The twin paradox

[michel123456]

Is there incomprehension here?

 

I was not asking how the traveler accelerates (decelerates) into my frame.

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

never mind.

 

I hope that someone around here understands my question.

[swansont]

Is there incomprehension here?

 

 

On this particular question any incomprehension is not mine.

 

I was not asking how the traveler accelerates (decelerates) into my frame.

 

And I didn't base my answer on that. I merely mentioned, parenthetically, that in the classic setup, the acceleration happens when you are co-located. Sorry if that extra information befuddled you. The change goes away when there is no travel time of the signal between you, i.e. when you are co-located. You could still be moving when this occurs.

 

As long as there is separation between you, there will be a delay in the signal. In proper physics, we account for that.

 

 

[DimaMazin]

We can't see a fast decay of fast muon.But this muon can see a fast decay of slow(relatively us) muon.

[Delta1212]

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

Very understandable.

Now what I cannot understand:

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

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

 

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

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

 

*exaggerating a bit for the sake of the argument.

To clarify, are you asking how something can hit you with the force of a subliminal object when you see it traveling many times faster than the speed of light, or are you asking what it would look like when that object reached you when it has looked like it was traveling at many times the speed of light?

[michel123456]

To clarify, are you asking how something can hit you with the force of a subliminal object when you see it traveling many times faster than the speed of light, or are you asking what it would look like when that object reached you when it has looked like it was traveling at many times the speed of light?

Something like that.

[swansont]

Something like that.

 

And it's been answered. There is no delay effect when the two objects are right next to each other. Why is that answer insufficient and being ignored?

[tar]

Iggy,

 

Oh. I didn't know I couldn't use the numbers, without admitting to length contraction and time dilation. I thought they could stand on their own.

 

Regards, TAR2

 

Made intuitive sense to me that the red shift behind would be the recipricol of the blue shift ahead.

[Iggy]

Iggy,

 

Oh. I didn't know I couldn't use the numbers, without admitting to length contraction and time dilation.

You can't. The numbers you're talking about come from the equation,

 

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

 

I can plug in the relevant numbers to show...

 

[math]f_\mathrm{obs} = f_\mathrm{rest}\sqrt{\left({1 - 0.5}\right)/\left({1 + 0.5}\right)}[/math]

 

[math]f_\mathrm{obs} = f_\mathrm{rest}\sqrt{1/3}[/math]

 

[math]f_\mathrm{obs} = f_\mathrm{rest} (0.577)[/math]

 

and 0.577 is the number you're talking about.

 

 

I can derive the equation I gave from the equation for time dilation plus classical Doppler shift if you like.

 

Point is that those numbers do take time dilation (and by extension, length contraction) into consideration

 

I thought they could stand on their own.

No number is an island.

 

I can tell you 4.

 

4 what? 4 apples? No number stands on its own. It either refers to something or it means nothing.

Made intuitive sense to me that the red shift behind would be the recipricol of the blue shift ahead.

It is. Until you can find the numerical value of red and blue shift without time dilation... well... yeah...

[DimaMazin]

And it's been answered. There is no delay effect when the two objects are right next to each other.

Well.But then who sees delaying time of neighbour?

[michel123456]

And it's been answered. There is no delay effect when the two objects are right next to each other. Why is that answer insufficient and being ignored?

There are 2 solutions.

Either Swansont did not understand the question either Michel does not understand the answer.

 

Actually Michel understands that the answer does not answer the question.

 

@Delta

Did you understand Swansont's answer?

You seem to have a good grasp at understanding other people problematics.

[md65536]

There are 2 solutions.

Either Swansont did not understand the question either Michel does not understand the answer.

The answer's straightforward. The question's nonsense.

 

What does it even mean for something to appear to travel faster than the speed of light, if you're not using the real definition of speed but instead a made up one?

 

If something arrives when or after it appears to leave, then it does not appear to travel faster than light. If it arrives before it is seen leaving, and yet is never out of sight, do you continuously see it appearing to move backward?

 

If you see something approach very fast, it appears to move faster but its time also appears to tick fast; it appears to cover greater distances but appears to take longer to do so.

 

Why use one measure for the speed of light (c, and the real definition of speed) and compare it to a different measure of an object (your mixed-frames measures).

 

Why are you wilfully ignoring any delay of light, yet comparing "apparent" speeds to c, now suddenly acknowledging a travel time of the same light?

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

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

Try working through your example in your imagination.

A traveler sends a letter by fast plane, and leaves simultaneously by slightly slower plane. The letter arrives, and the traveler shortly after (let's say 1 minute for example). Do you honestly not understand "how the hell" that is possible? Honestly?! Did the traveler really appear to come at you thousands of times faster than the plane by any reasonable measure, even a made up one?

Edited May 16, 2013 by md65536

[michel123456]

I feel really alone.

 

I need Spielberg to direct the scene.

 

An astronomer looks in his telescope and observe an E.T. coming to Earth.

 

In his telescope, the E.T. comes from a galaxy one thousand LY away and made the trip in one day, apparently , since the astronomer during 24h was really observing the travel of the E.T., each second of it, from the beginning when he embarked till the end, traveling 1000 LY in one day. (that is apparently - mistakenly 365000 times c) and aging about 400000 times faster than us.

 

Then, after 24h of observing, the astronomer sees the E.T. crashing on Earth at velocity less than c.

 

At which moment did the false impression disappear from the telescope and changed into the correct "less than c" velocity?

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

Michel's answer:

 

at no moment the astronomer observed the correct speed.

The astronomer observed the E.T. crashing upon Earth at velocity 365000 times c.

Only after calculating can the astronomer obtain the correct result.

[swansont]

I feel really alone.

 

I need Spielberg to direct the scene.

 

An astronomer looks in his telescope and observe an E.T. coming to Earth.

 

In his telescope, the E.T. comes from a galaxy one thousand LY away and made the trip in one day, apparently , since the astronomer during 24h was really observing the travel of the E.T., each second of it, from the beginning when he embarked till the end, traveling 1000 LY in one day. (that is apparently - mistakenly 365000 times c) and aging about 400000 times faster than us.

 

Slower than us, not faster. ET's clock runs slow.

 

 

Then, after 24h of observing, the astronomer sees the E.T. crashing on Earth at velocity less than c.

 

At which moment did the false impression disappear from the telescope and changed into the correct "less than c" velocity?

 

Once ET arrives. Asking the question again doesn't change the answer. The effect of time dilation is due to the motion. As long as ET is moving toward the astronomer, this effect of time dilation will be present. It will cease if ET stops, or the instant ET passes by (after which the signals will spread out in time, rather than compress)

 

The answer is not going to change if you re-form the question and ask it again.

 

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

Michel's answer:

 

at no moment the astronomer observed the correct speed.

 

Trivially true because one does not "observe" speed. One calculates it.

 

 

The astronomer observed the E.T. crashing upon Earth at velocity 365000 times c.

 

No, unless the astronomer makes an elementary mistake and uses an incorrect definition of speed.

 

Only after calculating can the astronomer obtain the correct result.

 

True under all scenarios. One must measure the change in the distance and the time in order to obtain the speed. The time one uses is NOT the time that the signal arrives.

[michel123456]

 

Slower than us, not faster. ET's clock runs slow.

 

 

 

Once ET arrives. Asking the question again doesn't change the answer. The effect of time dilation is due to the motion. As long as ET is moving toward the astronomer, this effect of time dilation will be present. It will cease if ET stops, or the instant ET passes by (after which the signals will spread out in time, rather than compress)

 

The answer is not going to change if you re-form the question and ask it again.

 

 

Trivially true because one does not "observe" speed. One calculates it.

 

 

No, unless the astronomer makes an elementary mistake and uses an incorrect definition of speed.

 

 

True under all scenarios. One must measure the change in the distance and the time in order to obtain the speed. The time one uses is NOT the time that the signal arrives.

(emphasis mine)

No?

One does not "observe" speed?

 

 

Slower than us, not faster. ET's clock runs slow.

(...)

You surely don't mean that the E.T. aged only 20h during the travel. For him the travel had a duration of many years. So the astronomer observed the E.T. living in fast motion. That's what I ment.

[swansont]

michel123456, on 16 May 2013 - 07:48, said:

(emphasis mine)

No?

One does not "observe" speed?

You observe motion. Speed is a quantification of that motion. It requires a (correct) calculation.

 

michel123456, on 16 May 2013 - 07:48, said:

You surely don't mean that the E.T. aged only 20h during the travel. For him the travel had a duration of many years. So the astronomer observed the E.T. living in fast motion. That's what I ment.

Yes, that's exactly what I mean.

 

If ET arrives one day after the photons were sent upon his departure, to him the trip took one day in his frame. If the trip for him took many years, he would not arrive one day after the signal arrived, he would arrive many years after. This is the same as the example I gave in post #120, except with a larger value of gamma. Now it's ~365,000. Meaning that 1000LY gets contracted down to ~1 light day in the moving frame. The astronomer does not see ET "living in fast motion".

 

edit: the stricken statement is ambiguous under the varying definitions being used.

[michel123456]

You observe motion. Speed is a quantification of that motion. It requires a (correct) calculation.

 

 

 

Yes, that's exactly what I mean.

 

If ET arrives one day after the photons were sent upon his departure, to him the trip took one day in his frame. If the trip for him took many years, he would not arrive one day after the signal arrived, he would arrive many years after. This is the same as the example I gave in post #120, except with a larger value of gamma. Now it's ~365,000. Meaning that 1000LY gets contracted down to ~1 light day in the moving frame. The astronomer does not see ET "living in fast motion".

My bad.

So if i understand clearly, it is physically possible to travel in one day to the other side of the universe at 0.9999999999999999999c

 

And at c there is no horizon problem.

[swansont]

My bad.

So if i understand clearly, it is physically possible to travel in one day to the other side of the universe at 0.9999999999999999999c

 

And at c there is no horizon problem.

 

No, for the reason that the horizon problem is an issue of expansion, which is tied in with general relativity. Which means you are contradicting the assumption of the twin paradox that we're dealing only with special relativity.

 

The solution applies only as far as you can ignore general relativity.

[michel123456]

You observe motion. Speed is a quantification of that motion. It requires a (correct) calculation.

 

(...)

O.K. the astronomer observes the motion of the E.T. coming from a galaxy 1000LY away in 1 day.

 

He observes this hilarously fast motion until the E.T. crashes on Earth.

The E.T. made the travel in one (of his)day.

He appeared (falsely) as if he made the travel in one of our days.

 

Is that it?

Edited May 16, 2013 by michel123456

[md65536]

If ET arrives one day after the photons were sent upon his departure, to him the trip took one day in his frame. If the trip for him took many years, he would not arrive one day after the signal arrived, he would arrive many years after. This is the same as the example I gave in post #120, except with a larger value of gamma. Now it's ~365,000. Meaning that 1000LY gets contracted down to ~1 light day in the moving frame. The astronomer does not see ET "living in fast motion".

I hope that I don't confuse michel123456, who seems to be struggling with even a classical notion of the notion of delay of light, but...

 

If gamma is ~365,000 then ET's clock *appears* to tick at such a high rate that its one day of travel is seen over about 0.1 seconds for the observer (relativistic Doppler). The observer measures the trip taking about 1000 years, but sees it in 0.1 seconds, leading to michel123456's intentional confusion (intentional because the mistake has been pointed out MANY TIMES NOW by many people but it is ignored).

 

The reduction in the delay of light is a first-order effect of v, and the slowing of its clock is a second-order effect... is that the right way to say it? So the observed frequency of ET's clock is faster than the astronomer's. I would call that appearing to live in fast motion, yet still not appearing to travel "at a fast velocity" as michel123456 might, because time has an inverse relationship to velocity (not to mention the proportional effect on distance here).

 

 

And michel123456, things appear to do what they're doing. If it travels at .99c then that's what it appears to do. You're saying "If something is really doing *this*, then it appears to do *that* (something different)" but then turning it around and saying "And if it appears to do *that*, then it's really doing *that*." Be consistent! Either speak of reality (and delayed observations), or acknowledge the difference between reality and appearance when converting from reality to appearance AND vice versa. You're also considering only what is real for light (v=c) and only what is apparent for ET (v=some made-up measure).

Edited May 16, 2013 by md65536

[Janus]

O.K. the astronomer observes the motion of the E.T. coming from a galaxy 1000LY away in 1 day.

 

He observes this hilarously fast motion until the E.T. crashes on Earth.

The E.T. made the travel in one (of his)day.

He appeared (falsely) as if he made the travel in one of our days.

 

Is that it?

Not quite. For one, 1000 ly away is still comfortably within our own galaxy.

 

But besides that, if we observe E.T. crossing 1000 ly in one of our days, This means that he is actually traveling at 0.999997260281479c relative to us.

 

Thus, due to length contraction according to him, and time dilation according to us. he will age 2.34 yr on the trip.

 

IOW, we will see him age 2.34 yrs in our one day of observing him make the trip.

 

This works because the Doppler shift factor for something approaching at 0.999997260281479c is 854.4, and 854.4 times 1 day is 2.34 yr.

 

 

 

 

 

 

 

 

0.99999726028147900000

[md65536]

But besides that, if we observe E.T. crossing 1000 ly in one of our days, This means that he is actually traveling at 0.999997260281479c relative to us.

 

Thus, due to length contraction according to him, and time dilation according to us. he will age 2.34 yr on the trip.

So let me get this straight:

ET traveled 365000 light-days in 365001 days (ignoring leap years) according to our measurements, equalling a velocity of 0.999997260281479c,

 

and was measured by us as taking 365001 days by our clocks, but appeared to take only 1 day because the image of it leaving is delayed by 365000 days due to light's finite speed

 

and that ET aged 854.4 days = 2.34 years during its trip according to its own clock, and appears to us to have aged 2.34 years during its trip according to its clock

 

and this all works out simply and consistently even if the theory behind the calculations might be unintuitive?

[swansont]

I hope that I don't confuse michel123456, who seems to be struggling with even a classical notion of the notion of delay of light, but...

 

If gamma is ~365,000 then ET's clock *appears* to tick at such a high rate that its one day of travel is seen over about 0.1 seconds for the observer (relativistic Doppler). The observer measures the trip taking about 1000 years, but sees it in 0.1 seconds, leading to michel123456's intentional confusion (intentional because the mistake has been pointed out MANY TIMES NOW by many people but it is ignored).

 

 

I should point out that because of some vague and varying definitions of things like apparent velocity and elapsed, there are now multiple scenarios being described. Thus, my example and the one given by Janus are somewhat different, and part of that was some confusion on my part in trying to reconstruct the scenario in terms of relativity, rather than not-relativity. Mea culpa.

[md65536]

I should point out that because of some vague and varying definitions of things like apparent velocity and elapsed, there are now multiple scenarios being described. Thus, my example and the one given by Janus are somewhat different, and part of that was some confusion on my part in trying to reconstruct the scenario in terms of relativity, rather than not-relativity. Mea culpa.

No problem... the original scenario presented by michel123456 (post #137, #143?) used 1 day for both the (proper) time that the traveler ages, AND the time that passes on Earth while the traveler is seen traveling, which doesn't work out. You presented the scenario where 1 day is the traveler's time, and Janus present the different scenario where 1 day is the Earth time while the traveler is watched, and the differences are now clear.

 

If that confuses anyone they could simply choose one scenario and ignore the other...

Edited May 16, 2013 by md65536

[Iggy]

Why are you wilfully ignoring any delay of light, yet comparing "apparent" speeds to c, now suddenly acknowledging a travel time of the same light?

 

I passed this when catching up in the thread, and I have to say it is the best worded summary of the situation I've encountered, or could imagine encountering.

 

 

Michel, velocity is travel distance divided by travel time. It has been that way for thousands of years. You can obsess yourself pink with the idea of travel distance divided by observation time, but that isn't velocity, and if you plug that thing you invented into relativity then you're going to get a nonsense answer.

 

I get why you're doing it. In post 54 you made a small mistake. You were arrogant about it. Two posts later you realized your mistake, but all that meant is that you would now have to spend 100 posts covering for your mistake though hell and high water, because that's what I've seen you do two or three times before. You err and forever more spending your time covering for your error despite *everyone* trying to explaining the error to you.

 

Ask yourself, is that the best approach? Someone who dissembles rather than admitting a mistake and learning from it?

 

The mere fact that you can't solve the 'apparent velocity' of an object the same way you solve it for light should be enough for you to say "Wait a minute, something is wrong with what I'm doing". The fact that you get a different 'apparent velocity' depending on the observer's position in space should be another.

 

But, you won't. You can't. You can't say that. You would rather write hundreds of nonsense posts than have someone correct you, and I don't even think you know you're doing it.

[Spyman]

Well said Iggy, +1.