest length between the two points is a straight line, and all other lines will be bent and have a longer length.

In spacetime a similar idea holds, except that now there are different kinds of paths, timelike paths whose lengths are measured by clocks and spacelike paths whose lengths are still measured by rods. The other difference is that for timelike paths the "straight line" statement gets flipped so the LONGEST time between two events is a straight line, and all other lines will be bent and have a SHORTER time.
 
  • #19
DaleSpam said:
For me, the best way to keep it straight is to think of it geometrically. Think about Euclidean geometry. Given two points you can connect them with an infinite number of paths, each of which will have a certain length. The shortest length between the two points is a straight line, and all other lines will be bent and have a longer length.

In spacetime a similar idea holds, except that now there are different kinds of paths, timelike paths whose lengths are measured by clocks and spacelike paths whose lengths are still measured by rods. The other difference is that for timelike paths the "straight line" statement gets flipped so the LONGEST time between two events is a straight line, and all other lines will be bent and have a SHORTER time.

hey,, that means...

An hour is closer to me than a minute?!
 
  • #21
ash64449 said:
hey,, that means...

An hour is closer to me than a minute?!

No. Consider three events ("event" is just another word for "point in space-time"):
1) You clap you hands right now. When you do, your wristwatch reads exactly noon. Call this event N.
2) You clap you hands again. This time your wristwatch reads 12:01. Call this event M.
3) You clap your hands a third time. This time your wristwatch reads 13:00. Call this event H.

Assume that you're just sitting still, no acceleration or other change of speed, for the entire hour. It might be best to assume that you're floating in empty space, so we don't have to worry about gravity, the rotation of the Earth, and the like.

Now N, M, and H all lie on the same straight line; M lies between N and H; and as you travel through space-time you are moving along that line, starting at point N, passing through M, and then reaching H.
Clearly M is closer to you than H because you had to pass through M to get to H on a straight-line path. Your path from N to M is one minute long, your path from M to H is 59 minutes long, and your path from N to H through M is one hour long.

However, there are paths through space-time from N to H that don't pass through M. Your path does, because all three events are you clapping your hands, so they're always going to happen where you are.
But consider a spaceship that starts out sitting right next to you, takes off when you clap your hands for the first time, flies for a while, turns around, and returns just in time to land and be sitting next to you when you clap your hands for the third time.
The spaceship's path through spacetime passes through N and H but not through M. It's also not a straight line.

DaleSpam is saying that the spaceship's path between N and H will be shorter than your path between N and H - not too surprising because they are different paths. That has nothing to do with the event M which is on your path but not the spaceship's. Your path from N to M is one minute long, your path from M to H is 59 minutes long, your path from N to H through M is one hour long, and for you H is further away than M.
 
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  • #22
nugatory,yes. You are right.. Makes sense..
I got exicted and searched in google and found a video just what i said. Unfortunately,i can't paste the url of the video.
Let me tell the description of the video and tell me what you feel from that video
video is from minutephysics.
Name of that video:Distance and Special Relativity:
How far away is tomorrow?
 
  • #23
DaleSpam said:
Huh?

i will tell you.
You said the shortest 'space like' intervel(straight line) is considered longest time between two events.(proper time-time like). Hour is longest time than a minute.so in space like intervel,hour is shortest distance than a minute so hour is closer to me than a minute! Dosen't that make sense.
I figured this out and got exicted and googled this thing and found a video.unfortunately,i am not able to paste the video's url.i will tell the description so that you can see it by yourself.
Video is from minutephysics
name of video:Distance and Special Relativity:
How far away is tomorrow?
 
  • #24
nugatory and dalespam:
can you guys explain me in simpler manner why shortest 'space like' intervel is considered as longest 'time like' intervel?
 
  • #25
and i think why we can get to hour faster than the minute is because whenever we move,we move through multiple dimensions!
That is we move through spatial dimension and time like dimension at the same time..
 
  • #26
ash64449 said:
nugatory and dalespam:
can you guys explain me in simpler manner why shortest 'space like' intervel is considered as longest 'time like' intervel?
This sentence is not correct in several ways. You need to know what the terms mean.
'spacetime' is just a word to show that we are talking about a concept in Einstein's relativity
'spacetime interval' Well, the interval in classical mechanics is just the distance between two points. This 'spacetime interval' is a similar concept in relativity. So it sort of represents a kind of 'distance', but it is not the same as distance as in classical mechanics. To be specific, this is the equation for the spacetime interval between two points (in 1d, with zero gravity):
\int \sqrt{c^2dt^2 - dx^2} = spacetime \ interval
so if you have dx=0 then the spacetime interval is going to be as great as possible. This corresponds to the person that does not move. So the person that does not move has a greater spacetime interval than the person that flies away on a ship and comes back again.
'time like' this is any path that is going at less than light-speed (i.e. all physical objects move on a time like path).
'space like' this is any path that is not 'time like', or moving at the speed of light. 'space like' paths are not important for the twin paradox problem.
'proper time' this means the same as the 'spacetime interval'
Also, the 'proper time' for a given path is the time which a person traveling on the path would measure on their watch. (or equivalently, how much they age). So from the above reasoning, the person that ages the most is the person who stays at home on the planet earth.

Finally, keep in mind that sometimes the spacetime interval is defined with the minus sign the other way around, and the units are sometimes defined differently.
 
  • #27
BruceW said:
To be specific, this is the equation for the spacetime interval between two points (in 1d, with zero gravity):
\int \sqrt{c^2dt^2 - dx^2} = spacetime \ interval

All I would add is that if someone prefers a calculus-free formula... as long as you're also working with velocities that are constant or change instantaneously (for example in the most common statement of the twin paradox we say that the traveling twin goes from 0 to v instantaneously at the start of the journey then travels at a constant v until the turnaround point) then this integral turns into:<br /> s^2 = (t_2-t_1)^2 - (x_2-x_1)^2<br />
where s is the spacetime interval and the t's and x's are the coordinates of the two events; it doesn't matter whose coordinates we use, the interval comes out the same.

This works for any spacetime diagram that only has straight lines, no curves, in it.
 
  • #28
ghwellsjr said:
Was the math of post #2 within your grasp?

It's not just the math of post #2; the OP needs to understand why the math of post #2, as you presented it, only applies to the traveling twin, not the stay-at-home twin. There is also a corresponding equation that applies to the stay-at-home twin, with the same ratio R in it, but it differs from the one you gave in post #2 in a crucial respect. What that crucial respect is is not a question of math, it's a question of physics. Post #2 gives a hint at the answer to the question of physics, but you didn't give it explicitly.
 
  • #29
ash64449 said:
You said the shortest 'space like' intervel(straight line) is considered longest time between two events.(proper time-time like).
No. I said the longest timelike interval between two events is a straight line. I never mixed up timelike and spacelike intervals.

The shortest spacelike interval between two events is a straight line.
The longest timelike interval between two events is a straight line.

Those are two completely separate statements about different pairs of events and different types of spacetime interval. I don't know why you are mixing them up like that.

ash64449 said:
Hour is longest time than a minute.so in space like intervel,hour is shortest distance than a minute so hour is closer to me than a minute! Dosen't that make sense.
No. You can certainly use geometrized units where time is measured in units of distance, but that doesn't make an hour closer than a minute. In such units an hour would be 1E9 km and a minute would be 1.8E7 km. So a minute is a shorter distance than an hour.

ash64449 said:
How far away is tomorrow?
2.6E10 km
 
  • #30
ash64449 said:
nugatory and dalespam:
can you guys explain me in simpler manner why shortest 'space like' intervel is considered as longest 'time like' intervel?
It isn't. You are mixing up spacelike and timelike intervals. You never compare timelike and spacelike intervals.
 
  • #31
BruceW said:
This sentence is not correct in several ways. You need to know what the terms mean.
'spacetime' is just a word to show that we are talking about a concept in Einstein's relativity
'spacetime interval' Well, the interval in classical mechanics is just the distance between two points. This 'spacetime interval' is a similar concept in relativity. So it sort of represents a kind of 'distance', but it is not the same as distance as in classical mechanics. To be specific, this is the equation for the spacetime interval between two points (in 1d, with zero gravity):
\int \sqrt{c^2dt^2 - dx^2} = spacetime \ interval
so if you have dx=0 then the spacetime interval is going to be as great as possible. This corresponds to the person that does not move. So the person that does not move has a greater spacetime interval than the person that flies away on a ship and comes back again.
'time like' this is any path that is going at less than light-speed (i.e. all physical objects move on a time like path).
'space like' this is any path that is not 'time like', or moving at the speed of light. 'space like' paths are not important for the twin paradox problem.
'proper time' this means the same as the 'spacetime interval'
Also, the 'proper time' for a given path is the time which a person traveling on the path would measure on their watch. (or equivalently, how much they age). So from the above reasoning, the person that ages the most is the person who stays at home on the planet earth.

Finally, keep in mind that sometimes the spacetime interval is defined with the minus sign the other way around, and the units are sometimes defined differently.
This is all wrong, unless somebody changed the definition of Spacetime Interval and didn't tell me. Where did you get that definition?

As far as I know, the Spacetime Interval is not an integral and therefore does not depend on the path between two points. A correct formula from wikipedia is:
3e6502fa4b2e9d9c0852e7a42b50e80f.png
(spacetime interval)

If s2 evaluates to a negative number, then the spacetime interval is timelike and can be measured by a clock at rest in an inertial frame. This is also the Proper Time on the clock but that is immaterial.

If s2 evaluates to a positive number, then the spacetime interval is spacelike and can be measured by a ruler at rest in an inertial frame. This is also the Proper Distance on the ruler but that is immaterial.

If s2 evaluates to zero, then the spacetime interval cannot be measured but is defined to be the propagation of light in any inertial frame.
 
  • #32
ghwellsjr said:
As far as I know, the Spacetime Interval is not an integral and therefore does not depend on the path between two points.

In flat spacetime, yes, you can define the "spacetime interval" as you have defined it; but the definition doesn't generalize to curved spacetime. (It also doesn't generalize to non-inertial frames in flat spacetime; your definition requires that the coordinates are defined relative to an inertial frame.)

The integral definition is more general in that it works in curved spacetime and in non-inertial frames in flat spacetime; but, as you note, it makes the "interval" path-dependent, because you have to choose a curve along which to integrate. If you choose a geodesic curve, the integral definition is equivalent to yours, but, as I said, it works in curved spacetime and in non-inertial frames in flat spacetime. If you choose some non-geodesic curve, the integral will give you a number, which may have a physical meaning (for example, it gives the elapsed proper time along a non-geodesic timelike curve), but will be different than the integral evaluated along a geodesic.

AFAIK the term "spacetime interval" is only used for the length along a geodesic in flat spacetime, which matches your definition; but that just means the term "spacetime interval" is of very limited application. The integral form, since it's more general, is much more useful, IMO.
 
  • #33
DaleSpam said:
It isn't. You are mixing up spacelike and timelike intervals. You never compare timelike and spacelike intervals.

yes. Sorry,i think i did mix them up...
 
  • #34
DaleSpam said:
It isn't. You are mixing up spacelike and timelike intervals. You never compare timelike and spacelike intervals.

But DaleSpam,you know what i meant...

Why is Longest Time like interval a straight line?
 
  • #35
ash64449 said:
Why is Longest Time like interval a straight line?

This follows from the structure of spacetime, where the interval between two events is given by {\Delta}s^2 = {\Delta}t^2 - {\Delta}x^2. Try plugging in some values for Δx and Δt (remember that we're talking about timelike intervals, so Δt is always greater than Δx, and I chose the sign convention to make it easy to work with timelike intervals) and you'll see that the straight line path between two points A and B is always longer than the unstraight path that passes through some third point C.

This is different from ordinary space where the interval between two points is given by {\Delta}s^2 = {\Delta}x^2 + {\Delta}y^2 (just the familiar Pythagorean theorem for calculating the distance between points in the x-y plane) where the straight line is always the shortest distance.
 
  • #36
Samshorn said:
ghwellsjr said:
I should have carried the explanation two steps further and applied the Principle of Relativity to point out that whatever Doppler ratio you see of the traveling observer's clock coming toward you compared to your clock, he sees the exact same Doppler ratio of your clock to his. And whatever Doppler ratio the traveling observer sees of the remote stationary clock to his, is the same that the remote observer sees of the traveling observer's clock to his own.
Yes, by invoking the principle of relativity along with the independence of light speed from the speed of the source, I agree that you can infer that the Doppler shift observed by the traveling twin on his outbound journey is the reciprocal of the shift on his return journey (with symmetrical speeds). And from this it follows that the total elapsed time for the traveling twin is less than for the stay-at-home twin. If I was to paraphrase your argument, I'd say you are considering three clocks a,b,c, with a and c mutually at rest, with clock b directly in between a and c, and moving away from a and toward c. Letting Da/Db denote the ratio of clock ticks from a reaching b to clock ticks of b, and so on, we have (Da/Db)(Db/Dc) = Da/Dc = 1, and by relativity we have Dc/Db = Db/Dc. Therefore, Da/Db is the reciprical of Dc/Db, and these represent the Doppler shift ratios for approaching and receding at the speed of b. Then, letting R denote the receding Doppler ratio, and 1/R the approaching Doppler ratio, you point out that for a symmetrical trip the average Doppler ratio for the whole trip is (R + 1/R)/2, which is strictly greater than 1 as R differs from 1. Hence the traveling twin ages less.
I'm glad you agree, thanks.

Samshorn said:
ghwellsjr said:
You must have understood my explanation enough to determine that there is no difference between Relativistic Doppler and Classical Doppler for what I described and I agree.
I don't follow that. Classical Doppler (for a theory in which the speed of the wave is independent of the speed of the source, like sound waves) would not lead to the same result, because classical Doppler gives different frequency ratio for the cases when the source or the receiver is moving.
I was referring just to the part in post #3, not #12. As long as the source, a, and the receiver, c, are moving identically so that they are in mutual rest, (Da/Db)(Db/Dc) = Da/Dc = 1. Do I need to draw a diagram?

Samshorn said:
Also I think this line of reasoning doesn't enable us to quantify the magnitude of the relativistic time dilation for any specific trip velocity, since it doesn't relate the speed to the Doppler ratios.
Correct, but it's irrelevant since everything applies at any velocity.

Samshorn said:
ghwellsjr said:
Keep in mind, I'm only trying to answer the OP's question: which twin ages more?
Yes, although I think it's debatable how much this explanation really clarifies why the twin who turns around ages less. It shows that if we impose the two principles, the turn-around twin must age less, but for a beginning student those two principles seem to be irreconcilable, and everyone knows we can derive infinitely many false conclusions from self-contradictory premises.
But these two principles are based on the history of experimental evidence and so they are not irreconcilable and so there is no danger of jumping to any false conclusions. I think it's important for a beginning student to grasp in a simple way that a traveling twin will return younger than the Earth twin and it's based on experimental evidence, facts, if you will and not merely theory. If we could easily do the experiment in a lab, they wouldn't find it so unusual or hard to grasp.

And especially when an OP asks the question the way this one did, we want to show him how the Principle of Relativity plus the second simple principle directly lead to the conclusion that the traveling twin ages less. He was thinking that the Principle of Relativity meant the twins would age the same.

Samshorn said:
So, for this reasoning to be persuasive, we need to explain why those two principles are not mutually contradictory (as they would be classically).
But the inverse nature of the two Dopplers is not classically contradictory as you have incorrectly concluded. Once you figure that out, maybe you'll have a different assessment.

But if that is not your concern, or even if it is, then this is one of those "why" questions that has no answer aside from "that's the way our world works".

Samshorn said:
And of course the light-speed principle applies only to inertial coordinates, so it comes back to the question of what makes one path inertial and the other not (if we imagine a purely relational world, for example, although the OP doesn't seem to be concerned about that - yet).
Yes, along with a slew of other assumptions that we take for granted but no one ever brings up when discussing other explanations of the Twin Paradox, so why bring them up for this one?
Samshorn said:
This, combined with the fact that this approach doesn't enable us to quantify the effect, may explain why this approach is not often followed.
Or, could it be that the approaches that are often followed are just repeated by people who don't really understand what they are talking about but this is one area of science where you're allowed to sound confusing and allowed to confuse your students and still be considered an expert?

This approach is not the end. It is just a beginning and needs to be followed up with more. I would follow it with a full explanation of Relativistic Doppler to state that what the twins observe is what a theory needs to explain and then show how Special Relativity fits perfectly (GR or gravity explanations are not needed) and use IRF's to show how they all explain all the observations and finally end with a non-inertial frame for the traveling twin based on radar measurements. All these frames identically support what each Twin sees, observes and measures and demonstrate the essential postulate of SR that light propagates at c. I would argue against the gravity, GR, multi-IRF or frame jumping explanations as promoting the Paradox instead of resolving it since they don't support what the twins see, observe or measure, nor do they demonstrate Einstein's second postulate of light propagating at c. At least that's my assessment and since I've not succeeded in getting anyone to prove me wrong, I'm sticking with it.

Again, thank you for your vote of confidence in this approach, even if it is only a beginning.
 
  • #37
ghwellsjr said:
But the inverse nature of the two Dopplers is not classically contradictory as you have incorrectly concluded. Once you figure that out, maybe you'll have a different assessment.

This seems to be the central point. The disagreement isn't over the fact that (Da/Db)(Db/Dc) = Da/Dc = 1, but rather over the claimed equality (by relativity) Dc/Db = Db/Dc, which you use to claim that Da/Db and Dc/Db are recipricals. In other words, you rely on the relativistic fact that two inertial observers both see the same Doppler shift of signals from the other. Classically this equality doesn't unambiguously hold. Here's why:

We have two inertial clocks A and B with some mutual velocity v, and we assert the two principles of (1) relativity between inertial coordinate systems, and (2) independence of light speed from the speed of the source (again in terms of any inertial coordinate system). On this basis we can examine the relationship between the frequencies of light signals transmitted and received by the clocks in terms of (for example) the rest frame of A, in which the other clock B has velocity v.

Now, by the classical Doppler effect (meaning no relativistic time dilation of B relative to the time coordinates of A's rest frame), it's trivial to see that the light-speed principle implies that a signal sent from A with frequency fA will arrive at B with frequency fB where fB/fA = 1-v, whereas a signal sent from B with frequency FB will arrive at A with frequency FA where FA/FB = 1/(1+v). These two ratios are not equal, but the relativity principle requires them to be equal, so classically the two principles are irreconcilable.

The only way of reconciling those principles is that clock B must not advance equally with the coordinate time of A's rest frame. Specifically, clock B must advance more slowly by the time-dilation factor sqrt(1-v^2). Accounting for this difference in the rate of B, we find that the two ratios are

fB/fA = (1-v)/sqrt(1-v^2) and FA/fB = sqrt(1-v^2)/(1+v)

Both of these equal the relativistic Doppler factor sqrt[(1-v)/(1+v)], so we have satisfied the relativity requirement. Note that the time dilation factor applies regardless of whether A and B are approaching or receding from each other, and of course the Doppler effects in those cases are reciprocals.

The relativistic time dilation factor, which is necessary to logically reconcile the principles, automatically enables us to compute (quantitatively, not just qualitatively) the different ages of the twins, so nothing more is needed, and nothing less suffices. I think a student would accept your version of the Doppler explanation only if he failed to realize that the premises are mutually contradictory unless we posit relativistic time dilation, and once we do that, nothing else is needed. Also note that no third clock is needed. At best the Doppler considerations provide just one more derivation of the time dilation factor, but of limited applicability, and without clearly defining how that fits with the Lorentz transformation. As such, it will certainly seem unsatisfactory to the student when he realizes that not only must B run slow in terms of the standard time coordinates of A's rest frame, but A must run slow in terms of the standard time coordinates of B's rest frame. The only way to really understand this is to understand how inertial coordinate systems are related to each other, in full (i.e., the Lorentz transformation).
 
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  • #38
PeterDonis said:
ghwellsjr said:
Was the math of post #2 within your grasp?
It's not just the math of post #2; the OP needs to understand why the math of post #2, as you presented it, only applies to the traveling twin, not the stay-at-home twin. There is also a corresponding equation that applies to the stay-at-home twin, with the same ratio R in it, but it differs from the one you gave in post #2 in a crucial respect. What that crucial respect is is not a question of math, it's a question of physics. Post #2 gives a hint at the answer to the question of physics, but you didn't give it explicitly.
I don't know if this is what you are referring to but if, or rather since, (R+1/R)/2 is the ratio of the stay-at-home twin's accumulated age compared to the traveling twin's accumulated age, then we can say its inverse, 2/(R+1/R), is the ratio of the traveling twin's accumulated age compared to the stay-at-home twin's age. In other words, its from the perspective of the stay-at-home twin looking at the traveling twin and we did this without knowing hardly anything at all.

But we do know something else which is that unlike the traveling twin who immediately sees a change from R to 1/R in the Doppler signal coming from the stay-at-home twin, the stay-at-home twin will not immediately see a change in the Doppler signal coming from the traveling twin but rather the Doppler will remain at R for a longer time than it remains at 1/R. And it's easy to calculate the exact fraction of time for R and 1/R.

Let's call F the fraction of the total trip time as measured by the stay-at-home twin that he observes a Doppler of R coming from the traveling twin. This means that the fraction of the total time that he observes a Doppler of 1/R must be (1-F) because the total time must be the sum of the two fractional parts. And if we multiple each of these fractions by their respective Doppler factors and add them together, they must equal the total of 2/(R+1/R). Now we can write an equation and solve for F as a function of R:

FR + (1-F)/R = 2/(R+1/R)
(FR2+1-F)/R = 2R/(R2+1)
FR2+1-F = 2R2/(R2+1)
F(R2-1)+1 = 2R2/(R2+1)
F(R2-1) = 2R2/(R2+1)-1
F(R2-1) = (2R2-R2-1)/(R2+1)
F(R2-1) = (R2-1)/(R2+1)
F = 1/(R2+1)

The other fraction is merely (1-F):

1-F = 1-1/(R2+1) = (R2+1-1)/(R2+1) = R2/(R2+1)

And of course the sum of the two bold terms is 1.

This was pure math so I don't know if it is what you were driving at. Could you enlighten me?
 
  • #39
ghwellsjr said:
Let's call F the fraction of the total trip time as measured by the stay-at-home twin that he observes a Doppler of R coming from the traveling twin. This means that the fraction of the total time that he observes a Doppler of 1/R must be (1-F) because the total time must be the sum of the two fractional parts.

This is what I was referring to; the fraction F for the stay-at-home twin is different than it is for the traveling twin. For the latter it's just 1/2, since the Doppler changes when he turns around halfway through the trip. That's the key item of physics that I was talking about; the traveling twin observes the change in Doppler as soon as he turns around (because his turning around is what changes it), but the stay-at-home twin doesn't observe the change in Doppler until light from the turnaround point reaches him. This is a key aspect of the scenario that is not symmetric between the twins.
 
  • #40
ghwellsjr said:
And it's easy to calculate the exact fraction of time for R and 1/R... And if we multiple each of these fractions by their respective Doppler factors and add them together, they must equal the total of 2/(R+1/R).

That isn't legitimate, because you're simply assuming your result, and then backing out whatever fraction is needed to give you that result. What you would really need to do is compute the result from the perspective of the stay-home twin from the same principles that are used to compute the result for the traveling twin, rather than simply assuming that the results are reciprocals of each other. This means you need to derive the fraction F from first principles, and then show this leads to the result 2/(R+1/R), rather than simply assuming the latter result and claiming that F just happens to have whatever value it needs to have in order to give that result. This will require you to actually quantify your reasoning about speeds, spatial distances, time intervals, and inertia, because ultimately that's the difference between the twins - one follows an inertial path and the other doesn't - so any valid explanation has to invoke the concept of inertia in some form.
 
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  • #41
Samshorn said:
ghwellsjr said:
But the inverse nature of the two Dopplers is not classically contradictory as you have incorrectly concluded. Once you figure that out, maybe you'll have a different assessment.
This seems to be the central point. The disagreement isn't over the fact that (Da/Db)(Db/Dc) = Da/Dc = 1, but rather over the claimed equality (by relativity) Dc/Db = Db/Dc, which you use to claim that Da/Db and Dc/Db are recipricals.
Good, I'm glad we got that detail ironed out.

Samshorn said:
In other words, you rely on the relativistic fact that two inertial observers both see the same Doppler shift of signals from the other.
And the relativistic fact that you are talking about here is the Principle of Relativity, not the Theory of Special Relativity, correct?

Samshorn said:
Classically this equality doesn't unambiguously hold.
If it doesn't, then classical concepts regarding the Principle of Relativity are inadequate, aren't they?

Samshorn said:
Here's why:

We have two inertial clocks A and B with some mutual velocity v, and we assert the two principles of (1) relativity between inertial coordinate systems, and (2) independence of light speed from the speed of the source (again in terms of any inertial coordinate system).
I make a distinction between inertial observers/clocks and inertial coordinate systems. If you're going to invoke inertial coordinate systems, then you need to assert another principle and I see at least three options:

1) Light propagates at c only in one detectable inertial frame.

2) Light propagates at c only in one nondetectable inertial frame.

3) Light propagates at c in any inertial frame.

If you pick option 1), then you have rejected the Principle of Relativity.

Samshorn said:
On this basis we can examine the relationship between the frequencies of light signals transmitted and received by the clocks in terms of (for example) the rest frame of A, in which the other clock B has velocity v.

Now, by the classical Doppler effect (meaning no relativistic time dilation of B relative to the time coordinates of A's rest frame), it's trivial to see that the light-speed principle implies that a signal sent from A with frequency fA will arrive at B with frequency fB where fB/fA = 1-v, whereas a signal sent from B with frequency FB will arrive at A with frequency FA where FA/FB = 1/(1+v). These two ratios are not equal, but the relativity principle requires them to be equal, so classically the two principles are irreconcilable.
No, the first principle is irreconcilable with one of the third optional principles, namely, 1), as I already pointed out. So don't pick a third principle and just leave the analysis as I have, concluding that the traveling twin will have accumulated less time than his brother, based only on the Principle of Relativity and the assertion that light propagates independently of the motion of its source (plus a slew of other assumptions that nobody ever bothers to bring up in other discussions about the Twin Paradox).

The rest of your post shows what you can do by adopting either option 2) or option 3) but is not required to answer the OP's question.

Samshorn said:
The only way of reconciling those principles is that clock B must not advance equally with the coordinate time of A's rest frame. Specifically, clock B must advance more slowly by the time-dilation factor sqrt(1-v^2). Accounting for this difference in the rate of B, we find that the two ratios are

fB/fA = (1-v)/sqrt(1-v^2) and FA/fB = sqrt(1-v^2)/(1+v)

Both of these equal the relativistic Doppler factor sqrt[(1-v)/(1+v)], so we have satisfied the relativity requirement. Note that the time dilation factor applies regardless of whether A and B are approaching or receding from each other, and of course the Doppler effects in those cases are reciprocals.

The relativistic time dilation factor, which is necessary to logically reconcile the principles, automatically enables us to compute (quantitatively, not just qualitatively) the different ages of the twins, so nothing more is needed, and nothing less suffices. I think a student would accept your version of the Doppler explanation only if he failed to realize that the premises are mutually contradictory unless we posit relativistic time dilation, and once we do that, nothing else is needed. Also note that no third clock is needed. At best the Doppler considerations provide just one more derivation of the time dilation factor, but of limited applicability, and without clearly defining how that fits with the Lorentz transformation. As such, it will certainly seem unsatisfactory to the student when he realizes that not only must B run slow in terms of the standard time coordinates of A's rest frame, but A must run slow in terms of the standard time coordinates of B's rest frame. The only way to really understand this is to understand how inertial coordinate systems are related to each other, in full (i.e., the Lorentz transformation).
Hopefully, a student would learn and understand that there is a difference between the observable differential aging between two observers/clocks and the unobservable time dilation explanation offered by a theory such as LET or SR. Maybe this difference might motivate a student to study the subject matter in a coherent way.
 
  • #42
Samshorn said:
ghwellsjr said:
And it's easy to calculate the exact fraction of time for R and 1/R... And if we multiple each of these fractions by their respective Doppler factors and add them together, they must equal the total of 2/(R+1/R).
That isn't legitimate, because you're simply assuming your result, and then backing out whatever fraction is needed to give you that result.
I didn't assume my result--I had no idea what it was going to be--and I was surprised at what it turned out to be.
Samshorn said:
What you would really need to do is compute the result from the perspective of the stay-home twin from the same principles that are used to compute the result for the traveling twin, rather than simply assuming that the results are reciprocals of each other. This means you need to derive the fraction F from first principles, and then show this leads to the result 2/(R+1/R), rather than simply assuming the latter result and claiming that F just happens to have whatever value it needs to have in order to give that result. This will require you to actually quantify your reasoning about speeds, spatial distances, time intervals, and inertia, because ultimately that's the difference between the twins - one follows an inertial path and the other doesn't - so any valid explanation has to invoke the concept of inertia in some form.
I don't understand your objection at all. If I can correctly determine that A sees B's clock accumulate Q times the amount of time on his own clock, then there can be no other answer than B sees A's clock accumulate 1/Q times the time on his own clock. Why is this controversial?
 
  • #43
ghwellsjr said:
Quote by ghwellsjr: "But the inverse nature of the two Dopplers is not classically contradictory as you have incorrectly concluded."
Quote by Samshorn: "By the classical Doppler effect (meaning no relativistic time dilation of B relative to the time coordinates of A's rest frame)... a signal sent from A with frequency fA will arrive at B with frequency fB where fB/fA = 1-v, whereas a signal sent from B with frequency FB will arrive at A with frequency FA where FA/FB = 1/(1+v). These two ratios are not equal, but the relativity principle requires them to be equal, so classically the two principles are irreconcilable."
Quote by ghwellsjr: "...then classical concepts regarding the Principle of Relativity are inadequate, aren't they?"

Indeed, which is why you were mistaken when you claimed above that the reciprocalness of the two Doppler effects is not classically contradictory. Hopefully that's clear now.

ghwellsjr said:
I make a distinction between inertial observers/clocks and inertial coordinate systems. If you're going to invoke inertial coordinate systems, then you need to assert another principle...

It isn't just me who is invoking inertial coordinate systems, you are invoking them too (albeit tacitly and with denials). The two principles that you explicitly cited as your foundation are both expressed in terms of inertial coordinate systems. For example, Einstein expresses the relativity principle by saying the laws of physics are not affected "whether they be referred to the one or the other of two systems of coordinates in uniform translatory motion". No other principle is needed for this, provided of course that you've given some operationally meaningful definition of "inertial coordinate system" (which Einstein did in the first sentence of the first paragraph of Part I of his EMB paper). Likewise the lightspeed principle (or any proposition about 'speeds') has meaning only in the context of operationally meaningful measures of space and time.

ghwellsjr said:
Just leave the analysis as I have, concluding that the traveling twin will have accumulated less time than his brother, based only on the Principle of Relativity and the assertion that light propagates independently of the motion of its source...

Again, those two principles are expressed in terms of inertial coordinate systems. So you can't claim to be invoking these principles while denying that you are treating with inertial coordinate systems. Also, even if we overlook the logical inconsistency of your analysis, it doesn't allow you to conclude the traveling twin ages less, because it fails to quantity Q, which could be 1.

ghwellsjr said:
Hopefully, a student would learn and understand that there is a difference between the observable differential aging between two observers/clocks and the unobservable time dilation explanation offered by a theory such as LET or SR. Maybe this difference might motivate a student to study the subject matter in a coherent way.

Special relativity is not incoherent, whereas the "explanation" that you propose as an alternative to special relativity is incoherent, for the reasons I've explained. Again, you invoke principles whose meanings are explicitly given in terms of inertial coordinates, and yet you claim to do without any reference to inertial coordinates. That isn't logically consistent. Also, you yourself admit that your approach can never be quantitative, and for the same reason, i.e., inability to connect with any operationally defined measures of space and time.

ghwellsjr said:
I didn't assume my result--I had no idea what it was going to be--and I was surprised at what it turned out to be.

No, you simply assumed your result, i.e., you assumed that the age ratios of the twins are reciprocal, and from this assumption you backed out what value F would need to have in order for your assumption to be true. That is backwards from what's needed to demonstrate why the twins get reciprocal results.

ghwellsjr said:
I don't understand your objection at all. If I can correctly determine...

Well, it's certainly possible to correctly determine the result according to special relativity, but one of my points is that you haven't correctly determined anything. Again, without providing any operational meaning to the concepts of speed (as in the 'speed' of light and the motions of the twins) you have no warrant to even expect any Doppler shift at all (e.g., your unquantified Q could always be 1), not to mention that you can't apply either of the principles you claim to be applying, which are based on the same operational meanings. All you've done is appropriated a result from special relativity and forfeited its quantitative content and claimed to be able to explain it qualitatively by some fuzzy hand-waving (with the mistaken notion that classical Doppler is all that's required) with no logical coherence. For example, you claim time dilation is not observable, which of course is false, and in fact time dilation is a necessary consequence of your premises, as explained previously.

ghwellsjr said:
...that A sees B's clock accumulate Q times the amount of time on his own clock, then there can be no other answer than B sees A's clock accumulate 1/Q times the time on his own clock. Why is this controversial?

It's called the twins "paradox" precisely because beginning students suspect that if we worked out the result from the other twin's perspective, applying the same principles, we would get the symmetrical (not reciprocal) result, typically because they think of relativity in purely kinematical terms, without understanding the dynamical foundations in the principle of inertia. Hence they suspect that special relativity is logically inconsistent. To explain why special relativity is logically consistent, and why the other twin sees 1/Q, it makes no sense to simply assume that special relativity is logically consistent and that therefore the twin must see 1/Q. This is the very thing that we need to show, by carrying out the analysis from the other twin's point of view. This requires us to explain why the two points of view are dynamically asymmetrical, even though they are kinematically symmetrical, and this requires us to define operationally meaningful measures of space and time - the very thing you fail to do.
 
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  • #44
Samshorn said:
Indeed, which is why you were mistaken when you claimed above that the reciprocalness of the two Doppler effects is not classically contradictory. Hopefully that's clear now.
I hope you are not purposely misrepresenting what I said. I will assume that you never understood what I said. So I will draw a diagram to show how even under classical Doppler the explanation that I offered in the second paragraph of post #3 works:

attachment.php?attachmentid=58642&stc=1&d=1368288714.png

The horizontal axis is distance and the vertical axis is time.

Here is that explanation:

ghwellsjr said:
Let's say that you are looking at two observers with clocks. One of them is a fixed distance away from you so you will see his clock ticking at the same rate as your own. They may have different times on them but we don't care about that, we only care about tick rates. The second observer is traveling at a constant speed from the first one directly toward you. Therefore, you will see his clock ticking faster than yours. You also know that you are seeing time progress on the traveler's clock compared to the first observer's clock just as the traveler is seeing them but delayed in time, provided that the light from both clocks travels towards you at the same speed. Therefore the ratio of the tick rates that you see of the distant fixed clock to the traveler's clock is some ratio that we called R. But the ratio of the tick rates that you see of the traveler's clock compared to your own (which is the same as the distant fixed clock) is the inverse of R.

You are the green observer C. You are looking at the clocks of the red observer A and the blue observer B. The red observer A is stationary with respect to you but both of you are traveling in the medium at the same speed. The blue observer, B, is traveling at some unknown, arbitrary speed from observer A towards you, the green observer C. The tick of each clock is represented by a colored dot.

In this example, you see blue's clock ticking three times faster than your own as depicted by the pale blue signal lines coming from his tick dots toward you. But you receive three of his for every one of your tick dots.

You can also see the same thing that the blue observer sees when he looks at the red observer's clock compared to his own, just delayed in time, namely that blue sees red's clock ticking at 1/3 the rate of his own clock. This is depicted by the black lines coming from each of red's tick dots and impinging on blue's tick dots but only every third one.

The two Doppler effects that I was talking about are your observations of B's clock compared to your own (3, in this example) and A's clock compared to B's clock (1/3, in this example). You can pick any other example and you will get the same result that the two Doppler factors are inverses of each other. And it doesn't matter if you consider classical Doppler or Relativistic Doppler or if you consider that the observers are moving with respect to a medium or stationary in it.

OK, got that? Please do not misrepresent what I said again. Of course if you think I am mistaken, then you're going to have to offer some proof--a diagram would be good.

Samshorn said:
It isn't just me who is invoking inertial coordinate systems, you are invoking them too (albeit tacitly and with denials). The two principles that you explicitly cited as your foundation are both expressed in terms of inertial coordinate systems. For example, Einstein expresses the relativity principle by saying the laws of physics are not affected "whether they be referred to the one or the other of two systems of coordinates in uniform translatory motion". No other principle is needed for this, provided of course that you've given some operationally meaningful definition of "inertial coordinate system" (which Einstein did in the first sentence of the first paragraph of Part I of his EMB paper). Likewise the lightspeed principle (or any proposition about 'speeds') has meaning only in the context of operationally meaningful measures of space and time.
The Principle of Relativity has experimental evidence which does not require the formulation of a specific set of laws and a transformation that embodies that principle. This is what Einstein was talking about in his 1920 book on relativity. For example, in at the end of chapter 5 which is about the Principle of Relativity, he says:

However, the most careful observations have never revealed such anisotropic properties in terrestrial physical space, i.e. a physical non-equivalence of different directions. This is a very powerful argument in favour of the principle of relativity.

Or in chapter 7 where he says:

Prominent theoretical physicists were therefore more inclined to reject the principle of relativity, in spite of the fact that no empirical data had been found which were contradictory to this principle.

So I'm talking about the "careful observations" and "empirical data" that support the Principle of Relativity without considering what specific laws or transformation is involved.

When MMX was performed, did they have to establish a reference frame or a coordinate system involving spatial coordinates and an establishment of coordinate time throughout that spatial coordinate system? In the same way, my thought experiment involving three observers and their clocks, which is separate from the twin scenario, does not establish any coordinate system. It does make a lot of assumptions which are always made in any discussion of this type but are rarely enunciated, so I don't know why we have to get any more specific than just stating the normal and usual way of discussing a thought experiment.

So in my three-observer/clock thought experiment, I stated that if the Principle of Relativity is assumed to be true, then whatever Doppler effect each observer determines for any other observer has a symmetrical relationship. This is the part that I said does not work under classical Doppler but that because classical Doppler does not conform to the Principle of Relativity in the way that we are concerned about here.

Samshorn said:
Again, those two principles are expressed in terms of inertial coordinate systems. So you can't claim to be invoking these principles while denying that you are treating with inertial coordinate systems. Also, even if we overlook the logical inconsistency of your analysis, it doesn't allow you to conclude the traveling twin ages less, because it fails to quantity Q, which could be 1.
Once again, you are misrepresenting what I said. What I said was:
ghwellsjr said:
If I can correctly determine that A sees B's clock accumulate Q times the amount of time on his own clock, then there can be no other answer than B sees A's clock accumulate 1/Q times the time on his own clock. Why is this controversial?
This is true even if Q = 1, isn't it? And it's true when Q ≠ 1, isn't it. It's not controversial, is it?

Samshorn said:
Special relativity is not incoherent, whereas the "explanation" that you propose as an alternative to special relativity is incoherent, for the reasons I've explained. Again, you invoke principles whose meanings are explicitly given in terms of inertial coordinates, and yet you claim to do without any reference to inertial coordinates. That isn't logically consistent. Also, you yourself admit that your approach can never be quantitative, and for the same reason, i.e., inability to connect with any operationally defined measures of space and time.
Again, a misrepresentation of what I said. I didn't say or imply that SR was incoherent. Why do you imply things like this? And if you properly understood my approach, you would see that it is perfectly coherent and logically consistent and has nothing to do with the fact that it has limitations.

Samshorn said:
No, you simply assumed your result, i.e., you assumed that the age ratios of the twins are reciprocal, and from this assumption you backed out what value F would need to have in order for your assumption to be true. That is backwards from what's needed to demonstrate why the twins get reciprocal results.
You make it sound like my assumption is invalid. How can it not be valid? You objection doesn't make any sense at all.

Samshorn said:
Well, it's certainly possible to correctly determine the result according to special relativity, but one of my points is that you haven't correctly determined anything. Again, without providing any operational meaning to the concepts of speed (as in the 'speed' of light and the motions of the twins) you have no warrant to even expect any Doppler shift at all (e.g., your unquantified Q could always be 1), not to mention that you can't apply either of the principles you claim to be applying, which are based on the same operational meanings. All you've done is appropriated a result from special relativity and forfeited its quantitative content and claimed to be able to explain it qualitatively by some fuzzy hand-waving (with the mistaken notion that classical Doppler is all that's required) with no logical coherence. For example, you claim time dilation is not observable, which of course is false, and in fact time dilation is a necessary consequence of your premises, as explained previously.
Another misrepresentation of what I said. I never said "that classical Doppler is all that's required". You need to read carefully all that I have written and understand it instead of misrepresenting it and accusing me of having mistake notions and characterizing my posts as "fuzzy hand-waving" and having "no logical coherence".

Samshorn said:
It's called the twins "paradox" precisely because beginning students suspect that if we worked out the result from the other twin's perspective, applying the same principles, we would get the symmetrical (not reciprocal) result, typically because they think of relativity in purely kinematical terms, without understanding the dynamical foundations in the principle of inertia. Hence they suspect that special relativity is logically inconsistent. To explain why special relativity is logically consistent, and why the other twin sees 1/Q, it makes no sense to simply assume that special relativity is logically consistent and that therefore the twin must see 1/Q. This is the very thing that we need to show, by carrying out the analysis from the other twin's point of view. This requires us to explain why the two points of view are dynamically asymmetrical, even though they are kinematically symmetrical, and this requires us to define operationally meaningful measures of space and time - the very thing you fail to do.
I made very clear that I was not talking about Special Relativity. I was talking about the Principle of Relativity. The OP did not ask about Special Relativity, he asked:

Moris526 said:
In the twin's paradox, why one of the brothers is older when they meet again?. if movement is relative. What determines which of them ends up older?

Movement being relative is an issue of the Principle of Relativity and I wanted to show that it's not because of any particular theory about relativity, but just because of that Principle and that light propagates independently of it the motion of its source, that we can determine which twin ends up older, just like he asked.

By the way, none of the ideas that I presented in this thread are unique to me. They have come up on this forum and in other references by other people in the past.
 

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  • #45
If this example is correct, it shows B seeing a slower clock rate for A than A sees for B. That would mean the classical doppler is not symmetrical.

https://www.physicsforums.com/attachments/58741
 
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  • #46
phyti said:
If this example is correct, it shows B seeing a slower clock rate for A than A sees for B. That would mean the classical doppler is not symmetrical.

https://www.physicsforums.com/attachments/58741
Actually, I didn't show A watching B but I have added it into this drawing:

attachment.php?attachmentid=58755&stc=1&d=1368667314.png

It turns out that A sees B's clock ticking at 1/3 his own just like B sees A's clock so in this case, it is symmetrical but that's because A and B are both traveling at the same speed in opposite directions with respect to the classical medium and so the ratio is 1 as your document states.

Later on, I'll upload some more examples where A and B are traveling at different speeds so the ratio of their Dopplers will not be 1 and yet B's Doppler observation of A is the inverse of C's Doppler observation of B. That, after all, was the point of the drawing, not any symmetry or non-symmetry anywhere else.
 

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  • #47
ghwellsjr said:
Please do not misrepresent what I said again. Of course if you think I am mistaken, then you're going to have to offer some proof--a diagram would be good.

I think your mistake has already been explained, and the exact equations were presented to prove it (i.e., the classical Doppler equations). Actually there are two classical Doppler models: The classical emission theory satisfies the relativity principle but not the independence of light-speed from the source, whereas the classical wave model satisfies the independence principle but not the relativity principle. Neither of these would imply any asymmetric ages for the re-united twins.

ghwellsjr said:
The Principle of Relativity has experimental evidence which does not require the formulation of a specific set of laws and a transformation that embodies that principle.

The point is that I quoted the statement of the relativity principle from Einstein's EMB paper, and it is explicitly expressed in terms of inertial coordinate systems. If that's the principle you are invoking, then you can't claim to not be referring to inertial coordinate systems. Conversely, if you claim your approach does not rely in any way on the concept of inertial coordinate systems, then you can't claim to be invoking the principle of relativity. If you want to propose some other principle, you're free to do so, but you shouldn't call it the principle of relativity, because that name is already taken.

ghwellsjr said:
I'm talking about the "careful observations" and "empirical data" that support the Principle of Relativity without considering what specific laws or transformation is involved.

Again, you need to clearly state your principles. My contention is that you can't - at least not without abandoning your approach and adopting special relativity.

ghwellsjr said:
When MMX was performed, did they have to establish a reference frame or a coordinate system involving spatial coordinates and an establishment of coordinate time throughout that spatial coordinate system?

If the experimenters had tried to do that themselves, they would have gotten it wrong, but nature did it for them. Physical entities in any uniform state of motion configure themselves in equilibrium in such a way that their descriptions are the same in terms of co-moving inertial coordinates. This is why the MMX returns null.

The coherent account provided by special relativity of all the available experimental results is based solidly on the empirical agreement between inertial isotropy (represented roughly by what Einstein called "coordinate systems in which Newton's laws of mechanics hold good") and light-speed isotropy. It is an empirical fact that the one-way speed of light (in vacuum) is c in terms of any such system of coordinates. Without understanding this, you can't really give a coherent account of special relativity.

ghwellsjr said:
This is the part that I said does not work under classical Doppler but that's because classical Doppler does not conform to the Principle of Relativity in the way that we are concerned about here.

No, you began by insisting that classical Doppler was adequate for your explanation. I'm glad to see you're not claiming that any more. (As a reminder, you said "there is no difference between Relativistic Doppler and Classical Doppler for what I described".)

ghwellsjr said:
I didn't say or imply that SR was incoherent. Why do you imply things like this?

Fortunately we have a verbatim record of what you said: "Hopefully, a student would learn and understand that there is a difference between the observable differential aging between two observers/clocks and the unobservable time dilation explanation offered by a theory such as LET or SR. Maybe this difference might motivate a student to study the subject matter in a coherent way."

Thus you claim your explanation differs from that offered by LET or SR because yours involves only observables, whereas LET or SR involve "unobservable (sic) time dilation", and you hope that when the student sees this he may be motivated to study the subject in a coherent way, by which you mean the way you have proposed based on "observables" rather than the way SR explains things in terms of "unobservable (sic) time dilation". You happen to be wrong, because time dilation IS observable, and the explanations given by special relativity ARE coherent, whereas your explanation is incoherent, for the reasons I've explained several times already. (You base yourself on a principle whose very terms you reject.)

ghwellsjr said:
This is true even if Q = 1, isn't it? And it's true when Q ≠ 1, isn't it. It's not controversial, is it?

Special relativity is valid and not controversial. As mentioned previously, what you've done is to appropriate one particular result from special relativity, denude it of its quantitative content (by disconnecting it from any physically meaningful conception of speed), and then proposed an inadequate and incoherent "explanation" of that result - and for good measure you've insinuated that special relativity is incoherent. And of course since your reasoning allows Q=1, it doesn't even unambiguously predict any asymmetric ages for the re-united twins. Yes, your claims are controversial.

ghwellsjr said:
The horizontal axis is distance and the vertical axis is time.

If so, then you are making use of a space-time coordinate system, and presumably you accept that your "distance" and "time" have some operational significance, since otherwise your drawings are meaningless. It's good that you now (at least tacitly) acknowledge that you are reasoning with space-time coordinates (contrary to your claim). If pressed, you would probably even admit (eventually) that you are dealing with a special class of space-time coordinates, specifically, a system in terms of which (as Einstein put it, somewhat imprecisely) "the equations of Newtonian mechanics hold good". But this contradicts your claim to be doing without inertial space-time coordinate systems, as well as your claim that time dilation is unobservable. Notice that you say the vertical axis is "time", and yet it cannot be "time" for the moving twin. (If it was, your principles are violated, as explained by the Doppler equations in a previous post.) Hopefully you can see that your diagram illustrates why "time" for the moving twin cannot be the "time" of your vertical axis.

ghwellsjr said:
You make it sound like my assumption is invalid. How can it not be valid?

Already explained.

ghwellsjr said:
Movement being relative is an issue of the Principle of Relativity and I wanted to show that it's not because of any particular theory about relativity, but just because of that Principle and that light propagates independently of it the motion of its source, that we can determine which twin ends up older, just like he asked.

Again, I quoted the statement of the principle of relativity from Einstein's EMB paper, which is explicitly expressed in terms of inertial coordinate systems, and yet you claim that you can apply this principle without any consideration of or reference to inertial coordinate systems. That is plainly illogical. It also explains why you think time dilation is unobservable, and why you don't understand the empirical content of the light-speed principle.

ghwellsjr said:
By the way, none of the ideas that I presented in this thread are unique to me. They have come up on this forum and in other references by other people in the past.

That's unfortunately true.
 
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  • #48
ghwellsjr said:
phyti said:
If this example is correct, it shows B seeing a slower clock rate for A than A sees for B. That would mean the classical doppler is not symmetrical.

https://www.physicsforums.com/attachments/58741
Actually, I didn't show A watching B but I have added it into this drawing:

attachment.php?attachmentid=58755&stc=1&d=1368667314.png

It turns out that A sees B's clock ticking at 1/3 his own just like B sees A's clock so in this case, it is symmetrical but that's because A and B are both traveling at the same speed in opposite directions with respect to the classical medium and so the ratio is 1 as your document states.

Later on, I'll upload some more examples where A and B are traveling at different speeds so the ratio of their Dopplers will not be 1 and yet B's Doppler observation of A is the inverse of C's Doppler observation of B. That, after all, was the point of the drawing, not any symmetry or non-symmetry anywhere else.
I have two more examples. Here's the first one:

https://www.physicsforums.com/attachment.php?attachmentid=58761&stc=1&d=1368690498​

In this example, A and C are stationary with respect to the medium and B is traveling at one-half the speed of a signal in the medium. You can see that B observes A's clock ticking at one-half the rate of his own and C can also observe this. You can see that C observes that B's clock is ticking at double the rate of his own. These two ratios are inverses of each other which is the point that I have been making, for all Doppler ratios, whether classical or relativistic, as long as the speed of the signal is independent of the source.

Now to the issue that you brought up, B observes A's clock ticking at one-half the rate of his own clock but A observes B's clock ticking at two-thirds the rate of his own clock.

A second example:

https://www.physicsforums.com/attachment.php?attachmentid=58762&stc=1&d=1368690498​

In this example, A is traveling to the left at 40% of the speed of signals in the medium and B is traveling to the right at 12.5%. As a result, B sees A's clock ticking at 5/8 of the rate of his own clock. And C sees B's clock ticking at 8/5 the rate of his own.

And A observes B's clock ticking at 8/15 of his own clock.

Does this all make sense? Can you see that the only time A and B have a symmetrical measurement of the other ones clock with respect to their own is when they are both traveling in the medium at the same speed.
 
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  • #49
ghwellsjr's first post is total bull, on the reverse trip each brother still sees the other's clock as slower, by exactly the same ratio if the speed difference is the same...

the key to this paradox is understanding the different definition of "now", or lines/planes/spaces of simultaneoty on a spacetime diagram, for different frames of reference. more specifically, the 'now' of the twin right before he reverses is much different than his 'now' after he reverses, from which comes the greater age of his distant twin despite he always seeing the rate of his clock slower.

draw a spacetime diagram and it is very obvious.
 
  • #50
georgir said:
ghwellsjr's first post is total bull, on the reverse trip each brother still sees the other's clock as slower, by exactly the same ratio if the speed difference is the same...
The two of you are just using the words "see" in different ways. You appear to be talking about the coordinate assignments made by the comoving inertial coordinate systems. He appears to be talking about the light that reaches the twins.
 

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