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The twin paradox


michel123456

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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.

Fair enough. Yes I committed errors but truly, genuinely, I don't understand, and I am sure now that you don't understand my question.

 

 

 

 

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

Just to put an end at this scenario so that i can learn something from it: and that other people reading this long thread get a final correct idea.

 

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

 

He observes this hilariously fast motion from the beginning till the end until the E.T. crashes on Earth.

The E.T. made the travel in 854.4 (of his) days.

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

The astronomer observed the E.T. live in fast motion.

 

Is that it?

 

Where is Delta?

Edited by michel123456
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Fair enough. Yes I committed errors but truly, genuinely, I don't understand, and I am sure now that you don't understand my question.

 

 

 

Just to put an end at this scenario so that i can learn something from it: and that other people reading this long thread get a final correct idea.

 

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

 

He observes this hilariously fast motion from the beginning till the end until the E.T. crashes on Earth.

The E.T. made the travel in 854.4 (of his) days.

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

The astronomer observed the E.T. live in fast motion.

 

Is that it?

 

Where is Delta?

 

Still not right, if ET traveled 1000LY in 854.4 (of his) days then just how fast was he travelling?

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Still not right, if ET traveled 1000LY in 854.4 (of his) days then just how fast was he travelling?

Ha.

back to the beginning.

He didn't travel in his own FOR, he was at rest. The Earth traveled to him in 854.4 days, and the distance was not 1000LY in his FOR (If I am not messing things again).

Edited by michel123456
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He appeared (falsely) as if he made the travel in one of our days.

Michel, you need to acknowledge that if the astronomer first sees the ET at a planet 1000 lightyears away and then later arrive on Earth the next day, then he knows that the journey took 1000 years + 1 day and not 24 hours, Earth time.

(Which is extremely fast but not impossible.)

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Michel, you need to acknowledge that if the astronomer first sees the ET at a planet 1000 lightyears away and then later arrive on Earth the next day, then he knows that the journey took 1000 years + 1 day and not 24 hours, Earth time.

(Which is extremely fast but not impossible.)

Yes, I acknowledge that the astronomer can calculate that, the same way he calculated the distance 1000LY in the first place. In his FOR.

 

But what he sees, what he lives, what he feels, the motion that crashes upon the Earth has nothing to do with a velocity "less than c".

As you said it takes 1000 years + 1 day ( or 854,4 days I have lost my mind on this), it does not take 1 day as the astronomer have seen.

I know that the 1 day figure is wrong, it is a mistake, it is false.

But that is what have been observed, all the way long, from the boarding till the crash.

Like the piano of George Clooney in the coffee commercial.

You see it coming very (very very) fast.

Then comes the astronomer and tells you that no, the piano was gently floating in the air and landed smoothly because that is what his calculations say.

 

That baffles my mind, forget it, I am afraid I will die with that misconception.

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But what he sees, what he lives, what he feels, the motion that crashes upon the Earth has nothing to do with a velocity "less than c".

As you said it takes 1000 years + 1 day ( or 854,4 days I have lost my mind on this), it does not take 1 day as the astronomer have seen.

I know that the 1 day figure is wrong, it is a mistake, it is false.

But that is what have been observed, all the way long, from the boarding till the crash.

Like the piano of George Clooney in the coffee commercial.

You see it coming very (very very) fast.

Then comes the astronomer and tells you that no, the piano was gently floating in the air and landed smoothly because that is what his calculations say.

The way to salvage understanding here is to use calculations that everyone can agree on (it seems to have been done, using the numbers Janus gave in post #145), and contemplate the meaning or the visuals around that.

 

The 1-day figure is a real observation of various events that occurred over a thousand years + a day for the astronomer, and the image of those events are just delayed different amounts.

 

If ET lands "gently" then it has somehow decelerated extremely quickly. This works fine if it has negligible mass, but... Consider that at race speeds, a car hitting a wall can disintegrate. At relativistic speeds there's no "gentle" landing here.

 

 

There are other relativistic effects that you'll see, too. If you can continuously watch the "hilarious fast motion" then there is light coming from ET for its 854.4 days, which you get all in one day, so it will appear brighter to you, also blue-shifted. To consider the hilarious motion, imagine it walking at relativistic speeds: Imagine that a light day is one step (of a giant)... the 1000 LY distance is 365000 steps for you, but the length-contracted distance is 854.4 steps for ET. That means that for every step you see it take, it covers 427.2 (=gamma) of your "rest" steps. In other words ET looks hilariously stretched, incoming (and hilariously squished if outgoing). The reason: Even though ET's lengths are actually length contracted according to you, the image of one step takes longer to reach you than the next step does (each step covers 472.2 of your light days so each next step is delayed 472.2 days less than the previous one). That is, it covered 1/854.4 of your rest distance in one step, taking 472.2 of your days to do it, but each step is seen over 101 observer seconds.

 

It's not moving hilariously quickly (the opposite actually), it just appears so due to reduction in the delay of light as it approaches, and it's not stretched but it appears so (what you call "false", but is perhaps just an "illusion" or simply its "appearance").

 

This is getting complicated now, but suffice to say: The hilarious image of ET will appear distorted in several ways. It's probably hard to image because we don't experience it in every day life, and so it would probably be weird to see.

 

(Note, they reduce c, while I scaled sizes way up with the giant ET, the result is the same.)

Edited by md65536
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Just to put an end at this scenario so that i can learn something from it: and that other people reading this long thread get a final correct idea.

 

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

 

He observes this hilariously fast motion from the beginning till the end until the E.T. crashes on Earth.

The E.T. made the travel in 854.4 (of his) days.

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

The astronomer observed the E.T. live in fast motion.

 

Is that it?

 

Where is Delta?

Assuming your numbers are correct, ET's velocity is v in:

 

[math]365242.199 = \frac{854.4}{\sqrt{1-v^2}}[/math]

 

v = 0.999997c

 

light would take 365242 days to make the trip and our long fingered friend would take 365243.09 days with that velocity (both from Earth's perspective).

 

So... your calculation for ET's proper time (854.4) was just about spot on.

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You see it coming very (very very) fast.

Then comes the astronomer and tells you that no, the piano was gently floating in the air and landed smoothly because that is what his calculations say.

0.99999726 c IS very (very very) very extremely fast... the calculations reflect that.

 

I was curious if it would be enough energy to destroy the entire Earth if ET crashed at that speed. Just using the google, I plugged the number into http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/releng.html, using the mass of Enterprise D, and got an answer of 1.524x10^28 joules. http://io9.com/5876473/how-much-energy-would-the-death-star-require-to-destroy-earth claims it would require 2.25x10^32 J to destroy the Earth using the Death Star's laser... almost 15000 times more.

 

So no, ET on the Enterprise at this speed would only destroy 1/15000th of the Earth. So that's like an area close to the size of Australia to a depth of 10km. Or vaporize 140m off the entire surface of the Earth.

 

Another way to put it, it's about 363 times the estimated energy of the asteroid that killed the dinosaurs.

 

These are just google-based calculations so I might be off by some orders of magnitude, but anyway: I assure you that the astronomer does not tell you this near-c impact will be gentle!!!

 

Edit: If it was just ET without any ship, it'd only be like 1/690th the impact of the Chicxulub impact... so probably nothing for most of us to worry about.

Edited by md65536
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Assuming your numbers are correct, ET's velocity is v in:

 

[math]365242.199 = \frac{854.4}{\sqrt{1-v^2}}[/math]

 

v = 0.999997c

 

light would take 365242 days to make the trip and our long fingered friend would take 365243.09 days with that velocity (both from Earth's perspective).

 

So... your calculation for ET's proper time (854.4) was just about spot on.

Now what I am wondering is this:

 

Calculation shows that E.T.'s velocity from Earth's perspective is 365243.09 days.

 

Maybe, I say "maybe" it is wrong to "correct" this number by substracting the time for light (365242 days)

and obtaining the "1 day" figure.

IOW maybe the astronomer does not "observe the ET coming in 1 day" but in 365243.09 days as the math suggest.

Edited by michel123456
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Now what I am wondering is this:

 

Calculation shows that E.T.'s velocity from Earth's perspective is 365243.09 days.

 

Maybe, I say "maybe" it is wrong to "correct" this number by substracting the time for light (365242 days)

and obtaining the "1 day" figure.

IOW maybe the astronomer does not "observe the ET coming in 1 day" but in 365243.09 days as the math suggest.

 

Right, the velocity would be (365242/365243.09)c... nearly c, and he arrives nearly as fast as light does. It all works out.

Edited by Iggy
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Right, the velocity would be (365242/365243.09)c... nearly c, and he arrives nearly as fast as light does. It all works out.

But if it is correct that the ET (or Tom) takes such time, then the diagram is wrong.

Edited by michel123456
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But if it is correct that the ET (or Tom) takes such time, then the diagram is wrong.

 

No one has yet drawn a diagram where the launch point is 1000 lightyears from landing. That said, no part of the description just given is incompatible with the typical minkowski-type diagrams that have been drawn in this thread.

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halfc_zps2f1f3da3.png

 

 

 

Following this graph

The astronomer at time 3 observes Tom making the turn.

At Time 4 the astronomer observes Tom landing.

The difference T4-T3 is a substraction that you seemed to agree cannot be done.

 

The astronomer should observe what the math says, in this graph it is the difference T4 minus T2.

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The astronomer should observe what the math says, in this graph it is the difference T4 minus T2.

Draw a horizontal line from Turn to the t-axis, and say "Light 'should' arrive instantaneously."

Why do you have light at a 45-degree angle, traveling at a velocity of c, if the astronomer really should observe T4-T2?

 

The astronomer does observe what the math says, including the observation of Turn being delayed by 1 unit of time due to the velocity of light.

 

The astronomer observes what the math says, what the graph says, what her eye says. The astronomer does not instantly observe the entire universe. I've never known anyone to have so much trouble with the concept of delayed observation due to speed of light.

 

I mean, how are you accounting for light in your diagram or description? You drew a line representing light... you realize that what the astronomer observes of Turn arrives at the astronomer in the form of light??? So what sort of calculation or reasoning could possibly convince you that the astronomer "should" observe Turn instantaneously?

Edited by md65536
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Following this graph

The astronomer at time 3 observes Tom making the turn.

At Time 4 the astronomer observes Tom landing.

The difference T4-T3 is a substraction that you seemed to agree cannot be done.

Sure it can be done. It just doesn't mean what you have said it means. Earth measures two years between the turn and the arrival. The traveler therefore measures,

 

[math]T_{\mathrm{ET}} = T_{\mathrm{earth}} \sqrt{1-v^2}[/math]

 

[math]T_{\mathrm{ET}} = 2 \sqrt{1-(0.5)^2}[/math]

 

[math]T_{\mathrm{ET}} = 1.732[/math]

 

That's how long it takes our traveler to make the trip, 1.732 years. To solve how long it takes our observer to view this 1.732 years of proper time use the equation I gave on the post your quoting the diagram from (post 68)

 

 

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

 

[math]f_\mathrm{obs} = 1_\mathrm{ET} \sqrt{\left( 1 + .5 \right) / \left( 1 - .5 \right)}[/math]

 

[math]1_\mathrm{obs} = 1.732_\mathrm{ET}[/math]

 

Each year on earth observes 1.732 years on ET's ship. That's the whole trip from turn to landing. The earthlings understand that ET traveled 2 years in Earth's frame since the turn, 1.732 years since the turn in its frame, and was observed over a period of 1 year on earth because of the Doppler shift caused by light delay.

 

The three numbers mean different things and they're all properly accounted for on the diagram and with the math. The proper equation says that earth should observe ET over a period of one year (between T=3 and T=4 on the diagram)

Edited by Iggy
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Sure it can be done. It just doesn't mean what you have said it means. Earth measures two years between the turn and the arrival. The traveler therefore measures,

What's funny, tho frustrating, is that the same diagram can be used to describe a classical version, where the traveler ages the same amount as the inertial twin. Or to be more precise: The diagram doesn't even specify the traveler's measure of time. Ignoring time dilation, the prediction of events according to only the inertial twin's clock can be made using only classical physics, and it still works out the same! Velocity = 0.5c, duration of trip = 4 units of time, Turn is seen at t=3, distance to turn is 1 unit of distance. Delay of observation of Turn, shown clearly on the diagram, is 1 unit of time.

 

Relativity isn't even needed to understand this diagram! It probably requires a grade-school level of understanding of math and physics to understand it??? (And highschool or likely greater to see it work in SR though.)

 

 

michel123456, this whole problem could be understood in terms of classical physics first, if that would make it easier!

Edited by md65536
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Increasing of distance increases interval of time between signals of ageing of escaping traveler.Because light, from each subsequent attribute of ageing, needs additional time for overcoming the increased distance.Reducing of distance reduces interval of time between signals of ageing of approaching traveler.Because light,from each subsequent attribute of ageing,needs less time for overcoming the reduced distance.

Edited by DimaMazin
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Following this graph

The astronomer at time 3 observes Tom making the turn.

At Time 4 the astronomer observes Tom landing.

The difference T4-T3 is a substraction that you seemed to agree cannot be done.

 

The astronomer should observe what the math says, in this graph it is the difference T4 minus T2.

T4-T3 yields the time between when the astronomer receives the light from Tom turning around (at a 1 year delay) and when the astronomer receives the light from Tom's return (at no delay). However, at T3, the astronomer sees Tom turn around at a point on the timeline that the astronomer believes is T2, because the astronomer knows there is a 1 year delay. Therefore to get the time that the astronomer observes Tom to take in a mathematical, rather than visual, sense, you must subtract the time that the astronomer calculates Tom turned around (T2) from his arrival.

 

So the astronomer observes Tom's turnaround as being at T2 when the astronomer sees the light from Tom's turnaround at T3. Part of the difficulty in communicating the concept here, I think, is simply that the term 'observe' can mean a lot of different things depending on how you're using it.

 

Actually, here's a good example of how this works in a manner that is more relatable to every day experience. Have you ever tried to figure out how far away lightning is by measuring the time between the flash and the thunder?

 

So let's use sound as a stand in for light, because both travel at a constant speed (for a given medium), but sound is slower so the effects of distance are more apparent in our every day perceptions whereas light is far too fast for the delay to be intuitively understandable based on every day experience.

 

So let's start with the lightning. In the case of Tom, you know how far away Tom is, and you know when the light from Tom arrives, so you need to know when Tom turned around. So say you're in the basement where you can't see the lightning (because frankly it's extraneous to the metaphor), but you hear the thunder.

 

At exactly 5 pm, you hear thunder. At exactly 5:01 lightning strikes your house. You later learn that the other lightning strike hit a church 12 miles away. Sound travels about a fifth of a mile per second, so it took 60 seconds for the sound to reach you. You "heard" one minute pass between the thunder from the first strike and the thunder from lightning striking your house, but you observed that two minutes passed between the strikes.

 

That's the same way the astronomer "sees" Tom turn around one year before arriving but observes that it was two years.

 

Now, you're wondering what it looks like when Tom arrives at Earth after traveling at a speed that looks ridiculously fast. Sound actually helps here, too. An object traveling at the speed of sound can be observed approaching but it will be entirely silent until it arrives, at which point all the sound that it created during the trip arrives simultaneously in the form of a sonic boom. Obviously, nothing can travel at light speed, but an object traveling just below the speed of sound will behave similarly to how something traveling just below the speed of light will.

 

So a slightly sub-sonic object will be Doppler shifted into very high frequencies, the way a siren sounds higher pitched while traveling toward you than when traveling away. The sound will also be sped up because the object emitting the sound is almost keeping pace with the waves it emits, so all the sound it emits arrives only slightly before it does and has to play in full.

 

If a jet with a loudspeaker blaring someone's voice was traveling at you at just below the speed of sound, the voice would be very loud, very high pitch and speak very quickly until the loud speaker reached you. When it was directly at your position (assuming it didn't hit you) you would briefly hear the voice at the exact speed, volume and pitch at which it was being broadcast. Then once it passed you, you would hear it at a lower pitch, more softly and slowed down slightly.

 

It's the same with light. An object traveling toward you will be bluer, appear to be traveling faster and be brighter than one traveling away from you. This is true of literally anything moving toward or away from you, but the speeds we're used to working with on a daily basis, even our fastest vehicles/weapons/whatever, move at such tiny fractions of the speed of light that the difference is literally imperceptible and we don't notice.

 

Once you are dealing with largest enough distances and speeds, though, light behaves more like how we experience sound.

 

So an object traveling toward you at 0.9999c will look like a movie that someone hit fast forward on in addition to being bluer and brighter, but once it reaches you, it will no longer be traveling toward you and it will look like it is moving at exactly the speed it is moving at, look exactly the color you'd normally see it at and be as bright as normal.

 

In fact, if it isn't on a collision course with you, but will pass right next to you instead, less of its speed will be pointed "at" you as it gets closer until it is parallel with your position and it momentarily isn't moving toward or away from you. In this case, it will look as if it is decelerating the whole way until it's traveling at it's actual velocity as it passes you, and then continues decelerating as it moves away. This is, again, only true of something moving toward you but not on a perfect collision course. In that case, you won't see it traveling at its actually velocity until it hits you, but it will hit you with the force of something traveling 0.967c, not 29c.

 

 

Hopefully something in there made some sense. If there's something specific confusing you, and some way you'd prefer it explained, let me know and I'll see if I can answer it in a more reasonable timeframe than this last time.

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In fact, if it isn't on a collision course with you, but will pass right next to you instead, less of its speed will be pointed "at" you as it gets closer until it is parallel with your position and it momentarily isn't moving toward or away from you. In this case, it will look as if it is decelerating the whole way until it's traveling at it's actual velocity as it passes you, and then continues decelerating as it moves away.

Sorry if I'm straying off-topic here: Ignore this post if you're only concerned with the twin paradox. But there's a curiosity here...

 

At the closest that the object comes, I think that light seen from this event will already be redshifted (http://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Transverse_Doppler_effect). So the change from blue to red appears to occur sometime earlier, when the object still appears to be approaching (transversely). I think?

 

This effect, and the time that the object appears to significantly change from blue to red, is negligible if it passes close enough.

Edited by md65536
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Sorry if I'm straying off-topic here: Ignore this post if you're only concerned with the twin paradox. But there's a curiosity here...

 

At the closest that the object comes, I think that light seen from this event will already be redshifted (http://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Transverse_Doppler_effect). So the change from blue to red appears to occur sometime earlier, when the object still appears to be approaching (transversely). I think?

 

This effect, and the time that the object appears to significantly change from blue to red, is negligible if it passes close enough.

That's actually interesting. I can sort of see why that would happen, but I need to look at it a bit more in-depth. Either way, thanks for the info.

 

Though yeah, it doesn't really change anything especially impactful about what I was trying to say. Still always good to learn new things.

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Sorry if I'm straying off-topic here: Ignore this post if you're only concerned with the twin paradox. But there's a curiosity here...

 

At the closest that the object comes, I think that light seen from this event will already be redshifted (http://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Transverse_Doppler_effect). So the change from blue to red appears to occur sometime earlier, when the object still appears to be approaching (transversely). I think?

 

This effect, and the time that the object appears to significantly change from blue to red, is negligible if it passes close enough.

Thank you for the link.

I understand the delay.

What I don't understand is wonderfully showed in this diagram from the link above.

 

LcprojectionS05n.gif

Diagram 3. The grey ellipse is a moving relativistic sphere, its oblate

shape due to Lorentz contraction. The colored ellipse is the visual

image of the sphere. Background curves are an xy-coordinates grid which

is rigidly linked to the sphere; it is shown only at one moment in time. (text in italic from Wiki)

 

 

I don't understand the abrupt change of "apparent velocity" (The motion of the colored ellipse) when the object reaches the observer.

I don't understand that because the object is supposed to have the same velocity when coming and leaving and here everything shows as if there was an abrupt deceleration acting.

 

--------

Or stating my question otherwise.

In this diagram, the grid is "linked to the sphere"

 

But the Earth is "at rest", so I suppose that there must be another regular orthogonal grid linked to the Earth not shown on the diagram (in this grid takes place the motion of the grey sphere)

 

I don't understand the abrupt change of apparent motion on one side and on the other side of the observer in relation with the not shown regular orthogonal grid.

Edited by michel123456
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That's actually interesting. I can sort of see why that would happen, but I need to look at it a bit more in-depth. Either way, thanks for the info.

 

Though yeah, it doesn't really change anything especially impactful about what I was trying to say. Still always good to learn new things.

 

You know...our Universal Space/Time Dimensional Geometry just will not allow certain things to happen as well as not allow certain things to be observed.

 

SR details a lot of the reasoning why behind this and it is folly to even make a chart using any concept of a craft and people within moving a any velocity faster than Light Speed or c.

 

But just because it is impossible for any craft using a form of PROPULSION to reach or exceed 186,282 Miles per Second...as at even a quarter of such velocities Time Dilation and Relativistic Issues become apparent as well as as a Space Crafts velocity get's closer to c...the Space Crafts Mass get's closer to being INFINITE IN MASS....it does not mean that it is not possible to use a Gravitic Drive to either Fold Space/Time or travel within a Warped Field or Subspace.

 

Let's look at Light or Photons...as a Light Wavelength travels through Space/Time...the Photons being a Quantum Particle/Wave Form have no mass and thus there is no increase in mass to the infinite....although Photons do have ANGULAR MOMENTUM.

 

Still...Photons are existing as both Particle and Wave and as such are existing in a Non-Linear Time as a Wave Function. This alone shows us that by either being to access or be interconnective to specific Universal or Multiversal Space/Time Dimensionality...as Photons as such Quanta...do...the rule comes forth that Depending upon Condition or State...it is possible to achieve Light Speeds Velocity.

 

Since this would mean understanding the UFT...Unified Field Theory...and thus meaning we would have to convert the Matter of a craft and the people within it...to it's very Quantum State and then transform such Quanta as Quarks, Gluons, Leptons, Mesons...etc....into Photons....then beam that to a specific point of travel...and this would still only be at a maximum velocity of c...and then at the reception point by the tenets of the UFT....convert the Photons back into the original Matter...we still have the problem of only achieving Light Speed as well as possibly killing everyone being beamed as conversion of the matter to energy and back does nothing to save the identity or soul if you will of a person.

 

The better and quicker way which would not have Time Dilation issues as well would not be breaking the laws of Physics as the graphs that detail obtaining faster than light speeds are but fantasy...would be to Fold Space/Time or travel within a Sub-Space.

 

Split Infinity

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!

Moderator Note

Please stay on topic - transverse doppler effect is marginal (but would still be better in its own thread) but unified field theory is definitely beyond the remit of this thread

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Thank you for the link.

I understand the delay.

What I don't understand is wonderfully showed in this diagram from the link above.

 

LcprojectionS05n.gif

Diagram 3. The grey ellipse is a moving relativistic sphere, its oblate

shape due to Lorentz contraction. The colored ellipse is the visual

image of the sphere. Background curves are an xy-coordinates grid which

is rigidly linked to the sphere; it is shown only at one moment in time. (text in italic from Wiki)

 

 

I don't understand the abrupt change of "apparent velocity" (The motion of the colored ellipse) when the object reaches the observer.

I don't understand that because the object is supposed to have the same velocity when coming and leaving and here everything shows as if there was an abrupt deceleration acting.

 

 

--------

Or stating my question otherwise.

In this diagram, the grid is "linked to the sphere"

 

But the Earth is "at rest", so I suppose that there must be another regular orthogonal grid linked to the Earth not shown on the diagram (in this grid takes place the motion of the grey sphere)

 

I don't understand the abrupt change of apparent motion on one side and on the other side of the observer in relation with the not shown regular orthogonal grid.

The abrupt change is a result of the difference in light delay. While the object is traveling toward you, each photon bouncing off of it is starting a little closer to the Earth, so it will have slightly less travel time than the last photon to bounce off of the object, and will arrive on Earth sooner after the previous one than it left.

 

When the object is traveling away from Earth, each successive photon is slightly farther away from Earth, so it has to travel farther and the gap between when the first arrives and the second arrives will be slightly longer than the gap between the first and second bouncing off the object.

 

This creates the illusion that objects moving toward you are faster than they really are, and objects moving away from you are slower than they really are, though it's only really noticeable at large fractions of light speed.

 

When the object reaches you, it transitions from the blue-shift "sped up" appearance to the red-shifted "slowed down" appearance, which creates the illusion of rapid deceleration even though the object is still actually traveling along an inertial path and doesn't feel any net forces.

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