The twin paradox

[swansont]

(emphasis mine)

Not your bookkeeping, your eyes.

 

Light takes 1 year to make the trip.

And you have seen in one year a spaceship making the trip.

 

But we are discussing relativity and we have to do bookkeeping. If you neglect how Einstein described how clocks are synchronized, you aren't discussing relativity, you're discussing something else. If you don't use the definition of speed that the rest of physics uses, you're describing a different variable.

 

At no point does the rocket keep pace with a photon. We get photons sent from the rocket all during the trip — we see the rocket approach.

[Delta1212]

If I use the same method I should use a part of time twice.

 

Explaining

 

The green triangle shows me what the earth observes till Tom reaches point R. I hope everyone agrees on that.

 

From this moment and after, Tom leaves point R.

 

I cannot use the time before he left (the orthogonal projection of R on the time axis) because this part of time has been already used in the green triangle.

 

 

(emphasis mine)

Not your bookkeeping, your eyes.

 

 

Light takes 1 year to make the trip.

And you have seen in one year a spaceship making the trip.

But what you see is not what happened, and c is only a speed limit for things that actually happen, not for things you see.

 

Let's say that instead of waving, he turns on a lightbulb. How long after you see him flick the switch will you see the light from the bulb reach you?

[Iggy]

If I use the same method I should use a part of time twice.

 

Explaining

 

The green triangle shows me what the earth observes till Tom reaches point R. I hope everyone agrees on that.

 

From this moment and after, Tom leaves point R.

 

I cannot use the time before he left (the orthogonal projection of R on the time axis) because this part of time has been already used in the green triangle.

 

I'm afraid that didn't make any sense to me.

 

Are you saying that the procedure for finding 'apparent velocity' changes depending on what you are measuring? Your idea of 'apparent velocity' sounds inconsistent in that case.

 

 

Actually, let's set that aside for a second, because you're going to end up with different 'apparent velocity' depending on where the observer's position is in space as well, and that's an even bigger inconsistency.

 

Imagine an observer at x = 1 and 1/3rd. That is to say, on your diagram, an observer is at x = 1.333 at rest (parallel to earth's world line). He is 2/3rds of a light-year from point R just like the earth observer is, except on the other side.

 

What does he measure as the 'apparent velocity'? It isn't going to be the same as earth.

[DimaMazin]

While on the outbound leg, the Earth ages more slowly than Tom, according to Tom

While on the inbound leg, the Earth ages more slowly than Tom, according to Tom.

 

However, Tom must change inertial frames when going from outbound to inbound, and this frame switch causes the time on Earth to advance ahead according to Tom. This is due to relativity of simultaneity. Thus, according to Tom's reckoning, the Earth ages 1.5 yr to his 1.732 yr during the outbound leg and ages 1.5 yr during the inbound leg, but ages 1 yr in the interval between the two legs.

 

Or let's put it this way. Assume we have a clock at the turn around point which is synced to the Earth. They both read a time of zero according to the Earth when Tom leaves (this of course means that the Earth won't actually see the turnaround clock read 0 until the Earth clock reads 1 yr)

 

According to Tom, traveling at 0.5c towards the Turnaround clock, the turnaround clock already reads 1/2 yr. During the 1.732 yrs it takes by his clock for the turmaround point to reach him, its clck advances 1.5 yr and reads 2 yr when they meet. This is the same 2 yr the Earth observer says the turnaround clock reads upon Tom's arrival.

 

At this moment, according to Tom the Earth clock reads 1.5 years.

 

Tom makes the switch to the return leg. The clock at turnaround stays at 2 yr, However, since Earth now has a relative velocity towards him, it is the Earth clock that must read ahead of the turnaround clock by 1/2 yr. Thus the Earth clock goes from reading 1.5 years to reading 2.5 years. The Earth clock then advances 1.5 years during the return to read 4 yr upon Tom's arrival.

 

So here is the gist of it. the Earth observer can consider himself at rest during the whole exercise. Tom can consider himself at rest on the outbound leg or he can consider himself at rest on the return leg. What he cannot do is consider himself at rest during the whole trip (if we restrict ourselves to SR) According to Tom outbound, Tom Inbound is moving, and According to Tom inbound, Tom outbound is moving.

 

Relativity does not contradict reality, it is reality.

Coordinate system of observer is motionless system concerning the observer. It is mathematical definition. You contradict math.

[tar]

md65536, on 14 May 2013 - 01:54, said:

It was a huge step forward in analysis that I saw, but I figured you gave up.

 

Well may I make a suggestion? Don't worry about all of those extra details and philosophical questions. Most of your questions would change as you learn anyway. Pretend you believe the predictions of SR are real, and work through the simplest twin experiment example from this thread or the web. First see what SR says is so, THEN begin questioning from there, or adding in all the details of reality vs. appearances etc, to see how everything fits. I think it would be a torturous task to try to intuitively understand the predictions of SR without first knowing what the predictions actually are.Certainly. That would be a Doppler analysis. I preferred figuring out the twin paradox using Doppler equations because they seemed so much more intuitive, describing it in terms of what observers actually see, however there are still a lot of complicated details of timing and frames etc and it's easy to make errors or get hopelessly lost. Knowing a Lorentz analysis of the same experiment lets you calculate and double-check any details you need, and makes a Doppler analysis easier.

MD65536.

 

Good advice. Although I think if I had the horsepower, I would have done it that way from the get go.

 

Iggy,

 

If I were to put numbers to the understanding, I think I would work with the peak frequency of black body radiation. Take that frequency and wavelength and figure how many actual waves there are, if one would take a static, instantanous, "now" picture of the space between here and the physical "turn around" point in our experiment. Counting the number of waves, I would have a number W. At the start of the experiment Earth would be at 0, the ship would be at 0 and the turn around point would be at W.

Then I would start the experiment and start all three beating at that frequency and put a label on each wave and plot its position in the static fast as thought stage we have set up. Everything should add up, and the ship should count Earths beats as longer wavelengths and lower frequency, at a .577 to normal rate, and should count the turnaround point's beats at shorter more frequent rate of 1.732 of normal. By the time the ship reaches the turn around point, he will be at his own beat count plus W and will find his count of the Earth's beats is at his own beat count, -W.

 

Regards, TAR2

[Iggy]

Iggy,

 

If I were to put numbers to the understanding, I think I would work with the peak frequency of black body radiation. Take that frequency and wavelength and figure how many actual waves there are, if one would take a static, instantanous, "now" picture of the space between here and the physical "turn around" point in our experiment.

 

How many waves there are over a distance in an instant?

 

That honestly doesn't make sense to me. I'm trying to understand what you mean, but I really don't follow.

 

 

Counting the number of waves, I would have a number W. At the start of the experiment Earth would be at 0, the ship would be at 0 and the turn around point would be at W.

Then I would start the experiment and start all three beating at that frequency and put a label on each wave and plot its position in the static fast as thought stage we have set up. Everything should add up, and the ship should count Earths beats as longer wavelengths and lower frequency, at a .577 to normal rate, and should count the turnaround point's beats at shorter more frequent rate of 1.732 of normal. By the time the ship reaches the turn around point, he will be at his own beat count plus W and will find his count of the Earth's beats is at his own beat count, -W.

 

Regards, TAR2

 

The numbers 0.577 and 1.732 have time dilation worked into them. That isn't just the classical Doppler effect. It is that plus time dilation. You can't assume that those numbers are correct while also assuming, like you said before, that the numbers can be found without time dilation.

 

We know through experiment that time dilation is a necessary factor. The only way to ignore time dilation is to call it by a different name.

[michel123456]

But we are discussing relativity and we have to do bookkeeping. If you neglect how Einstein described how clocks are synchronized, you aren't discussing relativity, you're discussing something else. If you don't use the definition of speed that the rest of physics uses, you're describing a different variable.

 

At no point does the rocket keep pace with a photon. We get photons sent from the rocket all during the trip — we see the rocket approach.

"Light takes 1 year to make the trip.

And you have seen in one year a spaceship making the trip."

 

That was with 0,5c

 

What about going at 0,99c?

 

In this case I observe in my telescope Tom waving his hand, embarking, and landing the next day!

 

If I follow this logic, I may observe in my telescope an E.T. coming from the other side of the Universe in just one day because he traveled at 0,99999999999 c.

And I would have observed something that we deny light can do.

 

I'm afraid that didn't make any sense to me.

 

Are you saying that the procedure for finding 'apparent velocity' changes depending on what you are measuring? Your idea of 'apparent velocity' sounds inconsistent in that case.

 

 

Actually, let's set that aside for a second, because you're going to end up with different 'apparent velocity' depending on where the observer's position is in space as well, and that's an even bigger inconsistency.

 

Imagine an observer at x = 1 and 1/3rd. That is to say, on your diagram, an observer is at x = 1.333 at rest (parallel to earth's world line). He is 2/3rds of a light-year from point R just like the earth observer is, except on the other side.

 

What does he measure as the 'apparent velocity'? It isn't going to be the same as earth.

(emphasis mine)

You can't do that. The observer is always upon his own world line, the observer is always on the Y axis.

[Iggy]

{emphasis mine)

 

You can't do that. The observer is always upon his own world line, the observer is always on the Y axis.

 

You can draw the Y axis wherever you want. The position of the y axis doesn't tell you (or imply anything about) where any particular world line is.

 

Do you want me to draw you a diagram where the Y axis is exactly where I said the observer was in my last question so that you can answer it?

[michel123456]

You can draw the Y axis wherever you want. The position of the y axis doesn't tell you (or imply anything about) where any particular world line is.

 

Do you want me to draw you a diagram where the Y axis is exactly where I said the observer was in my last question so that you can answer it?

No.

If you are the observer, you are on your world line. You cannot take the diagram and state that the observer is anywhere on it.

[Iggy]

No.

If you are the observer, you are on your world line. You cannot take the diagram and state that the observer is anywhere on it.

 

An observer can be anywhere on a diagram. You can imagine you are that observer if that somehow matters to you. You can make the Y axis the world line of that observer if that somehow matters to you. (by the way, those last two don't matter at all).

 

My question:

Imagine an observer at x = 1 and 1/3rd. That is to say, on your diagram, an observer is at x = 1.333 at rest (parallel to earth's world line). He is 2/3rds of a light-year from point R just like the earth observer is, except on the other side.

What does he measure as the 'apparent velocity'? It isn't going to be the same as earth.

Are you saying that this observer doesn't have an 'apparent velocity', because that alone would invalidate the velocity you've invented.

 

Are you saying that there can't be an observer there? Because, I'm pretty sure there can be an observer 1.333 light-years from earth and motionless relative to the earth.

 

Are you saying that "you" cannot be that observer? Because, I'm pretty sure you're just as capable of occupying a spacecraft as anyone else.

 

 

I think you're just refusing to answer the question because you know it gives you an inconsistent answer. Either that, or you don't know how to find the answer. If that is the case, let me know and I'll solve it and explain the solution.

[michel123456]

An observer can be anywhere on a diagram. You can imagine you are that observer if that somehow matters to you. You can make the Y axis the world line of that observer if that somehow matters to you. (by the way, those last two don't matter at all).

 

My question:

Imagine an observer at x = 1 and 1/3rd. That is to say, on your diagram, an observer is at x = 1.333 at rest (parallel to earth's world line). He is 2/3rds of a light-year from point R just like the earth observer is, except on the other side.

What does he measure as the 'apparent velocity'? It isn't going to be the same as earth.

Are you saying that this observer doesn't have an 'apparent velocity', because that alone would invalidate the velocity you've invented.

 

Are you saying that there can't be an observer there? Because, I'm pretty sure there can be an observer 1.333 light-years from earth and motionless relative to the earth.

 

Are you saying that "you" cannot be that observer? Because, I'm pretty sure you're just as capable of occupying a spacecraft as anyone else.

 

 

I think you're just refusing to answer the question because you know it gives you an inconsistent answer. Either that, or you don't know how to find the answer. If that is the case, let me know and I'll solve it and explain the solution.

(emphasis mine)

i didn't even read the question. from the moment you propose to put the observer out of his world line, it stops to make sense.

You can't make a diagram for observer Michel in Greece describing what Iggy observes where he is.

either you will make a diagram for Michel and the Y axis will be his world line, or make another diagram for Iggy and then Iggy will be on the Y axis.

Putting the observer out of the Y axis is wrong.

Edited May 15, 2013 by michel123456

[Iggy]

I didn't even read the question.

 

I think we've found the problem.

 

 

from the moment you propose to put the observer out of his world line

 

I proposed no such thing. The observer is 1.333 light-years from earth. The world line is 1.333 light-years from earth.

 

The world-line is not on the axis of the diagram you had previously drawn, but an axis is not a world-line. An axis is just a label. It labels the lines running horizontal on the diagram. You can put it wherever you want and it makes no difference.

 

*I did not put the observer off his world line*

 

I don't know how more plainly to say it.

 

 

, it stops to make sense.

 

My question makes perfect sense. What 'apparent velocity' does an observer 1.333 light-years from earth measure for the blue line?

 

If that question doesn't make sense then there is something seriously wrong with your 'apparent velocity'.

 

You can't make a diagram for observer Michel in Greece describing what Iggy observes where he is.

 

I'm not asking you the 'apparent velocity' observed from earth by the observer 1.333 lightyears from earth.

 

I'm asking you the 'apparent velocity' observed 1.333 lightyears from earth by the observer 1.333 lightyears from earth.

 

 

either you will make a diagram for Michel and the Y axis will be his world line, or make another diagram for Iggy and then Iggy will be on the Y axis.

Putting the observer out of the Y axis is wrong.

 

Putting the observer off the y axis is not wrong. But, it doesn't matter. Put the axis on the observer if you want. I want to know the 'apparent velocity' of the blue line from the perspective of an observer 1.333 light-years from earth. Put the y-axis right on top of him if you want. I don't want to know anything about any other observer.

 

How is this not clear?

[swansont]

"Light takes 1 year to make the trip.

And you have seen in one year a spaceship making the trip."

 

That was with 0,5c

 

What about going at 0,99c?

 

In this case I observe in my telescope Tom waving his hand, embarking, and landing the next day!

 

If I follow this logic, I may observe in my telescope an E.T. coming from the other side of the Universe in just one day because he traveled at 0,99999999999 c.

And I would have observed something that we deny light can do.

Not at all. You're ignoring how physics has defined things. You can't ignore physics to try and show that physics is wrong.

[michel123456]

I think we've found the problem.

 

 

 

I proposed no such thing. The observer is 1.333 light-years from earth. The world line is 1.333 light-years from earth.

 

The world-line is not on the axis of the diagram you had previously drawn, but an axis is not a world-line. An axis is just a label. It labels the lines running horizontal on the diagram. You can put it wherever you want and it makes no difference.

 

*I did not put the observer off his world line*

 

I don't know how more plainly to say it.

 

 

 

My question makes perfect sense. What 'apparent velocity' does an observer 1.333 light-years from earth measure for the blue line?

 

If that question doesn't make sense then there is something seriously wrong with your 'apparent velocity'.

 

 

I'm not asking you the 'apparent velocity' observed from earth by the observer 1.333 lightyears from earth.

 

I'm asking you the 'apparent velocity' observed 1.333 lightyears from earth by the observer 1.333 lightyears from earth.

 

 

 

Putting the observer off the y axis is not wrong. But, it doesn't matter. Put the axis on the observer if you want. I want to know the 'apparent velocity' of the blue line from the perspective of an observer 1.333 light-years from earth. Put the y-axis right on top of him if you want. I don't want to know anything about any other observer.

 

How is this not clear?

It is clear that you want to know the 'apparent velocity' of the blue line from the perspective of an observer 1.333 light-years from earth.

 

This diagram shows what earthlings observe. It does not show what an observer 1.333 light-years from earth observes.

It does not show what Tom observes either. Nor any other observer in any other FOR.

 

_You want me to draw a diagram that shows an observer 1.333 light-years from Earth, in the same FOR with the Earth, observing Tom that leaves the Earth at 0,5c (as measured in Earth's FOR), then turning back to Earth at 0,5c (Earths FOR).

I suppose that you want Tom to approach the observer.

 

That's a bit tuff. Working on it.

 

Not at all. You're ignoring how physics has defined things. You can't ignore physics to try and show that physics is wrong.

Which of my statements was wrong?

 

This one?

What about going at 0,99c?

 

In this case I observe in my telescope Tom waving his hand, embarking, and landing the next day!

[swansont]

Which of my statements was wrong?

 

This one?

 

 

"I may observe in my telescope an E.T. coming from the other side of the Universe in just one day because he traveled at 0,99999999999 c.

And I would have observed something that we deny light can do."

 

One will only "observe" something that can't be done by being ignorant of physics. In this case, willfully ignorant, because the observer is aware of relativity.

[michel123456]

--------------------
Here you are (hoping not being toooo wrong)

I needed to introduce a "help observer" at point R. The velocity that observes the observer at o is the same as observed at R with a delay caused by c. (words in italic edited)





It is clear like a crystal.

"I may observe in my telescope an E.T. coming from the other side of the Universe in just one day because he traveled at 0,99999999999 c.
And I would have observed something that we deny light can do."

One will only "observe" something that can't be done by being ignorant of physics. In this case, willfully ignorant, because the observer is aware of relativity.

So the statement:

What about going at 0,99c?

In this case I observe in my telescope Tom waving his hand, embarking, and landing the next day!

Is correct.

and this part

"I may observe in my telescope an E.T. coming from the other side of the Universe in just one day because he traveled at 0,99999999999 c.

Is correct too.

Edited May 15, 2013 by michel123456

[Iggy]

--------------------

Here you are (hoping not being toooo wrong)

 

I needed to introduce a "help observer" at point R. The velocity that observes the observer at o is the same as observed at R with a delay caused by c. (words in italic edited)

 

 

 

 

It is clear like a crystal.

 

"clear like a crystal"

 

he he That was good.

 

I can't tell what's going on with that. There are no world lines where I would expect them. Let me draw the scenario...

 

I have to run a couple errands this morning first. It'll take a sec.

Edited May 15, 2013 by Iggy

[swansont]

!

Moderator Note

Moved, because we haven't been talking about accepted physics for a while in this thread.

[Delta1212]

--------------------

Here you are (hoping not being toooo wrong)

 

I needed to introduce a "help observer" at point R. The velocity that observes the observer at o is the same as observed at R with a delay caused by c. (words in italic edited)

 

 

 

 

It is clear like a crystal.

 

 

 

 

 

 

So the statement:

 

Is correct.

 

 

and this part

Is correct too.

Yes. As long as you realize that it didn't actually take one day to travel all that way from Earth's FOR because you have to take Doppler shift into account when calculating actual speed.

 

Light is like a postcard. It tells you where someone was when they sent it. It doesn't tell you where they are now. If someone sends you a postcard saying they just arrived in Hawaii for their vacation, and the card doesn't reach you for 5 days, you don't assume that their vacation was two days long when they return two days after you receive the card. You look at the postmark to see when the card was sent, and can then determine they spent a week away, even though it "looks" like they spent two days.

 

With light, you have to measure from the time the light was sent, rather than the time you received the light, in order to measure something's actual velocity. If something is keeping pace with the original light that reflected toward Earth, they can arrive very soon after it, and it will "look" like they moved much more quickly than is physically possible.

 

That doesn't mean they actually did, of course.

[swansont]

So the statement:

 

Is correct.

 

 

and this part

Is correct too.

 

Sure. If someone 1 LY away is traveling at some ridiculously fast speed such that gamma is ~365, and they send a signal at departure w/ negligible acceleration time, you will get the signal and they will show up a day after that signal arrives. They will think the trip took them a day, because length contraction makes the trip 1 light-day long. They will be one day older than when they left.

 

BUT if you then claim that the earth observer measures their trip to be one day long, then this is not a physics/relativity discussion. Relativity defines how clocks are synchronized, and physics defines what speed is. Clock synchronization accounts for the time of a light signal travel being c. Thus, if ET's clock said Jan 1, 2012 when he left, and that message was sent, you would be getting a message on Jan 1, 2013 with the 2012 timestamp on it.

 

This is functionally no different than if you send a letter to a friend who is 1000 miles away, saying you will come visit, and you immediately hop in your car. The letter arrives two days later at noon, and then you show up at 1PM, and the friend deciding that you must have traveled 1000 miles per hour.

 

edit: xpost w/ Delta. Funny we used similar examples.

[michel123456]

 

Sure. If someone 1 LY away is traveling at some ridiculously fast speed such that gamma is ~365, and they send a signal at departure w/ negligible acceleration time, you will get the signal and they will show up a day after that signal arrives. They will think the trip took them a day, because length contraction makes the trip 1 light-day long. They will be one day older than when they left.

 

BUT if you then claim that the earth observer measures their trip to be one day long, then this is not a physics/relativity discussion. Relativity defines how clocks are synchronized, and physics defines what speed is. Clock synchronization accounts for the time of a light signal travel being c. Thus, if ET's clock said Jan 1, 2012 when he left, and that message was sent, you would be getting a message on Jan 1, 2013 with the 2012 timestamp on it.

 

This is functionally no different than if you send a letter to a friend who is 1000 miles away, saying you will come visit, and you immediately hop in your car. The letter arrives two days later at noon, and then you show up at 1PM, and the friend deciding that you must have traveled 1000 miles per hour.

 

edit: xpost w/ Delta. Funny we used similar examples.

O.K. Good example (good to Delta too)

Very understandable.

Now what I cannot understand:

In the letter example, one can imagine the letter traveling on the back of a donkey and the writer traveling with an airplane. In such a a way that the traveler can knock at your door and enter gently even before the letter.

With light and massive objects, there are some conditions: 1. the "letter" takes the faster existing airplane and the writer can take only a slightly slower airplane and 2. the straight path of the letter is the same as the straight path of the writer.

 

When the letter arrives, it crashes upon your house because its velocity is c.

I don't understand how the hell the traveler will knock your door gently* (traveling at less than c) when you have seen him with your eyes coming to you at "apparent velocity" thousands of times that of the airplane.

 

*exaggerating a bit for the sake of the argument.

[Iggy]

--------------------

Here you are (hoping not being toooo wrong)

 

I needed to introduce a "help observer" at point R. The velocity that observes the observer at o is the same as observed at R with a delay caused by c. (words in italic edited)

 

 

 

 

It is clear like a crystal.

 

 

This was the diagram you had before:

 

 

 

 

 

 

I recreated it and added the scenario I asked about:

 

 

 

 

 

 

 

The two different observers, earth and B, measure a different apparent velocity (given the method you proposed for finding apparent velocity).

 

I know you're going to object for reasons I can't quite unravel, and that's fine. I look forward to it. But, I think this particular line of questions has played itself out. We've added confusion on top of confusion on top of confusion and I don't think it's helping anything.

Edited May 15, 2013 by Iggy

[swansont]

O.K. Good example (good to Delta too)

Very understandable.

Now what I cannot understand:

In the letter example, one can imagine the letter traveling on the back of a donkey and the writer traveling with an airplane. In such a a way that the traveler can knock at your door and enter gently even before the letter.

With light and massive objects, there are some conditions: 1. the "letter" takes the faster existing airplane and the writer can take only a slightly slower airplane and 2. the straight path of the letter is the same as the straight path of the writer.

 

When the letter arrives, it crashes upon your house because its velocity is c.

I don't understand how the hell the traveler will knock your door gently* (traveling at less than c) when you have seen him with your eyes coming to you at "apparent velocity" thousands of times that of the airplane.

 

*exaggerating a bit for the sake of the argument.

 

Two reasons.

 

We are ignoring acceleration as an unnecessary complication of the concept.

 

"Apparent velocity" is completely bogus.

 

That's why.

[michel123456]

Two reasons.

 

We are ignoring acceleration as an unnecessary complication of the concept.

 

"Apparent velocity" is completely bogus.

 

That's why.

 

 

That is not acceleration. We are talking about the transit between an "apparent velocity" (that dubious unphysical effect you have seen with your own eyes looking into your telescope) and "real physical velocity" when the writer knocks your door.

When did that change happen?

[swansont]

That is not acceleration. We are talking about the transit between an "apparent velocity" (that dubious unphysical effect you have seen with your own eyes looking into your telescope) and "real physical velocity" when the writer knocks your door.

When did that change happen?

 

That transition occurs when the traveler is co-located with you (and has accelerated into your frame, in the classic scenario). As has been described, this "apparent velocity" artifact is not relativistic in origin. It comes from improper clock synchronization and ignoring the definition of velocity.