1. Originally Posted by Jeff Root
No, that isn't how it works. You don't redefine the
unit of length and you don't alter the speed of light.
You just apply the appropriate corrections to get the
correct values. It's similar to the difference between
air speed and ground speed in flying. If what you want
to know is how far you have travelled, but all you can
measure is your direction and air speed, then you need
to find out from someone else what are the direction
and speed of the wind that you are flying through, and
your ground speed, and from that, the distance.

-- Jeff, in Minneapolis
Are you talking to me or Webbo? He/she is the one that has a picture in his/her head that would require the speed of light not to be constant and the length of a meter to have different values depending on the orientation with respect to the direction of acceleration.

If I'm travelling to the moon or any other object I can record my speed relative to that object just fine and through similar processes record the distance to the object at any given time along my journey. My speed can be deduced by simple Doppler measurements. My distance can be deduced by the total time it takes to do a single measurement of my speed. Why do you think you need a 3rd observer to do this?

Take for instance an AEW&C plane. Their radar lets the operators know the vector of the object which includes the distance to an object and that object's direction and speed relative to the AEW&C plane. Sure the radar system can do further calculations and say how fast, far and what direction objects are travelling with regard to a point on the ground but that is an additional transformation that the system performs. The measurements of AEW&C always start out relative to the AEW&C plane itself.

2. Order of Kilopi
Join Date
Dec 2004
Posts
11,219
Originally Posted by Webbo
According to all the previous postings the distances
between the stars ahead are supposed to contract.
Why has the angle increased and therefore caused
an increase of observed distance between them?
Because they appear to be closer.

fingers wide apart. Look at them through one eye.
appearing to move closer to the spacecraft due to
length contraction. You see the angular distance

-- Jeff, in Minneapolis

3. Order of Kilopi
Join Date
Dec 2004
Posts
11,219
Originally Posted by WayneFrancis
Originally Posted by Jeff Root
No, that isn't how it works. You don't redefine the
unit of length and you don't alter the speed of light.
You just apply the appropriate corrections to get the
correct values. It's similar to the difference between
air speed and ground speed in flying. If what you want
to know is how far you have travelled, but all you can
measure is your direction and air speed, then you need
to find out from someone else what are the direction
and speed of the wind that you are flying through, and
your ground speed, and from that, the distance.
Are you talking to me or Webbo?
You. Although I'm agreeing with "Webbo" on this
specific point, I'm not sure he understands what he
is saying, or whether he could follow the argument.

Originally Posted by WayneFrancis
He/she is the one that has a picture in his/her head
that would require the speed of light not to be constant
and the length of a meter to have different values
depending on the orientation with respect to the
direction of acceleration.
He didn't indicate that in the post you were replying to.
As far as I can see, it was your idea that the speed of
light and the length of a meter would need to have
different values depending on the orientation with
respect to the direction of acceleration. After all, you
were telling him that -- he wasn't telling you. So it is
to correct.

Originally Posted by WayneFrancis
If I'm travelling to the moon or any other object I can
record my speed relative to that object just fine and
through similar processes record the distance to the
object at any given time along my journey. My speed
can be deduced by simple Doppler measurements.
My distance can be deduced by the total time it takes
to do a single measurement of my speed.
Oh, lord!

You can do that with radar, at low speeds (say, a
million kilometers per hour), but the faster you go,
the less accurate your measurements will be.
Getting a good value goes from very complex to
terribly complex as you approach the speed of light
relative to the object. And the value becomes less
your distance is from the Moon when the pulse is
sent out when you are 100,000 km from the Moon
and the echo received back a few milliseconds later
when you are just 1 km from the Moon? How do
you interpret the value?

Originally Posted by WayneFrancis
Why do you think you need a 3rd observer to do this?
That wasn't intended to be an essential feature of the
analogy. Different kinds of information are needed in
the two scenarios. The other observer in the analogy
provides the speed of the wind relative to the ground.
The interstellar traveller needs to know his integrated
acceleration, or the Doppler shift of signals from ahead
or behind, or the like. Sorry I made that sound like it
was supposed to mean more than it did.

Originally Posted by WayneFrancis
Take for instance an AEW&C plane. Their radar lets the
operators know the vector of the object which includes
the distance to an object and that object's direction and
speed relative to the AEW&C plane. Sure the radar system
can do further calculations and say how fast, far and what
direction objects are travelling with regard to a point on
the ground but that is an additional transformation that
the system performs. The measurements of AEW&C
always start out relative to the AEW&C plane itself.
Well, yeah. Same thing. They don't need to redefine
the meter or speed of light to translate measurements
into other coordinate systems. Of course, they aren't
moving at nearly the speed of light relative to what
they look at, but the translation is similar.

-- Jeff, in Minneapolis

4. Originally Posted by Jeff Root
You. Although I'm agreeing with "Webbo" on this
specific point, I'm not sure he understands what he
is saying, or whether he could follow the argument.
So you think he/she is arguing based on ignorance. You are arguing I'm hoping on a rejection of semantics and your statements that indicate that you think there is a preferred frame everyone can agree is best instead of just using the frame your in and using coordinate transformations if you want to relate your frame with another frame.

Originally Posted by Jeff Root
He didn't indicate that in the post you were replying to.
As far as I can see, it was your idea that the speed of
light and the length of a meter would need to have
different values depending on the orientation with
respect to the direction of acceleration. After all, you
were telling him that -- he wasn't telling you. So it is
to correct.
Well maybe let webbo clarify them self because I'm not the only one that thinks webbo's comments indicate that webbo doesn't understand what is going on even when pictures are drawn for them. I'm not sure you do either by your comments recently but then it could be that you are in one of your modes of arguing semantics and claiming the way you think about something is the best way. What I'm presenting is the mainstream view about coordinate system transformations.

Originally Posted by Jeff Root
Oh, lord!

You can do that with radar, at low speeds (say, a
million kilometers per hour), but the faster you go,
the less accurate your measurements will be.
Getting a good value goes from very complex to
terribly complex as you approach the speed of light
relative to the object. And the value becomes less
your distance is from the Moon when the pulse is
sent out when you are 100,000 km from the Moon
and the echo received back a few milliseconds later
when you are just 1 km from the Moon? How do
you interpret the value?
You are raising a straw man. Given you know your speed via doppler shift it isn't hard to work out what distance you where at when you started the measurement and when you finished it.
Thought experiment 1: note these are not the only way to do this but just simple methods I can spout off the top of my head
1) obtain relative speed via Doppler shift. This doesn't care if the object is stationary relative to you or .999c relative to you. The Doppler shift will indicate the relative motion between you and your target.
2) time the measurement.
Given the following
a) you know your relative speed with respect to the object (from step #1)
b) you are heading straight towards the object
c) the signal took x amount of time to go out and bounce back
d) you are not accelerating during this time period
Then there is only 1 distance you could have been at the beginning of the measurement and only 1 at the end.
Hell if you know your angle and amount you accelerated during that time you can still figure out where you where at the start and finish of the measurement.

Thought experiment 2
1) obtain relative speed via Doppler shift. This doesn't care if the object is stationary relative to you or .999c relative to you. The Doppler shift will indicate the relative motion between you and your target.
2) obtain distance measurement using parallax

Originally Posted by Jeff Root
That wasn't intended to be an essential feature of the
analogy. Different kinds of information are needed in
the two scenarios. The other observer in the analogy
provides the speed of the wind relative to the ground.
The interstellar traveller needs to know his integrated
acceleration, or the Doppler shift of signals from ahead
or behind, or the like. Sorry I made that sound like it
was supposed to mean more than it did.
So you are now agreeing that a ship in transit can make measurements on its distance and speed at any point during its trip to another body without outside help....I'm really confused on what you are complaining about.

Originally Posted by Jeff Root

Well, yeah. Same thing. They don't need to redefine
the meter or speed of light to translate measurements
into other coordinate systems. Of course, they aren't
moving at nearly the speed of light relative to what
they look at, but the translation is similar.

-- Jeff, in Minneapolis
Good, we both agree that in any frame 1 second=1 second and 1m=1m in that frame and that other frames measure the the same time and distance different.
IE ship is travelling at ~.866c They measure time and distances normally on their ship 1s/1s and 1m=1m . An observer from where that ship took off would watch the ship's clock and say it was ticking half as fast and that the metre stick on the ship looks like it is only 50cm long if the stick is oriented in the direction of motion. Again my post was kind of a reply to more then one of Webbo's posts and its my mistake for not requoting what he was saying multiple times when I pointed out the implications of some of his statements.

It doesn't matter how fast they are moving! They are using photons to do all the measurements. All they have to worry about is that their receiver is capable of receiving the appropriate frequency range. Sure a radar gun police use can't measure the speed of an object travelling at .999c but that isn't a physical barrier. Its just stupid for a police radar gun or an AEW&C airplane to have record things that go that fast. But if we built a ship capable of relativistic speeds it wouldn't be stupid and the receiver would be able to record very high energy photons.

Just because AEW&C don't do it doesn't mean it is physically impossible. And as I've said that isn't the only way to get distance measurements. It is only 1 of many ways.

But we are getting off the OP of this thread now. If you have issue of any of my diagrams please raise it with me. If you have some other topic you want to discuss please point me to a thread of your own.

Go read Webbo's posts. He's indicated multiple times that if an observer measures 2 objects in front of them as having an angular separation of 20 degrees between them then accelerate towards them that no matter the speed the 2 objects should still appear to only have 20 degrees of angular separation between them which is FALSE.

5. Originally Posted by Webbo
I dont follow.
Since that post, I see a lot of better illustrations and explainations of it. So; I won't even try. In fact, I erred in not elongating the rear part in my haste to come up with a simple diagram.

Originally Posted by Webbo
And you don't find it just a little odd that there are currently 2 entities (Earth and craft B) at essentially the same point in space and one (Earth) measures the distance to our target star as 22.3 ly and has the knowledge that even if a craft were to pass at 99.9% of c the fastest it could get there is 1 ly of time (as measured by the passing craft itself), and the other (craft B) has measured the destination to be just a couple of weeks away?
Not at all.
One problem that most people have, and I suspect you too, is that it is hard to imagine the your speed relative to another speed rather than a fixed object. The root of that is that we experience our speed linearly in our daily lives. (ie. 10mph = 2* 5mph, 100mph = 2* 50mph, etc). Speed is not linear. It is a curve.
At the very slow relative speeds (our movment to the Earth) we are on such a small arc of that curve that it appears as a line because the changes (that define the shape of that curve) are imperceptable to us.

It was not clear to me until I understood calculus with equations where "x approaches a limit".

6. Established Member
Join Date
Aug 2009
Posts
753
Originally Posted by NEOWatcher
Not at all.
Then what is stopping the observer on Earth building an identical vehicle to craft B, and reaching the target star 22.3 ly away in say 3 weeks (to allow for acceleration) of on board travel time?

7. Originally Posted by WayneFrancis
It is a fact that the distance contracts in front of the travelling observer and elongates behind them.
Not quite correct. Lengths in the direction of relative movement contract, regardless of whether they are in front or behind. Optically, objects appear to elongate "in front", approaching the observer, and contract while receding from the observer, appearing skewed/rotated in the transition (Penrose-Terrell effect/Terrell rotation).

Webbo is still indisputably wrong in his apparent claim that there should be no aberration, though, you just got the nature of the aberrations a bit turned around:
http://www.fourmilab.ch/cship/aberration.html
http://demonstrations.wolfram.com/Re...dDopplerShift/

8. Originally Posted by Webbo
Then what is stopping the observer on Earth building an identical vehicle to craft B, and reaching the target star 22.3 ly away in say 3 weeks (to allow for acceleration) of on board travel time?
3 weeks to those on board the craft? Well currently technology because we don't know how to get any craft up to the speed of 0.999996653453784c.
Think of it this way. The fastest particles we've created are hadrons at the LHC and their speed is 0.999999991c which is only 0.000003337546246% faster. That is about 3/1,000,000 of 1% faster

If we could speed a ship up to that speed, and slow it back down at the other end those aboard would have only aged 3 weeks while we here on Earth would have aged 22.3 years

9. Originally Posted by cjameshuff
Not quite correct. Lengths in the direction of relative movement contract, regardless of whether they are in front or behind. Optically, objects appear to elongate "in front", approaching the observer, and contract while receding from the observer, appearing skewed/rotated in the transition (Penrose-Terrell effect/Terrell rotation).

Webbo is still indisputably wrong in his apparent claim that there should be no aberration, though, you just got the nature of the aberrations a bit turned around:
http://www.fourmilab.ch/cship/aberration.html
http://demonstrations.wolfram.com/Re...dDopplerShift/
That first link explicitly states that
we've set the main viewer to compensate for Lorentz contraction and Doppler shift, leaving only aberration to affect our view
What my diagrams are showing is the the Lorentz contraction alone and its effect on what we see.

10. Originally Posted by WayneFrancis
What my diagrams are showing is the the Lorentz contraction alone and its effect on what we see.
Lorentz contraction doesn't cause elongation...just contraction. It depends only on relative velocity, not on whether an object is approaching or receding. Other pages on the fourmilab.ch site demonstrate length contraction and combined effects.

11. Originally Posted by Webbo
Then what is stopping the observer on Earth building an identical vehicle to craft B, and reaching the target star 22.3 ly away in say 3 weeks (to allow for acceleration) of on board travel time?
Technology, energy consumption and G forces.

12. Established Member
Join Date
Aug 2009
Posts
753
Of course there is another consideration. If space is contracted in the direction of travel then so is ship, the crew and any apparatus used to measure the parallax/distance. Therefore would the observers on board even be able to measure any change?

13. Originally Posted by Webbo
Of course there is another consideration. If space is contracted in the direction of travel then so is ship, the crew and any apparatus used to measure the parallax/distance.
No, they aren't. Why would they be contracted in their own frame of reference? Velocity with respect to that frame is zero.

14. Established Member
Join Date
Aug 2009
Posts
753
Originally Posted by cjameshuff
No, they aren't. Why would they be contracted in their own frame of reference? Velocity with respect to that frame is zero.
So where does the point of contraction start?

15. Order of Kilopi
Join Date
Jan 2010
Posts
3,582
Originally Posted by Webbo
So where does the point of contraction start?
It doesn't start or stop anywhere, an object is contracted if it is in relative motion with the observer irrespective of where the object is. The spaceship isn't contracted because it is not in relative motion with the observer, not because it is closer to him.

16. Banned
Join Date
Mar 2012
Posts
211
These drawings are completely inappropriate, because they ignore the aberration of light (Bradley).

For velocity v = 0.99c perpendicular rays aberration is over 81 degrees!

17. Ok then lets get back to the original post and I'm sorry for any confusion my diagrams may have introduced

to cjameshuff, Hetman and anyone else that has relevant knowledge:

1) When using parallax to obtain the distance to an object is the distance based on the observer's frame of reference?
If the answer to Question #1 is yes what is the answer to the following senario.
Point A and Point B are not moving relative to each other.
Point A and Point B are not in a significantly different gravity well.
Point A and B given those 2 criteria are 10ly appart.

Observer Q accelerates instantly away from Point A towards Point B at ~.866c

2)What distance does Observer Q think Object B is at?

3)Would Observer Q's parallax measurement change, for Object B, from before and after Observer Q acceleration.

I'm understanding now where I went wrong as far as the what happened behind the observer. The distance has to be contracted in both the forward and reverse direction otherwise looking back the accelerated observer would calculate that they've travelled faster then they actually have within their own frame of reference.

Now My answers for the above are
1) Yes
2) 5ly
3) Yes

If I'm wrong please tell me why.

18. Banned
Join Date
Mar 2012
Posts
211
Aberration distort the image.
http://en.wikipedia.org/wiki/Penrose-Terrell_rotation

I think that the distances determined in Minkowski space would be incorrect, because would depend on the direction.

19. Order of Kilopi
Join Date
Sep 2004
Posts
5,445
Originally Posted by Hetman
Aberration distort the image.
http://en.wikipedia.org/wiki/Penrose-Terrell_rotation

I think that the distances determined in Minkowski space would be incorrect, because would depend on the direction.
They are not incorrect, they are different. That is the whole point.

The fact that the object looks different is also irrelevant to the parallax measurement. So the prolate spheroidal star looks like a different prolate spheroid.

Also, you should look at the link you provided. It is a rotation of an object passing the observer. The observation of something nearly head on will have a minimal rotation.

20. Banned
Join Date
Mar 2012
Posts
211
Perhaps, but these distortions have nothing to do with transformations of space (distances).
This is ordinary optics - theory of waves.

Moreover, this aberration is clearly not symmetric, thus reveals that the frames of reference are not equivalent.

21. Established Member
Join Date
Aug 2009
Posts
753
Originally Posted by caveman1917
It doesn't start or stop anywhere, an object is contracted if it is in relative motion with the observer irrespective of where the object is. The spaceship isn't contracted because it is not in relative motion with the observer, not because it is closer to him.
Ok. So what if there are 2 ships travelling towards the star with one placed in front of the other at a fixed distance of 2 light years between them. Does that mean as the 2 ships accelerate to 0.99 c the space between the 2 ships remains 2 light years (as it must in the same reference frame) yet the star that was originally 22.3 ly away (as measured from the ship placed behind) is now only 1 ly away and space has somehow contracted and pulled the star though the first ship. Of course the first ship I assume would still measure the star to be about 0.8 light years ahead. Please explain how this is possible.

22. Established Member
Join Date
Aug 2009
Posts
753
Originally Posted by WayneFrancis
I'm understanding now where I went wrong as far as the what happened behind the observer. The distance has to be contracted in both the forward and reverse direction otherwise looking back the accelerated observer would calculate that they've travelled faster then they actually have within their own frame of reference.
So would that mean that for our moving ship a star that is behind would also have its distance contracted from 22.3 ly to 1 ly?

23. Originally Posted by Webbo
Ok. So what if there are 2 ships travelling towards the star with one placed in front of the other at a fixed distance of 2 light years between them. Does that mean as the 2 ships accelerate to 0.99 c the space between the 2 ships remains 2 light years (as it must in the same reference frame) yet the star that was originally 22.3 ly away (as measured from the ship placed behind) is now only 1 ly away and space has somehow contracted and pulled the star though the first ship. Of course the first ship I assume would still measure the star to be about 0.8 light years ahead. Please explain how this is possible.
No, the trailing ship will not see the star contract through the other ship. Your basic setup is impossible, you can't observe what's happening 2 light years away in real time...the ships can't start accelerating simultaneously in all frames. If each accelerates simultaneously according to an observer at their midpoint, each will accelerate, and then see the other ship accelerate starting from a contracted distance of about 0.09 light years, with the leading ship reaching the star first.

24. Originally Posted by Webbo
Ok. So what if there are 2 ships travelling towards the star with one placed in front of the other at a fixed distance of 2 light years between them. Does that mean as the 2 ships accelerate to 0.99 c the space between the 2 ships remains 2 light years (as it must in the same reference frame) yet the star that was originally 22.3 ly away (as measured from the ship placed behind) is now only 1 ly away and space has somehow contracted and pulled the star though the first ship. Of course the first ship I assume would still measure the star to be about 0.8 light years ahead. Please explain how this is possible.
No, the 2 ships are in similar frames but just like the SR Thought experiment of what happens when a train 200m long travels at .99c through a 100m long tunnel.

What will happen is that the ships distances will get closer because you won't be able to get both ships to start accelerating at the same time in all reference frames. While their will be a frame in which they appear to accelerate at the same time there will be other frames where 1 will accelerate before the other and close the gap. Nice thought experiment but it is just like asking about the front and back of a train that accelerates. To those on board the train nothing strange happens. But what people outside the train see is very different. They see the train contracting in length from 1 end and that contraction propagates to the other end.

25. Originally Posted by Webbo
So would that mean that for our moving ship a star that is behind would also have its distance contracted from 22.3 ly to 1 ly?
Yea I got that wrong and I'm still working it all out in my head. The light would be red shifted but the distance is still contracted. If it wasn't then you'd have situation where you've moved forward 1ly but when you look back you'd think that you travelled 2ly.

I haven't been able to sit down and play with the maths to get everything straight in my head. My son's formal is tonight and he just told me at 9pm that he didn't get the money to one of his friends quick enough for the limo and there is no room for him. He's known this for over 3 weeks yet tells me last night 22 hours before he's supposed to be at the formal. so I'v been running around fixing his stuff ups

26. It's all pretty darn complicated: e.g. http://en.wikipedia.org/wiki/Penrose-Terrell_rotation

The main thing I learn, is "it" all can't simply be dismissed with a simple bit of what appears to be logic.

27. Established Member
Join Date
Aug 2009
Posts
753
Originally Posted by cjameshuff
No, the trailing ship will not see the star contract through the other ship. Your basic setup is impossible, you can't observe what's happening 2 light years away in real time...the ships can't start accelerating simultaneously in all frames. If each accelerates simultaneously according to an observer at their midpoint, each will accelerate, and then see the other ship accelerate starting from a contracted distance of about 0.09 light years, with the leading ship reaching the star first.
You say it's impossible because they cannot observe, but it is not theoretically impossible to accelerate at the same time and stay in the same frame. They could certainly check each other was still in the same frame by transmitting a signal to each other. As long as there is no doppler shift they are in the same reference frame and maintain a 2 ly distance.

28. Established Member
Join Date
Aug 2009
Posts
753
Originally Posted by WayneFrancis
Yea I got that wrong and I'm still working it all out in my head. The light would be red shifted but the distance is still contracted. If it wasn't then you'd have situation where you've moved forward 1ly but when you look back you'd think that you travelled 2ly.
Ok. Now we have established that space has contracted behind as well, what if the ship were to rotate 180 degrees on its axis, declare itselt at rest (which I assume it can do without issue considering it's new reality), and accelerate towards that star. As it's only 1ly away it should be able to reach that star in just a few light weeks as it accelerates to 0.99% c. Of course once this is achieved and the space is contracted to 2 light weeks it could repeat the manouver and rotate and switch back to the original star which is only 2 weeks away, declare itself at rest, and accelerate towards the original star which it should reach in about 12 hours. We have now reached our original star in a matter of hours without apparently ever leaving the vicinity of Earth. Am I correct?

29. Originally Posted by Webbo
Ok. Now we have established that space has contracted behind as well, what if the ship were to rotate 180 degrees on its axis, declare itselt at rest (which I assume it can do without issue considering it's new reality), and accelerate towards that star. As it's only 1ly away it should be able to reach that star in just a few light weeks as it accelerates to 0.99% c. Of course once this is achieved and the space is contracted to 2 light weeks it could repeat the manouver and rotate and switch back to the original star which is only 2 weeks away, declare itself at rest, and accelerate towards the original star which it should reach in about 12 hours. We have now reached our original star in a matter of hours without apparently ever leaving the vicinity of Earth. Am I correct?
No because at some point it will not be moving with respect to the star and so the distance will be back at 22 (or whatever) light years. When he gets back up to speed, it will shrink back to 1 ly.

30. Established Member
Join Date
Aug 2009
Posts
753
Originally Posted by Strange
No because at some point it will not be moving with respect to the star and so the distance will be back at 22 (or whatever) light years. When he gets back up to speed, it will shrink back to 1 ly.
I dont understand why the ship cannot be at rest and accelerate toward a star that is 1 ly in distance. The distance I am assured is not an illusion/distortion but an actual distance.

From the other thread (for some reason this was not transfered over);

Originally Posted by korjik
It isnt a distortion to the moving observer. It is a realistic result, not some made up thing. While the ship is moving that fast, space along the direction of motion is shorter. It isnt a distortion, it is actually shorter.

Why when the ship accelerates towards the second star which was behind are you now saying that the space between expands from 1 ly to 22.3 ly?

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•