The twin paradox

[md65536]

"All travel for a fixed time" sounded for sure like you meant by earth's clock. Made perfect sense to me that way. You should have run with it

Thanks! It didn't make perfect sense to me!

 

I spend too much time trying to figure out the details, and rarely get to the point of understanding enough to know which details are important, and which don't need to be mentioned at all!

OK, the equations work out, and agree with experiment, Still not sure though what proper time is.

Proper time is the time between two events measured by a clock that passes through those two events. In other words it's the time according to a given single clock. In the twin paradox it's "aging"; you age along with a local clock and you can't separate yourself from that clock and age according to some distant clock.

 

Each clock can have its own proper time, which depends on its path. For example, between the events "twins separate" and "twins reunite", the paths and the aging of the two twins are different, even though they both record a proper time between the same pair of events.

If red makes a turnaround, would that not mean she would have to expend a great deal of energy to "brake" back into our FOR for that moment where she is no longer traveling at a v away, and is then traveling at a v toward us.

Not necessarily. Red could have very low mass (eg. a neutrino) and the Lorentz transformation would still apply.

The light from the rest of the galaxy is extremely red shifted hitting her stern and extremely blue shifted hitting her bow. Off at right angles to her motion things would look about right, but although light is hitting both her bow and stern at C, the frequency is likely doubled in front and halved in back. This unbalance of energy hitting the craft, should have a slowing effect. As well as imbalance of heat. She would KNOW which way she was headed, and could not fool herself into thinking she was stationary.

But she could be stationary in a system that is moving around her; yes she is accelerated by stuff that hits her but that can still happen with a moving system that moves past her. By the equivalence principle, she is stationary in her frame of reference. Edited May 2, 2013 by md65536

[michel123456]

The earth distance= 7.5 ly. Speed =.6c. Expected travel time= 7.5/.6= 12.5 y.

gamma =1.25. Time dilation for Pam = 12.5/1.25=10 y. Since her clock and her sense of time indicate 10y, she concludes the interval of 7.5 ly has contracted to 7.5/1.25= 6 ly.

 

(...)

As Iggy stated, that is the straight path.

See the followings

 

 

On the left is the A diagram as observed from the Earth, on the right the B diagram as observed by the moving one. The axis are orthogonal in both cases because both observe things as inertial observers. In fact the right diagram is almost a perfect scaling of the left one at 1,25 scale factor. The main difference between the 2 graphs is that on diagram A the spacing of the dots on the diagonal correspond to nothing, while on diagram B the spacing correspond to the time axis.

 

Now, I tried to put the B diagram upon the A without great success. Here below is a try where the new axes are not orthogonal anymore, the bisect line is not 45 degrees. But the projection on the axes is perpendicular.

 

 

Then I tried as explained in Iggy's link but it doesn't work well.

 

The bisect of the new axes intersects point R and is never at 45 degrees (the bold line)

Edited May 2, 2013 by michel123456

[Iggy]

As Iggy stated, that is the straight path.

See the followings

 

 

On the left is the A diagram as observed from the Earth, on the right the B diagram as observed by the moving one. The axis are orthogonal in both cases because both observe things as inertial observers. In fact the right diagram is almost a perfect scaling of the left one at 1,25 scale factor. The main difference between the 2 graphs is that on diagram A the spacing of the dots on the diagonal correspond to nothing, while on diagram B the spacing correspond to the time axis.

 

Now, I tried to put the B diagram upon the A without great success. Here below is a try where the new axes are not orthogonal anymore, the bisect line is not 45 degrees. But the projection on the axes is perpendicular.

 

 

Then I tried as explained in Iggy's link but it doesn't work well.

 

The bisect of the new axes intersects point R and is never at 45 degrees (the bold line)

 

 

---

 

 

You're quite close.

 

The person at rest in a reference frame is moving straight up, so if earth is at rest in one frame while the ship moves to the right:

 

 

 

then the ship is at rest in the other while earth moves left,

 

 

 

 

 

A Minkowski diagram with both coordinate systems would be,

 

 

 

Or, it could alternatively be done the other way:

 

 

 

 

The Lorentz transformations can help with drawing it too. If you didn't have the red part drawn you could, for example...

 

 

if you know where those black events are in green coordinates you can solve them for red with the lorentz transforms

 

For (4,12) x = 4, t = 12, and v = 0.6 (all in light units). In the red coordinate system:

 

[math]x' = \frac{x-v \cdot t}{\sqrt{1 - v^2}} = \frac{4-0.6 \cdot 12}{\sqrt{1 - 0.6^2}} = -4[/math]

 

[math]t' = \frac{t-v \cdot x}{\sqrt{1 - v^2}} = \frac{12-0.6 \cdot 4}{\sqrt{1 - 0.6^2}} = 12[/math]

 

so that event (4,12) is at (-4,12) in the red coordinate system. Or, the one on the light ray (4,4)

 

x = 4, t = 4

 

[math]x' = \frac{x-v \cdot t}{\sqrt{1 - v^2}} = \frac{4-0.6 \cdot 4}{\sqrt{1 - 0.6^2}} = 2[/math]

 

[math]t' = \frac{t-v \cdot x}{\sqrt{1 - v^2}} = \frac{4-0.6 \cdot 4}{\sqrt{1 - 0.6^2}} = 2[/math]

 

green(4,4) = red(2,2)... Locating events along the light ray would help draw it.

 

By the way, the above coordinates mean that it would take a ray of light four years to go from our sun to another star 4 lightyears away, but if someone chased that same light ray at 0.6c then they would find it takes 2 years for the ray to reach the star two lightyears away. Fun

Edited May 2, 2013 by Iggy

[tar]

Well,

 

I am sort of seeing what your saying, or should I say, what spacetime diagrams are saying.

 

Thanks.

 

Regards, TAR2

[michel123456]

(...)

 

A Minkowski diagram with both coordinate systems would be,

 

 

(...)

I don't get it.

 

Where you label "Earth" I suppose you mean "Earth's time"

Where you label "ship" I suppose you mean "Ship's time"

 

1. I don't see a line joining "ship's time" with "ship's distance".

2. I don't understand the thin diagonal lines joining point 8 on Earth's time with point 10 on Ship's time.

 

I don't even get this one below.

You stated that "the ship is at rest", but that happens only when the ship stops at the return point, when both the ship and Earth are in the same FOR. While the ship is moving the things as observed by the ship should show differently. For the moving ship, distance is less and time is less too. No?

(...)

 

then the ship is at rest in the other while earth moves left,

 

It should show 10 years on the time axis instead of 12.5, and 6 lightyears on the distance axis instead of 7.5 Edited May 3, 2013 by michel123456

[Iggy]

 

(...)

 

A Minkowski diagram with both coordinate systems would be,

 

 

(...)

I don't get it.

 

Where you label "Earth" I suppose you mean "Earth's time"

Where you label "ship" I suppose you mean "Ship's time"

 

 

I was thinking more that the green line marked 'earth' indicates earth's world line and the same with the ship. But, I agree, the green one is the T axis in the green coordinate system and the red one is the T' axis in the red coordinate system.

 

 

1. I don't see a line joining "ship's time" with "ship's distance".

2. I don't understand the thin diagonal lines joining point 8 on Earth's time with point 10 on Ship's time.

 

 

It is difficult to explain over the internet.

 

Hum... First, consider an ordinary coordinate system, nothing tilted or skewed:

 

 

NASA launches a high speed projectile at a comet to destroy it. The black dot marks the event where the comet is destroyed. You measure velocity by measuring the change in X divided by the change in T. The projectile moves from x = 0 to x = 6 light-minutes and from 10 = 0 to 10 = 10 minutes. According to the clocks and measuring devices here on earth it went 6 light-minutes in 10 minutes:

 

 

 

If you do the same procedure for earth, you see earth is at rest in this frame. It is always at x = 0. Its velocity is 0c.

 

Now let's ignore Earth's frame of reference for a sec and look at the projectile's frame of reference:

 

 

Let's look at Earth first. This is the projectile's frame of reference, so earth should be moving in this frame. That is to say, if you were sitting on this projectile, and it was at rest relative to you, then you would see earth moving relative to you. We do the same procedure as before:

 

 

According to the projectile, earth recedes at 0.6c. Notice earth measures the projectile at 0.6c and the projectile measures earth at -0.6c. They call that the principle of reciprocity of velocity. Whatever velocity you think something moves relative to you, so too does it think you move that fast relative to it.

 

Now, if you do the same procedure we just did for earth in the above image for the projectile you will see that it moves zero light-minutes over any period of time. In this frame it has a velocity of 0c. In other words, the projectile does not move relative to itself.

 

Now hopefully answers your question,

 

 

The bold red lines are the axes for the red coordinate system. The measurements that the ship makes will coincide with that coordinate system.

 

 

I don't even get this one below.

You stated that "the ship is at rest", but that happens only when the ship stops at the return point, when both the ship and Earth are in the same FOR.

 

 

The ship is at rest in its own frame of reference. Sort of like how a person is at rest in an airplane when they take their seat.

 

 

No?It should show 10 years on the time axis instead of 12.5, and 6 lightyears on the distance axis instead of 7.5

 

Yes, that was confusing. I didn't mean for the black dot on both images to represent the same event. I was trying to show that each sees the other time dilated. The black dot in these two would be the same event:

 

 

 

 

And, again, all of the measurements for any event you could mark on the left image corresponds with the green parts of the green and red image above, and the image on the right correspond with the red parts of the image.

 

The left image is Earth's frame of reference and the right image is the ship's frame of reference. BTW, "earth distance" means "distance from earth" and "ship distance", "distance from ship"

 

EDIT:

 

 

2. I don't understand the thin diagonal lines joining point 8 on Earth's time with point 10 on Ship's time.

 

It occurs to me that I should also answer this in physical terms. The green line joining Earth's proper time 8, and the ship's proper time 10 means that when the ship's clock says 10, the ship will figure that Earth's clocks say 8. It is a 'line of simultaneity', also known as a 'present instant'. The ship will say that red-T = 10 and green-T = 8 are simultaneous.

 

Earth will, on the other hand, say that green-T = 12.5 and red-T = 10 are simultaneous.

 

Simultaneity is relative, which is why the ship's x axis is tilted relative to Earth's.

Edited May 4, 2013 by Iggy

[michel123456]

I am grateful that you took so much time in explaining those things. Thank you.

 

I was thinking more that the green line marked 'earth' indicates earth's world line and the same with the ship. But, I agree, the green one is the T axis in the green coordinate system and the red one is the T' axis in the red coordinate system.

1. o.k.

It is difficult to explain over the internet.

2. You are doing great!

Hum... First, consider an ordinary coordinate system, nothing tilted or skewed:



NASA launches a high speed projectile at a comet to destroy it. The black dot marks the event where the comet is destroyed. You measure velocity by measuring the change in X divided by the change in T. The projectile moves from x = 0 to x = 6 light-minutes and from 10 = 0 to 10 = 10 minutes. According to the clocks and measuring devices here on earth it went 6 light-minutes in 10 minutes:









If you do the same procedure for earth, you see earth is at rest in this frame. It is always at x = 0. Its velocity is 0c.

3. I am ok with your new example but the first diagrams were with 12,5 time and 7.5 distance from Earth. Something that you reintroduce in your last diagram at the end (see point 6). To me what the Earth observes in term of distances & time is different from what the projectile observes. The diagrams should not be symmetric, see point 4 & point 6.

Now let's ignore Earth's frame of reference for a sec and look at the projectile's frame of reference:



Let's look at Earth first. This is the projectile's frame of reference, so earth should be moving in this frame. That is to say, if you were sitting on this projectile, and it was at rest relative to you, then you would see earth moving relative to you. We do the same procedure as before:



According to the projectile, earth recedes at 0.6c. Notice earth measures the projectile at 0.6c and the projectile measures earth at -0.6c. They call that the principle of reciprocity of velocity. Whatever velocity you think something moves relative to you, so too does it think you move that fast relative to it.

4. I understand the principle of reciprocity of velocity but maybe not the same way as you do.
Velocity 6LY/10Y is the same as 7.5LY/12.5Y = 0,6C.
observers on Earth and on the projectile observe different times and distances but according to the same ratio. Since the ratio in question is what we call velocity, both observers measure the same velocity.

It does not mean that they observe the same distances and the same times. Also I do not understand why the black point (junction between comet & projectile) does not correspond to point 10 on Earth's line. I made a slight change to your diagram, see below.

Now, if you do the same procedure we just did for earth in the above image for the projectile you will see that it moves zero light-minutes over any period of time. In this frame it has a velocity of 0c. In other words, the projectile does not move relative to itself.

5. I understand that the observer on the projectile does not move relative to itself. But as he sees the Earth receding, he sees the comet approaching at exactly the same rate. The Earth and the comet are in the same FOR in the example.

 

6. here below your diagram with Earth at 12.5 time and 7.5 distance. One can draw a vertical line from 6 red to 10 red, showing the Comet. what observes the projectile is different from what the Earth observes, the velocity is the same though.

Now hopefully answers your question,



The bold red lines are the axes for the red coordinate system. The measurements that the ship makes will coincide with that coordinate system.




The ship is at rest in its own frame of reference. Sort of like how a person is at rest in an airplane when they take their seat.



Yes, that was confusing. I didn't mean for the black dot on both images to represent the same event. I was trying to show that each sees the other time dilated. The black dot in these two would be the same event:





And, again, all of the measurements for any event you could mark on the left image corresponds with the green parts of the green and red image above, and the image on the right correspond with the red parts of the image.

The left image is Earth's frame of reference and the right image is the ship's frame of reference. BTW, "earth distance" means "distance from earth" and "ship distance", "distance from ship"


EDIT:



It occurs to me that I should also answer this in physical terms. The green line joining Earth's proper time 8, and the ship's proper time 10 means that when the ship's clock says 10, the ship will figure that Earth's clocks say 8. It is a 'line of simultaneity', also known as a 'present instant'. The ship will say that red-T = 10 and green-T = 8 are simultaneous.

Earth will, on the other hand, say that green-T = 12.5 and red-T = 10 are simultaneous.

Simultaneity is relative, which is why the ship's x axis is tilted relative to Earth's.

Your last diagrams are not symmetric so I think that at the end we do not disagree.

 

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

 

Here below are your diagrams again with some of my intervention. Hoping not to be too mistaken:

 

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

7. Here I made a scaling of 2 4 6 labeling in order to obtain a correspondance between the intersection points. The result shows as a deformed rectangle (a rhombus).

 

 

 

 

 

 

 

8. Here I simply added the Comet line, which is parallel to Earth line.

 

 

 

 

 

 

9. Here it gets tricky:

The left diagram is the same as above.

The right is the precedent right one, rotated so that the Earth & Comet lines become vertical, and scaled so that the 2 diagrams can superpose.

And the 2 diagrams do superpose without any almost trouble: the ship line at left corresponds to the ship line at right. Same inclination, same dimension.

Except that the "Light" line does not correspond anymore. And that is ennoying.

Edited May 4, 2013 by michel123456

[Iggy]

the ship line at left corresponds to the ship line at right. Same inclination, same dimension.

 

That's the principle of reciprocity again. If the earth thinks the ship recedes at .6c then the ship thinks the same of earth.

 

 

Except that the "Light" line does not correspond anymore. And that is ennoying.

 

The perspective would have to change on the image on the right. Remember it gets skewed and squished. You're trying to end up with this one:

 

 

You can't rotate your right hand image alone in order to make it look like the red parts above. The t lines get rotated differently than the x lines. Your x axis is tilted the wrong way.

Edited May 4, 2013 by Iggy

[xyzt]

False, I wrote this section for wiki that clearly disproves your claim.

The twin paradox again.

The paradox is not that the twins have a different age at the end, the paradox is that both twins should observe the other aging less because motion is relative.

 

In all examples the one who moves observes the one at rest aging more than him (instead of less) and that does not solve the paradox, that IS the paradox.

Edited May 5, 2013 by xyzt

[michel123456]

False, I wrote this section for wiki that clearly disproves your claim.

Well the twin paradox is usually presented as a situation where all observers are in an inertial Frame Of Reference. Or maybe that's me who understand things that way, but as it looks in the other thread about acceleration, I am not alone reading the twin paradox that way.

 

Your section of Wiki begins with this:

 

Δτ represents the time of the non-inertial (travelling) observer K' as a function of the elapsed time Δt of the inertial (stay-at-home) observer K for whom observer K' has velocity v(t) at time t.

where evidently you consider that the traveler is NOT in an Inertial FOR. That is all the question. And if you are correct on this, you are correct all the way long.

After much thinking and following the other thread, yes you must be right. IMHO of course.

 

 

 

[xyzt]

Well the twin paradox is usually presented as a situation where all observers are in an inertial Frame Of Reference. Or maybe that's me who understand things that way, but as it looks in the other thread about acceleration, I am not alone reading the twin paradox that way.

 

Your section of Wiki begins with this:

 

where evidently you consider that the traveler is NOT in an Inertial FOR. That is all the question. And if you are correct on this, you are correct all the way long.

After much thinking and following the other thread, yes you must be right. IMHO of course.

 

 

 

The twin paradox does not exist if there is no acceleration. Acceleration is what introduces the asymmetry between the twins. Besides, you complained that the result in not the same as viewed from the traveling twin. I proved that no matter from what frame you are judging the "paradox", the observers agree.

[michel123456]

The twin paradox does not exist if there is no acceleration. Acceleration is what introduces the asymmetry between the twins. Besides, you complained that the result in not the same as viewed from the traveling twin. I proved that no matter from what frame you are judging the "paradox", the observers agree.

They agree what? That they disagree about their age in comformity with Relativity?

 

(Edit)

 

I am really glad to hear that "The twin paradox does not exist if there is no acceleration".

Edited May 5, 2013 by michel123456

[xyzt]

They agree what? That they disagree about their age in comformity with Relativity?

 

(Edit)

 

I am really glad to hear that "The twin paradox does not exist if there is no acceleration".

They agree which twin is younger.

[tar]

xyzt,

 

What confuses me about the experiment, is the question of why does not the opposite aging effect happen during the return trip.

 

Let's say for example that the traveler keeps track of two similar pulsars, 180 degrees from each other, one in the direction of the outbound trip, and one in the direction of the Earth. On the way out, she counts pulses from both, and finds the one she is headed toward is ticking faster then it did when she was stationary on Earth (where the similar ticks were established) and the one in the direction of the Earth is ticking slower. On the way back however, the pulsar behind the Earth is now blue shifted and ticking faster, and the one behind her is red shifted and ticking slower. When she gets home, and adds up the ticks she has experienced exactly the same number of ticks from each, and this number agrees with the count of the stay at home twin. She would not consider herself a different age then her twin.

 

To solidify this argument, consider the fact that we see the light from a star 4 lys away, 4 years after the light is emitted, however there is only one instance of that star. We cannot consider it both existing as we see it, and existing presently sending out light which we will not see for 4 years. Well we can, but we have to define carefully which "instance" of the star, we are considering.

 

As the traveling twin goes out, she remains as only one instance of a person, but her increasing distance from Earth causes her to be continually separated from Earth's point of view, and continually closer to the "imaginary" now of the star we know is emitting light that takes four years to get to us. If she should reach the star, she would be in its present moment for real, and it would then be Earth, which to her would appear as it was four years before. Still took her the appropriate time to get there, and will still take her, the appropriate time to get back. She should be as much older as the time it took to make the trip.

 

Regards TAR2

[xyzt]

xyzt,

 

What confuses me about the experiment, is the question of why does not the opposite aging effect happen during the return trip.

 

 

Because, at the turn around point the line of simultaneity makes a "jump". Half of the value of the jump can be attributed to the "before" the turnarounf and the other half to the "after" the turnaround. The situation you describe is depicted here.

[Dekan]

During the outward trip, going away from Earth, the light from the starship gets red-shifted.

Then during the return trip, the ship's light gets blue-shifted, by exactly the same amount.

So when the ship lands back on Earth, the blue/red shift has been normalised. Nature has maintained equilibrium. Everything's back the same as it was.

 

So will the ship's crew, as they disembark, really astonish us, by looking unnaturally young?

 

 

 

 

 

 

 

[xyzt]

During the outward trip, going away from Earth, the light from the starship gets red-shifted. Then during the return trip, the ship's light gets blue-shifted, by exactly the same amount. So when the ship lands back on Earth, the blue/red shift has been normalised. Nature has maintained equilibrium. Everything's back the same as it was. So will the ship's crew, as they disembark, really astonish us, by looking unnaturally young?

They will be younger. How much younger depends on the length of the trip and how fast they were going. Do not confuse frequency shift up/down with elapsed proper time, they are different entities.

Edited May 11, 2013 by xyzt

[michel123456]

During the outward trip, going away from Earth, the light from the starship gets red-shifted. Then during the return trip, the ship's light gets blue-shifted, by exactly the same amount. So when the ship lands back on Earth, the blue/red shift has been normalised. Nature has maintained equilibrium. Everything's back the same as it was. So will the ship's crew, as they disembark, really astonish us, by looking unnaturally young?

Yes I had the same question.

It has been answered by (valuable) members in another thread that when the traveler remains inertial there is no paradox, everything is fine and everybody has the same age. I still wonder (and I am not alone) how one can figure out a twin experiment entirely based on inertial observers. See the interminable other thread.

 

When someone "forget' what happens at the turning point the result is that would be observed for a twin that would not have turned back and finally reached a point very far away.

[DimaMazin]

During the outward trip, going away from Earth, the light from the starship gets red-shifted. Then during the return trip, the ship's light gets blue-shifted, by exactly the same amount. So when the ship lands back on Earth, the blue/red shift has been normalised. Nature has maintained equilibrium. Everything's back the same as it was. So will the ship's crew, as they disembark, really astonish us, by looking unnaturally young?

If they with our galaxy travel relatively of us then they will be older.

Edited May 11, 2013 by DimaMazin

[tar]

xyzt,<br /><br />But if the "jump" is only in the turn around, how do you account for the traveling twin "aging" differently? Is she the same age as the stay at home, all the way out, then becomes a different age at the turn around, and then continues aging at the same pace as the stay at home, all the way back?<br /><br />And what is happening to the stay at home, that would account for the gap in his black dots? Should not his black dots be equally spaced for the whole trip?<br /><br />I am no longer as clear on the ts and the taus, as I though I was getting.<br /><br />I am still considering that what happens on the way out, in term of the traveling twin getting farther from the stay at home's now, and therefore closer to the now of the turnaround point, must be the opposite (time relationshipwise) as what happens on the way back.<br /><br />Also, I do not comprehend the meaning of v x/c2.<br /><br />What is that supposed to represent? I have a hard time understanding the speed of light, if time and distance are not already understood as invariant. And if they are to vary, vary from what fixed understanding?<br /><br />And one other little problem I have. What exactly is a square year? Wouldn't you need a fifth dimension to hold such a thing? Can you cube a year? What would you have, if you did? Or what quantity does C squared represent?<br /><br />Regards, TAR2

[Iggy]

Yes I had the same question.

It has been answered by (valuable) members in another thread that when the traveler remains inertial there is no paradox, everything is fine and everybody has the same age.

 

There is *never* any paradox when relativity is used correctly.

 

"everybody has the same age" is not a correct interpretation of inertial relative motion. See the diagram on the previous page. I believe you understand it.

 

 

 

This is inertial relative motion.

 

From the perspective of green, when green is 10 years older, red is 8 years older. When green is celebrating his 10th birthday since the separation, he says "red is *currently* celebrating his 8th birthday since the separation"

 

From the perspective of red, when red is 10 years older, green is 8 years older. When red is celebrating his 10th birthday since the separation, he says "green is *currently* celebrating his 8th birthday since the separation"

 

"currently" means something different to the two of them. That is why the red lines are skewed relative to the green lines. The symmetrical situation between them cannot be summarized as "they age at the same rate". The symmetry is broken when either a third observer is introduced, or when one of the twins accelerates.

[Janus]

During the outward trip, going away from Earth, the light from the starship gets red-shifted. Then during the return trip, the ship's light gets blue-shifted, by exactly the same amount. So when the ship lands back on Earth, the blue/red shift has been normalised. Nature has maintained equilibrium. Everything's back the same as it was. So will the ship's crew, as they disembark, really astonish us, by looking unnaturally young?

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.

[Iggy]

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.

 

exceedingly well written and intuitive post. I actually surprised myself at how ferociously I attacked the +1 arrow

[tar]

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

[DimaMazin]

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.

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