The twin paradox

[michel123456]

You must agree that velocity is relative.

Each observer, from his own FOR, measures a different velocity. So each observer measures what I call an "apparent velocity" which is different for another observer in another FOR.

 

The "apparent velocity" of light is the same for all observers.

 

In this diagram velocity of light is a line at 45 degrees (or 135 degrees) oriented from the past to the future, always going from down to up in the diagram. Never going up to down. For all observers.

 

In this diagram, from Earth's FOR, the apparent velocity is distance divided by time.

It is not necessarily the angle of a line. It is the distance as read by earthlings divided by time as read by earthlings.

 

ONLY for light the angle represents accurately velocity.

[Delta1212]

The issue that you're running into is that, 'apparent' velocity doesn't have a cap of c the way actual velocity does. Things are allowed to look like they are traveling up to a speed approximately equivalent to instantaneous.

 

Let's say there's a supernova 10,000 light years away. It releases neutrinos, which travel at very close to the speed of light. Earth sees the supernova, and an hour later, the neutrinos arrive, having fallen slightly behind the photons over the course of the trip.

 

Well, it had only been an hour since Earth saw the release of the neutrinos, so their apparent velocity from Earth must have been 10,000 ly per hour.

[DimaMazin]

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.

Then in Tom's frame the Earth should be younger.Here you and relativity contradict with reality.

[Janus]

Then in Tom's frame the Earth should be younger.Here you and relativity contradict with reality.

 

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.

[michel123456]

The issue that you're running into is that, 'apparent' velocity doesn't have a cap of c the way actual velocity does. Things are allowed to look like they are traveling up to a speed approximately equivalent to instantaneous.

 

Let's say there's a supernova 10,000 light years away. It releases neutrinos, which travel at very close to the speed of light. Earth sees the supernova, and an hour later, the neutrinos arrive, having fallen slightly behind the photons over the course of the trip.

 

Well, it had only been an hour since Earth saw the release of the neutrinos, so their apparent velocity from Earth must have been 10,000 ly per hour.

I am very glad that you raised this argument.

 

The situation you describe is different, you have moved the goal post.

 

The common understanding of the twin experiment consists in drawing a spacetime diagram as one would have drawn the path of an object bouncing on a wall, a kind of billiard game.

 

My diagram consists simply in stretching the diagram and in such a way that the deformation equal to c.

This stretching reflects the fact that earthlings observe all along the path the velocity of Tom being 0,5c. That includes also the instant where Tom is landing back on Earth. Note again that the 2 triangles have the same area, the same basis, the same height: they represent the same velocity.

 

The stretching also reflects the fact that earthlings observe the turning point with a delay of c, although the turning appears to them at half the trip (as observed by them).

 

In you argument, the earthlings did not send anything to make the supernova explode. There is no delay to include before the explosion.

 

But if you whish, I engage you to make the diagram of the twin experiment with neutrinos sended from Earth and back and provide here the results of the standard solution. At what speed will the earthlings observe the (real) neutrinos?

Edited May 13, 2013 by michel123456

[Delta1212]

I am very glad that you raised this argument.

 

The situation you describe is different, you have moved the goal post.

 

The common understanding of the twin experiment consists in drawing a spacetime diagram as one would have drawn the path of an object bouncing on a wall, a kind of billiard game.

 

My diagram consists simply in stretching the diagram and in such a way that the deformation equal to c.

This stretching reflects the fact that earthlings observe all along the path the velocity of Tom being 0,5c. That includes also the instant where Tom is landing back on Earth. Note again that the 2 triangles have the same area, the same basis, the same height: they represent the same velocity.

 

The stretching also reflects the fact that earthlings observe the turning point with a delay of c, although the turning appears to them at half the trip (as observed by them).

 

In you argument, the earthlings did not send anything to make the supernova explode. There is no delay to include before the explosion.

 

But if you whish, I engage you to make the diagram of the twin experiment with neutrinos sended from Earth and back and provide here the results of the standard solution. At what speed will the earthlings observe the (real) neutrinos?

Here, let's say that I've got a flashlight that is pointed towards Earth and traveling along the path of the accelerated twin in the twin paradox at neutrino speeds, so just barely under c. Let's say it falls 1 light day behind c per month (which is actually considerably slower than neutrinos) for the sake of math. The flashlight flicks on and then off once a month.

 

After the first month, the flashlight is 29 light days away from Earth when it flickers. The photons take 29 days to reach Earth, so 59 days after the flashlight leaves, the Earth sees the flashlight 29 light days away. This gives the flashlight an apparent velocity of just under 0.5c from Earth.

 

Let's put up a quick timeline of dates:

Day 30: Flashlight flickers 29 light days from Earth (velocity: 29/30c or 0.967c)

Day 59: Earth sees flashlight flicker 29 light days away (apparent velocity: 29/59c or 0.492c)

Day 60: Flashlight flickers 58 light days away from Earth (velocity: 0.967c)

Day 90: Flashlight flickers 87 light days away. (velocity: 0.967c)

Day 118: Earth sees flashlight flicker 58 light days away. (apparent velocity: 0.492c)

Day 120: Flashlight flickers at 116 light days. (apparent velocity: 0.967c)

Day 150: Flashlight flickers at 145 light days. (apparent velocity: 0.967c)

Day 180: Flashlight flickers at 174 light days. (apparent velocity: 0.967c)

Day 177: Earth sees flashlight flicker 87 light days away. (apparent velocity: 0.492c)

Day 210: Flashlight flickers at 203 light days. (apparent velocity: 0.967c)

Day 236: Earth sees flashlight flicker 116 light days away. (apparent velocity: 0.492c)

Day 240: Flashlight flickers at 232 light days. (apparent velocity: 0.967c)

Day 270: Flashlight flickers at 261 light days. (apparent velocity: 0.967c)

Day 295: Earth sees flashlight flicker 145 light days away. (apparent velocity: 0.492c)

Day 300: Flashlight flickers at 290 light days. (apparent velocity: 0.967c)

Day 330: Flashlight flickers at 319 light days. (apparent velocity: 0.967c)

Day 354: Earth sees flashlight flicker 174 light days away. (apparent velocity: 0.492c)

Day 360: Flashlight flickers at 348 light days, then turns around and heads back to Earth at the same velocity. (velocity: 0.967c)

Day 390: Flashlight flickers at 319 light days away. Covered 29 light days in 30 days. (velocity: 0.967c)

Day 413: Earth sees flashlight flicker 203 light days away. (apparent velocity: 0.492c)

Day 420: Flashlight flickers at 290 light days. Covered 29 light days in 30 days. (velocity: 0.967c)

Day 450: Flashlight flickers at 261 light days away. Covered 29 light days in 30 days. (velocity: 0.967c)

Day 472: Earth sees flashlight flicker 232 light days away. (apparent velocity: 0.492c)

Day 480: Flashlight flickers at 232 light days away. Covered 29 light days in 30 days. (velocity: 0.967c)

Day 510: Flashlight flickers at 203light days away. Covered 29 light days in 30 days. (velocity: 0.967c)

Day 531: Earth sees flashlight flicker 261 light days away. (apparent velocity: 0.492c)

Day 540: Flashlight flickers at 174 light days away. Covered 29 light days in 30 days. (velocity: 0.967c)

Day 570: Flashlight flickers at 145 light days away. Covered 29 light days in 30 days. (velocity: 0.967c)

Day 590: Earth sees flashlight flicker 290 light days away. (apparent velocity: 0.492c)

Day 600: Flashlight flickers at 116 light days away. Covered 29 light days in 30 days. (velocity: 0.967c)

Day 630: Flashlight flickers at 87 light days away. Covered 29 light days in 30 days. (velocity: 0.967c)

Day 649: Earth sees flashlight flicker 319 light days away. (apparent velocity: 0.492c)

Day 660: Flashlight flickers at 58 light days away. Covered 29 light days in 30 days. (velocity: 0.967c)

Day 690: Flashlight flickers at 29 light days away. Covered 29 light days in 30 days. (velocity: 0.967c)

Day 708: Earth sees flashlight flicker at 348 light days. (apparent velocity: 0.492c)

Day 709: Earth sees flashlight flicker at 319 light days. Covered 29 light days in 1 day. (apparent velocity: 29c)

Day 710: Earth sees flashlight flicker at 290 light days. Covered 29 light days in 1 day. (apparent velocity: 29c)

Day 711: Earth sees flashlight flicker at 261 light days. Covered 29 light days in 1 day. (apparent velocity: 29c)

Day 712: Earth sees flashlight flicker at 232 light days. Covered 29 light days in 1 day. (apparent velocity: 29c)

Day 713: Earth sees flashlight flicker at 203 light days. Covered 29 light days in 1 day. (apparent velocity: 29c)

Day 714: Earth sees flashlight flicker at 174 light days. Covered 29 light days in 1 day. (apparent velocity: 29c)

Day 715: Earth sees flashlight flicker at 145 light days. Covered 29 light days in 1 day. (apparent velocity: 29c)

Day 716: Earth sees flashlight flicker at 116 light days. Covered 29 light days in 1 day. (apparent velocity: 29c).

Day 717: Earth sees flashlight flicker at 87 light days. Covered 29 light days in 1 day. (apparent velocity: 29c)

Day 718: Earth sees flashlight flicker at 58 light days. Covered 29 light days in 1 day. (apparent velocity: 29c)

Day 719: Earth sees flashlight flicker at 29 light days. Covered 29 light days in 1 day. (apparent velocity: 29c)

Day 720: Flashlight reaches Earth. Covered 29 light days in 30 days (velocity: 0.967c). Earth sees flashlight reach Earth. Covered 29 light days in 1 day. (Apparent velocity: 29c).

 

 

Now, can you point to one of the dates marking Earth's observations that is incorrect, and tell me what the correct day on which Earth would observe the flashlight at that distance would be?

[Iggy]

You must agree that velocity is relative.

Each observer, from his own FOR, measures a different velocity. So each observer measures what I call an "apparent velocity" which is different for another observer in another FOR.

 

The "apparent velocity" of light is the same for all observers.

 

In this diagram velocity of light is a line at 45 degrees (or 135 degrees) oriented from the past to the future, always going from down to up in the diagram. Never going up to down. For all observers.

 

In this diagram, from Earth's FOR, the apparent velocity is distance divided by time.

It is not necessarily the angle of a line. It is the distance as read by earthlings divided by time as read by earthlings.

 

ONLY for light the angle represents accurately velocity.

Degrees are not units of speed. One doesn't measure the speed of light with a protractor. Saying that the 'apparent velocity' of light is a medium sized angle makes no sense.

 

In post 54 you accidentally divided the distance traveled by the time observed trying to find velocity. That was a very small mistake. You needed to divide the distance traveled by the time traveled.

 

It is better to go back and recognize what happened than trying to invent a new kind of velocity and a new kind of physics to work with it.

[tar]

#66 md65536,<br /><br />You are right to take back any suggestion that I am seeing this correctly. We would not want to mislead anybody actually trying to learn.<br /><br />I did, myself, think I was getting it, earlier on in the thread, but I lost it.<br /><br />I am thinking my on again off again attitude toward this, has to do with something I am figuring in the same manner as Delta1212. Perhaps it has something to do with the concept of simultaneity in terms of "when" something happens, and the shift one has to make between the instance of a thing, and the image of a thing.<br /><br />From a pure reality point of view, the most recent events in the universe are happening here and now, and everything else we see is old news, images of events that happened a millisecond ago, a lightmillisecond away, things that happened a second ago, a lightsecond away, and so forth, out to the Cosmic Backgroung radiation, that happened 13.8 billion years ago, 46 billion lightyears away.<br /><br />In addition, as best as I can figure it, there is only one instance of the universe, each and every piece and part of it, actually no older and no younger than 13.8 billion years old. And in this sense, a star 4.2 lightyears from us, is true, in two ways. One, we know it is happening now in a way we have no immediate access to, but that we can imagine is occuring in a way that will become "true" to our senses, in 4.2 years, when the light from that current event reaches our equipment. And two, we know the star is true, because we look up, and see the darn thing shining up there.<br /><br />If there are these two true instances of that star, which there is, then it is critical that when we speak of it, we declare which of the two we are refering to as real, or when we are referring to both the actual instance of the thing, and when we are referring to its presence in our sky and its effect on our senses and the world around us. Both "instances" of the thing, are very real to us, and neither is imaginary. Neither is "just a thought", or "just an image". There is only one instance of the thing, and it is happening now in both senses.<br /><br />Where I think Delta1212 and I are in agreement, is in the knowledge, that when something moves away from Earth at velocities approaching C, which is the speed at which the two instances or the two senses of a vastly distant thing are connected to each other, that something is actually making the trip, between the image type sense of the thing, and the actual thing. Staying in its own now, as something the exact same age, as the rest of the universe, but separating itself positionwise, and timewise, from Earth, and approaching positionwise and timewise, the actual now of the distant thing.<br /><br />No doubt, the math of relativity works out. But it seems to me, that you can explain the same numbers with the above understanding, and the red and blue shifts of what actual nows you are moving toward, staying at rest in relation to, or moving away from.<br /><br />Allowing one to perhaps explain a thing, without time dilation, length contraction and such things.<br />Perhaps.<br />Perhaps not.<br /><br />Regards, TAR2

 

Is the Mars rover doing something now? Is that thing its doing now the thing we see it doing now, or the thing we will see it doing in 14 minutes? One has to be specific about these uses of now.

[md65536]

#66 md65536,<br /><br />You are right to take back any suggestion that I am seeing this correctly. We would not want to mislead anybody actually trying to learn.

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.

No doubt, the math of relativity works out. But it seems to me, that you can explain the same numbers with the above understanding, and the red and blue shifts of what actual nows you are moving toward, staying at rest in relation to, or moving away from.

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.

[michel123456]

I cut out your post for the sake of clarity and emphasis mine:

Here, let's say that I've got a flashlight that is pointed towards Earth and traveling along the path of the accelerated twin in the twin paradox at neutrino speeds, so just barely under c. Let's say it falls 1 light day behind c per month (which is actually considerably slower than neutrinos) for the sake of math. The flashlight flicks on and then off once a month.

After the first month, the flashlight is 29 light days away from Earth when it flickers. The photons take 29 days to reach Earth, so 59 days after the flashlight leaves, the Earth sees the flashlight 29 light days away. This gives the flashlight an apparent velocity of just under 0.5c from Earth.

Let's put up a quick timeline of dates:
(...)


Now, can you point to one of the dates marking Earth's observations that is incorrect, and tell me what the correct day on which Earth would observe the flashlight at that distance would be?

"29 light days" as seen from which FOR? Are this the words of Tom or the words of Earth? This is a distance you are talking about and distances are measured differently from one FOR to another.

"the photon take 29 days to reach the Earth" as seen from which FOR? This is a time. Is this the time as seen from Tom? (as seen by a neutrino?)

 

Let's put up a quick timeline of dates:

Day 30: Flashlight flickers 29 light days from Earth (velocity: 29/30c or 0.967c)

Day 59: Earth sees flashlight flicker 29 light days away (apparent velocity: 29/59c or 0.492c)

 

(...)

 

"Day 30": _that is clearly Earth's time.

 

"Flashlight flickers 29 light days from Earth (velocity: 29/30c or 0.967c)" _that is clearly a distance in Earth's FOR. (or Am I wrong? Tom-flash-light didn't traveled 29 days of his time.)

 

"Day 59": _that is clearly Earth's time.

 

"Earth sees flashlight flicker 29 light days away (apparent velocity: 29/59c or 0.492c)" _Here you have supposed that Tom-flash-light traveled 29 light-days-earths-FOR in 29 days-earths-FOR, But for the flash to click, one has to wait a month in Tom's FOR. Or was that each month in Earth's FOR?

 

It is a mess. I prefer working with diagrams.

 

Degrees are not units of speed. One doesn't measure the speed of light with a protractor. Saying that the 'apparent velocity' of light is a medium sized angle makes no sense.

That is the basic of a spacetime diagram. A velocity is a ratio, and an angle is a ratio too.

In post 54 you accidentally divided the distance traveled by the time observed trying to find velocity. That was a very small mistake. You needed to divide the distance traveled by the time traveled.

Oh? I thought velocity was frame dependent. There is no one "distance traveled" and "time traveled". There are an infinity of "distances observed" and "time observed", each one for each FOR which divided give a "velocity observed".

It is better to go back and recognize what happened than trying to invent a new kind of velocity and a new kind of physics to work with it.

That is not my invention. You are speaking like there was an absolute velocity.

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

(edit)

 

I could understand the following:

 

Tom leaves the Earth and for 2 years he travels to a point.

Then he comes back, traveling 2 years and landing on Earth.

 

In this case, Time and distances are both in Tom's FOR. Not in Earth's FOR.

The diagram is the same as in a conventional diagram, the labeling is different.

[Iggy]

And in this sense, a star 4.2 lightyears from us, is true, in two ways. One, we f there are these two true instances of that star,

 

There are *not* two instances of the star. There is one instance of "tar observes star" (this is an event that happens on earth), and there is a different instance of "star emits light", (which, believe it or not, happens on the star).

 

They are two different events. Time separates the events. This has almost nothing to do with relativity. It is just common sense. "star emits" happens before "earth observes".

 

 

No doubt, the math of relativity works out. But it seems to me, that you can explain the same numbers with the above understanding... without time dilation, length contraction and such things.

 

Can you please punch numbers into "the above understanding", so that I may have an example to show "where length contraction and time dilation are unneeded"? Thank you.

 

 

That is the basic of a spacetime diagram. A velocity is a ratio, and an angle is a ratio too.

 

Please tell me in meters per second what the 'apparent velocity' of light is. Thank you.

[michel123456]

Day 360: Flashlight flickers at 348 light days, then turns around and heads back to Earth at the same velocity. (velocity: 0.967c)

You see, "day 360" is Earth's time.

 

In fact Tom makes the turn at half his time.

 

What happens is that Tom travels X days forth and X days back in his FOR, because he travels at the same velocity. So the symmetric diagram is from Tom's FOR and not from Earth's FOR. IOW distances & times on the conventional diagram are Tom's, and not Earth's.

 

(edited)

Day 708: Earth sees flashlight flicker at 348 light days. (apparent velocity: 0.492c)

 

Day 709: Earth sees flashlight flicker at 319 light days. Covered 29 light days in 1 day. (apparent velocity: 29c)

 

Day 710: Earth sees flashlight flicker at 290 light days. Covered 29 light days in 1 day. (apparent velocity: 29c)

 

Day 711: Earth sees flashlight flicker at 261 light days. Covered 29 light days in 1 day. (apparent velocity: 29c)

 

Day 712: Earth sees flashlight flicker at 232 light days. Covered 29 light days in 1 day. (apparent velocity: 29c)

 

Day 713: Earth sees flashlight flicker at 203 light days. Covered 29 light days in 1 day. (apparent velocity: 29c)

 

Day 714: Earth sees flashlight flicker at 174 light days. Covered 29 light days in 1 day. (apparent velocity: 29c)

 

Day 715: Earth sees flashlight flicker at 145 light days. Covered 29 light days in 1 day. (apparent velocity: 29c)

 

Day 716: Earth sees flashlight flicker at 116 light days. Covered 29 light days in 1 day. (apparent velocity: 29c).

 

Day 717: Earth sees flashlight flicker at 87 light days. Covered 29 light days in 1 day. (apparent velocity: 29c)

 

Day 718: Earth sees flashlight flicker at 58 light days. Covered 29 light days in 1 day. (apparent velocity: 29c)

 

Day 719: Earth sees flashlight flicker at 29 light days. Covered 29 light days in 1 day. (apparent velocity: 29c)

 

Day 720: Flashlight reaches Earth. Covered 29 light days in 30 days

(velocity: 0.967c). Earth sees flashlight reach Earth. Covered 29 light

days in 1 day. (Apparent velocity: 29c).

You see, at the end, your calculation show that the flashlight is observed flashing once a day (Earth's FOR) although at the beginning it was assumed that the flashlight was flashing once a month (Earth's FOR). There has been a switching somewhere.

 

In fact the "once a month" is always in Toms-flash-light FOR and never in Earth's FOR.

 

If that is the case that means that at day 59 the Earth does not observe the first flash.

(...)

 

 

 

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

 

(...)

In Iggy's graph, time is Tom's, and distances are Tom's.

 

Quoting myself:

I could understand the following:

 

Tom leaves the Earth and for 2 years he travels to a point.

Then he comes back, traveling 2 years and landing on Earth.

 

In this case, Time and distances are both in Tom's FOR. Not in Earth's FOR.

The diagram is the same as in a conventional diagram, the labeling is different.

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

.

.

.

(...)

Please tell me in meters per second what the 'apparent velocity' of light is. Thank you.

299 792 458 m / s

It is the red triangles on my diagram (see below).

 

Edited May 14, 2013 by michel123456

[swansont]

Note again that the 2 triangles have the same area, the same basis, the same height: they represent the same velocity.

 

 

Area is not the same as velocity. Area has units of distance*time

[michel123456]

Area is not the same as velocity. Area has units of distance*time

Correct.

 

The green triangle has height D and basis T (numbers are not important)

The blue triangle has height D and basis T

 

The formula for velocity is V=D/T

 

So the 2 rectangles represent the same value V=D/T

 

 

Meaning simply that the ratio of D/T for the 2 rectangles is the same.

Of course the product DT/2 is the same too. It is the triangle area but it is irrelevant here.

Edited May 14, 2013 by michel123456

[LaurieAG]

Why wouldn't you just use two clocks at points one light year apart, that both start when they detect the first type 1a supernova from an equidistant galaxy, and run a race between the two clocks/points?

 

You could then send out a series of staged light pulses every hour from the second clock location and delay the departure from the first location by a year so that the spaceship can start when the first pulse hits the detector on its nose and can compare its own time clock with the absolute time at both start and end points during the trip, regardless of whatever speed it was travelling.

[swansont]

Correct.

 

The green triangle has height D and basis T (numbers are not important)

The blue triangle has height D and basis T

 

The formula for velocity is V=D/T

 

So the 2 rectangles represent the same value V=D/T

 

 

Meaning simply that the ratio of D/T for the 2 rectangles is the same.

Of course the product DT/2 is the same too. It is the triangle area but it is irrelevant here.

 

D/T is the slope of a line. Your use of "basis" is not valid for an evaluation of a speed.

[michel123456]

D/T is the slope of a line. Your use of "basis" is not valid for an evaluation of a speed.

That's what blocked my mind so many times.

Then I thought that we interpret the velocity as the slope of a line. The diagram does not speak. We interpret the diagram as we wish.

 

So in this diagram, the slope of a line is velocity only for c. But it is a special case.

 

In this diagram, as a generalization the slope is not velocity. But it is always what the equation tells us: it is the ratio height/basis.

[swansont]

The diagram does not speak. We interpret the diagram as we wish.

 

We also set up definitions of the terminology we use, so while we can interpret things as we wish, those interpretations can be consistent with the definitions (i.e. correct) or inconsistent with the definitions (i.e. wrong)

 

In the paradigm of wanting to be correct, then, we cannot interpret the diagram as we wish.

[michel123456]

We also set up definitions of the terminology we use, so while we can interpret things as we wish, those interpretations can be consistent with the definitions (i.e. correct) or inconsistent with the definitions (i.e. wrong)

 

In the paradigm of wanting to be correct, then, we cannot interpret the diagram as we wish.

What is the terminology of velocity if it is not V=D/T? Edited May 14, 2013 by michel123456

[swansont]

What is the terminology of velocity if it is not V=D/T?

 

It is V=D/T But you have to use the distance it traveled and the time of the travel. The time of travel from point R to point E is not the 2 years you are using, i.e. it is not the "base". It's the projection of that line onto the time axis. Similarly, the time of travel to point R is the projection of that point on to the time axis, which is less than 2 years.

[michel123456]

It is V=D/T But you have to use the distance it traveled and the time of the travel. The time of travel from point R to point E is not the 2 years you are using, i.e. it is not the "base". It's the projection of that line onto the time axis. Similarly, the time of travel to point R is the projection of that point on to the time axis, which is less than 2 years.

What i present is that the projection onto the time axis is through the 45 degrees diagonal. It is not possible to observe velocity in "real time" (with a horizontal projection).

 

At time 2, the Earth observes Tom reaching point R.

The velocity is 0,5 c (the slope of the line as you said).

So earthlings calculate that after 2 years earths time, Tom has reached a point 2 LY away (Earths distance) minus the time (earths time) the light had to reach the Earth. That is point R.

So i believe until point R we should not have any disagreement, except the fact that I call "apparent velocity" the ratio D/T. I will stop doing that.

 

The question is:

_is that the point R where Tom makes the turn?

or

_at what point should Tom make the turn so that his velocity (as measured from Earths FOR) is the same when coming back?

 

I mean: in the standard explanation it blows my mind to think about observing Tom waving hand on a planet 1 LY away and one year after shaking my hand after landing at "regular" velocity. The astronomer looking in his telescope would be incapable of seeing Tom traveling back. Tom would be apparently moving as a photon! In other words the astronomer would see Tom landing on Earth popping out of nowhere.

 

On the other hand the velocity of Tom landing must be the same as the velocity when he left.

In my diagram, the time lapse between the waving and shaking is 2 years, because Tom comes from a planet 1LY away and travels at 0,5c. Which is the regular definition of velocity V=D/T

 

Imagine an E.T. that you observe in your telescope. The E.T. is on a planet 1 LY away. He waves his hands and embark a spaceship that travels at 0,5c. In how much time do you expect to see him landing?

(edit)

If you answer "next year" (counting 2 years of travel minus 1 year delay of the waving hand image), that means from earth the E.T. is seen as a photon.

Edited May 14, 2013 by michel123456

[md65536]

Imagine an E.T. that you observe in your telescope. The E.T. is on a planet 1 LY away. He waves his hands and embark a spaceship that travels at 0,5c. In how much time do you expect to see him landing?

(edit)

If you answer "next year" (counting 2 years of travel minus 1 year delay of the waving hand image), that means from earth the E.T. is seen as a photon.

So he travels at the speed of light? Or is this photon "slow"?

 

Imagine you're watching the ET through your telescope and you see it flip a switch on a laser. Do you expect to receive the signal from the laser a year later?

[Iggy]

 

What is the 'apparent velocity' of light in m/s

 

.299 792 458 m / s

It is the red triangles on my diagram (see below).

 

 

Yes. Or, one light-year per year. That is the normal velocity of light because it is the x distance between M and R divided by the t distance between M and R.

 

M is 2/3rds of a year above R and 2/3rds of a light-year to the left of R.

 

Please find the 'apparent velocity' of the blue line from R to E using the same method, and let me know what it is. I really do appreciate this. Thank you.

[swansont]

What i present is that the projection onto the time axis is through the 45 degrees diagonal. It is not possible to observe velocity in "real time" (with a horizontal projection).

 

Which is not consistent with the formulation of relativity and the definition of velocity. You don't get to call this the velocity.

 

 

 

 

I mean: in the standard explanation it blows my mind to think about observing Tom waving hand on a planet 1 LY away and one year after shaking my hand after landing at "regular" velocity. The astronomer looking in his telescope would be incapable of seeing Tom traveling back. Tom would be apparently moving as a photon! In other words the astronomer would see Tom landing on Earth popping out of nowhere.

 

I must have missed where anyone was claiming that such a thing happens. At no point does any traveler keep pace with a photon.

 

 

On the other hand the velocity of Tom landing must be the same as the velocity when he left.

In my diagram, the time lapse between the waving and shaking is 2 years, because Tom comes from a planet 1LY away and travels at 0,5c. Which is the regular definition of velocity V=D/T

 

Imagine an E.T. that you observe in your telescope. The E.T. is on a planet 1 LY away. He waves his hands and embark a spaceship that travels at 0,5c. In how much time do you expect to see him landing?

(edit)

If you answer "next year" (counting 2 years of travel minus 1 year delay of the waving hand image), that means from earth the E.T. is seen as a photon.

 

If ET waves his hand and hops in his spaceship and travels at 0.5c, he will arrive 1 year after you see the wave, i.e. 1 year after the photon left, because the hand-wave signal is a year old. If you think that he is traveling as fast as a photon then your bookkeeping is messed up. You will get photons from his ship all through the trip. He doesn't just pop into existence.

[michel123456]

Yes. Or, one light-year per year. That is the normal velocity of light because it is the x distance between M and R divided by the t distance between M and R.

 

M is 2/3rds of a year above R and 2/3rds of a light-year to the left of R.

 

Please find the 'apparent velocity' of the blue line from R to E using the same method, and let me know what it is. I really do appreciate this. Thank you.

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.

 

Which is not consistent with the formulation of relativity and the definition of velocity. You don't get to call this the velocity.

 

 

 

 

I must have missed where anyone was claiming that such a thing happens. At no point does any traveler keep pace with a photon.

 

 

 

If ET waves his hand and hops in his spaceship and travels at 0.5c, he will arrive 1 year after you see the wave, i.e. 1 year after the photon left, because the hand-wave signal is a year old. If you think that he is traveling as fast as a photon then your bookkeeping is messed up. You will get photons from his ship all through the trip. He doesn't just pop into existence.

(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.