The twin paradox

[Janus]

Janus,

 

Judging by Iggys approval I suppose you are right.

 

However, running the same numbers from the travelers point of view, she sees the Earth age at .557 per year rate, for two years and then at a 1.732 per year rate for another two, on her way back. Making the stay at home and the Earth, 4.618 (1.154 plus 3.464) years older, which they are not.

 

Although 4.618 x .866 would give you four years and I suppose that is what time dialation is for, it seems to add up for the wrong reasons.

 

Regards, TAR2

No. Due to length contraction, the distance between Earth and the Turn around point is 0.866 ly according to Tom. He takes 1.732 years to reach the turn around and sees Earth age 1 year (After you account for the fact that 0.577 and 1.732 are rounded off values.) He then sees Earth age at a rate of 1.732 for another 1.732 years on the return trip for another 3 yrs. (again accounting for rounding.)

 

One thing to keep in mind is that both Earth and Tom have to agree as what time Tom's clock reads at turn-around. The Earth reads 1.732 yr due to red-shift and light travel delay, and Tom reads 1732 yrs because that is how long it takes to travel 0.866 ly at 0.5c.

According to relativity Tom is observer then Earth travels and comes back,according to relativity Earth is younger in end of the experiment.

 

Check my answer to tar. Tom sees Earth red shifted and blue shifted for equal amounts of time and thus sees Earth age more. Earth has to wait for a year before it gets the information that Tom turnes around and sees the transistion for Red to blue shift. Tom, on the other hand is the one that turns around and this causes an immediate transistion of red to blue shift as far as he is concerned.

Edited May 12, 2013 by Janus

[DimaMazin]

No. Due to length contraction, the distance between Earth and the Turn around point is 0.866 ly according to Tom. He takes 1.732 years to reach the turn around and sees Earth age 1 year (After you account for the fact that 0.577 and 1.732 are rounded off values.) He then sees Earth age at a rate of 1.732 for another 1.732 years on the return trip for another 3 yrs. (again accounting for rounding.)

 

One thing to keep in mind is that both Earth and Tom have to agree as what time Tom's clock reads at turn-around. The Earth reads 1.732 yr due to red-shift and light travel delay, and Tom reads 1732 yrs because that is how long it takes to travel 0.866 ly at 0.5c.

 

Check my answer to tar. Tom sees Earth red shifted and blue shifted for equal amounts of time and thus sees Earth age more. Earth has to wait for a year before it gets the information that Tom turnes around and sees the transistion for Red to blue shift. Tom, on the other hand is the one that turns around and this causes an immediate transistion of red to blue shift as far as he is concerned.

Then Tom is a moving observer in own system.Agen you contradict relativity.

[md65536]

However, running the same numbers from the travelers point of view, she sees the Earth age at .557 per year rate, for two years and then at a 1.732 per year rate for another two, on her way back. Making the stay at home and the Earth, 4.618 (1.154 plus 3.464) years older, which they are not.

This is an excellent analysis demonstrating correct application of the principle of relativity with respect to the relativistic Doppler effect, with a small error corrected by Janus, in which you're describing the traveler's point of view using the Earth's clock (two years each way) instead of the traveler's own clock (1.732 years each way). Edited May 12, 2013 by md65536

[michel123456]

The red and blue shifts do not cancel out.

 

Let's say that Tom leaves Earth at 0.5c, travels for 2 yrs (according to the Earth), turns around and comes back to Earth at 0.5c. This means that he travels a distance of 1 light year as measured by the Earth.

 

What does Earth observer see, taking red and blue shift into account?

 

On the outbound trip, he will see Tom red shift by a factor of 0.577 ( for every year that passes for him, he sees Tom age 0.577 years) He will see Tom age at this rate unitl he sees Tom turn around and start to come back. This will take 3 yrs (two years for Tom to travel 1 light year plus 1 year caused by the time it takes light to travel 1 light year. Tom arrives in two years, But the Earth Observer doesn't see him doing so for another year.)

 

3 yrs *0.577 = 1.732 yrs, which is how much the Earth sees Tom age on the outbound leg.

 

The Earth then sees Tom blue shifted for the return leg at a rate of 1.732. Since the total trip take 4 yrs and the Earth has spent 3 of them seeing a red shift, this leaves 1 year to see a blue shift. (From the Earth's point of view, Tom is chasing his own light signal.)

 

This means that the Earth see Tom age 1.732 years during the return leg. Thus in total, the Earth sees Tom age 3.464 years while 4 yrs have passed on the Earth. All in all, Tom has aged 0.886 as much as the Earth.

 

This is exactly the age difference predicted by time dilation.

This is wrong.

1_What is the velocity of Tom at the begin of the travel, when leaving the Earth: it is 0,5 c as observed by the Earth

 

2_What is the velocity of Tom at the end of the travel, when coming back to Earth: it is 0,5 c as observed by the Earth

 

The travel is totally symmetric as observed by the Earth. The Earth observes after 2 years that Tom has reached a point 1 LY away because Tom travels at 0,5 c as observed from the Earth. And not after 3 years.

The symmetric diagram is made from the FOR of the Earth. (edited)

 

_In your concept, when leaving, what is the velocity of an object that, as observed from the Earth, travels 1LY in 3 years? It is 0,33 C and not 0,5c It is contrary to what was stated in the beginning.

_In your concept,when returning, what is the velocity of an object that as observed from the Earth travels 1LY in 1 year? that is c, that is not 0,5c and it is not allowed.

 

At no moment the traveler is "chasing its own light signal", that is BS.

Edited May 12, 2013 by michel123456

[tar]

This is an excellent analysis demonstrating correct application of the principle of relativity with respect to the relativistic Doppler effect, with a small error corrected by Janus, in which you're describing the traveler's point of view using the Earth's clock (two years each way) instead of the traveler's own clock (1.732 years each way).

 

md65536,

 

So Tom does not think he has reached the turnaround point, when he turns around? He has not followed the plan, and turned around before he has gotten to the designated turnaround point. He has neither gone far enough nor long enough, or he has not traveled at the designated speed. I think we should take Tom's license away, for failure to follow instructions, and messing up the experiment.

 

Regards, TAR2

 

P.S. Over the last 10 days or so, I have personally been dealing with a double vision issue. Most of the time I can fuse the two images, but my eye doctor has determined that my eyes are out of vertical alignment (have not yet grasped the concept, but have verified the issue by looking at an object through two paper towel roll tubes, and the image my right eye focuses on is "higher" and a little to the left from the image presented by my left eye).

 

I mention this, because the fusing of the image presented to us by the Earth, and the image presented to us by Tom, is a difficult task. We can not put ourselves in a position where we have two actual images that are presented simultaneously for fussion, during the entire experiment. Our binocular observer has to have one eye in one reference frame, and the other eye in the other reference frame.

I am sure that Minkowski space and Fourier transforms, and four vector analysis and such, have been designed for this task. However, I am woefully lacking in such skills, and keep hoping for a view that adds up and fuses, without the math.

[michel123456]

Explaining my post #54

 

what does it mean "a velocity of 0,5c"?

 

_it means that, as observed from Earth, the object travels an observed distance of 0,5 LY in an observed time of 1 Year.

 

The common explanation of the twin experiment says otherwise. The common explanation says that the velocity as put on the diagram is 0,5 c and as observed from the Earth is different.

That;s the nail of the question.

Edited May 12, 2013 by michel123456

[Delta1212]

This is wrong.

1_What is the velocity of Tom at the begin of the travel, when leaving the Earth: it is 0,5 c as observed by the Earth

 

2_What is the velocity of Tom at the end of the travel, when coming back to Earth: it is 0,5 c as observed by the Earth

 

The travel is totally symmetric as observed by the Earth. The Earth observes after 2 years that Tom has reached a point 1 LY away because Tom travels at 0,5 c as observed from the Earth. And not after 3 years.

The symmetric diagram is made from the FOR of the Earth. (edited)

 

_In your concept, when leaving, what is the velocity of an object that, as observed from the Earth, travels 1LY in 3 years? It is 0,33 C and not 0,5c It is contrary to what was stated in the beginning.

_In your concept,when returning, what is the velocity of an object that as observed from the Earth travels 1LY in 1 year? that is c, that is not 0,5c and it is not allowed.

 

At no moment the traveler is "chasing its own light signal", that is BS.

The Earth will not see something that is 1LY away until a year later. So anything the Earth sees at 1LY distant, the Earth concludes to have happened one year previous. The calculates the ship to be traveling at 0.5c and going 1LY, hence taking 2 years. Upon arrival however, it will then take 1 year for the light from the spaceship to reach Earth, for a total of 3 years between the ship leaving and the light from its arrival reaching Earth, but 2 years between the time the ship leaves and the arrival as measured by Earth, because Earth knows to correct for the time delay.

 

So the image of the ship would recede at less than 0.5c, but Earth is smart enough to know that the location of the ship and the location of the ship's image as seen from Earth are not the same thing at any given moment, and the difference can be calculated based upon the distance and velocity.

 

On the return trip, any light emitted from the shit at 1LY will take 1 year to reach Earth. 6 months after the ship turns around, the light showing that it has turned around will be 6 months from Earth, while the ship will be six months closer. In effect, it is "chasing its own light" because they are both moving in the same direction. The ship will therefore be much closer to Earth than the turnaround point by the time the Earth actually sees the ship turn around.

[michel123456]

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

IOW the problem states:

 

Tom is leaving the Earth at T=0 at v=0,5c

At t=2 (Earth time) observers from Earth look in their telescope and observe Tom making a turn.

At t=4 (Earth time) observers on Earth welcome back Tom landing at velocity 0,5 c.

[Delta1212]

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

IOW the problem states:

 

Tom is leaving the Earth at T=0 at v=0,5c

At t=2 (Earth time) observers from Earth look in their telescope and observe Tom making a turn.

At t=4 (Earth time) observers on Earth welcome back Tom landing at velocity 0,5 c.

Actually, they'd look out of their telescopes at t=3 and observe that Tom turned around 1 year ago at t=2, but otherwise yes.

[michel123456]

Actually, they'd look out of their telescopes at t=3 and observe that Tom turned around 1 year ago at t=2, but otherwise yes.

Let's say that observers from Earth look in their telescope and 1 year ago look at Tom making the turnaround.

in this case observers from Earth will be astonished to see in their telescope Tom coming in 1 year from a point 1 LY away. That is Tom is observed traveling at c.

I am pretty sure that changing slightly the numbers in this scenario observers from Earth could observe Tom coming back faster than c! and that is not allowed.

Edited May 12, 2013 by michel123456

[Delta1212]

The blue-shifted image of Tom's ship will appear to be approaching faster than c, but c is a restriction on massive objects, not images. If Tom's ship turned around and suddenly accelerated to the speed of light traveling towards Earth (in violation of the laws of physics) he would arrive simultaneously with the light from his turnaround (seeing as he is traveling at the same speed). The Earth would therefore see Tom arrive back at Earth at the same time they see him turn around 1LY away.

 

Because Tom takes 1 year longer than his light to travel the same lightyear, the people on Earth will know that he was, in reality, traveling at half the speed of light.

[michel123456]

The blue-shifted image of Tom's ship will appear to be approaching faster than c, but c is a restriction on massive objects, not images. If Tom's ship turned around and suddenly accelerated to the speed of light traveling towards Earth (in violation of the laws of physics) he would arrive simultaneously with the light from his turnaround (seeing as he is traveling at the same speed). The Earth would therefore see Tom arrive back at Earth at the same time they see him turn around 1LY away.

 

Because Tom takes 1 year longer than his light to travel the same lightyear, the people on Earth will know that he was, in reality, traveling at half the speed of light.

that makes no sense.

The Earth does not welcome the image of Tom, but Tom himself.

[Delta1212]

Ok, let's do this in increments.

 

At the moment Tom turns around, he is 1ly away from Earth. At the same time Tom starts traveling back toward Earth, a photon bounces off Tom and travels toward Earth as well. This is one of the photons that will allow Earth to see Tom turnaround when it reaches Earth. The photon is traveling at c. Tom is traveling at 0.5c.

 

6 months after beginning the trip back to Earth, the photon has traveled half a light year towards Earth. Tom has traveled a quarter of a light year. The photon has half a light year remaining. Tom has three quarters of a light year remaining.

 

6 months after that, the photon has traveled another half a light year and reaches Earth. Earth now sees Tom turn around. During the same 6 months, Tom travels another quarter of a lightyear towards Earth, and is now half a light year away.

 

6 months later, the Earth saw Tom turn around 6 months ago. In the intervening period, Tom has traveled another quarter of a light year and is now a quarter of a light year from Earth.

 

6 months after that, the Earth saw Tom turn around one year ago, and Tom travels another quarter of a light year, ending up at Earth.

 

Thus two years have elapsed since Tom turned around, and one year has elapsed since the light from Tom turning around has reached Earth.

[michel123456]

Ok, let's do this in increments.

 

At the moment Tom turns around, he is 1ly away from Earth. At the same time Tom starts traveling back toward Earth, a photon bounces off Tom and travels toward Earth as well. This is one of the photons that will allow Earth to see Tom turnaround when it reaches Earth. The photon is traveling at c. Tom is traveling at 0.5c.

 

6 months after beginning the trip back to Earth, the photon has traveled half a light year towards Earth. Tom has traveled a quarter of a light year. The photon has half a light year remaining. Tom has three quarters of a light year remaining.

 

6 months after that, the photon has traveled another half a light year and reaches Earth. Earth now sees Tom turn around. During the same 6 months, Tom travels another quarter of a lightyear towards Earth, and is now half a light year away.

 

6 months later, the Earth saw Tom turn around 6 months ago. In the intervening period, Tom has traveled another quarter of a light year and is now a quarter of a light year from Earth.

 

6 months after that, the Earth saw Tom turn around one year ago, and Tom travels another quarter of a light year, ending up at Earth.

 

Thus two years have elapsed since Tom turned around, and one year has elapsed since the light from Tom turning around has reached Earth.

Re-read your own post and you will conclude that the earth actually observes Tom making the trip from 1LY away in 1 year.

 

After that, replace 0,5c with 0.6c and tell everybody what the earthlings observe in their telescopes.

[Delta1212]

Re-read your own post and you will conclude that the earth actually observes Tom making the trip from 1LY away in 1 year.

 

After that, replace 0,5c with 0.6c and tell everybody what the earthlings observe in their telescopes.

They "see" him travel 1 ly in 1 year, but they measure him traveling 1 ly in 2 years. There is no prohibition on something looking like it is traveling FTL. There is only a prohibition on actually traveling FTL.

 

They see the image of his ship turn around and say "This happened 1 lightyear away, therefore it took 1 lightyear for the light to reach us. Tom turned around 1 year ago." One year after that, Tom arrives and Earth says "One year ago, it was one year since Tom turned around, therefore it took two years for Tom to return 1 ly. He must have been traveling at 0.5c." Those two years of reflected light will arrive over the course of one year because it is being blue shifted by Tom's approach velocity.

[md65536]

This is wrong.

1_What is the velocity of Tom at the begin of the travel, when leaving the Earth: it is 0,5 c as observed by the Earth

 

2_What is the velocity of Tom at the end of the travel, when coming back to Earth: it is 0,5 c as observed by the Earth

 

The travel is totally symmetric as observed by the Earth. The Earth observes after 2 years that Tom has reached a point 1 LY away because Tom travels at 0,5 c as observed from the Earth. And not after 3 years.

You're mixing what is measured (Lorentz analysis) and what is seen (Doppler analysis). Account for the delay of light, and you get a consistent answer.

 

The Earth measures Tom has reached that point after 2 years. It is 1 LY away in Earth's rest frame, so the observation reaches Earth one year later.

 

So Tom does not think he has reached the turnaround point, when he turns around? He has not followed the plan, and turned around before he has gotten to the designated turnaround point. He has neither gone far enough nor long enough, or he has not traveled at the designated speed. I think we should take Tom's license away, for failure to follow instructions, and messing up the experiment.

My mistake about your analysis. I thought you were so close to getting it, but I think you may have only presented the math because it (incorrectly) confirmed your belief that it shouldn't work out. When you accidentally got the right answer (by dividing by gamma) you reasoned out why the answer should be dismissed.

 

Tom reaches the turnaround point when he does, when he thinks he does, when everyone thinks he does, only---as Janus has pointed out---the distance to that point, which is a rest length in Earth's frame, is length-contracted in Tom's frame, all according to plan.

[Delta1212]

Maybe flipping it will make this more intuitive:

 

On Tom's return trip, a photon leaves his ship heading toward Earth. One second later, another leaves his ship and heads toward Earth. During that 1 second, the first photon will have traveled 1 light second. The ship, moving at 0.5c will have traveled half a light second. Therefore the second photon, released 1 second later, is released only half a light second behind the first photon. It will arrive on Earth half a second behind the first photon, but will represent 1 second's worth of Tom's travel time.

 

In effect, Earth will be receiving Tom's light more frequently than it was released (I.e. blue shifted) which will make Tom appear to be moving faster than he is if you don't correct for Doppler shift.

[Iggy]

Explaining my post #54

 

what does it mean "a velocity of 0,5c"?

 

_it means that, as observed from Earth, the object travels an observed distance of 0,5 LY in an observed time of 1 Year.

 

The common explanation of the twin experiment says otherwise. The common explanation says that the velocity as put on the diagram is 0,5 c and as observed from the Earth is different.

That;s the nail of the question.

I remember us talking about signal delay before and you understood it very well. If something happens on the sun now then we will see it roughly 8 minutes from now. We heard the Apollo astronauts say "touchdown" 1.3 seconds after they touched down... etc..

 

One year after the turnaround, earth would see the turnaround. I remember you knowing that.

 

 

 

It takes Tom 2 years to travel 1 light year. Half a light-year per year. v = 0.5c

 

The light from the event marked "turn" reaches earth at t = 3 years.

 

 

---

 

 

If anyone was curious, the equations for these:

...On the outbound trip, he will see Tom red shift by a factor of 0.577 ( for every year that passes for him, he sees Tom age 0.577 years)...

 

 

...The Earth then sees Tom blue shifted for the return leg at a rate of 1.732...

are,

 

[math]f_\mathrm{obs} = f_\mathrm{rest}\sqrt{\left({1 - v/c}\right)/\left({1 + v/c}\right)}[/math]

 

when the source is moving away, and,

 

[math]f_\mathrm{obs} = f_\mathrm{rest}\sqrt{\left({1 + v/c}\right)/\left({1 - v/c}\right)}[/math]

 

when it is approaching

[Janus]

Then Tom is a moving observer in own system.Agen you contradict relativity.

 

No, in Tom's frame it is the Earth and turnaround point that is moving relative to him and that is why the distance between the two undergoes length contraction according to Tom. There is no contradiction with Relativity as this is exactly what Relativity says happens.

[michel123456]

I remember us talking about signal delay before and you understood it very well. If something happens on the sun now then we will see it roughly 8 minutes from now. We heard the Apollo astronauts say "touchdown" 1.3 seconds after they touched down... etc..

 

One year after the turnaround, earth would see the turnaround. I remember you knowing that.

 

 

 

It takes Tom 2 years to travel 1 light year. Half a light-year per year. v = 0.5c

 

The light from the event marked "turn" reaches earth at t = 3 years.

 

 

---

 

 

If anyone was curious, the equations for these:

 

are,

 

[math]f_\mathrm{obs} = f_\mathrm{rest}\sqrt{\left({1 - v/c}\right)/\left({1 + v/c}\right)}[/math]

 

when the source is moving away, and,

 

[math]f_\mathrm{obs} = f_\mathrm{rest}\sqrt{\left({1 + v/c}\right)/\left({1 - v/c}\right)}[/math]

 

when it is approaching

You said

"One year after the turnaround, earth would see the turnaround. I remember you knowing that"

 

Yes I know that. But i cannot swallow that. I changed my mind.

Let me explain.

And tell me where I am wrong:

 

a. On your graph, earthlings observe the turnaround when they are at time 3.

 

b. After that, earthlings observe Tom landing back. That happens at time 4.

 

c. the difference between the 2 events is 4-3=1 year

.

d. Earthlings have observed Tom coming from a planet 1 LY away in 1 year

.

e. Thus, the apparent velocity of Tom as observed by earthlings is c.

 

 

 

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

(edit)

take some time and make a similar diagram with 0.66c (1 by 1.5) and you will be surprised of what the apparent velocity of Tom becomes.

 

Here i did it for you:

 

 

Tom is observed from Earth at point 2.5

Tom lands on earth at point 3

 

Earthlings have observed in their telescopes Tom coming from a planet 1 LY away in half a year. That is twice c!

 

How can one observe an object traveling faster than light?

It is wrong.

(edit) I mean the diagram is wrong. Relativity is correct.

Edited May 12, 2013 by michel123456

[Delta1212]

You said

"One year after the turnaround, earth would see the turnaround. I remember you knowing that"

 

Yes I know that. But i cannot swallow that. I changed my mind.

Let me explain.

And tell me where I am wrong:

 

a. On your graph, earthlings observe the turnaround when they are at time 3.

 

b. After that, earthlings observe Tom landing back. That happens at time 4.

 

c. the difference between the 2 events is 4-3=1 year

.

d. Earthlings have observed Tom coming from a planet 1 LY away in 1 year

.

e. Thus, the apparent velocity of Tom as observed by earthlings is c.

 

 

e. holds if you fail to account for Doppler shift, but that's fine. The ban on faster than c travel applies to objects actually traveling at c. Things are allowed to look like they're traveling faster than c, as long as they don't actually travel faster than c.

 

In this case, it's an optical trick caused by Tom's direction of travel. You can't take an accurate measurement if Tom's velocity until that optical effect has been accounted for.

 

Likewise, you could point a laser at the moon and move it such that the dot on the moon's surface would travel across the lunar surface at faster than c. You can do this because it is merely a succession of images that create apparent motion that is faster than c. Nothing is actually moving faster than c.

 

-----

 

If you're still not convinced, try reversing your assumed position. Let's say that Earth looks through its telescopes at t=2 and sees Tom turning around. Where is Tom actually at that point, and when did the light that Earth is seeing leave Tom's ship?

[Iggy]

e. Thus, the apparent velocity of Tom as observed by earthlings is c.

 

It seems like you're trying to invent a new kind of velocity... an 'apparent velocity'.

 

To help me understand the concept better, could you tell me the 'apparent velocity' of light?

 

 

I mean the diagram is wrong.

 

Ordinary velocity (as it has been defined for thousands of years) is run over rise on a graph (change in x divided by change in y is the procedure for finding velocity).

 

Are you proposing that we need a new kind of graph for 'apparent velocity', or a new procedure on a normal graph for finding 'apparent velocity'? I can't tell what exactly you're trying to accomplish.

 

[md65536]

Earthlings have observed in their telescopes Tom coming from a planet 1 LY away in half a year. That is twice c!

 

How can one observe an object traveling faster than light?

It is wrong.

(edit) I mean the diagram is wrong. Relativity is correct.

You keep asserting that things you don't understand are wrong.

 

And if Earth at year 3 sees Tom sending a light signal, and Earth at year 3 receives that signal, how fast has that signal traveled? Infinite speed? Could the observation of Tom sending the signal, or leaving the planet, not be delayed by the travel time of light?

 

Edit: Oops, I see Iggy has already covered this...

Edited May 12, 2013 by md65536

[michel123456]

With all the risks, this below is what i think happens in the twins experiment:

 

Tom is heading at 0,5c to point P.

 

 

When Earth is at time M =2 years (half the waiting time), then earthlings observe Tom making the U-turn at point R.

After that, earthlings wait for another 2 years before Tom comes back.

 

Analysis:

 

-notice that the blue and green triangles have the same area. Same basis & same height.

 

_the velocity of Tom during the outbound is distance/time (the height of the green triangle divided by its basis)

 

_ the velocity of Tom during the return travel is distance/time (the height of the blue triangle divided by its basis)

 

The 2 velocities are equal to each other.

Tom never reached point P.

Edited May 12, 2013 by michel123456

[Iggy]

So that I can understand the diagram, what is the 'apparent velocity' of light?

 

edit:

 

or, are you no longer thinking in those terms?

Edit...

In any case, the outbound velocity shown is .5c and the inbound velocity shown is .25c. Outbound is difference in x (2/3) divided by difference in y (4/3) between point A and R... v = .5c

Inbound is difference in x (2/3) divided by difference in y (8/3) between point R and E... v = .25c

The ship travels 2/3rds of a light-year from earth. It turns around at R and earth views it turn at M (2/3rds of a year later). Edited May 12, 2013 by Iggy