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Webbo
2012-Jun-26, 11:15 AM
Surely a parallax measurement would be the same for either observer whether stationary or travelling near c. Why would movement change the triangulation?



Moved from here: http://www.bautforum.com/showthread.php/135745-maximum-speed

Strange
2012-Jun-26, 11:21 AM
Surely a parallax measurement would be the same for either observer whether stationary or travelling near c. Why would movement change the triangulation?

Because the distance is shorter (because of relativity).

Webbo
2012-Jun-26, 12:05 PM
Because the distance is shorter (because of relativity).Which distance? The only distance that is measured is the distance between the 2 viewpoints and these are accross the direction of movement. My understanding was that the shortened distance was only in the direction of movement. Am I incorrect?

Strange
2012-Jun-26, 12:26 PM
Which distance? The only distance that is measured is the distance between the 2 viewpoints and these are accross the direction of movement. My understanding was that the shortened distance was only in the direction of movement. Am I incorrect?

You are correct. It is the distance to the distant object which is less, therefore the measured angles would be different.

Webbo
2012-Jun-26, 12:37 PM
You are correct. It is the distance to the distant object which is less, therefore the measured angles would be different.The angles are used to derive the distance not the other way around. The distance is unknown; it's a calculation based on the baseline and the angles.

Strange
2012-Jun-26, 01:42 PM
The angles are used to derive the distance not the other way around. The distance is unknown; it's a calculation based on the baseline and the angles.

Yes. That is what I said (or intended to :)). You measure the baseline and the angles, calculate the distance: the "stationary" observe calculates 22 ly; the moving one calculates 1 ly. They measure the same baseline but different angles (because the distance to the object is different).

Webbo
2012-Jun-26, 01:58 PM
Yes. That is what I said (or intended to :)). You measure the baseline and the angles, calculate the distance: the "stationary" observe calculates 22 ly; the moving one calculates 1 ly. They measure the same baseline but different angles (because the distance to the object is different).But why would the angles be different? They are either determined by the direction that the light is received or by the target stars lateral movement against the background stars. I can't see how either is affected.

Strange
2012-Jun-26, 02:02 PM
But why would the angles be different? They are either determined by the direction that the light is received or by the target stars lateral movement against the background stars. I can't see how either is affected.

We seem to be going round in circles. Because the distance to the star is less (because of relativity).

Webbo
2012-Jun-26, 02:37 PM
We seem to be going round in circles. Because the distance to the star is less (because of relativity).But like I said, for that to happen either the lateral movement against the background stars would need to contract or the distance between the viewpoints, but that is not possible as that is not the direction of travel.

Look at the method used.

1.Distance is measured between viewpoints = real number
2.Lateral movement is measured = real number
3.Angles determined = derived from 1 & 2
4.Distance calculated = derived from 1, 2 & 3

1 & 2 are the only real measurement hence only they can cause differences in the angle and distance. How can forward movement affect those first 2 measurements?

Strange
2012-Jun-26, 02:39 PM
But like I said, for that to happen either the lateral movement against the background stars would need to contract or the distance between the viewpoints, but that is not possible as that is not the direction of travel.

Look at the method used.

1.Distance is measured between viewpoints = real number
2.Lateral movement is measured = real number
3.Angles determined = derived from 1 & 2
4.Distance calculated = derived from 1, 2 & 3

1 & 2 are the only real measurement hence only they can cause differences in the angle and distance. How can forward movement affect those first 2 measurements?

The amount of lateral movement seen depends on how far away the star is, no?

(I'll leave it there as I am not sure quite which bit of this you are not clear on)

NEOWatcher
2012-Jun-26, 02:59 PM
But like I said, for that to happen either the lateral movement against the background stars would need to contract or the distance between the viewpoints, but that is not possible as that is not the direction of travel.
The problem is anything but an exact 90 degrees has a contraction as part of its vector.
Let me try to explain what I see: (difficult because of my lack of correct terminology)
To judge speed, you are taking a time measurement between two objects or positions at two different times. No matter which object you choose as your 90 degree angle, the measurement at the other time point has changed it's angle and introduces contraction.

Webbo
2012-Jun-26, 03:01 PM
The amount of lateral movement seen depends on how far away the star is, no?

Which means that the stationary observer would have the target star in one particlular position on a photographic plate when compared to the background, and the moving observer would have the target star in a different position on the photographic plate when compared to the background as it passed virtualy the same point in space. Not sure how that is possible.

Strange
2012-Jun-26, 03:01 PM
Good point NEOWatcher.

Webbo, are you thinking that length contraction only happens in exactly the direction of travel?

Strange
2012-Jun-26, 03:06 PM
Which means that the stationary observer would have the target star in one particlular position on a photographic plate when compared to the background, and the moving observer would have the target star in a different position on the photographic plate when compared to the background as it passed virtualy the same point in space. Not sure how that is possible.

As a crude analogy, think of taking two photographs from different distances but with different focal length cameras such that the main object in the picture appears the same size in both. Other objects will have a different spatial relationship in the two images because one is "compressed" in the direction the picture is taken.

Similarly, the travelling observer will see everything "squished" which will change their apparent relationships.

Webbo
2012-Jun-26, 03:08 PM
The problem is anything but an exact 90 degrees has a contraction as part of its vector.
Let me try to explain what I see: (difficult because of my lack of correct terminology)
To judge speed, you are taking a time measurement between two objects or positions at two different times. No matter which object you choose as your 90 degree angle, the measurement at the other time point has changed it's angle and introduces contraction.We're not trying to determine speed, we're trying to determine distance and therefore doesn't need to be a time difference between measurements. It's currently required for parallax measurement because our baseline is the diameter of the Earths orbit.

Webbo
2012-Jun-26, 03:10 PM
Good point NEOWatcher.

Webbo, are you thinking that length contraction only happens in exactly the direction of travel?No but it shouldn't be evident at all at right angles to direction which are the only measurements we can make.

Strange
2012-Jun-26, 03:18 PM
No but it shouldn't be evident at all at right angles to direction which are the only measurements we can make.

Ah, I think I see (maybe). You are considering the displacement and the baseline as both being at right angles to the direction of travel? But the displacement is a result of the trigonometry of the setup; i.e. the angle between two different viewpoints looking at the same object.

The angles to the star (which you are measuring as a displacement) are between lines in the direction of travel (which will be shorter) and a line orthogonal to the direction of travel (which will not be shorter).

Do I need to draw a diagram, I wonder... (ETA: no need!)

For the travelling observer, the distance being measured (indirectly) is shorter and therefore the angle/displacement must be greater.

WayneFrancis
2012-Jun-26, 03:23 PM
Surely a parallax measurement would be the same for either observer whether stationary or travelling near c. Why would movement change the triangulation?

Since length contraction is only in the direction of movement the angles are different.
17143

A=B,C>D \therefore F>E
Actually a better representation would be
A=B, F>E \therefore C>D as the observers would be using the angles F & E to derive the distance of C & D

I'll add that both ships at the tips can tell that they are travelling parallel to each other and trade data to get the parallax (purple lines is the direction of motion of the observers)

Webbo
2012-Jun-26, 03:27 PM
As a crude analogy, think of taking two photographs from different distances but with different focal length cameras such that the main object in the picture appears the same size in both. Other objects will have a different spatial relationship in the two images because one is "compressed" in the direction the picture is taken.

Similarly, the travelling observer will see everything "squished" which will change their apparent relationships.I don't see why it would be squished. It's an image reproduced on a plate at a particular point in space in either case. Forward movement shouldn't have the same effect as changing focus.

Strange
2012-Jun-26, 03:29 PM
I don't see why it would be squished. It's an image reproduced on a plate at a particular point in space in either case. Forward movement shouldn't have the same effect as changing focus.

[changing focal length]

Look at Wayne's diagram. Everything is foreshortened from the moving observer's point of view. Things are nearer him and hence nearer to each other.

Webbo
2012-Jun-26, 03:32 PM
Since length contraction is only in the direction of movement the angles are different.
17143

A=B,C>D \therefore F>E
Actually a better representation would be
A=B, F>E \therefore C>D as the observers would be using the angles F & E to derive the distance of C & D

I'll add that both ships at the tips can tell that they are travelling parallel to each other and trade data to get the parallax (purple lines is the direction of motion of the observers)Parallax is not measured using that method. Only lateral measurements are made. Angles are derived not observed/measured.

WayneFrancis
2012-Jun-26, 03:33 PM
I don't see why it would be squished. It's an image reproduced on a plate at a particular point in space in either case. Forward movement shouldn't have the same effect as changing focus.

But parallax is the use of 2 images. With a really wide ship or 2 ships or as we do it here 2 points separated by 6 months and ~2AU.

Webbo
2012-Jun-26, 03:39 PM
Just to clarify. If everything is squished then the arc second must also be squished hence measurements would be exacly the same. What's required is for just the target star to be slightly shifted compared to the background.

WayneFrancis
2012-Jun-26, 03:46 PM
Just to clarify. If everything is squished then the arc second must also be squished hence measurements would be exacly the same. What's required is for just the target star to be slightly shifted compared to the background.

No
17144

Those 2 triangles where identical before I squashed one of them. Angle A ≠ B. This is basic euclidean geometry

Ah wait. I think I know where you are getting confused. An obserer will not agree with an observer, in a different inertial frame, on the angular separation of 2 objects. This is why the accelerated observer will come up with a different parallax measurement then the non accelerated obsserver.

Webbo
2012-Jun-26, 03:53 PM
No
17144

Those 2 triangles where identical before I squashed one of them. Angle A ≠ B. This is basic euclidean geometryIf the camera takes an image of say the forward hemisphere and everything is squished then there will be an empty band around the outside. Only an inept scientist would continue to use the empty band as part of the total arc measurement. If an area of the photo is squished, then the unit of measurement must also be squished.

WayneFrancis
2012-Jun-26, 04:09 PM
If the camera takes an image of say the forward hemisphere and everything is squished then there will be an empty band around the outside. Only an inept scientist would continue to use the empty band as part of the total arc measurement. If an area of the photo is squished, then the unit of measurement must also be squished.

There is no empty band.

In this image the left one is a stationary observer.
The right one is an accelerated observer.

17146

The accelerated observer will see a larger angular distance between the 2 objects then the non accelerated observer. The closer an object is to the direction of travel the more it will have its distance contracted. They can still see a continuous 360 degrees around them. The objects in front of them look a lot closer, the objects perpendicular to their direction of motion seem to be the same, objects directly behind them seem to be very far away. Through out the entire 360 degrees it is a completely smooth transition

WayneFrancis
2012-Jun-26, 04:17 PM
Perhaps you could do some simple drawing on what you think is happening. Notice I've squished the 2 red dots but I haven't squished the sphere around the observer. This is why the parallax angle changes.

Webbo
2012-Jun-26, 04:22 PM
There is no empty band.

In this image the left one is a stationary observer.
The right one is an accelerated observer.

17146


The accelerated observer will see a larger angular distance between the 2 objects then the non accelerated observer. The closer an object is to the direction of travel the more it will have its distance contracted. They can still see a continuous 360 degrees around them. The objects in front of them look a lot closer, the objects perpendicular to their direction of motion seem to be the same, objects directly behind them seem to be very far away. Through out the entire 360 degrees it is a completely smooth transitionSo you are saying the stars in front are squashed together but perpendicular are the same. Which means at somewhere between the 2 points there must be larger than expected gaps. I would argue that such a distortion should not be measured using fixed arcseconds. Any attempt to do so would be scientifically inept.

NEOWatcher
2012-Jun-26, 04:28 PM
We're not trying to determine speed, we're trying to determine distance...
Fair enough. this thread topic is about speed, and to determine speed you need a distance, so the lines got a little blurred. But the concept still applies...

...and therefore doesn't need to be a time difference between measurements. It's currently required for parallax measurement because our baseline is the diameter of the Earths orbit.
Even removing time from the equation, you are still talking about 3 points in space. Unless they are a straight line, there will be at least one angle that is not 90 degrees from the direction of travel. Therefore, contraction can not be removed from the equation.


If the camera takes an image of say the forward hemisphere and everything is squished then there will be an empty band around the outside. Only an inept scientist would continue to use the empty band as part of the total arc measurement. If an area of the photo is squished, then the unit of measurement must also be squished.
No, there is no empty band until you reach c itself where a photon can't "catch up" to you. The "squish" is non-uniform around the field of view. It gets more compressed as you look forward, and less compressed as you look back. The distortion does not exist at 90 degrees and gets more "squished" as you look forward. As you go from 90 to 180, the field of view is stretched.

korjik
2012-Jun-26, 04:32 PM
So you are saying the stars in front are squashed together but perpendicular are the same. Which means at somewhere between the 2 points there must be larger than expected gaps. I would argue that such a distortion should not be measured using fixed arcseconds. Any attempt to do so would be scientifically inept.

Why? A spherical coordinate system is a spherical coordinate system. There isnt any requirement that the distribution of stars be distributed in any specific way. The universe is distorted at relativistic velocities. Changing the coordinate system to make it look 'normal' wouldnt be useful

Webbo
2012-Jun-26, 04:33 PM
Perhaps you could do some simple drawing on what you think is happening. Notice I've squished the 2 red dots but I haven't squished the sphere around the observer. This is why the parallax angle changes.I dont think anything is happening. I believe the images would be the same for both observers. If there is a distortion in the image as described then a corresponding distortion should be made for the arcsecond unit.

Lets assume the stars ahead are squished. Well we know that when stationary the distant fixed background, made up of say 2 distant galaxies, are 10 arseconds apart and 2 billion ly away. Why because we accelerate toward our local star would they move closer together and thereby reduce the distance between them by hundreds or millions of ly?

Webbo
2012-Jun-26, 04:41 PM
Fair enough. this thread topic is about speed, and to determine speed you need a distance, so the lines got a little blurred. But the concept still applies...

Even removing time from the equation, you are still talking about 3 points in space. Unless they are a straight line, there will be at least one angle that is not 90 degrees from the direction of travel. Therefore, contraction can not be removed from the equation.


No, there is no empty band until you reach c itself where a photon can't "catch up" to you. The "squish" is non-uniform around the field of view. It gets more compressed as you look forward, and less compressed as you look back. The distortion does not exist at 90 degrees and gets more "squished" as you look forward. As you go from 90 to 180, the field of view is stretched.Contraction is not an issue because no angle is measured. Distance is the reciprocal of the parallax which is measured in arcseconds (It's movement accross the sky compared to the background).

If its compressed in one part it must be expanded in another part.

Webbo
2012-Jun-26, 04:57 PM
Why? A spherical coordinate system is a spherical coordinate system. There isnt any requirement that the distribution of stars be distributed in any specific way. The universe is distorted at relativistic velocities. Changing the coordinate system to make it look 'normal' wouldnt be usefulIt's to account for a known distortion not to make it normal, however, if it's accepted and expected that the universe is distorted at relativistic velocities, and any measurements taken are known to be distorted then fine. If so, then perhaps it should be stated that "although the moving observer measures the distance as 1 ly this is a known distortion and with correction the distance would be 22.3 ly".

Personally if I were in such a position knowing that my observations were distorted I would make an adjusment to correct for said distortion to acquire a realistic result.

NEOWatcher
2012-Jun-26, 05:02 PM
Contraction is not an issue because no angle is measured. Distance is the reciprocal of the parallax which is measured in arcseconds (It's movement accross the sky compared to the background).
What? Arcseconds IS the unit of measure for an angle. You can't use arcseconds (or minutes, degrees etc) without an angle.


If its compressed in one part it must be expanded in another part.
To what are you referring? I already explained where it was compressed and where it is expanded.

NEOWatcher
2012-Jun-26, 05:06 PM
Personally if I were in such a position knowing that my observations were distorted I would make an adjusment to correct for said distortion to acquire a realistic result.
And that correction for distortion is exactly what we are talking about. The correction is the measurement for the non-moving frame of reference, and the non correction is for the moving frame of reference. Unfortunately, in space there is no non-moving frame of reference. It's all relative (thus the name relativity).
The person on Earth is only non-moving because you say that they are.

Webbo
2012-Jun-26, 05:29 PM
What? Arcseconds IS the unit of measure for an angle. You can't use arcseconds (or minutes, degrees etc) without an angle.But for parallax no angle is used only it's movement against the background.



To what are you referring? I already explained where it was compressed and where it is expanded.Which was not logical. You stated that there is no distiortion at 90 degrees but how can all stars contract ahead if the field of view is exactly the same (half a sphere), and how can all the stars spread out behind you (in the other half a sphere), with the same number of stars in each as they had before?

Webbo
2012-Jun-26, 05:36 PM
And that correction for distortion is exactly what we are talking about. The correction is the measurement for the non-moving frame of reference, and the non correction is for the moving frame of reference. Unfortunately, in space there is no non-moving frame of reference. It's all relative (thus the name relativity).
The person on Earth is only non-moving because you say that they are.So for arguments sake if the person on the space craft could argue he is at rest, then I could introduce his friend on craft B which is passing him at 99.9% of c, heading towards the same star, and he will measure the distance to the star as a couple of light weeks? What would our person on Earth have say about that?

NEOWatcher
2012-Jun-26, 05:59 PM
But for parallax no angle is used only it's movement against the background.
How is that possible?
Parallax is EVERYTHING about angles. Items further away have less of an angle difference than those in the foreground.
But; even if I ignore that. YOU said measure in arcseconds with no angle involved.

Which was not logical. You stated that there is no distiortion at 90 degrees but how can all stars contract ahead if the field of view is exactly the same (half a sphere), and how can all the stars spread out behind you (in the other half a sphere), with the same number of stars in each as they had before?
Because the distortion is perpendicular to the sphere.

Let me illustrate:
The black circle is the person considered to be moving. (should probably be a point, or at least the distance between eyeballs)
The red circle is what the person who is not considered moving views a consistant distance "x" from the center.
The blue circle is what the moving person considers "x" distance to be.
They agree at 90, and progressively disagree as it goes away from 90.
(the illustration is from the stationary person's point of view, from the mover's point of view, the circle to ellipse would be reversed and off 90 degrees)
17147

So for arguments sake if the person on the space craft could argue he is at rest, then I could introduce his friend on craft B which is passing him at 99.9% of c, heading towards the same star, and he will measure the distance to the star as a couple of light weeks?
Yep...

What would our person on Earth have say about that?
That this third person's "circle" is squished even more than the blue one.

Webbo
2012-Jun-26, 11:53 PM
Because the distortion is perpendicular to the sphere.

Let me illustrate:
The black circle is the person considered to be moving. (should probably be a point, or at least the distance between eyeballs)
The red circle is what the person who is not considered moving views a consistant distance "x" from the center.
The blue circle is what the moving person considers "x" distance to be.
They agree at 90, and progressively disagree as it goes away from 90.
(the illustration is from the stationary person's point of view, from the mover's point of view, the circle to ellipse would be reversed and off 90 degrees)
17147I dont follow. Lets say that there are just 360 stars surrounding Earth and the ship on a single plane spaced by 1 degree. From Earth we therefore see that there are 180 stars ahead and 180 behind. Lets say our target star is number 360 straight ahead which places star 90 to our right, star 180 directly behind and star 270 to our left. As we know that there is no distortion to 90 and 270 which means these are fixed in both observers views, how can all spaces between the stars ahead (270 through to 90) move closer together and all stars behind (90 through to 270) move further apart? This cannot make any logical sense when we know that 90 & 270 are fixed directly to our left and right respectively.

Webbo
2012-Jun-27, 12:02 AM
Yep...

That this third person's "circle" is squished even more than the blue one.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?

WayneFrancis
2012-Jun-27, 12:08 AM
It's to account for a known distortion not to make it normal, however, if it's accepted and expected that the universe is distorted at relativistic velocities, and any measurements taken are known to be distorted then fine. If so, then perhaps it should be stated that "although the moving observer measures the distance as 1 ly this is a known distortion and with correction the distance would be 22.3 ly".

Personally if I were in such a position knowing that my observations were distorted I would make an adjusment to correct for said distortion to acquire a realistic result.

But then you wouldn't say that the speed of light is constant because if you are going .999c and want to say that the distance you are observing is still 22.3 light years then you also have to say that the speed of light observed by you is not 299,792,458m/s but ~6,685,371,813m/s because you've now redefined what a meter is to keep your distances reading the same.

It is a fact that the distance contracts in front of the travelling observer and elongates behind them. Yes they see a distorted view but what you are asking them to do is have a compass that has almost every degree actually measure a different arc length based on the speed you are travelling. But then what are you comparing that speed to? Since SR tells us there is no preferred frame who's base line do you pick?

This is why you just pick the spherical coordinate systems based on your frame of reference. If you want to compare it with someone in a different frame then you do the calculations to get an answer that converts one of the 2 coordinate systems to the other.

WayneFrancis
2012-Jun-27, 12:10 AM
But for parallax no angle is used only it's movement against the background.


Which was not logical. You stated that there is no distiortion at 90 degrees but how can all stars contract ahead if the field of view is exactly the same (half a sphere), and how can all the stars spread out behind you (in the other half a sphere), with the same number of stars in each as they had before?

Time for another diagram....give me a few.

Webbo
2012-Jun-27, 12:23 AM
But then you wouldn't say that the speed of light is constant because if you are going .999c and want to say that the distance you are observing is still 22.3 light years then you also have to say that the speed of light observed by you is not 299,792,458m/s but ~6,685,371,813m/s because you've now redefined what a meter is to keep your distances reading the same. No. I would say that IF the stars ahead are distorted then the observer on board would be aware of that fact and adjust his readings accordingly.


It is a fact that the distance contracts in front of the travelling observer and elongates behind them. Yes they see a distorted view but what you are asking them to do is have a compass that has almost every degree actually measure a different arc length based on the speed you are travelling. But then what are you comparing that speed to? Since SR tells us there is no preferred frame who's base line do you pick?

This is why you just pick the spherical coordinate systems based on your frame of reference. If you want to compare it with someone in a different frame then you do the calculations to get an answer that converts one of the 2 coordinate systems to the other.If it contracts ahead and elogates behind what is the effect on my scenario with the 360 surrounding stars? If star 90 & 270 are fixed at the angles of 90 & 270 how can 180 stars contract is one sphere and elongate in the other when there mutual 2 end points are fixed?

Webbo
2012-Jun-27, 12:48 AM
Look at it this way. Surround yourself with 360 stars all at 100 ly distance from yourself and separated by 1 degree. Now move them all closer to 50 ly distance. What's the degree separation? It's stil 1 degree. Now move half in one hemisphere back out to 100 ly. What's the degree separation. It's still 1 degree. Now move them randomly to various positions from 50 ly to 1 billion yr. What's the degree separation. It's still 1 degree. It is not possible to change the degree separation by physically moving their distances nor is it possible by moving toward any particular star. To condense in one whole hemisphere would require that some of the stars migrate from the other hemisphere which means in our scenario stars that were behind now need to be ahead.

WayneFrancis
2012-Jun-27, 12:53 AM
Ok here we go.

17149

The left is a normal spherical coordinate system.
The middle shows what happens to that same sphere with acceleration
The right shows what Webbo would want us to do so that 2 objects would have the same angular measurement between them in both frames.
Note that the left and right obviously don't share the same angles the left one does.

Also to get everything to work out the meter would have to be redefined an would actually be a variable so that it was shorter in the direction of travel, the same distance perpendicular to the direction of travel and longer in the opposite direction of travel.

Webbo
2012-Jun-27, 12:59 AM
Ok here we go.

17149

The left is a normal spherical coordinate system.
The middle shows what happens to that same sphere with acceleration
The right shows what Webbo would want us to do so that 2 objects would have the same angular measurement between them in both frames.
Note that the left and right obviously don't share the same angles the left one does.

Also to get everything to work out the meter would have to be redefined an would actually be a variable so that it was shorter in the direction of travel, the same distance perpendicular to the direction of travel and longer in the opposite direction of travel.And just as I have been trying to say, from the observation point in the centre of the green middle sphere, there is no observable contraction nor elongation anywhere. All stars are in the same directions/angles/parallax so both observers would measure the parallax distance as 22.3 ly.

WayneFrancis
2012-Jun-27, 01:03 AM
No. I would say that IF the stars ahead are distorted then the observer on board would be aware of that fact and adjust his readings accordingly.


Then you are picking a reference frame of some other observer. No one claims you wouldn't be able to tell you are accelerated. We know we are accelerated with regards to the CMBR by this but we don't adjust our spherical coordinate system because of it unless we want to know something specific in another frame of reference. The effect is small because the amount of acceleration is relatively small but that doesn't matter. The effect is still there. What you want is a none spherical coordinate system which is fine as long as you make that clear to others you are actually using a coordinate transformation from another frame of reference. This is all well and good but if you want to say that the distance to some object stays at 22.3ly while accelerated to .999c then you have to also say it still took you a bit over 22.3 years to get there but your body will still not have aged 22.3 years. Your watch on your arm would not show 22.3 years have past.



If it contracts ahead and elogates behind what is the effect on my scenario with the 360 surrounding stars? If star 90 & 270 are fixed at the angles of 90 & 270 how can 180 stars contract is one sphere and elongate in the other when there mutual 2 end points are fixed?

Look at my diagram in post 45

This is really one of those things that playing around with the maths actually helps. Don't get offended but your gut feel of what is going on is not correct. It isn't easy to conceptualise if you haven't been playing around with the concepts much and your intuition is probably going to fail you.

Webbo
2012-Jun-27, 01:05 AM
Then you are picking a reference frame of some other observer. No one claims you wouldn't be able to tell you are accelerated. We know we are accelerated with regards to the CMBR by this but we don't adjust our spherical coordinate system because of it unless we want to know something specific in another frame of reference. The effect is small because the amount of acceleration is relatively small but that doesn't matter. The effect is still there. What you want is a none spherical coordinate system which is fine as long as you make that clear to others you are actually using a coordinate transformation from another frame of reference. This is all well and good but if you want to say that the distance to some object stays at 22.3ly while accelerated to .999c then you have to also say it still took you a bit over 22.3 years to get there but your body will still not have aged 22.3 years. Your watch on your arm would not show 22.3 years have past.



Look at my diagram in post 45

This is really one of those things that playing around with the maths actually helps. Don't get offended but your gut feel of what is going on is not correct. It isn't easy to conceptualise if you haven't been playing around with the concepts much and your intuition is probably going to fail you.Place yourself at the centre of your middle sphere. Where is the observed contraction?

WayneFrancis
2012-Jun-27, 01:16 AM
And just as I have been trying to say, from the observation point in the centre of the green middle sphere, there is no observable contraction nor elongation anywhere. All stars are in the same directions/angles/parallax so both observers would measure the parallax distance as 22.3 ly.

What? The accelerated observer most definitely sees contraction and elongations.

17150

Look at the points You can't say that they are the same. The objects in space appear to move.

In the end you've had multiple people try to tell you about coordinate transformations. You keep claiming that nothing would change. The facts are they do change. The distances to all objects that are not perpendicular to the direction of motion changes. The angle between them and the direction of travel also changes. People have explained this. I've drawn little pictures for you but you still some how refuse to accept the facts. If the pictures don't work for you I'm afraid that the maths behind the pictures will be completely lost on you.

Webbo
2012-Jun-27, 01:28 AM
What? The accelerated observer most definitely sees contraction and elongations.

17150

Look at the points You can't say that they are the same. The objects in space appear to move.

In the end you've had multiple people try to tell you about coordinate transformations. You keep claiming that nothing would change. The facts are they do change. The distances to all objects that are not perpendicular to the direction of motion changes. The angle between them and the direction of travel also changes. People have explained this. I've drawn little pictures for you but you still some how refuse to accept the facts. If the pictures don't work for you I'm afraid that the maths behind the pictures will be completely lost on you.Which direction is the craft moving in the middle sphere?

WayneFrancis
2012-Jun-27, 01:51 AM
Look at the following diagram
17151

You are sitting in your ship.
You measure the distance to A as 1 light year
You measure the angle to the side at 22.5 degrees
You then accelerate to ~.866c towards D
A has shifted to appear to be at B
You measure the distance as .5999 light years
You measure the angle to now be ~39.64 degrees off from your direction of motion

The distance to D went from 1ly to .5ly

If you stopped you'd see the object shift back to the same points before you accelerated, as long as you didn't stay accelerated to long.

Webbo
2012-Jun-27, 01:55 AM
Look at the following diagram
17151

You are sitting in your ship.
You measure the distance to A as 1 light year
You measure the angle to the side at 22.5 degrees
You then accelerate to ~.866c towards D
A has shifted to appear to be at B
You measure the distance as .5999 light years
You measure the angle to now be ~39.64 degrees off from your direction of motion

The distance to D went from 1ly to .5ly

If you stopped you'd see the object shift back to the same points before you accelerated, as long as you didn't stay accelerated to long.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?

WayneFrancis
2012-Jun-27, 02:06 AM
Which direction is the craft moving in the middle sphere?

The circle is before accelerating. The egg shape is after accelerating. The points shift as the observer accelerates. If the observer accelerates in a different direction, like in the complete opposite direction, then the points will again shift according to the direction and amount of accelerating.

Webbo
2012-Jun-27, 02:10 AM
The circle is before accelerating. The egg shape is after accelerating. The points shift as the observer accelerates. If the observer accelerates in a different direction, like in the complete opposite direction, then the points will again shift according to the direction and amount of accelerating.The reason I ask is because both hemispheres (top and bottom) have sections where angles increase (expand) and decrease (contract). How can that be possible?

pzkpfw
2012-Jun-27, 02:16 AM
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?

The contraction is in the direction C-D.
The horizontal distance between the line C-D and the point A isn't changing.
However, A "appears" at B due to length contraction (in the direction of motion, only).
So the angle has to increase, and that's what those on board would measure.

(The diagram isn't showing where D would appear - I'd guess because it's still directly ahead so that doesn't matter in measuring the angle.)

Webbo
2012-Jun-27, 02:23 AM
The contraction is in the direction C-D.
The horizontal distance between the line C-D and the point A isn't changing.
However, A "appears" at B due to length contraction (in the direction of motion, only).
So the angle has to increase, and that's what those on board would measure.

(The diagram isn't showing where D would appear - I'd guess because it's still directly ahead so that doesn't matter in measuring the angle.)So should all the stars angles in the forward sphere increase?

pzkpfw
2012-Jun-27, 02:25 AM
So should all the stars angles in the forward sphere increase?

Isn't that what post #49 shows?

WayneFrancis
2012-Jun-27, 02:54 AM
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?

Ok, lets deal with just objects directly "ahead" Before you accelerate you measure the distance to 2 object. One, object A, is 1ly away the other, object B, is 3ly. Thus A→B is 2ly.

You accelerate to ~.866c.
You measure the distance to you and A is now only .5ly away and B is 1.5ly away. A→B is now 1ly.
All distances are reduces by half.

Now for object ahead but off centre the distance will all decrease but since the contraction is only in the direction of motion the ratio of X,Y is skewed. All the distances will decrease but the distance perpendicular to the direction of motion are less effected. This is why the angular distance increases from the direction of motion. Not all angles increase just the ones measured from straight ahead

17153

So
DCB>DCA
DCG>DCB
but, its hard to see here.
ACF>BCG, ie the angle between those 2 AF,BG objects got smaller, ACF>BCG. But this is ok because we are not measuring the angle from straight ahead.

C→A transformed to C→B
C→F transformed to C→G
C→A-C→B > C→F-C→G b

The DC[Insert 3rd point here] angle will always be larger except for the 90 degree angle which remains the same.

Jeff Root
2012-Jun-27, 04:22 AM
We know we are accelerated with regards to the
CMBR by this but we don't adjust our spherical
coordinate system because of it unless we want
to know something specific in another frame of
reference.
The large majority of images of the CMBR data
have been adjusted in just this way, so that they
show what would be seen by an observer at our
position but "at rest" relative to the spherical
shell from which the CMBR was emitted. When
you see an image of the CMBR that is almost all
one color in one hemisphere and the opposite
color in the opposite hemisphere, you are looking
at an image which has not been so adjusted.

-- Jeff, in Minneapolis

Jeff Root
2012-Jun-27, 04:54 AM
Personally if I were in such a position knowing that my
observations were distorted I would make an adjusment
to correct for said distortion to acquire a realistic result.
But then you wouldn't say that the speed of light is
constant because if you are going .999c and want to
say that the distance you are observing is still 22.3
light years then you also have to say that the speed
of light observed by you is not 299,792,458m/s but
~6,685,371,813m/s because you've now redefined
what a meter is to keep your distances reading the
same.
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
vectorally add that to your wind speed to determine
your ground speed, and from that, the distance.

-- Jeff, in Minneapolis

WayneFrancis
2012-Jun-27, 05:08 AM
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
vectorally add that to your wind speed to determine
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.

Jeff Root
2012-Jun-27, 05:12 AM
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.

Hold your arm out in front of you and spread your
fingers wide apart. Look at them through one eye.
Now move your hand closer to your face, like stars
appearing to move closer to the spacecraft due to
length contraction. You see the angular distance
between your fingertips increase.

-- Jeff, in Minneapolis

Jeff Root
2012-Jun-27, 06:15 AM
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
vectorally add that to your wind speed to determine
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.



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
you I am addressing and your statement that I hope
to correct.



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
and less meaningful. What does your radar tell you
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?



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.



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

WayneFrancis
2012-Jun-27, 07:30 AM
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.




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
you I am addressing and your statement that I hope
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.




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
and less meaningful. What does your radar tell you
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



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.




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.

NEOWatcher
2012-Jun-27, 12:19 PM
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.


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".

Webbo
2012-Jun-27, 02:51 PM
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?

cjameshuff
2012-Jun-27, 03:53 PM
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/RelativisticAberrationAndDopplerShift/
http://www.youtube.com/watch?v=9E1WmP4fGCo

WayneFrancis
2012-Jun-27, 03:55 PM
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

WayneFrancis
2012-Jun-27, 04:17 PM
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/RelativisticAberrationAndDopplerShift/
http://www.youtube.com/watch?v=9E1WmP4fGCo

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.

cjameshuff
2012-Jun-27, 04:50 PM
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.

NEOWatcher
2012-Jun-27, 05:55 PM
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.

Webbo
2012-Jun-27, 10:33 PM
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?

cjameshuff
2012-Jun-27, 11:36 PM
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.

Webbo
2012-Jun-27, 11:47 PM
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?

caveman1917
2012-Jun-28, 12:03 AM
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.

Hetman
2012-Jun-28, 01:14 AM
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!

WayneFrancis
2012-Jun-28, 05:31 AM
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.

Hetman
2012-Jun-28, 02:52 PM
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.

korjik
2012-Jun-28, 03:27 PM
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.

Hetman
2012-Jun-28, 10:45 PM
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.

Webbo
2012-Jun-28, 10:56 PM
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.

Webbo
2012-Jun-28, 11:05 PM
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?

cjameshuff
2012-Jun-29, 04:38 AM
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.

WayneFrancis
2012-Jun-29, 05:12 AM
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.

WayneFrancis
2012-Jun-29, 05:47 AM
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

pzkpfw
2012-Jun-29, 07:22 AM
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.

Webbo
2012-Jun-29, 12:10 PM
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.

Webbo
2012-Jun-29, 12:21 PM
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?

Strange
2012-Jun-29, 12:43 PM
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.

Webbo
2012-Jun-29, 12:57 PM
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);


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?

Strange
2012-Jun-29, 01:05 PM
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);

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?

You suggested that the spaceship turn round and head towards another star. In order to do this it will have to slow down, stop, then speed up again in the opposite direction (all relative to the destination star). While it is travelling at 0.99c the destination star will be 1 ly away. When it stops, the star will be 22.3 ly away. When it gets back up to 0.99c the star will be 1 ly away.

Webbo
2012-Jun-29, 01:12 PM
You suggested that the spaceship turn round and head towards another star. In order to do this it will have to slow down, stop, then speed up again in the opposite direction (all relative to the destination star). While it is travelling at 0.99c the destination star will be 1 ly away. When it stops, the star will be 22.3 ly away. When it gets back up to 0.99c the star will be 1 ly away.Are you saying the ship cannot declare itself at rest and the contracted distance behind the ship is an illusion?

Hetman
2012-Jun-29, 01:16 PM
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.

You can also reach this by eliminating the optical anomalies:

http://home.scarlet.be/leo.gooris/mich/page5.html

Anomalous reflection:
according to Michelson we should observe the blurring in many optical instruments, in which there are multiple light reflection.

http://home.scarlet.be/leo.gooris/mich/img71.gif
Ellipsoidal wavefront of a moving source.

Strange
2012-Jun-29, 01:27 PM
Are you saying the ship cannot declare itself at rest and the contracted distance behind the ship is an illusion?

I don't really understand what you are getting at. When the ship is travelling at 0.99c (relative to the distant stars) it will see distance ahead and behind compressed from 22 to 1 ly. When it is stationary (relative to those distant stars) it will see the distance as 22 ly.

That is true whether it is heading from star A to star B or vice versa.

korjik
2012-Jun-29, 01:47 PM
Are you saying the ship cannot declare itself at rest and the contracted distance behind the ship is an illusion?

The ship can declare itself stationary. Then the rest of the universe is moving at .999c in the opposite direction and the distance is 1ly

caveman1917
2012-Jun-29, 01:57 PM
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.

The problem here is that they are not in an inertial reference frame (which is a frame moving at constant velocity) because they are accelerating. That complicates things quite a lot. Your example is related to Bell's spaceship paradox. Look at the wiki page on Bell's spaceship paradox and look at the second diagram (with the lines connecting the events). The dashed diagonal line represents the distance between the ships in their own frame, the horizontal line the distance between them in the rest frame. Bell's paradox is stated in terms that the rest frame distance remains the same, but you can see what happens if we say that the distance in their own frame has to remain the same, the spaceships will have different acceleration curves. What will happen is that by the time they reach 0.99c the first ship will indeed have passed the star already (perhaps the second ship too). For a more mathematical treatment look at reference 7 of that wiki page.

Webbo
2012-Jun-29, 02:08 PM
The problem here is that they are not in an inertial reference frame (which is a frame moving at constant velocity) because they are accelerating. That complicates things quite a lot. Your example is related to Bell's spaceship paradox. Look at the wiki page on Bell's spaceship paradox and look at the second diagram (with the lines connecting the events). The dashed diagonal line represents the distance between the ships in their own frame, the horizontal line the distance between them in the rest frame. Bell's paradox is stated in terms that the rest frame distance remains the same, but you can see what happens if we say that the distance in their own frame has to remain the same, the spaceships will have different acceleration curves. What will happen is that by the time they reach 0.99c the first ship will indeed have passed the star already (perhaps the second ship too). For a more mathematical treatment look at reference 7 of that wiki page.The are stationary with respect to each other. You said;

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.
Whether the observer is in the ship, standing ten feet behind it or 2 ly behind it in another ship is irrelevant. They are not in relative motion with each other therefore space is not contracted between them.

Webbo
2012-Jun-29, 02:15 PM
The ship can declare itself stationary. Then the rest of the universe is moving at .999c in the opposite direction and the distance is 1lySo if the ship rotates 180 degrees and accelerate toward the star that was behind, how long will it take to reach it and will it at any time measure the distance to increase to 22.3ly?

Strange
2012-Jun-29, 02:22 PM
So if the ship rotates 180 degrees and accelerate toward the star that was behind, how long will it take to reach it and will it at any time measure the distance to increase to 22.3ly?

Yes. At the half way point between moving in one direction and moving in the other - at that point it will be stationary with respect to the stars.

But maybe you are thinking it can go instantly from 0.99c in one direction to 0.99c in the other, without slowing down ...

caveman1917
2012-Jun-29, 02:23 PM
The are stationary with respect to each other. You said;

Whether the observer is in the ship, standing ten feet behind it or 2 ly behind it in another ship is irrelevant. They are not in relative motion with each other therefore space is not contracted between them.

Yes, so the front spaceship will accelerate a lot less than the back spaceship in order to keep their proper distance the same. Realistically with the numbers you are imposing both will have long passed the star by the time they reach 0.99c.

Strange
2012-Jun-29, 02:24 PM
The are stationary with respect to each other. You said;

Whether the observer is in the ship, standing ten feet behind it or 2 ly behind it in another ship is irrelevant. They are not in relative motion with each other therefore space is not contracted between them.

That is only true if they are in inertial motion (not accelerating). Once you bring acceleration into it, that simple view no longer applies. Work though this: http://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox

Webbo
2012-Jun-29, 02:30 PM
Yes. At the half way point between moving in one direction and moving in the other - at that point it will be stationary with respect to the stars.

But maybe you are thinking it can go instantly from 0.99c in one direction to 0.99c in the other, without slowing down ...Korjik, do you have the same answer?

Webbo
2012-Jun-29, 02:37 PM
Yes, so the front spaceship will accelerate a lot less than the back spaceship in order to keep their proper distance the same. Realistically with the numbers you are imposing both will have long passed the star by the time they reach 0.99c.There has been no acceleration. The original ship was passing at 0.99c, I have just added a companion. They are at rest with each other and always have been.

EDIT - I think I added acceleration at one point but there was no need for it. I am just trying to ascertain why the space between two ships contracts but not the ship itself.

Strange
2012-Jun-29, 02:40 PM
There has been no acceleration. The original ship was passing at 0.99c, I have just added a companion. They are at rest with each other and always have been.

But you said:


... as the 2 ships accelerate to 0.99 c ...

caveman1917
2012-Jun-29, 02:41 PM
There has been no acceleration. The original ship was passing at 0.99c, I have just added a companion. They are at rest with each other and always have been.

So basically you just have a 2 lightyear long spaceship passing by at 0.99c and everything else remains the same?

caveman1917
2012-Jun-29, 02:45 PM
So what you're saying is: if the observer on the 2 lightyear long spaceship says that the distance between the earth and the star is 1 lightyear then the front of his spaceship is behind the star. However to an observer on earth the distance to the star is 22 lightyears and the length of the spaceship is contracted to say 0.1 lightyears then the front of the spaceship is in front of the star.

So the question is, does the spaceship fit between the earth and the star or not? This is equivalent to the ladder paradox (http://en.wikipedia.org/wiki/Ladder_paradox) where instead of asking whether a moving ladder fits in a garage you're asking whether a moving spaceship fits between the earth and the star.

Strange
2012-Jun-29, 02:56 PM
There has been no acceleration. The original ship was passing at 0.99c, I have just added a companion. They are at rest with each other and always have been.

EDIT - I think I added acceleration at one point but there was no need for it. I am just trying to ascertain why the space between two ships contracts but not the ship itself.

I think you are making the common mistake of being unclear which reference frame you are considering.

From the point of view of a "stationary" observer on Earth, the spaceship gets shorter (and in the case of two ships travelling at the same velocity, the distance between them gets shorter by the same proportion).

From the point of view of the "stationary" spaceship(s), they stay the same length and the distance between them stays the same but the distance to (and between) the "moving" stars decreases.

Webbo
2012-Jun-29, 03:05 PM
So what you're saying is: if the observer on the 2 lightyear long spaceship says that the distance between the earth and the star is 1 lightyear then the front of his spaceship is behind the star. However to an observer on earth the distance to the star is 22 lightyears and the length of the spaceship is contracted to say 0.1 lightyears then the front of the spaceship is in front of the star.

So the question is, does the spaceship fit between the earth and the star or not? This is equivalent to the ladder paradox (http://en.wikipedia.org/wiki/Ladder_paradox) where instead of asking whether a moving ladder fits in a garage you're asking whether a moving spaceship fits between the earth and the star.Indeed. An observer on Earth sees a ship start to go past at 0.999c heading towards a star that is 22.3 ly away. When the star finally passes 2 years later the Earth observer would calculate (and possibly detect) that the front of the ship must still be 20 ly from its destination. According to the ship observer the star is halfway between the front and back of the ship. As it's created a paradox my suspicion is that as stated at the begining the ship does not calculate the distance to be contracted at 1 ly, probably because the ship itself is contracted along with its instruments.

caveman1917
2012-Jun-29, 03:10 PM
Indeed. An observer on Earth sees a ship start to go past at 0.999c heading towards a star that is 22.3 ly away. When the star finally passes 2 years later the Earth observer would calculate (and possibly detect) that the front of the ship must still be 20 ly from its destination. According to the ship observer the star is halfway between the front and back of the ship. As it's created a paradox my suspicion is that as stated at the begining the ship does not calculate the distance to be contracted at 1 ly, probably because the ship itself is contracted along with its instruments.

It's an apparent paradox, not an actual paradox. It results from thinking in terms of absolute simultaneity. Have you read the wikipedia article on the ladder paradox that i linked to? It includes a full explanation of the situation. It's the same question, just think of the front door of the garage as the earth and the back door of the garage as the star, and the ladder being the spaceship.

Webbo
2012-Jun-29, 03:12 PM
I think you are making the common mistake of being unclear which reference frame you are considering.

From the point of view of a "stationary" observer on Earth, the spaceship gets shorter (and in the case of two ships travelling at the same velocity, the distance between them gets shorter by the same proportion).

From the point of view of the "stationary" spaceship(s), they stay the same length and the distance between them stays the same but the distance to (and between) the "moving" stars decreases.Yes I think I confused the accelerating ship with a non-accelerating ship.

In light of that I would go back to what I originally stated about the ship. Originally I said it would have its instruments contracted because it's moving at 0.999 c but what I should have said is that the instruments are contracted when it accelerates to 0.999c. I think that would mean that it could not detect the contraction of space in front or behind and the distance to the star would remain 22.3 ly.

Strange
2012-Jun-29, 03:19 PM
Yes I think I confused the accelerating ship with a non-accelerating ship.

Let's forget acceleration. I was pointing out what is seen from two different points of view when they consider themselves stationary.


In light of that I would go back to what I originally stated about the ship. Originally I said it would have its instruments contracted because it's moving at 0.999 c but what I should have said is that the instruments are contracted when it accelerates to 0.999c.

That is so confusing (or confused) I'm not sure what you are trying to say.


I think that would mean that it could not detect the contraction of space in front or behind and the distance to the star would remain 22.3 ly.

They would see lengths ahead (and behind) contracted; from their point of view, the distance to the star is 1 ly.

Webbo
2012-Jun-29, 03:21 PM
It's an apparent paradox, not an actual paradox. It results from thinking in terms of absolute simultaneity. Have you read the wikipedia article on the ladder paradox that i linked to? It includes a full explanation of the situation. It's the same question, just think of the front door of the garage as the earth and the back door of the garage as the star, and the ladder being the spaceship.What the difference between an actual and apparent paradox. What actual paradox's are there? If it's just an apparent paradox what's the correct answer?

Webbo
2012-Jun-29, 03:28 PM
That is so confusing (or confused) I'm not sure what you are trying to say.If two ships, one in front of the other, accelerate then the space between them contracts as pointed out earlier. The same must be true of front and back of the ship and therefore everything on board, hence I believe that the instruments would not be able to detect the contraction of space ahead and would continue to measure the star as 22.3 ly away.[/quote]

Strange
2012-Jun-29, 03:30 PM
What the difference between an actual and apparent paradox.

An apparant paradox is where people use the word for something that is counter-intuitive but perfectly well understood (the single-photon version of the two slit experiment, the twin paradox in relativity, etc).


What actual paradox's are there?

Not many. Mainly in logic I think; "this statement is false" sort of thing.


If it's just an apparent paradox what's the correct answer?

The answer comes down to the fact that the two observers (moving and stationary) do not agree what "at the same time" means. See the Wikipedia page for the full description.

Strange
2012-Jun-29, 03:38 PM
If two ships, one in front of the other, accelerate then the space between them contracts as pointed out earlier. The same must be true of front and back of the ship and therefore everything on board, hence I believe that the instruments would not be able to detect the contraction of space ahead and would continue to measure the star as 22.3 ly away.

I think you are mixing up two different things.

If we say that, from the Earth's reference frame, the two ships start accelerating at the same time and stay the same distance apart then, from the reference frame of one of the spaceships, they will start accelerating at different times and so the distance between them will change.

This is separate from the fact that they will see the distance to the the destination star contract.

The first is caused by relativity of simultaneity ("at the same time" means something different to the Earth as it does on the moving spaceships); the latter is caused by the the velocity of the star relative to the spaceship.

caveman1917
2012-Jun-29, 03:43 PM
What the difference between an actual and apparent paradox.

An apparent paradox is if you get contradictory answers because you made a mistake in your analysis and the paradox is resolved by a more careful consideration of the problem at hand. Usually these get to be well-known paradoxes because the mistake being made is very intuitive and the resolution lies in somewhat subtle aspects of the situation.

In this example the mistake is by thinking that two events (front of spaceship being at point x and back of spaceship being at point y) being simultaneous in one frame (the earth-observer frame) means they are simultaneous in another frame (the spaceship-observer frame).

Another example of an apparent paradox is the so-called twin paradox. If time dilation is symmetric, ie twin A sees twin B's time as going slow and twin B sees twin A's time as going slow, then how come only the twin that took the journey is younger? Shouldn't either think the other is younger? The resolution lies in the fact that time dilation is only symmetric between inertial frames (that don't accelerate), however the journey-taking twin must at some point start to return, at which time he accelerates (he slows down and then reverses course), so we can unambiguously say which twin is younger and which is older.


What actual paradox's are there?

In relativity none that i know of. It would be a disaster if there were because an actual paradox would be getting contradictory answers without making a mistake which would mean that the theory is inconsistent.

An example of an actual paradox in logic would be the liar paradox, ie "this sentence is false". If the sentence is true then it is false, but if it is false then it is true. (note that there are resolutions to the paradox, but in classical logic it is an actual paradox)


If it's just an apparent paradox what's the correct answer?

The correct answer is that the spaceship fits between the earth and the star. Because in the frame of the spaceship the events "back of spaceship at earth" and "front of spaceship behind star" are not simultaneous, therefor they do not represent an actual "state" for the spaceship to be in at some time.

ETA: i must learn to type faster than Strange :)

Webbo
2012-Jun-29, 03:45 PM
An apparant paradox is where people use the word for something that is counter-intuitive but perfectly well understood (the single-photon version of the two slit experiment, the twin paradox in relativity, etc).



Not many. Mainly if logic I think, "this statement is flase" sort of thing.In that case, in the physical world, as no actual paradox exists, to describe the difference beween actual and apparent is erroneous. It's a paradox and that's it. In the physical world there shouldn't be one.



The answers comes down to the fact that the two observers (moving and stationary) do not agree what "at the same time" means. See the Wikipedia page for the full description.There is no "at the same time" issue. The observers on Earth watch the 2ly ship go by and spend the next 20 years observing it approach the star. The ship observers watch the front of the ship pass the star before the back of the ship has passed Earth. Both cannot physically be correct in their observations so something is wrong with one of them. One of our theoretical observations is incorrect. I believe that is because the ship observers always see the star as 22.3 ly distant.

Webbo
2012-Jun-29, 03:48 PM
I think you are mixing up two different things.

If we say that, from the Earth's reference frame, the two ships start accelerating at the same time and stay the same distance apart then, from the reference frame of one of the spaceships, they will start accelerating at different times and so the distance between them will change.

This is separate from the fact that they will see the distance to the the destination star contract.

The first is caused by relativity of simultaneity ("at the same time" means something different to the Earth as it does on the moving spaceships); the latter is caused by the the velocity of the star relative to the spaceship.Do objects contract as they approach the speed of light?

caveman1917
2012-Jun-29, 03:50 PM
I think you are mixing up two different things.

If we say that, from the Earth's reference frame, the two ships start accelerating at the same time and stay the same distance apart then, from the reference frame of one of the spaceships, they will start accelerating at different times and so the distance between them will change.

This is separate from the fact that they will see the distance to the the destination star contract.

The first is caused by relativity of simultaneity ("at the same time" means something different to the Earth as it does on the moving spaceships); the latter is caused by the the velocity of the star relative to the spaceship.

The resolution in the case in which the distance between them must stay constant in their own frame comes from considering the Rindler horizon. If we look at the front spaceship and say that it must accelerate to 0.99c before passing the star, it must have a very high proper acceleration. This means that the back spaceship is behind its Rindler horizon, so it would have to have infinite acceleration to keep their proper distance the same. The best we can do is put the proper acceleration of the first spaceship such that the back spaceship is barely in front of the Rindler horizon of the first, but this acceleration is a lot lower than what we required. The result is that both spaceships will have long passed the star before they reach 0.99c (because there is a limit on their acceleration).

Strange
2012-Jun-29, 03:52 PM
In that case, in the physical world, as no actual paradox exists, to describe the difference beween actual and apparent is erroneous. It's a paradox and that's it. In the physical world there shouldn't be one.

Indeed. If you find one in a physical theory it is either an example of reductio ad absurdum (and the theory is wrong) or you have made a mistake.


The ship observers watch the front of the ship pass the star before the back of the ship has passed Earth.

But you are talking about the relationship in time between two events. The observers on Earth will disagree about when these events occur. Therefore no paradox. You need to work out the detail, rather than relying on intuition (as caveman says).

Strange
2012-Jun-29, 03:52 PM
ETA: i must learn to type faster than Strange :)

Yeah, but you wrote more than me! (it's all relative)

Webbo
2012-Jun-29, 03:55 PM
In relativity none that i know of. It would be a disaster if there were because an actual paradox would be getting contradictory answers without making a mistake which would mean that the theory is inconsistent.Indeed. Which is what this is about.



The correct answer is that the spaceship fits between the earth and the star. Because in the frame of the spaceship the events "back of spaceship at earth" and "front of spaceship behind star" are not simultaneous, therefor they do not represent an actual "state" for the spaceship to be in at some time.So from the observers point of view there is a distortion of reality hence my belief they would indeed still see the star as 22.3 ly away.

Strange
2012-Jun-29, 03:56 PM
Do objects contract as they approach the speed of light?

Yes. As seen by an external "stationary" observer; i.e. the one they are approaching the speed of light relative to. They do not feel themselves getting shorter.

Actually, your wording ("approach") is ambiguous. They are contracted when their velocity is near the speed of light (actually at any speed, it only becomes significant at high speed; GPS systems, which are much slower than light speed, have to take this into account).

Webbo
2012-Jun-29, 04:08 PM
Yes. As seen by an external "stationary" observer; i.e. the one they are approaching the speed of light relative to. They do not feel themselves getting shorter.

Actually, your wording ("approach") is ambiguous. They are contracted when their velocity is near the speed of light (actually at any speed, it only becomes significant at high speed; GPS systems, which are much slower than light speed, have to take this into account).Yes I agree by the external observer or any external event such as the light from stars interacting with the contracted instruments. On board they would assume everything is normal, however, due to their contracted instruments, the contracted space ahead of them would appear unchanged and still read as if the star was 22.3 ly away.

cjameshuff
2012-Jun-29, 04:15 PM
Yes I agree by the external observer or any external event such as the light from stars interacting with the contracted instruments. On board they would assume everything is normal, however, due to their contracted instruments, the contracted space ahead of them would appear unchanged and still read as if the star was 22.3 ly away.

Their instruments aren't contracted, their surroundings are.

caveman1917
2012-Jun-29, 04:17 PM
Indeed. Which is what this is about.

There is no paradox. Different observers will have different "stories" about what happens, but the results will be the same.

Another example, suppose two supernovae happen with two stars 1 lightyear apart. If you are at star A you immediately see supernova A happen and a year later supernova B, the situation is reversed if you are at star B. So observer A will say that star A blew a year earlier than star B and observer B will say that star B blew a year earlier than star A. That's not a paradox either, even though both observers disagree about the sequence of events.

Now let's make our spaceship example physical. Suppose that both at earth and at the star we have a "spaceship detector". A light that gives of green light if a spaceship part is at that location and a red light if there is no spaceship part at that location. Now suppose that in the middle between earth and the star we have a device (stationary with the earth and star) that will explode if and only if it receives a green photon from both sides at the same time.

From the earth observer frame it is obvious that the device should never explode because the spaceship is never at both places at the same time.
Do you think that from the spaceship observer frame the device should explode or not? If yes then we would have an actual paradox, if no then we just have two observers telling different stories about how events unfold without changing the "physical world" (ie an exploded vs a non-exploded device).

Webbo
2012-Jun-29, 04:22 PM
So if the ship rotates 180 degrees and accelerate toward the star that was behind, how long will it take to reach it and will it at any time measure the distance to increase to 22.3ly?

Yes. At the half way point between moving in one direction and moving in the other - at that point it will be stationary with respect to the stars.

But maybe you are thinking it can go instantly from 0.99c in one direction to 0.99c in the other, without slowing down ...While we await Korjik's answer could I ask another question in relation to this?

Let's look at another identical ship somewhere else in the universe. This ship spots an object moving directly away from it at 0.999 c and currently it's distance is 1 light year. I assume we all agree that as the ship observes this object and does nothing it would take about another 21.3 years before it reaches a distance of 22.3 light years.

Now, if instead the ship accelerates towards the object when it spots it at 1 ly distance roughly how long would it take to reach and would the distance ever increase to 22.3 ly.

Let's assume it takes all these ships 1 hour to reach maximum speed.

Webbo
2012-Jun-29, 04:24 PM
Their instruments aren't contracted, their surroundings are.According to themselves, but not according to the light outside approaching the intruments.

Webbo
2012-Jun-29, 04:32 PM
There is no paradox. Different observers will have different "stories" about what happens, but the results will be the same.

Another example, suppose two supernovae happen with two stars 1 lightyear apart. If you are at star A you immediately see supernova A happen and a year later supernova B, the situation is reversed if you are at star B. So observer A will say that star A blew a year earlier than star B and observer B will say that star B blew a year earlier than star A. That's not a paradox either, even though both observers disagree about the sequence of events.

Now let's make our spaceship example physical. Suppose that both at earth and at the star we have a "spaceship detector". A light that gives of green light if a spaceship part is at that location and a red light if there is no spaceship part at that location. Now suppose that in the middle between earth and the star we have a device (stationary with the earth and star) that will explode if and only if it receives a green photon from both sides at the same time.

From the earth observer frame it is obvious that the device should never explode because the spaceship is never at both places at the same time.
Do you think that from the spaceship observer frame the device should explode or not? If yes then we would have an actual paradox, if no then we just have two observers telling different stories about how events unfold without changing the "physical world" (ie an exploded vs a non-exploded device).Then one of the observers is not observing an actual reality which IMO are the ones based on the ship. They will always see the star as 22.3 ly distant and never trigger the device. No paradox.

caveman1917
2012-Jun-29, 04:39 PM
Then one of the observers is not observing an actual reality which IMO are the ones based on the ship. They will always see the star as 22.3 ly distant and never trigger the device. No paradox.

I wasn't asking who is observing an "actual reality" :). My question was, do you think that in the frame of the observer on the spaceship the device would explode? In other words, would they trigger the device if they see the earth-star distance as contracted?

Strange
2012-Jun-29, 05:00 PM
Yes I agree by the external observer or any external event such as the light from stars interacting with the contracted instruments. On board they would assume everything is normal, however, due to their contracted instruments, the contracted space ahead of them would appear unchanged and still read as if the star was 22.3 ly away.

From their point of view, their instruments are not contracted. So no.

On the other hand, in their frame of reference they are stationary and the universe is rushing past at 0.99c. Therefore they will see everything external contracted in the direction of travel.

Strange
2012-Jun-29, 05:02 PM
According to themselves, but not according to the light outside approaching the intruments.

The light is approaching their instruments at c. Just as it would if they were still on earth. The speed that light hits you does not tell you anything because it is always the same.

Strange
2012-Jun-29, 05:03 PM
Let's look at another identical ship somewhere else in the universe. This ship spots an object moving directly away from it at 0.999 c and currently it's distance is 1 light year. I assume we all agree that as the ship observes this object and does nothing it would take about another 21.3 years before it reaches a distance of 22.3 light years.

Now, if instead the ship accelerates towards the object when it spots it at 1 ly distance roughly how long would it take to reach and would the distance ever increase to 22.3 ly.

Let's assume it takes all these ships 1 hour to reach maximum speed.

You don't provide enough information to answer this. And even if you did, I would suggest you work it out yourself as that is the only way you will learn.

1) At what rate does this second ship accelerate?
2) What is the maximum speed?

Hetman
2012-Jun-29, 05:16 PM
That is only true if they are in inertial motion (not accelerating). Once you bring acceleration into it, that simple view no longer applies. Work though this: http://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox

22 ly / gamma = 1 => 22 ly * gamma = 484 ly

the result obtained according to the calculations of this paradox.
It is calculated directly from the Lorentz transformation:
x'(t=0) = x.gamma

there is the same, but for any t:
dx'= x'_1 - x'_2 = dx.gamma

This is the distance determined from the two-way speed of light, because the average c + v and c - v is equal to c / gamma^2, plus time dilation, which reduces it to c / gamma.

Grey
2012-Jun-29, 07:32 PM
There is no "at the same time" issue. The observers on Earth watch the 2ly ship go by and spend the next 20 years observing it approach the star. The ship observers watch the front of the ship pass the star before the back of the ship has passed Earth. Both cannot physically be correct in their observations so something is wrong with one of them. One of our theoretical observations is incorrect. I believe that is because the ship observers always see the star as 22.3 ly distant. [my emphasis added]No, actually, what you describe happening here is pretty close to correct. From the point of view of observers on Earth, the ship is contracted to a length of about 0.1 light years. So it takes about 0.1 years to pass by Earth, and then another 22.2 years for the front of the ship to reach the star (plus another 0.1 years after that for the back of the ship to reach the star). For observers on Earth, there is a significant amount of time when the spaceship is entirely between the Earth and the star.

On the other hand, from the point of view of the observers on the spaceship, the Earth passes by the front of the ship, and then a year later (when the Earth is at about the midpoint of the ship), the star passes the front of the ship. After another year, the Earth passes the back of the ship (and the star reaches the midpoint), and a year after that, the star passes the back of the ship. From the point of view of those aboard the spaceship, there is never any point when the ship is entirely between the Earth and the star. Instead, there's a period of time when the Earth and the star are both between the front and back of the ship.

The problem with any of this is your intuitive claim that "Both [observations] cannot physically by correct" (and your claim that there is no simultaneity issue here). Those claims are false. They seem natural, but only because we spend our whole lives at sub-relativistic speeds where these issues don't come up, so we never experience relativistic effects directly. But letting go of this notion that there's an absolute space and time is what's required to make relativity work. Physicists didn't like that idea either, really; we don't accept relativity because we want to, we accept it because it fits the experimental evidence.

BioSci
2012-Jun-29, 09:58 PM
(My emphasis) No, actually, what you describe happening here is precisely correct. From the point of view of observers on Earth, the ship takes about 2 years to pass by Earth, and then another 20 years for the front of the ship to reach the star (plus another 2 years after that for the back of the ship to reach the star). For observers on Earth, there is a significant amount of time when the spaceship is entirely between the Earth and the star.


Just a nitpick on your excellent summation:

If the spaceship is measured to be 2 ly in the frame of the ship, then in earth's frame it would only be ~ 1/10 light year...

If it was 2 ly in the earth frame , then it would be ~40 LY in the ship's frame!

;)

Grey
2012-Jun-29, 10:01 PM
Just a nitpick on your excellent summation:

If the spaceship is measured to be 2 ly in the frame of the ship, then in earth's frame it would only be ~ 1/10 light year...

If it was 2 ly in the earth frame , then it would be ~40 LY in the ship's frame!

;)Heh. While you were typing that, I noticed that mistake, too, and corrected it. It doesn't change the substantive argument, but I thought it best to remove so as to avoid any confusion.

Strange
2012-Jun-29, 10:02 PM
22 ly / gamma = 1 => 22 ly * gamma = 484 ly

I'm afraid I don't follow what you are trying to say. Could you be a little more explict / less cryptic; for example, what is the 484 ly supposed to refer to?

Hetman
2012-Jun-30, 09:29 PM
I'm afraid I don't follow what you are trying to say. Could you be a little more explict / less cryptic; for example, what is the 484 ly supposed to refer to?

This is just the result directly from the Lorentz transformation.
x '(t = t' = 0) = gamma * x; t = t '= 0 - synchronization of clocks.

This is the instantaneous distance.

The same is here:
http://en.wikipedia.org/wiki/Bell%27s_spaceship_paradox

This is a direct consequence of c = const, and time dilation.

When you measure less time (because the clock slows down), then you must increase the distance, in order to maintain speed, including the c.

In GR is the same: dtdxdydz = inv.

ctcoker
2012-Jul-01, 05:03 AM
No, that's not true. If you dilate time, then the distance must decrease as well. A spacetime diagram will show this pretty easily, independent of any math. Mathematically it's easiest to show this using spacetime intervals. Light follows null geodesics, so the spacetime interval along any path a photon takes is zero. If you decrease the amount of time light takes to get somewhere (whether through a boosted frame or anything else), it must have traveled a shorter distance.


In GR is the same: dtdxdydz = inv.That is not the invariant in GR. ds2 = gmndxmdxn is.

WayneFrancis
2012-Jul-01, 08:29 AM
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. You accelerate up to .99c. A star that was 7ly behind you now appears to be 1ly. You turn around and that star is racing away from you a .99c While you accelerate to .99c towards it you'll see that you are actually bringing yourself and the star back into a similar frame of reference and that means the distances will now expand back to 7ly.

Can't have your cake and eat it too.

WayneFrancis
2012-Jul-01, 10:44 AM
Are you saying the ship cannot declare itself at rest and the contracted distance behind the ship is an illusion?
You can but then you have to remember that the star is racing away from you at .99c or .999c if you want the ~22x contraction.

note i'm using .999c which is a ~22x dilation
So if you accelerate to .999c towards the star, 1ly away, all you've done is put yourself back into a similar reference frame as that star.
Remember you started in a frame where you and the star are at rest with respect to each other and separated by ~22ly. You accelerated to .999c and that distance appears to be 1ly. But now you and that star are not in the same reference frame. Either you or the star is moving at .999c. If you want to consider yourself at rest you can't ignore that the star is racing away from you at .999c. So when you turn around and start your acceleration to .999c your really just bringing yourself back to the rest frame common between you and that star.

Mathematically you have to add your two speeds. It is racing away from you at .999c minus your speed of n (.999c-n) and put that into the formula
\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}

c= 1 and v is a faction of 1
So step 1. You and star relative velocity = 0.

= \frac{1}{\sqrt{1-\frac{0^{2}}{1^{2}}}}
\frac{1}{\sqrt{1-\frac{0}{1}}}
\frac{1}{\sqrt{1-0}}
\frac{1}{1}

You accelerate away from the star at .999c
\frac{1}{\sqrt{1-\frac{.999^{2}}{1^{2}}}}
\frac{1}{\sqrt{1-0.998001}}
\frac{1}{\sqrt{0.001999}}
\frac{1}{0.04471017781221631419961342300205}
22.366272042129221710662042522285 is the amount of time and space dilation.
Now either you or the star are travelling at .999c
You turn your ship around and accelerate towards the star at .999c
Star's velocity away from you. .999c Your velocity towards the star .999c
total amount of velocity between you and the star. .999c-.999c = 0.0c

WayneFrancis
2012-Jul-01, 11:33 AM
While we await Korjik's answer could I ask another question in relation to this?

Let's look at another identical ship somewhere else in the universe. This ship spots an object moving directly away from it at 0.999 c and currently it's distance is 1 light year. I assume we all agree that as the ship observes this object and does nothing it would take about another 21.3 years before it reaches a distance of 22.3 light years.

Now, if instead the ship accelerates towards the object when it spots it at 1 ly distance roughly how long would it take to reach and would the distance ever increase to 22.3 ly.

Let's assume it takes all these ships 1 hour to reach maximum speed.

This sounds counter intuitive but here is what is going on
17182

A and B do NOT measure the distance the same.

If A is the one that accelerated then and A measures the distance as 1ly then B will measure the distance as 22ly.
If B measures the distance as 1ly the A will measure the distance as 1/22ly

If B starts accelerating towards A and it takes B 1 hour to reach .999c then to object A it would take B 22hours to reach .999c. But this isn't 1 hour of B's time. But B's initial reference frame's time. Let me call that observer C. At which time Object A & B will be in roughly the same frame and be separated by about 1/22ly

If you want the acceleration to be one hour in B's reference frame then what will happen is that it will take a lot longer then 1 hour according to C.

WayneFrancis
2012-Jul-01, 12:17 PM
Then one of the observers is not observing an actual reality which IMO are the ones based on the ship. They will always see the star as 22.3 ly distant and never trigger the device. No paradox.

Simultaneous events are not possible across all reference frames. They can be worked out using the appropriate formulas and the consequences will be the same but what 2 different observers sees can be in different orders.

Do you agree that the faster you go the less time passes for you locally?

Say I have 2 stations, A and B, separated by 1ly in the same inertial reference frame. Each is blinking a laser once every day at a frequency of 450nm
If I accelerate towards from A from B at .866c what do I see?
I observe the stations are only .5ly appart.
I measure the amount of time it takes me to get from A to B as .557367 years or about 7 months
I see Station A & B's lasers blink 421 times.
I'll see stations B's lasers pulses less frequently then A's until the end of the journey when I slow down.
I see A's laser light blue shift both do to me my time dilation and a doppler effect by approaching the laser
I see B's laser with a combination of blue shift because of my time dilation and red shift do to me travelling away from the laser.

Everyone will agree that 421 laser pulses where flashed during the trip. We'll not agree on how fast they pulsed and what frequency of light they where at.

Hetman
2012-Jul-01, 04:12 PM
No, that's not true. If you dilate time, then the distance must decrease as well. A spacetime diagram will show this pretty easily, independent of any math. Mathematically it's easiest to show this using spacetime intervals. Light follows null geodesics, so the spacetime interval along any path a photon takes is zero. If you decrease the amount of time light takes to get somewhere (whether through a boosted frame or anything else), it must have traveled a shorter distance.

Mathematically, to keep c = x/t = const, you need to transform t opposite to x, which means xt = const, paradoxically.


That is not the invariant in GR. ds2 = gmndxmdxn is.

four volume = dtdxdydz = inv.
In other coordinates, such as spherical, we have to take into account the Jacobian.
The space is preserved.

ctcoker
2012-Jul-01, 09:44 PM
Mathematically, to keep c = x/t = const, you need to transform t opposite to x, which means xt = const, paradoxically.No, this is wrong. If c = x/t, then c = (5x)/(5t), not 5x/(1/5)t = 25(x/t) or vice versa. If you want to keep c constant, x and t must transform the same way.


four volume = dtdxdydz = inv.
In other coordinates, such as spherical, we have to take into account the Jacobian.
The space is preserved.Never mind, you're right about that, sorry.

Webbo
2012-Jul-02, 04:47 PM
This sounds counter intuitive but here is what is going on
17182

A and B do NOT measure the distance the same.

If A is the one that accelerated then and A measures the distance as 1ly then B will measure the distance as 22ly.
If B measures the distance as 1ly the A will measure the distance as 1/22ly

If B starts accelerating towards A and it takes B 1 hour to reach .999c then to object A it would take B 22hours to reach .999c. But this isn't 1 hour of B's time. But B's initial reference frame's time. Let me call that observer C. At which time Object A & B will be in roughly the same frame and be separated by about 1/22ly

If you want the acceleration to be one hour in B's reference frame then what will happen is that it will take a lot longer then 1 hour according to C.What is the answer considering we don't know who accelerated away. The ship spotted the object constantly moving away at 0.999 so;

a) from the ships own frame how long would it take to reach assuming it could accelerate to 0.999 c in 1 hour?
b) from the ships own frame does the objects distance ever increase to more than 22.3ly away?

Webbo
2012-Jul-02, 04:53 PM
The light is approaching their instruments at c. Just as it would if they were still on earth. The speed that light hits you does not tell you anything because it is always the same.I was thinking more about the angle rather than the speed.

Webbo
2012-Jul-02, 05:00 PM
Now either you or the star are travelling at .999c
You turn your ship around and accelerate towards the star at .999c
Star's velocity away from you. .999c Your velocity towards the star .999c
total amount of velocity between you and the star. .999c-.999c = 0.0cSo how long until the ship reaches it?

Webbo
2012-Jul-02, 05:18 PM
Interestingly from the maximum speed thread Grey provides a link that explains why relativistic length contraction is invisible.

http://faraday.physics.utoronto.ca/PVB/Harrison/SpecRel/Flash/ContractInvisible.html

Grey
2012-Jul-02, 07:35 PM
Interestingly from the maximum speed thread Grey provides a link that explains why relativistic length contraction is invisible.That same link also makes it clear that you will measure length contraction as real. It's just that light travel time delays will make it look as if the object is rotated to your eye (or a camera), rather than contracted. But a measurement of the size of the object will show it to be smaller.

Webbo
2012-Jul-02, 11:52 PM
That same link also makes it clear that you will measure length contraction as real. It's just that light travel time delays will make it look as if the object is rotated to your eye (or a camera), rather than contracted. But a measurement of the size of the object will show it to be smaller.But the whole point is what is observed. No contraction is observed therefore parallax would be the same in both cases; at rest or moving at 0.999c.

Grey
2012-Jul-03, 02:21 AM
But the whole point is what is observed. No contraction is observed therefore parallax would be the same in both cases; at rest or moving at 0.999c.Only because of the light delay time. You have to take that into account when working out parallax. It's been known that the light time delay affects the apparent position of an object for hundreds of years, well before relativity theory was introduced. Again, the angles involved are always very small, but only because we're never moving very fast.

WayneFrancis
2012-Jul-03, 07:50 AM
What is the answer considering we don't know who accelerated away. The ship spotted the object constantly moving away at 0.999 so;


Don't mistake "you can always consider yourself at rest" with "I don't know if I accelerated or not" The 2 are not the same.



a) from the ships own frame how long would it take to reach assuming it could accelerate to 0.999 c in 1 hour?


How long would it take according to an observer at rest with the initial frame of the ship before accelerating? The time, t, on Earth would be 1hr < t < 22hr
It isn't a simple calculation. Is the acceleration constant with respect to the ship or the stationary observer. If it is constant with respect to the stationary observer then those on the ship would experience an an exponential increase in acceleration at the end of the trip. If it is with respect to the ship then it you'd have the stationary observer see the ship slow down over time.




b) from the ships own frame does the objects distance ever increase to more than 22.3ly away?

Can you ever get a distance greater then the proper distance? Not that I can think of. This is because contraction is always contraction. Its kind of like asking can you go before the beginning.

Webbo
2012-Jul-04, 10:03 AM
Only because of the light delay time. You have to take that into account when working out parallax. It's been known that the light time delay affects the apparent position of an object for hundreds of years, well before relativity theory was introduced. Again, the angles involved are always very small, but only because we're never moving very fast.Does it matter what the reason is? My OP suggested that both the at rest observer on Earth and the 0.999c moving observer would both measure the distance to be 22.3ly. The ships instruments do not know that the contraction is offset by the light delay time.

Webbo
2012-Jul-04, 10:13 AM
Can you ever get a distance greater then the proper distance? Not that I can think of. This is because contraction is always contraction. Its kind of like asking can you go before the beginning.Yet in the other scenario that is exactly what happens. It seems to me that prior knowledge of any movement/acceleration needs to be known before an effect can be determined, which in turn leads me to believe that, contrary to relativity, there must be some prefered reference frame.

caveman1917
2012-Jul-04, 01:34 PM
Does it matter what the reason is? My OP suggested that both the at rest observer on Earth and the 0.999c moving observer would both measure the distance to be 22.3ly. The ships instruments do not know that the contraction is offset by the light delay time.

The contraction is only offset by the light delay time for an extended object, such as the disk in the example. When you're doing a parallax measurement you're considering angles to point sources, so the ship will measure 1 ly, not 22.3 ly.

caveman1917
2012-Jul-04, 01:35 PM
Yet in the other scenario that is exactly what happens.

What other scenario? Measuring more than the proper distance should never happen.

caveman1917
2012-Jul-04, 01:43 PM
Can you ever get a distance greater then the proper distance? Not that I can think of. This is because contraction is always contraction. Its kind of like asking can you go before the beginning.

You can't. A lorentz transformation is equivalent to a rotation in 4d spacetime, so an object that is in its rest frame aligned with say the x-axis, will in another frame be rotated into the time dimension, so a size measurement in that frame (which is taking a slice of constant time) will always get a lower size because the object has "rotated away".

Perhaps an example, just think in 3d. Think of a cylinder aligned with the x-axis. When you measure its length you will get the full length. Now rotate it so that it is standing diagonal. When you now measure the length along the x-axis you will get a lower number. (i know it's a bit of an imperfect example but that's what you get with 3d instead of 4d :))

Webbo
2012-Jul-04, 02:03 PM
The contraction is only offset by the light delay time for an extended object, such as the disk in the example. When you're doing a parallax measurement you're considering angles to point sources, so the ship will measure 1 ly, not 22.3 ly.Why would only objects be extended and not space between the objects?

WayneFrancis
2012-Jul-04, 02:09 PM
Yet in the other scenario that is exactly what happens. It seems to me that prior knowledge of any movement/acceleration needs to be known before an effect can be determined, which in turn leads me to believe that, contrary to relativity, there must be some prefered reference frame.

True, I didn't think if both objects where moving together but at relativistic speeds. Ah but then the proper distance is still the distance between those 2 objects. So while there are frames that can measure a distance longer then the proper distance between 2 points that will indicate that those 2 points are both accelerated with respect to the point doing the measurements.

Webbo
2012-Jul-04, 02:13 PM
What other scenario? Measuring more than the proper distance should never happen.See below;

No. You accelerate up to .99c. A star that was 7ly behind you now appears to be 1ly. You turn around and that star is racing away from you a .99c While you accelerate to .99c towards it you'll see that you are actually bringing yourself and the star back into a similar frame of reference and that means the distances will now expand back to 7ly.So it would appear that prior knowledge of the movement between the observer and the object needs to be known.

The question I suppose is, who and when is it decided what the proper distance is?

WayneFrancis
2012-Jul-04, 02:21 PM
See below;
So it would appear that prior knowledge of the movement between the observer and the object needs to be known.

The question I suppose is, who and when is it decided what the proper distance is?
proper distance is the distance between 2 objects that are not moving with respect to each other that those 2 object would measure.

So if there is movement relative to each other then on of the objects would have to apply some amount of acceleration in an opposite direction to bring the 2 objects to a point where they are stationary with respect to each other. Doesn't matter which one does it. Either can do it and the results would be the same.

Webbo
2012-Jul-04, 02:35 PM
proper distance is the distance between 2 objects that are not moving with respect to each other that those 2 object would measure.

So if there is movement relative to each other then on of the objects would have to apply some amount of acceleration in an opposite direction to bring the 2 objects to a point where they are stationary with respect to each other. Doesn't matter which one does it. Either can do it and the results would be the same.Then I guess the OP cant really be answered. The ship cannot measure the proper distance, however, the suggestion that contraction cannot be observed still needs to be addressed.

WayneFrancis
2012-Jul-04, 03:37 PM
Then I guess the OP cant really be answered. The ship cannot measure the proper distance, however, the suggestion that contraction cannot be observed still needs to be addressed.

why can't it be measured? Any observer can measure a distance. When the distance to an object is measured and that object is not in relative motion to the observer then we call that "proper distance".

No one has presented a valid reason why the contraction can not be observed. The closest anyone has come was spherical objects because of their motion will still have an optical illusion that they are spherical. That doesn't effect distance measurement as I understand it.

Webbo
2012-Jul-06, 01:30 AM
why can't it be measured? Any observer can measure a distance. When the distance to an object is measured and that object is not in relative motion to the observer then we call that "proper distance".The moving ship is always in motion relative to the target star hence it doesn't have a proper distance to measure. How can we compare what they measure? The comparison is meaningless.


No one has presented a valid reason why the contraction can not be observed. The closest anyone has come was spherical objects because of their motion will still have an optical illusion that they are spherical. That doesn't effect distance measurement as I understand it.So would there be any contraction evident instead of a disc, there was a perfect circle of stars. I see no reason why the physics would be different. A disc was used merely to demostrate the effect. Why would the effect be unique to discs?

WayneFrancis
2012-Jul-06, 05:46 AM
The moving ship is always in motion relative to the target star hence it doesn't have a proper distance to measure. How can we compare what they measure? The comparison is meaningless.

So would there be any contraction evident instead of a disc, there was a perfect circle of stars. I see no reason why the physics would be different. A disc was used merely to demostrate the effect. Why would the effect be unique to discs?

So what if the ship is in motion. If I measure the distance to my destination as 1ly how is this meaningless to me? It gives me an indication on how long I have left on my trip. Just like passing a mile marker on the highway lets you know how far you are from a certain point. If I want to compare it to the distance someone else in a different frame is in I can do that to. I'm not sure what you're issue.

Grey
2012-Jul-06, 02:34 PM
I'm wondering if it might clear things up to suggest a concrete method of measuring the distance, using the method Einstein suggested, and see how it plays out. So we're going to measure the distance to the star using radar, assuming we have a really large dish antenna, both to send out a strong signal, and to receive the weak return signal. Suppose I send out a radar pulse to an object, and it takes 10 years to get back. I now know that it's 5 light years away. Or, more precisely, that it was 5 light years away 5 years ago (this is where taking the light travel time delay is important). If the object is at rest with respect to me, it's still 5 light years away. If it's moving, I could calculate how far away it is right now (assuming my own frame of reference for simultaneity), if I assume it hasn't changed velocity since the last signal I received from it. Sending multiple radar pulses will serve to get a series of position measurements, which can then be used to calculate velocity. For our test here, I'm going to assume that nobody accelerates; everyone is moving at constant velocity. I'm going to talk about a single radar pulse and see what happens, but I'm assuming that in each case this was actually a series of pulses, so that I get a velocity measurement as well as an instantaneous position measurement.

Now, from the point of view of Earth, we're essentially at rest with respect to the star, so we can send out a pulse whenever we feel like it. We get a return signal 44.6 years later, so we know that the star is 22.3 light years away (obviously, we're not concerned with how long any of this takes). Now, suppose we notice that the ship send a radar pulse toward the star just as it passes us. Here's what we'd expect to happen from our frame of reference. The pulse travels toward the star, taking about 22.3 years to reach it. But the ship is travelling toward the star during that time, at 99.9% the speed of light. By the time the light pulse arrives, the ship should be only about 8 light days away from the star. The pulse bounces back, and since the ship is continuing to move at nearly light speed, the ship and the pulse arrive in the middle of that remaining 8 light days about 4 days later. The total time elapsed for the beam of light is 22.3 years plus 4 days. However, since I also notice that the clock on the ship is running slowly by a factor of about 22.3, I'll expect that the ship clock will only say that about 1 year has passed during that time. Since the ship clock is running slow, that would be just a little over 4 hours measured on the ship clock, so the observers on Earth will expect that the ship clock will read 1 year and 4 hours when the ship receives the return radar pulse. The ship is now only 4 light days away from the star, though, so the observers on Earth would expect the ship to pass the star in another 4 days. That would again be about 4 hours on the ship clock, so observers on Earth would expect the ship clock to read 1 year and 8 hours when it reaches the star, even though it "really" took a little over 22.3 years.

Let's switch to the ship's perspective. From their point of view, the ship is at rest, and the Earth is zipping past them at 99.9% the speed of light. If they've sent out earlier radar pulses toward the star to calculate its position and velocity, they'll know that it is also moving toward them at 99.9% the speed of light, and that it's 1 light year away from them as they pass the Earth. Since it's moving toward them, the radar pulse will meet the star about half way after just 6 months (actually, since the beam of light is a bit faster, the star won't have made it quite halfway; the radar pulse will make it to the halfway point first, and then reach the star when the star is about 2 light hours away from the halfway point). Now the radar pulse starts on its journey home. Since it took 6 months and 2 hours to go out, it will take another 6 months and 2 hours to come back, and the ship will receive the return pulse when their clock reads 1 year and 4 hours. During that time, of course, the observers on the ship would expect the star to continue moving toward them, and they'll calculate it to be only 4 light hours away by the time they receive the return radar pulse. Sure enough, 4 hours later by the ship clock, the star will go zooming past, with the ship clock reading 1 year and 8 hours as it passes.

Note that the observers disagree with each other about things like the distance to the star and the elapsed time. Note also that from Earth's perspective, most of the radar pulse's journey is spent on the outward trip, with only a short time spent on the return trip, since from Earth's perspective, the ship is almost keeping pace with the pulse, and is almost to the star by the time the radar pulse reaches the star. On the other hand, from the ship's perspective, they're standing still and the star is racing toward them, so the radar pulse meets the star about halfway, and spends the same amount of time on both legs of the journey. However, note that they'll agree with each other about specific measurable events, like what time the ship's clock displays when it receives the return radar pulse, or what time it displays when the ship passes the star. Everything remains consistent from either observer's point of view, but they differ in how they interpret those events.

Okay, so even if it's consistent, isn't it all unnecessarily convoluted? Can't we have everyone's clocks moving at the same rate and everyone measuring distances to be the same? Well, we could, at least in principle, but then there would be no way for everyone to measure a beam of light to have the same speed. People moving at different speeds relative to the beam of light would measure light speed to be different. And we don't observe that. Einstein's relativity is the simplest system that keeps light speed a universal constant, regardless of how an observer is moving. And when we take the relativistic transformations to distances and clocks and use them in places where the measurements are accurate enough for it to matter (for example, the GPS satellites), we discover that the system works. If we don't use those relativistic transformations, it doesn't work. I agree that it seems completely bizarre and counterintuitive that people who are moving at different speeds will measure distances and time differently from each other, but in experiment after experiment, it's been shown to be true.

Webbo
2012-Jul-10, 02:37 PM
I agree that it seems completely bizarre and counterintuitive that people who are moving at different speeds will measure distances and time differently from each other, but in experiment after experiment, it's been shown to be true.Surely you mean it's assumed to be true. What experiments do we have for observers moving at near c speeds so we can compare their records to observers that are not?

caveman1917
2012-Jul-10, 02:49 PM
Surely you mean it's assumed to be true. What experiments do we have for observers moving at near c speeds so we can compare their records to observers that are not?

Muons produced in the atmosphere that reach the ground before decaying. Which means the distance for them must be smaller than for us.

Hornblower
2012-Jul-10, 02:52 PM
Surely you mean it's assumed to be true. What experiments do we have for observers moving at near c speeds so we can compare their records to observers that are not?

It is predicted to be true by calculations based on the same theory that is in good agreement with high energy physics experiments we are able to perform. Merely contradicting what seems like common sense to those of us who are not experts in modern physics is not sufficient reason to reject it. The universe is what it is and does what it does, and it does not care whether or not we think it makes sense.

cjameshuff
2012-Jul-10, 02:56 PM
Surely you mean it's assumed to be true. What experiments do we have for observers moving at near c speeds so we can compare their records to observers that are not?

Such experiments are unnecessary. We do have a wide variety of particle accelerator experiments (relativistic effects become important for something as simple and low energy as a large CRT display), spacecraft, etc, which are sufficient to make it quite clear that time dilation and length contraction are real.

NEOWatcher
2012-Jul-10, 05:22 PM
What experiments do we have for observers moving at near c speeds...
It doesn't take "near c" speeds. It takes any difference. "near c" just makes it obvious.

Plus; we are way past the experimental stage. Relativity is already a part of our technology. Relativistic adjustments are necessary for GPS to work (http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit5/gps.html).

Grey
2012-Jul-10, 08:20 PM
Surely you mean it's assumed to be true. What experiments do we have for observers moving at near c speeds so we can compare their records to observers that are not?It's only at near light speed that the effects become obvious, but even at moderate speeds the effects are clearly measurable if your equipment is sensitive enough. Hafele-Keating was an early test of both special and general relativity, done with airplanes flying around the world. There are places where relativistic corrections are needed every day, such as the routine operation of GPS equipment. The predictions of relativity have been tested in too many ways to count since Einstein's day, and in every single case, they have been borne out.

Moreover, any theory that might replace relativity some day will have to produce exactly the same results as relativity in all of the areas where it has been tested, just as relativity itself reduces to Newtonian mechanics when speeds are sufficiently low. So, even if relativity is overturned tomorrow, things like time dilation and Lorentz contraction will remain part of any new theory.

Webbo
2012-Jul-12, 01:07 AM
Muons produced in the atmosphere that reach the ground before decaying. Which means the distance for them must be smaller than for us.I don't believe there were any observers in the muons frame to report on their obsevations during that journey.

Webbo
2012-Jul-12, 01:10 AM
It is predicted to be true by calculations based on the same theory that is in good agreement with high energy physics experiments we are able to perform. Merely contradicting what seems like common sense to those of us who are not experts in modern physics is not sufficient reason to reject it. The universe is what it is and does what it does, and it does not care whether or not we think it makes sense.But that still doesn't show it to be true as Grey stated. Its truthfulness is expected/inferred.

Webbo
2012-Jul-12, 01:12 AM
Such experiments are unnecessary. We do have a wide variety of particle accelerator experiments (relativistic effects become important for something as simple and low energy as a large CRT display), spacecraft, etc, which are sufficient to make it quite clear that time dilation and length contraction are real.So the statement that "experiment after experiment, it's been shown to be true" is incorrect. Such experiments have never been performed.

Webbo
2012-Jul-12, 01:13 AM
It doesn't take "near c" speeds. It takes any difference. "near c" just makes it obvious.

Plus; we are way past the experimental stage. Relativity is already a part of our technology. Relativistic adjustments are necessary for GPS to work (http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit5/gps.html).But what has that got to do with observed length contraction? Obvious or otherwise.

Webbo
2012-Jul-12, 01:16 AM
It's only at near light speed that the effects become obvious, but even at moderate speeds the effects are clearly measurable if your equipment is sensitive enough. Hafele-Keating was an early test of both special and general relativity, done with airplanes flying around the world. There are places where relativistic corrections are needed every day, such as the routine operation of GPS equipment. The predictions of relativity have been tested in too many ways to count since Einstein's day, and in every single case, they have been borne out.

Moreover, any theory that might replace relativity some day will have to produce exactly the same results as relativity in all of the areas where it has been tested, just as relativity itself reduces to Newtonian mechanics when speeds are sufficiently low. So, even if relativity is overturned tomorrow, things like time dilation and Lorentz contraction will remain part of any new theory.However, length contraction has not been observed by experiment which is all I was stating. It's theorised so it has not been shown in experiment after experiment.

cjameshuff
2012-Jul-12, 01:54 AM
I don't believe there were any observers in the muons frame to report on their obsevations during that journey.

Are you denying that we have measurements of the half life of muons?



So the statement that "experiment after experiment, it's been shown to be true" is incorrect. Such experiments have never been performed.

I was referring specifically to your proposed experiment, not to experiments testing relativity in general. Due to practical reasons in accelerating instrumentation to a large fraction of c, and the complete lack of a need for doing so in order to test relativity, it hasn't been and almost certainly won't ever be done. However, my statement was somewhat incorrect, as it's actually been done in principle by the muon observations, and similar observations in particle accelerators.



However, length contraction has not been observed by experiment which is all I was stating. It's theorised so it has not been shown in experiment after experiment.

Saying it hasn't doesn't change the fact that it has. The universe doesn't care how uncomfortable you are with reality. Experiments have verified relativity time and again, time dilation and length contraction are indisputably real no matter how much you dislike them.

Grey
2012-Jul-12, 03:01 AM
However, length contraction has not been observed by experiment which is all I was stating. It's theorised so it has not been shown in experiment after experiment.Well, there have been no experiments where we take a meter stick, accelerate it to near light speed, and then measure the length. That's not feasible with our technology (or with any foreseeable in the near future). However, that's certainly not the only way to test Lorentz contraction. Indeed, the original Michelson-Morley and Kennedy-Thorndike experiments require both time dilation and Lorentz contraction to explain the results. There are other ways to measure contraction as well. For example, we can't directly measure the length of heavy ions moving at relativistic speeds as they zip through particle accelerators. However, from relativity we'd expect them to be shaped something like pancakes rather than little spheres from the perspective of the lab frame. And indeed, if you try to treat them as spheres, you get the wrong scattering results, but if you apply Lorentz contraction and treat them as compressed in the direction of motion, you get results that match observation.

Relativity makes a very long list of predictions, some of which involve time dilation alone, some of which involve Lorentz contraction alone, and some of which involve both. Some of those predictions cannot be tested easily, but many of them have been. I stand by my statement that Lorentz contraction has been clearly shown by experiment many times in the century since Einstein introduced relativity.

WayneFrancis
2012-Jul-12, 06:55 AM
I don't believe there were any observers in the muons frame to report on their obsevations during that journey.

This type of argument shows a very basic lack of understanding about science and observations. It is like saying no one was around a 150 million years ago so their for the Jarassic period is suspect.
We know muons decay very rapidly
We know how they are generated in the upper atmosphere
We know their speeds when the reach the surface of the Earth
We know that given their decay rates that without SR they wouldn't reach the ground before they decay.
We know SR is correct by multiple independent methods to a very high degree of accuracy.
Therefore we can reasonably conclude what is happening to the muons in their frame of reference and that matches observation.

Jeff Root
2012-Jul-12, 09:55 AM
I don't believe there were any observers in the muons
frame to report on their observations during that journey.
Are you denying that we have measurements of the half
life of muons?
It doesn't look like it to me. He's denying that any
measurements have been made by somebody moving
at sufficient speed to measure length contraction of
everything else in the Universe. I agree with Webbo.
I think that such a measurement is so difficult as to
be practically impossible. It will never be done.

Length contraction due to the observer's motion can
only be inferred from observations of other things,
such as half-lives of muons.

-- Jeff, in Minneapolis

WayneFrancis
2012-Jul-12, 11:29 AM
It doesn't look like it to me. He's denying that any
measurements have been made by somebody moving
at sufficient speed to measure length contraction of
everything else in the Universe. I agree with Webbo.
I think that such a measurement is so difficult as to
be practically impossible. It will never be done.

Length contraction due to the observer's motion can
only be inferred from observations of other things,
such as half-lives of muons.

-- Jeff, in Minneapolis

OK for a second say we didn't have any tests to show length contraction. How do you and webbo then propose that Special Relativity has been confirmed with such a high degree of precision if there is no length contraction?

You can't throw out half of a theory and then just say the other half gives the exact same results as the whole theory combined. I'm sorry Jeff but this sounds like more of your "I don't think we'll ever be able to measure it so I doubt it" without you seemingly thinking through all the things that already confirm it.

It just boggles my mind that you can dismiss away half the theory and all its implications and expect to get the same result. Let me use a cooking analogy.

Here are the basic ingredients for a very basic lasagne


flour
egg
olive oil
garlic
beef mince
tomatoes

cheese
milk
parsley
butter
salt
pepper

Now remove half of them. I'll divide the ingredients up 3 different ways and you tell me how you'd get a lasagne from just those ingredients


Just first half
Just 2nd half
Alternating


flour
cheese
flour


egg
milk
olive oil


olive oil
parsley
beef mince


garlic
butter
milk


beef mince
salt
butter


tomatoes
pepper
pepper



Tell me in what way any of those three recipes would make a lasagne. Then when you are done with that. Please reformulate SR so that it works without length contraction. It isn't something that is just a by-product of the theory. So if you want to remove it you have to do it in such a way that your modified version of SR would still get all the same results as has been confirmed by numerous observations.

NEOWatcher
2012-Jul-12, 12:02 PM
But what has that got to do with observed length contraction? Obvious or otherwise.
I suggest you read some history on all the experiments leading up to GPS. Even Sputnik had an element of reletivity experimentation.
Then, I would like you to explain how the measurements made can be done without measuring any kind of movement between two known differences.
I see no way to measure time without involving a known distance, whether that be the length of a pendulum swing, or the difference between waves in an atomic emission.

cjameshuff
2012-Jul-12, 01:26 PM
It doesn't look like it to me. He's denying that any
measurements have been made by somebody moving
at sufficient speed to measure length contraction of
everything else in the Universe. I agree with Webbo.
I think that such a measurement is so difficult as to
be practically impossible. It will never be done.

But it was done, with the muon observations. The figure for muon half life is for an observer in the muon's rest frame. Webbo's claim is equivalent to claiming such things are unknown.



Length contraction due to the observer's motion can
only be inferred from observations of other things,
such as half-lives of muons.

Time dilation and length contraction are the same phenomena. What we measure as time dilation in the frame of a detector counting muons at the bottom of Earth's atmosphere is length contraction in the frame of the muons. Both are required for relativity to have any internal consistency, you can't get one without the other. Time dilation is just the easier aspect to measure directly, though as Grey said, some experiments do directly depend on length contraction as well.

Jeff Root
2012-Jul-12, 04:19 PM
He's denying that any measurements have been made
by somebody moving at sufficient speed to measure
length contraction of everything else in the Universe.
I agree with Webbo. I think that such a measurement
is so difficult as to be practically impossible. It will
never be done.
But it was done, with the muon observations.
No it was not.



The figure for muon half life is for an observer in the
muon's rest frame. Webbo's claim is equivalent to
claiming such things are unknown.
Not if my expression of his claim is accurate. We can
call it my claim. I am responsible for prompting Webbo
to post (in a different thread) the original post here.
I disagreed with korjik when he claimed that someone
travelling for one year at .999 c relative to the nearby
stars would think they travelled a distance of about
one light-year. I still disagree with that.

Let's see if these new links work.

http://cosmoquest.org/forum/showthread.php/135745-maximum-speed?p=2031033#post2031033

My claim, for the current discussion, is that no
measurements have been made by anyone moving
at sufficient speed to measure length contraction of
everything else in the Universe. Such a measurement
is so difficult as to be practically impossible. It will
never be done.



Time dilation and length contraction are the same
phenomena.
Different aspects of the same phenomenon.



What we measure as time dilation in the frame of a
detector counting muons at the bottom of Earth's
atmosphere is length contraction in the frame of the
muons. Both are required for relativity to have any
internal consistency, you can't get one without the
other. Time dilation is just the easier aspect to
measure directly, though as Grey said, some
experiments do directly depend on length
contraction as well.
I have never disagreed with any of that, or implied
that I disagreed with any of it.

No measurements have been made by anyone moving
at sufficient speed to measure length contraction of
everything else in the Universe. Such a measurement
is so difficult as to be practically impossible. It will
never be done.

-- Jeff, in Minneapolis

Grey
2012-Jul-12, 04:50 PM
No measurements have been made by anyone moving
at sufficient speed to measure length contraction of
everything else in the Universe. Such a measurement
is so difficult as to be practically impossible. It will
never be done.Even this isn't true. That is, you're correct that such a measurement has not been done, but it's not "practically impossible". The Space Interferometry Mission (http://en.wikipedia.org/wiki/Space_Interferometry_Mission), designed to measure the position of stars very precisely, would be sufficiently sensitive to detect changes in apparent positions due to the changing Lorentz contraction effects as Earth travels in it's orbit. Sadly, budget cuts at NASA have postponed this mission indefinitely. However, there's a lot of focus these days on finding exoplanets, and many methods for doing that involve measuring the positions of stars precisely enough to detect the wobble from an orbiting planet. We're very close now to the point where improvements in resolution of any such mission will demonstrate Lorentz contraction of the rest of the universe as a side effect of those precise measurements. Even if SIM never launches, some other mission will probably give similar results within a decade or so.

So, suggesting that the experiment is beyond technical feasibility is false. We could do it now if we had the funding. But there's no real doubt of the result, even so. As I noted above, there are plenty of other instances where we have to take Lorentz contraction of fast moving objects into account, or we get clear disagreement with experimental results. Lorentz contraction is a real result, just like time dilation, confirmed by numerous experiments. If you seriously want to propose a replacement to relativity that does not include Lorentz contraction, the place to do that would be the AtM section of the board, but honestly, I don't think you'd be able to do so and still have a consistent theory that also matches all the experimental evidence to date.

caveman1917
2012-Jul-12, 07:24 PM
No it was not.

But the muon is the instrument. What difference does it make if an observer uses a swiss watch or a batch of muons as a clock? What difference does it make if the observer is human or robotic (ie consisting of instruments only, ie consisting of simply the muons).

Suppose that you accept time dilation but not length contraction. Suppose that we want to test length contraction.

We put an observer with a watch somewhere and let him go at relativistic speed. We agree that the observer will give us a signal after he has been some predefined time in his own frame. Because we know time dilation and thus the rate on his clock and his speed we can predict where that will be in both cases, with length contraction and without.

Do you accept that this is an experiment that directly tests length contraction?

Now suppose that instead of an observer we simply use a timebomb with a set timer. Now suppose that instead of a normal watch, we use a bunch of muons (which constitutes a clock by their known decay time). And suppose that the signal is simply "decay".

Do you accept that this is still essentially the same experiment? Because that is exactly what the muon experiments do. The muons should decay before reaching the ground if length contraction in their frame is false, they should not if it is real. The experiment shows them hitting the ground*, ergo length contraction is real.

strictly speaking we're talking about which fraction of them hit the ground before decaying

Webbo
2012-Jul-13, 01:04 AM
Are you denying that we have measurements of the half life of muons?No. Are you implying that the muons had observers travelling with them?



I was referring specifically to your proposed experiment, not to experiments testing relativity in general. Due to practical reasons in accelerating instrumentation to a large fraction of c, and the complete lack of a need for doing so in order to test relativity, it hasn't been and almost certainly won't ever be done. However, my statement was somewhat incorrect, as it's actually been done in principle by the muon observations, and similar observations in particle accelerators.I was also talking about my experiment. Grey's quote was "I agree that it seems completely bizarre and counterintuitive that people who are moving at different speeds will measure distances and time differently from each other, but in experiment after experiment, it's been shown to be true.". And you stated "Such experiments are unnecessary" which did not refute what I stated, and now you are agreeing with me.




Saying it hasn't doesn't change the fact that it has. The universe doesn't care how uncomfortable you are with reality. Experiments have verified relativity time and again, time dilation and length contraction are indisputably real no matter how much you dislike them.And now you introduce a straw man. I am very specifically only talking about observations of length contraction. What experiments do we have where observers have observed length contraction while travelling?

Webbo
2012-Jul-13, 01:14 AM
I stand by my statement that Lorentz contraction has been clearly shown by experiment many times in the century since Einstein introduced relativity.Your claim was;

I agree that it seems completely bizarre and counterintuitive that people who are moving at different speeds will measure distances and time differently from each other, but in experiment after experiment, it's been shown to be true.What experiment can demonstrate your claim? You didn't claim that experiments show Lorentz contraction, you claimed travellers moving at different speeds will measure distances differently.

Webbo
2012-Jul-13, 01:17 AM
This type of argument shows a very basic lack of understanding about science and observations. It is like saying no one was around a 150 million years ago so their for the Jarassic period is suspect.
We know muons decay very rapidly
We know how they are generated in the upper atmosphere
We know their speeds when the reach the surface of the Earth
We know that given their decay rates that without SR they wouldn't reach the ground before they decay.
We know SR is correct by multiple independent methods to a very high degree of accuracy.
Therefore we can reasonably conclude what is happening to the muons in their frame of reference and that matches observation.You are reasonably concluding that. Grey did not resonably conclude that a moving observer would measure distances differenty, he stated "experiment after experiment" has shown it to be true.

Webbo
2012-Jul-13, 01:24 AM
I suggest you read some history on all the experiments leading up to GPS. Even Sputnik had an element of reletivity experimentation.
Then, I would like you to explain how the measurements made can be done without measuring any kind of movement between two known differences.
I see no way to measure time without involving a known distance, whether that be the length of a pendulum swing, or the difference between waves in an atomic emission.What element of the system experiences length contraction that it needs to account for or correct?

cjameshuff
2012-Jul-13, 03:17 AM
No. Are you implying that the muons had observers travelling with them?

The muons are the observers.



And now you introduce a straw man. I am very specifically only talking about observations of length contraction. What experiments do we have where observers have observed length contraction while travelling?

Once again...the muon observations, among others.

NEOWatcher
2012-Jul-13, 11:53 AM
What element of the system experiences length contraction that it needs to account for or correct?
Did you read the post. I point it out very clearly at the end.

WayneFrancis
2012-Jul-16, 06:04 PM
You are reasonably concluding that. Grey did not resonably conclude that a moving observer would measure distances differenty, he stated "experiment after experiment" has shown it to be true.

Well either length contraction is 'true' or you need to explain how SR comes up with accurate prediction despite half of it being wrong. I mean you might as well look at this 2 + 4 = 6 then say yea that is right but I don't believe in the number 4 and ignore that 2 != 6.

Jeff Root
2012-Jul-17, 01:11 AM
Wayne,

Do you understand what Webbo said in the quote you
just replied to? It doesn't look like it.

Webbo said that you reasonably conclude what is happening
to the muons. I agree. But nobody has ever measured any
length or distance in the muon's frame, and that is the
disagreement. Length contraction has never been observed.
It has only been calculated using relativity theory. In your
math analogy, 2 + 4 = 6 is easy to calculate given 2 and 6,
but 4 has never been observed, only calculated.

-- Jeff, in Minneapolis

WayneFrancis
2012-Jul-17, 03:53 AM
Wayne,

Do you understand what Webbo said in the quote you
just replied to? It doesn't look like it.

Webbo said that you reasonably conclude what is happening
to the muons. I agree. But nobody has ever measured any
length or distance in the muon's frame, and that is the
disagreement. Length contraction has never been observed.
It has only been calculated using relativity theory. In your
math analogy, 2 + 4 = 6 is easy to calculate given 2 and 6,
but 4 has never been observed, only calculated.

-- Jeff, in Minneapolis

I realise that is what webbo said but webbo, and you, go beyond just saying the nobody has ever measured the distances in the muon's frame. Webbo has been saying that if you could and did do measurements that they don't believe the distance would not be length contracted. You keep using arguments from incredulity that we can't even in principal do measurements while in the frames at relativistic speeds. Both of you seemingly trying to raise doubt on something that has more evidence for it then against it.

You've missed my point. If you even if you don't believe in the number 4 and never observed it you have to ask why it fits the equation perfectly. If you want to claim that length contraction isn't real then you HAVE to replace it with something else that would cause us to get the EXACT same results we have observed.

Honestly saying we don't know because we've never been in the muon's frame is on par with creationist saying that the Earth isn't 4.5 billion years old because none of us were around 4.5 billion years ago to see it first hand.

Now if you want to continue on your claim that length contraction isn't real and that you can't measure distances in accelerated frames then back it up with more then hand waving. Give us a valid reason beyond the staw men and hand waving you've done up to this point. Show me I'm wrong. Show me how the maths of SR still produce such accurate predictions and results, in so many cases, if you remove half of the theory.

Length contraction is currently the ONLY answer that fits and makes SR work. Claiming it isn't real without providing a replacement that is just as bad as creationist saying "I don't like evolution! Intelligent design explains it." but constantly refusing to say what ID actually "is"

Jeff Root
2012-Jul-17, 12:28 PM
Wayne,

The problem is that you keep arguing for length contraction.
Over and over and over and over again, you argue for length
contraction.

But I haven't argued against length contraction at all. I've
said several times that I have no disagreement of any kind
with length contraction. I just say that length contraction
due to motion of the observer has never been measured.
As Grey said at the beginning of post #181 (at the top of
this page).

More specifically, I claim that the relativistic traveller in
the thread which spawned this one would not think that the
distance from the Solar System to the target star changed
from 22.3 light-years to about 1 light-year. He would know
that the distance was still 22.3 light-years, even though he
gets there in just one year of his time, in part because he
would understand how special relativity works.

-- Jeff, in Minneapolis

Strange
2012-Jul-17, 12:41 PM
But I haven't argued against length contraction at all. I've
said several times that I have no disagreement of any kind
with length contraction. I just say that length contraction
due to motion of the observer has never been measured.

And when examples of length contraction happening are provided, you say "but it hasn't been directly measured". So what? How is this "no true scotsman" argument relevant? GPS has to take into account the difference in distance observed by the satellite and the receiver due to their relative velocities. But I assume that is not "direct" enough for you.


More specifically, I claim that the relativistic traveller in
the thread which spawned this one would not think that the
distance from the Solar System to the target star changed
from 22.3 light-years to about 1 light-year.

And yet you have provided no reason, evidence or justification of this denial of the basics of relativity.


He would know
that the distance was still 22.3 light-years, even though he
gets there in just one year of his time, in part because he
would understand how special relativity works.

No. He would know, from special relativity, that a different observer would think it was 22.3 ly, even though he sees, measures and knows it as 1 ly.

Jeff Root
2012-Jul-17, 03:49 PM
Strange,

You breathed yesterday.

How could I know that? I don't even know what continent
you are on, or even whether you are on a continent.

Is it a mystery to you how I did it?

No, of course it isn't a mystery. You know that I didn't
observe you breathing yesterday. I deduced that you
breathed yesterday, from other information.

Observing something and deducing it are two entirely
different things.




But I haven't argued against length contraction at all. I've
said several times that I have no disagreement of any kind
with length contraction. I just say that length contraction
due to motion of the observer has never been measured.
And when examples of length contraction happening are
provided, you say "but it hasn't been directly measured".
So what?
So it hasn't been directly measured. It has never been
observed. It has only been deduced from other things
which can be and have been observed.



How is this "no true scotsman" argument relevant?
It isn't an argument. It is a statement. You and others
here have been trying to argue against it. All I have
done is repeated the statement and asked for a better
argument against it. I have not made any argument
relating to it. The statement stands by itself:
Length contraction due to motion of the observer has
never been measured. It has only been deduced from
observations of other things.



GPS has to take into account the difference in distance
observed by the satellite and the receiver due to their
relative velocities. But I assume that is not "direct"
enough for you.
The satellite doesn't observe any distances.

Neither do the ground stations.




More specifically, I claim that the relativistic traveller in
the thread which spawned this one would not think that the
distance from the Solar System to the target star changed
from 22.3 light-years to about 1 light-year.
And yet you have provided no reason, evidence or
justification of this denial of the basics of relativity.
What denial? I didn't deny anything!

:P

I certainly didn't deny any "basics of relativity".




He would know that the distance was still 22.3 light-years,
even though he gets there in just one year of his time, in
part because he would understand how special relativity
works.
No. He would know, from special relativity, that a
different observer would think it was 22.3 ly, even
though he sees, measures and knows it as 1 ly.
He would be smarter than that. He would understand
special relativity better than that. He would know that
the actual distance is 22.3 light-years, and that if he
could measure the distance, he would get a distorted
value. And he would know how to deduce from that
the value that everyone agrees is the correct value.

-- Jeff, in Minneapolis

Strange
2012-Jul-17, 04:20 PM
The satellite doesn't observe any distances.

The GPS receiver triangulates using the broadcast positions of the satellites. These have to be adjusted to take into account the fact that the satellite and the receiver have different frames of reference and therefore observe distances as being different.


What denial? I didn't deny anything!

You deny that someone travelling at significant speed would see length contraction.


He would be smarter than that. He would understand
special relativity better than that. He would know that
the actual distance is 22.3 light-years, and that if he
could measure the distance, he would get a distorted
value. And he would know how to deduce from that
the value that everyone agrees is the correct value.

There is no "actual" distance or "correct" value. There is one value seen from one frame of reference and a different value in another frame of reference. These are equally correct. By insisting that only one of these is correct and real, you are claiming that there is a special (absolute) frame of reference.

Hornblower
2012-Jul-17, 04:37 PM
I don't know what Jeff is trying to argue or not argue as the case may be, and I am not going to wade through his posts throughout the thread to try to find out. We all acknowledge that we have no way of getting a passing spacecraft to fly by fast enough to visually observe the length contraction that relativity theory predicts. We have inferred the occurrence of such contraction from the observations of high-energy particles we are able to observe and evaluate, and we have confidence in the validity of these inferences. The success of relativistic adjustments to the GPS system further strengthen that confidence. The fact that Webbo apparently does not have such confidence is of no adverse consequence to the advancement of physics. Physicists do not need unanimous consent from the public at large to proceed with their work.

korjik
2012-Jul-17, 06:58 PM
I don't know what Jeff is trying to argue or not argue as the case may be, and I am not going to wade through his posts throughout the thread to try to find out. We all acknowledge that we have no way of getting a passing spacecraft to fly by fast enough to visually observe the length contraction that relativity theory predicts. We have inferred the occurrence of such contraction from the observations of high-energy particles we are able to observe and evaluate, and we have confidence in the validity of these inferences. The success of relativistic adjustments to the GPS system further strengthen that confidence. The fact that Webbo apparently does not have such confidence is of no adverse consequence to the advancement of physics. Physicists do not need unanimous consent from the public at large to proceed with their work.

Slight amplification: The behavior we see of high energy particles does not match theory without the length contraction. The particles cannot behave the way they do if they have to travel the lab frame distances. This means we have an observation that cannot work in the lab frame, yet matches the theory nearly exactly.

This also applies to the GPS system. Without the frame corrections mandated by GR, the GPS system would drift to tens of meters of error in weeks. Again, the observation does not match classical theory yet matches what GR says quite well.

Strange
2012-Jul-17, 07:13 PM
This also applies to the GPS system. Without the frame corrections mandated by GR, the GPS system would drift to tens of meters of error in weeks. Again, the observation does not match classical theory yet matches what GR says quite well.

It is not even a matter of drift (which is mainly due to the time dilation due to the different height in the gravitational field) it is the instantaneous corrections that have to be made because of the relative velocity of the satellite and receiver; without this the calculated position would be many meters out.

grapes
2012-Jul-17, 07:47 PM
So it hasn't been directly measured. It has never been
observed. It has only been deduced from other things
which can be and have been observed.

I think this may be the central point to your argument.

However, in the same sense, it is valid to say that no measurements have ever been directly observed. All measurements are deduced, in some fashion. The distance to the moon has not been measured, it has been deduced from time of flight observations. My BMI has not been measured, just deduced from other calculations.

Such an objection can be frustrating, and the general level of discussion here reflects that. We should move off that point.

But measurements are repeated all the time--using the same, as well as different, techniques (and deductions). Which new (old) measurements should be made (repeated)? Why not discuss that?

Strange
2012-Jul-17, 08:46 PM
However, in the same sense, it is valid to say that no measurements have ever been directly observed.

Exactly. That is why it is such an .... <bites tongue> ... "unhelpful" comment. It adds nothing at all to the discussion.

pzkpfw
2012-Jul-17, 10:05 PM
... He would know that
the actual distance is 22.3 light-years ...

Are these two objects that are "actually" 22.3 ly apart, motionless with respect to each other?

Jeff Root
2012-Jul-18, 07:05 AM
Are these two objects that are "actually" 22.3 ly
apart, motionless with respect to each other?
Yes.

I think I made that clear in reply to you in the other
thread, and agreed that if the objects are in relative
motion, then there is no "actual" distance that can
be generally agreed on, because there can be no
agreement that the observations of the positions
of the two objects were simultaneous.

The proper distance between two objects which are
motionless relative to each other will be agreed by all
observers. It is the simplest and easiest measurement,
the only one which is not distorted, requiring correction.

-- Jeff, in Minneapolis

Jeff Root
2012-Jul-18, 07:07 AM
The satellite doesn't observe any distances.
The GPS receiver triangulates using the broadcast
positions of the satellites. These have to be adjusted
to take into account the fact that the satellite and
the receiver have different frames of reference and
therefore observe distances as being different.
It is still the case that neither the GPS satellites nor
the GPS receivers observe any distances.




What denial? I didn't deny anything!
You deny that someone travelling at significant
speed would see length contraction.
Okay, I did deny that. But I didn't deny "the basics of
relativity", which is what I was denying that I denied.

I'm not the least bit sure that the length contraction
couldn't be observed. But I asked korjik over and
over again, many, many times, to suggest a technique
by which it might be done, and he refused every time.
Nobody else suggested anything useful either. Maybe
no useable technique was put forward because no
technique was possible, or at least practical. So that
became my stance. I asserted that measurement of
distances by relativistically-moving observers were so
difficult as to be practically impossible. I never once
claimed that length contraction doesn't happen.




He would be smarter than that. He would understand
special relativity better than that. He would know that
the actual distance is 22.3 light-years, and that if he
could measure the distance, he would get a distorted
value. And he would know how to deduce from that
the value that everyone agrees is the correct value.
There is no "actual" distance or "correct" value.
Sure there is. It is the proper distance. The distance
measured when at rest relative to the objects being
measured, which every observer will agree is correct.



There is one value seen from one frame of reference
and a different value in another frame of reference.
These are equally correct.
It is easy as pie to distinguish the correct value from
an incorrect one.



By insisting that only one of these is correct and real,
Which I do.



you are claiming that there is a special (absolute)
frame of reference.
No way. That doesn't even make sense. I am most
definitely not claiming that there is an absolute frame
of reference of any kind, nor that any one frame is in
any way superior to others in general.

I'm just saying that the proper measurement of a
length or distance gives the correct value, while
a measurement made in motion relative to what is
being measured cannot give the correct value.
Everyone will agree with the results of the proper
measurement. In general, everyone disagrees with
results of measurements made in relative motion.

-- Jeff, in Minneapolis

WayneFrancis
2012-Jul-18, 07:50 AM
Say I have Observer A & B in a nearly empty universe.

Observer A & B believe they are at rest relative to each other. A & B both measure the distance to each other as 1 light minute.

Now This universe has 2 other observers C & D who are also at rest relative to each other. They also measure the distance between themselves as 1 light minute.

Observer A & B measure the distance between C & D and believe that the distance is 22.3 light minutes and that C & D are travelling almost straight at them at .999c

Observer C & D measure the distance between A & B and believe that the distance is 22.3 light minutes and that A & B are travelling almost straight at them at .999c

So who's frame is the correct frame? According to Jeff both sets of observers should have some way to figure out which one of them have the real frame that they both can agree on.

SR says that either frame is valid and the final frame will be dependant on how much each one accelerates.

If A & B don't do anything and C & D accelerate to become at rest with A & B then C & D will see that the distance between them will expand to 22.3 light minutes and that the distance between A & B seems to shrink to just 1 light minute.

If C & D don't do anything and A & Baccelerate to become at rest with C & D then A & B will see that the distance between them will expand to 22.3 light minutes and that the distance between C & D seems to shrink to just 1 light minute.

If A & B and C & D both accelerate at the same rate until they come to rest with respect to each other then do you realise that none of the observers will measure the distance between anyone at 1 light minute or 22.3 light minutes?!?!
Not hard to calculate what they will measure.

korjik
2012-Jul-18, 07:59 AM
It is still the case that neither the GPS satellites nor
the GPS receivers observe any distances.


Okay, I did deny that. But I didn't deny "the basics of
relativity", which is what I was denying that I denied.

I'm not the least bit sure that the length contraction
couldn't be observed. But I asked korjik over and
over again, many, many times, to suggest a technique
by which it might be done, and he refused every time.
Nobody else suggested anything useful either. Maybe
no useable technique was put forward because no
technique was possible, or at least practical. So that
became my stance. I asserted that measurement of
distances by relativistically-moving observers were so
difficult as to be practically impossible. I never once
claimed that length contraction doesn't happen.


Sure there is. It is the proper distance. The distance
measured when at rest relative to the objects being
measured, which every observer will agree is correct.


It is easy as pie to distinguish the correct value from
an incorrect one.


Which I do.


No way. That doesn't even make sense. I am most
definitely not claiming that there is an absolute frame
of reference of any kind, nor that any one frame is in
any way superior to others in general.

I'm just saying that the proper measurement of a
length or distance gives the correct value, while
a measurement made in motion relative to what is
being measured cannot give the correct value.
Everyone will agree with the results of the proper
measurement. In general, everyone disagrees with
results of measurements made in relative motion.

-- Jeff, in Minneapolis

Actually, Korjik repeatedly said that any technique would work. That is slightly different from what you said. It is up to the person who believes there is an error to show the error.

Proper is not real Jeff. You do not measure the proper distance, ever. You measure the real distance, which if the observer isnt moving, is the same as the proper distance.

You are just plain wrong. There have been numerous mentions of observations that cannot work without length contraction. You are also just plain wrong that the proper value is the 'correct' value. Whether you are making the stationary frame special is a bit debatable, because technically making everyone measure the same distances throws all of SR out the window and forces Galliean relativity/transforms which throws frames right out the window.

Jeff Root
2012-Jul-18, 08:45 AM
When I use a tape measure to measure the length of
my hallway, I am making a direct measurment of length.

When some ancient Greek guys found the distance of
the Moon using geometry and Earth's diameter, the angle
of Earth's umbra, the angular diameter of the Moon, and
the size of Earth's shadow on the Moon, they measured
angles, which are just ratios of lengths. For the purposes
of this thread, I'd say they measured the distance. Not
directly, for sure, but they measured distance, not time.

Similarly when measurement of the Moon's parallax
became possible. Although widely-separated observers
on Earth might synchronize their observations, time is
not a factor in the measurement. What is actually
measured are angles and distances. The result is a
measured distance to the Moon, using geometry.
Not direct, but a measurement of length as opposed
to something else.

My objection in the other thread was to korjik's
assertion that the traveller would think the distance
from the Solar System to the distant star was only
about 1 light-year. That is what I disagree with, and
what I would argue against if I knew how. The 1 ly
figure, even if it could actually be measured, is an
obvious distortion of the true distance. That is the
core of my assertion.

-- Jeff, in Minneapolis

Jeff Root
2012-Jul-18, 09:14 AM
I asked korjik over and
over again, many, many times, to suggest a technique
by which it might be done, and he refused every time.
Actually, Korjik repeatedly said that any technique would work.
Yes, and you refused to specify one.



That is slightly different from what you said.
Yes, it is slightly different from what I said, and it is
perfectly true, but what I said is also perfectly true.



It is up to the person who believes there is an error to
show the error.
Yes. In this case I asserted that you were in error in
saying that some technique could be used by a relativistic
traveller to measure the distance between stars. I wanted
you to specify a technique so that we could analyze it.
But you refused, over and over and over. If you actually
knew of a technique that would work, I would think you
would suggest it.



Proper is not real Jeff. You do not measure the proper
distance, ever. You measure the real distance, which if
the observer isnt moving, is the same as the proper
distance.
Does that make sense to anyone?

Does it make sense to you, korjik?



You are just plain wrong.
Then show it.



There have been numerous mentions of observations that
cannot work without length contraction.
Three, actually. Cosmic-ray generated muons, particles
in accelerators, and GPS measurements. However, none
of them involve measurement of distance by an observer
in relativistic motion, which is what is in question.



You are also just plain wrong that the proper value is
the 'correct' value.
If there is a 'correct' value, then it is the proper value.
No other value can claim that status.



Whether you are making the stationary frame special is
a bit debatable,
I'll belabor the point that I'm not talking about a
"stationary frame", which of course is a fiction, but
a frame which is common to observer and observed.



because technically making everyone measure the same
distances throws all of SR out the window and forces
Galliean relativity/transforms which throws frames right
out the window.
I don't believe that for a millisecond.

-- Jeff, in Minneapolis

pzkpfw
2012-Jul-18, 09:15 AM
... The 1 ly
figure, even if it could actually be measured, is an
obvious distortion of the true distance. That is the
core of my assertion.

-- Jeff, in Minneapolis

The core is that you think one of the measures, in one situation, is the "true distance", and the others are "distortions".

Why?

Strange
2012-Jul-18, 09:25 AM
Okay, I did deny that. But I didn't deny "the basics of
relativity", which is what I was denying that I denied.

You deny length contraction would be seen by someone travelling at high speed; that is a basic fact of relativity. You are denying it.


I'm not the least bit sure that the length contraction couldn't be observed.

I used to be doubtful about this; but having recently seen a camera that can film things moving at the speed of light (it is quite spooky to see a flash of light passing slowly through a bottle) the only thing we need to do now is accelerate a macroscopic object to a significant proportion of the speed of light. How hard can that be...

The trouble is, even if we did that and showed something being length contracted, at least some people would say that this wasn't a "direct measurement" (there was a digital camera involved, after all). That is the problem with the "no true scotsman" argument: the person making it can just keep moving the goalposts.

But, as the "direct" (however you choose to define that) measurement of this is totally and utterly irrelevant (even if an interesting engineering challenge) can we just drop it?


But I asked korjik over and
over again, many, many times, to suggest a technique
by which it might be done, and he refused every time.

No. He quite reasonably said you can use any technique that you would use if you were stationary. You waffled that this wouldn't work because the results would be "unceratain" for some unspecified reason.


Nobody else suggested anything useful either.

Wayne [and Grey] produced quite a detailed descriptions of techniques using triangulation and radar. I rather lost interest for a while and didn't see your responses to those.


No way. That doesn't even make sense. I am most
definitely not claiming that there is an absolute frame
of reference of any kind, nor that any one frame is in
any way superior to others in general.

Even though only one of them is correct? So the "correct" frame of reference is not superior to the "incorrect" ones?

The thing is, Jeff, I'm sure you don't mean to sound like a relativity-denier, but you do. You sound as if you are denying length contraction ("but there is no evidence for it ... ok, there is no direct evidence for it"). You sound as if you are insisting on a special frame of reference ("only one is correct"). You seem to ignore arguments you don't like ("no one has shown me how it could be measured"; "OK here are three ways ..."; "no one has shown me how it could be measured").

Jeff Root
2012-Jul-18, 10:57 AM
Say I have Observer A & B in a nearly empty universe.

Observer A & B believe they are at rest relative to each
other. A & B both measure the distance to each other
as 1 light minute.

Now This universe has 2 other observers C & D who are
also at rest relative to each other. They also measure
the distance between themselves as 1 light minute.

Observer A & B measure the distance between C & D
and believe that the distance is 22.3 light minutes and
that C & D are travelling almost straight at them at .999c

Observer C & D measure the distance between A & B
and believe that the distance is 22.3 light minutes and
that A & B are travelling almost straight at them at .999c
You got the numbers backward. The last two measured
distances you refer to need to be shorter, not longer than
the others. The easiest way to fix this would be to make
the proper distance between each pair of observers 22.3
light-minutes, and the distance as measured by the other,
relatively-moving pair, 1 light-minute.



So who's frame is the correct frame? According to Jeff
both sets of observers should have some way to figure
out which one of them have the real frame that they
both can agree on.
Not hardly. I never said or implied anything about a
"real frame". The "correct frame" for a measurement
of a thing is the frame of the thing being measured.

Simple. Obvious. Universally applicable.

All of the observers know that proper measurements
give correct values, while the measurements in relative
motion give incorrect values.

Since you established the distance between each pair
of observers to be the same for both pairs, you limited
potential dramatic effect of your point.



SR says that either frame is valid and the final frame
will be dependant on how much each one accelerates.

If A & B don't do anything and C & D accelerate to become
at rest with A & B then C & D will see that the distance
between them will expand to 22.3 light minutes and that
the distance between A & B seems to shrink to just
1 light minute.
Ouch. At least two mistakes there. First, the same
mistake as earlier: The distance between A & B seems
to increase, not shrink.

Second, the acceleration throws everything out of synch.
You are talking about a really enormous amount of
acceleration. And the distance between C & D is not
trivial. This means that C & D will not be able to stay
a constant distance apart even in their own frame. If
the distance in their own frame can change to anything
(depending on how they choose to synchronize their
acceleration), then the distance in anyone else's frame
is at least equally unpredictable.

Third, I'm not sure that the reason you think "C & D will
see that the distance between them will expand" is the
reason I just gave. You seem to be saying it is due to
the cessation of length contraction. Ouch ouch ouch.



If C & D don't do anything and A & B accelerate to become
at rest with C & D then A & B will see that the distance
between them will expand to 22.3 light minutes and that
the distance between C & D seems to shrink to just
1 light minute.
I agree that the situations are completely symmetric.



If A & B and C & D both accelerate at the same rate until
they come to rest with respect to each other then do you
realise that none of the observers will measure the distance
between anyone at 1 light minute or 22.3 light minutes?!?!
Not hard to calculate what they will measure.
The distances will vary depending on how the accelerations
are synchronized.

Edit to add:

I'm thinking about this aspect a bit more. You still have
the numbers backward, but you may be right in principle
about the change due to acceleration.

-- Jeff, in Minneapolis

Jeff Root
2012-Jul-18, 11:09 AM
The core is that you think one of the measures, in
one situation, is the "true distance", and the others
are "distortions".

Why?
Because they are.

I'm sorry, I understand how pathetic that answer is.
I don't know how to explain what is self-evident.

-- Jeff, in Minneapolis

Grey
2012-Jul-18, 01:43 PM
I'm not the least bit sure that the length contraction
couldn't be observed. But I asked korjik over and
over again, many, many times, to suggest a technique
by which it might be done, and he refused every time.
Nobody else suggested anything useful either. Maybe
no useable technique was put forward because no
technique was possible, or at least practical. So that
became my stance. I asserted that measurement of
distances by relativistically-moving observers were so
difficult as to be practically impossible.I suggested radar, which is a remarkably common method for measuring distances to objects which are too far away to use a ruler, and showed how two observers moving relative to each other would get different results. I also pointed out that we're remarkably close to having the needed sensitivity to be able to see the movement of stellar positions from relativistic effects due to the motion of the Earth. It's far from practically impossible.


But I didn't deny "the basics of
relativity", which is what I was denying that I denied.
...
I'm just saying that the proper measurement of a
length or distance gives the correct value, while
a measurement made in motion relative to what is
being measured cannot give the correct value.This is "denying the basics of relativity". One of the postulates of special relativity is that all inertial reference frames are equally valid. Measurements made in one frame are not preferred over any other. You can't simultaneously claim that this postulate is false and also insist that you aren't "denying the the basics of relativity". (Well, at least you can't be consistent and claim both of those things at once; you are apparently trying to claim both of those things nevertheless).

Hornblower
2012-Jul-18, 02:11 PM
As an exercise in physics, my inclination is simply to perform the triangulation measurements and make the corresponding calculation of the distance to the target, making sure to specify the frame of reference unambiguously. Then as a thought exercise I could do an exact transformation to the frame of reference of another observer who is moving relative to me. I would just let the language of mathematics do the talking and not get bogged down in quibbling with words concerning what is "correct" or "true".

Once again, this thread has become so labored and convoluted that I cannot tell what is being asserted or denied as the case may be, and I really do not care.

Grey
2012-Jul-18, 02:33 PM
Because they are.

I'm sorry, I understand how pathetic that answer is.
I don't know how to explain what is self-evident.I'd say that it is self-evident that there is one correct measurement for any given distance and time. Unfortunately, Einstein discovered that what is "self-evident" turns out to not actually be true! :) In most of physics, I think we've realized that many of the obvious things we take for granted have turned out to be wrong.

NEOWatcher
2012-Jul-18, 02:47 PM
Because they are.
I'm sorry, I understand how pathetic that answer is.
I don't know how to explain what is self-evident.
Perhaps you could tell us how to determine which person is "true" as opposed to the others that are "distorted".

caveman1917
2012-Jul-18, 03:30 PM
Why would you consider proper distance to be "true"? We already have a term to differentiate it from the rest, ie "proper". Isn't that enough?

Also your reasoning falls apart somewhere. You say that all observers will agree on the length of an object in its rest frame, ie its proper length. But the question is, how would those observers, that are themselves not in that rest frame, come to agree upon that? Because they apply the lorentz transformations from their respective frames to the rest frame, and each of them will arrive at the same answer. But that can be said about every frame, not just the rest frame. If some observer A is moving relative to the object, then all observers will also agree on the length as measured by A (as opposed to the rest frame of the object), by doing the exact same thing, applying the lorentz transformations from their respective frames to this observer A's frame. Again all will get the same result and agree upon it.

Strange
2012-Jul-18, 03:39 PM
That's a good point: in Jeff's world it seems that the moving observer cannot measure the distance and therefore cannot use the Lorentz transformation to convert it to the distance that the "stationary" observer would measure. And so they cannot agree on the distance...

korjik
2012-Jul-18, 04:34 PM
Because they are.

I'm sorry, I understand how pathetic that answer is.
I don't know how to explain what is self-evident.

-- Jeff, in Minneapolis

It isnt, You are wrong. You are replacing belief with evidence. You miss the entire point of relativity. The whole reason why it is called relativity as opposed to absloutity.

WayneFrancis
2012-Jul-19, 09:10 AM
I regret I messed up my thought experiments numbers and Jeff focused on that instead on admitting that the basic premiss of the thought experiment was the "correct measurement" would depend completely on how the 4 observers eventually come to rest with respect to each other. And since there was 3 different solution that I can think of that would achieve that end there would be 3 different answers that could be deemed "correct" or more exactly the would be no way to determine which one was "correct" which is the very point of SR.

Jeff Root
2012-Jul-19, 10:32 AM
You deny length contraction would be seen by someone
travelling at high speed; that is a basic fact of relativity.
You are denying it.
That is not a fact of relativity.

Relativity says that length contraction occurs. It does
not say that the contraction will be seen.

I said that I think the measurement is "so difficult as
to be practically impossible." Under the relentless hours
of interrogation I may have sometimes expressed it more
strongly than that, but that is what I meant. And as I
admit in the next quote, even that is too strong. But so
far no evidence has been put forward to refute it.




I'm not the least bit sure that the length contraction
couldn't be observed.
...
The trouble is, even if we did that and showed something
being length contracted, at least some people would say
that this wasn't a "direct measurement" (there was a digital
camera involved, after all). That is the problem with the
"no true scotsman" argument: the person making it can
just keep moving the goalposts.
I haven't moved any goalpost. My original point in the
other thread was that the traveller would not think the
distance between the Solar System and the distant star
would change from 22.3 ly to about 1 ly. In order to try
to meet the objections to that assertion, and based on
all the evidence I had (and still have), I further claimed
that the traveller simply would not be able to measure
the distance, because the conditions would make it
practically impossible.

Since then several ideas have been suggested about
measuring the times of radar pulses, muons or whatnot.
Those are not measurements of distance. They are
measurements of time. Distance can only be derived
from the time measurement through theory-dependent
calculations. Yes, I accept the theory as correct. But
that doesn't mean I won't object when you call a
measurement of time a measurement of distance.



But, as the "direct" (however you choose to define that)
measurement of this is totally and utterly irrelevant (even
if an interesting engineering challenge) can we just drop it?
I don't remember if I'm the one who first used the term
"direct" or if someone else used it first. In any case I
agree that it is irrelevant. A measurement of time is not
a measurement of distance. My objection is that the
traveller would not measure the distance as about 1 ly,
but even if I'm wrong about that, he still certainly would
not think that the distance actually was about 1 ly.




But I asked korjik over and
over again, many, many times, to suggest a technique
by which it might be done, and he refused every time.
No. He quite reasonably said you can use any technique
that you would use if you were stationary. You waffled
that this wouldn't work because the results would be
"unceratain" for some unspecified reason.
He refused to specify technique, over and over and over.

He wants me to choose one. I would choose a technique
which I think will not work. That is unavoidable, because
I have not been able to think of a technique which would
work. So if I were to end up showing that the technique
I picked wouldn't work, korjik and you and everyone else,
even including myself, would complain that I set the
situation up incorrectly-- I used the wrong technique.
So korjik, or you, or some other proponent of the idea
that the distance can be measured, has to specify what
technique should be analyzed. By asking me to choose
a technique, korjik is requiring me to create a strawman
argument. There is no need for that. Just specify a
technique that can be analyzed, and we can go forward.



Wayne [and Grey] produced quite a detailed descriptions
of techniques using triangulation and radar.
Radar measures time. The triangulation possibility was
extremely vague. It wasn't even made clear what was
supposed to be measured. The descriptions were not
detailed at all.

My own reply was not waffling, but it was at least as
vague as Wayne's description of the measurement
technique: I said that the distance measurement would
be difficult or impossible because of the problem of
determining simultaneity. When the two ends of the
tape measure are located is as important as where they
are located, and there will be a huge unresolveable
discrepency in this case.

Grey did give some good descriptions of things, but not
of how the traveller could measure the distance from the
Solar System to the distant star.




I am most definitely not claiming that there is an
absolute frame of reference of any kind, nor that any
one frame is in any way superior to others in general.
Even though only one of them is correct? So the
"correct" frame of reference is not superior to the
"incorrect" ones?
There is no absolute frame of reference. No one frame
of reference is superior to others in general. The two
statements amount to the same thing.

The frame of a thing being measured is the only frame
which allows correct measurement of the thing's length.



The thing is, Jeff, I'm sure you don't mean to sound like
a relativity-denier, but you do. You sound as if you
are denying length contraction ...
This is what I said to korjik back on June 24:



I am not saying there is any problem with the
observations or the theory. I am saying there is a
problem with your description.



... ("but there is no evidence for it ... ok, there is no
direct evidence for it"). You sound as if you are
insisting on a special frame of reference ("only one is correct").
Only a measurement made in an object's frame can give
a correct length measurement of that object. If the
object's length is changing, then no measurement of
its length can be correct because there can be no
general agreement on the simultaneity of the two
endpoint measurements.



You seem to ignore arguments you don't like
("no one has shown me how it could be measured";
"OK here are three ways ..."; "no one has shown me
how it could be measured").
No one has shown me how the distance between the
Solar System and a distant star could be measured by
a relativistic traveller.
"Okay, here is a way that the half-life of relativistic
muons is measured, a way that light travel times
might be measured given a magical radar, and a way
involving triangulation that might work if I could say
exactly what it was."

-- Jeff, in Minneapolis

Strange
2012-Jul-19, 10:49 AM
I haven't moved any goalpost.

I very carefully worded my comment to avoid saying that you had moved any goal posts.


The frame of a thing being measured is the only frame
which allows correct measurement of the thing's length.

So you are simply using the word "correct" as a synonym for "proper" rather than "true" or "right"?


a way involving triangulation that might work if I could say exactly what it was.

I am curious why you think it works from earth but not from a moving spaceship ...

Jeff Root
2012-Jul-19, 11:48 AM
I suggested radar, which is a remarkably common
method for measuring distances to objects which
are too far away to use a ruler, and showed how
two observers moving relative to each other would
get different results.
As I said to Strange, radar measures time, not dstance.
In mundane situations, where relative speeds are low,
such as traffic monitoring or detecting ICBMs, that is
not a problem. At the very high speed of the scenario
in question, it is.



I also pointed out that we're remarkably close to having
the needed sensitivity to be able to see the movement
of stellar positions from relativistic effects due to the
motion of the Earth. It's far from practically impossible.
I agree that we may be close to having the sensitivity
needed to see the movement of stellar positions due to
a relativistic effect of the motion of the Earth. I do not
agree that the effect is a change in distance.



This is "denying the basics of relativity". One of the
postulates of special relativity is that all inertial reference
frames are equally valid. Measurements made in one frame
are not preferred over any other. You can't simultaneously
claim that this postulate is false and also insist that you
aren't "denying the the basics of relativity".
I agree completely with the postulate. What I disagree
with is your interpretation of it.

All inertial reference frames are equally valid. All inertial
reference frames equally allow observers in those frames
to accurately measure the lengths of objects in their own
frame. All inertial reference frames equally require that
observers in those frames will be unable to accurately
measure the lengths of objects moving relative to their
own frame.

Special relativity has no postulate or equation stating
that an observer can accurately measure the lengths of
things moving relative to him.

Can you pet a kitten without killing it?

How about if the relative speed between you and the
kitten is .999 c ? Yet you can safely pet a kitten
that you take with you aboard your spaceship nomatter
what your speed relative to anything else. Apparently
"all inertial reference frames are equally valid" doesn't
mean that what is possible in one frame is also possible
between two frames. It just means that it is possible
in any frame.

-- Jeff, in Minneapolis

Jeff Root
2012-Jul-19, 12:00 PM
Once again, this thread has become so labored and
convoluted that I cannot tell what is being asserted
or denied ...
I'm denying that a relativistic traveller would think
that everything outside his spaceship shrank because
his speed changed.

Everyone else seems to be stuck in interpreting that
as denying the basics of relativity.

-- Jeff, in Minneapolis

Jeff Root
2012-Jul-19, 12:09 PM
Perhaps you could tell us how to determine which
person is "true" as opposed to the others that are
"distorted".
I think I've said about a dozen times, now:

An observer not in motion relative to the thing
being measured can get an accurate measurement
of true length or distance. An observer in motion
relative to the thing being measured will always
get a distorted measurement.

-- Jeff, in Minneapolis

Jeff Root
2012-Jul-19, 01:08 PM
I regret I messed up my thought experiments numbers
and Jeff focused on that instead on admitting that the
basic premiss of the thought experiment was the
"correct measurement" would depend completely on
how the 4 observers eventually come to rest with
respect to each other. And since there was 3 different
solution that I can think of that would achieve that
end there would be 3 different answers that could be
deemed "correct" or more exactly the would be no way
to determine which one was "correct" which is the
very point of SR.
Can we try again? That bit at the end surprised me,
and I think you were basically right: The two ships
travelling together, one ahead of the other, would
see the distance between them increase when they
match speed with the other pair of ships. I thought
at first that they'd stay the same distance apart.

That is part of the simultaneity problem which I say
would make distance measurements so difficult as
to be practically impossible.

Please do try again.

-- Jeff, in Minneapolis

korjik
2012-Jul-19, 04:30 PM
I regret I messed up my thought experiments numbers and Jeff focused on that instead on admitting that the basic premiss of the thought experiment was the "correct measurement" would depend completely on how the 4 observers eventually come to rest with respect to each other. And since there was 3 different solution that I can think of that would achieve that end there would be 3 different answers that could be deemed "correct" or more exactly the would be no way to determine which one was "correct" which is the very point of SR.

He isnt interested in what SR actually says, or what evidence there is for it. That is why every time someone tries to give a concrete example, he ignores it, picks at an irrelevant error, or give a non-sequitur reason why it would not work.

Hornblower
2012-Jul-19, 04:41 PM
Here is another take on this topic with a thought exercise. Suppose we have two identical spacecraft, A and B. A is on the ground and B flies by just over it at nearly the speed of light, while we observe from off to the side. With the camera and flash gun of our dreams we take a picture at just the right instant, as B is directly over A and at the same distance from us. From this picture we can get a visual measurement of the apparent length of B in comparison with A. In principle we get an exact visual observation of B's length contraction and can compare it with what is predicted by the SR theory. We also can make an exact calculation of what B's crew would see, which should be the appearance that A is length-contracted. In principle we could ask them what they saw and compare their report with our calculations.

What sort of uncertainties or inaccuracies does anyone think I am missing in my remarks?

Strange
2012-Jul-19, 04:58 PM
Relativity says that length contraction occurs. It does not say that the contraction will be seen.

But there is absolutely no reason to think it won't be (other than your personal beliefs).

Strange
2012-Jul-19, 04:59 PM
I'm denying that a relativistic traveller would think that everything outside his spaceship shrank because his speed changed.

Everyone else seems to be stuck in interpreting that as denying the basics of relativity.

Perhaps that is because it is. Unless you can show, using the mathematics of relativity, that he wouldn't?

korjik
2012-Jul-19, 05:20 PM
Here is another take on this topic with a thought exercise. Suppose we have two identical spacecraft, A and B. A is on the ground and B flies by just over it at nearly the speed of light, while we observe from off to the side. With the camera and flash gun of our dreams we take a picture at just the right instant, as B is directly over A and at the same distance from us. From this picture we can get a visual measurement of the apparent length of B in comparison with A. In principle we get an exact visual observation of B's length contraction and can compare it with what is predicted by the SR theory. We also can make an exact calculation of what B's crew would see, which should be the appearance that A is length-contracted. In principle we could ask them what they saw and compare their report with our calculations.

What sort of uncertainties or inaccuracies does anyone think I am missing in my remarks?

lets see: B would have to be travelling through the atmosphere at close to the speed of light, so it could not happen. The Flash gun of your dreams would have to be too expensive so it would not actually get built. B's crew would be too busy hanging on for dear life to actually use the radio, and the redshift would make it too hard to use anyway.

At least, that is the quality of the argument so far....

Hornblower
2012-Jul-19, 05:49 PM
lets see: B would have to be travelling through the atmosphere at close to the speed of light, so it could not happen. The Flash gun of your dreams would have to be too expensive so it would not actually get built. B's crew would be too busy hanging on for dear life to actually use the radio, and the redshift would make it too hard to use anyway.

At least, that is the quality of the argument so far....

This is a thought exercise with no atmosphere or other mechanical impediments, which I perhaps did not make clear. Cost of the flash is not an issue. I am not asking about real world technical or cost issues. Let's put us in free fall out in space alongside A, while B flies by. We in principle have the means to measure B's relative speed and apparent length exactly. As I think I understand the theory that should make it possible, in principle, to calculate exactly what B's crew would see. Does anyone think I still am missing something that would introduce and inherent uncertainty?

tusenfem
2012-Jul-19, 08:07 PM
okay I think this has been going round in circles long enough, there is no added value to continuing this thread.
I have the impression that Jeff Root is somehow arguing for a perfect rest frame where everything should be measured true. There is no such frame in relativity.
If you would like to discuss this then take it to ATM.

As usual, if there are any good reasons why this should be re-opened, report this post.