Creating a Target with 3 Photons in SR Propagating Sphere of Light

[Dale]

I place my right angle tubes over any two photons moving at right angles to each other.

In which frame are the two photons moving at right angles to each other?

 

[Dale]

Are you really saying all photons will return to the start -event point of moving frame tubes after your frame as moved on during sphere propagation.

You mean, e.g. if you have a mirror at the end of each tube? Then yes, I am really saying that.

The "right angle" photons in any frame will meet back in the middle of the tubes at exactly the same time as the "forward" and "backward" photons. This is exactly what was proven by the Michelson Morely experiment.

Your idea has been experimentally shown to be incorrect for more than 100 years now.

 

[Reff]

Hi DaleSpam
Originally Posted by Reff
Are you really saying all photons will return to the start -event point of moving frame tubes after your frame as moved on during sphere propagation.

You mean, e.g. if you have a mirror at the end of each tube? Then yes, I am really saying that
Yes I can see part of that but the right angle photon in your moving frame has not done a 180 degree reversal, it is reflected back in a state of constant tube intersection The true reflected angle would be seen doing my geometry.
I am saying a frame cannot drag the event point with it. It is instantly independent of propagation of photons which are all on their own heading-- direction. I locate myself central to a sphere of photons and remain there inertialy now pass by me with a frame of any speed and create an event adjacent to me and others in sequence from as many frames as you wish right at that same point. All 180 degree photons will return to that same point and all the moving frames have gone.

What are your thoughts on two at rest frames as I described in a previous post, not ever being able to meet. I further qualify them in this post.

You say
The "right angle" photons in any frame will meet back in the middle of the tubes at exactly the same time as the "forward" and "backward" photons. This is exactly what was proven by the Michelson Morely experiment.
You say
Your idea has been experimentally shown to be incorrect for more than 100 years now.

I still don't think you have quite grasped my geometry.

c is c is c whatever the frame speed and photons will sit in a same heading sequence whatever speed or direction the frames were doing when the photons were generated.


You ask
In which frame are the two photons moving at right angles to each other?

During propagation of photons I take two photons moving at right angles to each other. whilst remaining centered on the sphere and place my right angle tubes over those specific photons and allow them to pass through without touching the inside of the tubes. Right angle photons and right angle tubes.
In any moving frame one photon at right angles to the direction of travel must be substituted for another photon moving at less than 90 degrees to the directional photon, thus a time dilated frame. That is a big difference. My frame is not time dilated and cannot move to a similar frame, All moving frames can meet any other on an intersecting course.
Take a moving frame with a right angle tube aimed at a distant galaxy and allow a photon to pass through. will it hit the distant galaxy or travel on a line from my marked event through the point it exits the tube which is actualy that photons sphere edge. There is a big difference.

 

[Dale]

It would be nice if you would use the quote feature to more easily identify what you are quoting from whom.

but the right angle photon in your moving frame has not done a 180 degree reversal

It has done a 180º reversal, in the moving frame.

The true reflected angle would be seen doing my geometry.

What makes the angle in one frame the "true" angle? What experiment can be done to distinguish the "true" angle in one frame from the non-"true" angles in another frame? I mean, if you simply are allowed to pick the stationary frame as the "true" frame why cannot I pick the moving frame as the "true" frame. In which case the right angle photon in the stationary frame has not done a "true" 180º reversal.

I still don't think you have quite grasped my geometry.

Then draw a picture as I requested so long ago. I have read your confusing and disjointed descriptions carefully, and I think that I have grasped your geometry now. If you doubt it then it is up to you to communicate more clearly.

During propagation of photons I take two photons moving at right angles to each other.

In the stationary frame.

In any moving frame one photon at right angles to the direction of travel must be substituted for another photon moving at less than 90 degrees to the directional photon

Incorrect. In the moving frame the other photon is moving at 90º. It is only moving at less than 90º in the stationary frame.

That is a big difference. My frame is not time dilated

I have news for you. Your frame is time dilated according to the moving frame. If you do not understand that then you do not understand one of the most basic parts of relativity.

Take a moving frame with a right angle tube aimed at a distant galaxy and allow a photon to pass through. will it hit the distant galaxy or travel on a line from my marked event through the point it exits the tube which is actualy that photons sphere edge. There is a big difference.

There is no difference. Do the same thing from the moving frame and you get the same result.

 

[Reff]

It would be nice if you would use the quote feature to more easily identify what you are quoting from whom.

Hope this works.

It has done a 180º reversal, in the moving frame.

What makes the angle in one frame the "true" angle? What experiment can be done to distinguish the "true" angle in one frame from the non-"true" angles in another frame? I mean, if you simply are allowed to pick the stationary frame as the "true" frame why cannot I pick the moving frame as the "true" frame. In which case the right angle photon in the stationary frame has not done a "true" 180º reversal.

I think this is another step in the right direction.
And I believe I can answer this constructively. I believe we have to determine what is the true direction-heading of a photon in both frames.
Take the moving frame tube pointing in the direction of travel--The photon within it is on a true heading. Where the tube is pointing, the photon will go, confirmed by the photons collision with a distant object and the tube still sighting the photon up to the collision.
True heading works fine for me.
Now a slight variation slightly reversed engineered. Consider a photons collision at a distant-- galaxy considered Stationary like the Earth is for a dilation exercise. Rewind the exercise and use a taught line from the collision point out to that photon and well past it. Can we say that the taught line scribes the photons heading-- its direction-- its true heading. Can we also say that somewhere on the line past that photon from the galaxy, there must have occurred the event which created the photon.
Now we look at the moving frame with your right angle tube and the photon moving within it.
If we consider the True heading of your photon I am saying it does not perform a 180 degree reversal from a mirror at the end of the tube. It has two vectors. One moving through the tube at right angles and one moving fwd with the frame. Within the expandng
sphere. the event point-- start point at the confluence of the tubes has moved so as I have said before, in a sphere of 200mm the moving frame tube confluence at .8c has moved 80mm in the direction of travel. The photon in the tube will have crossed 60 mm. Now I am saying that the true heading of the photon is on a taught line passing through the right angle 60mm point and through a point 80 mm behind the progress of the directional tube.

And yes you are somewhat correct when you say

Then draw a picture as I requested so long ago. I have read your confusing and disjointed descriptions carefully, and I think that I have grasped your geometry now. If you doubt it then it is up to you to communicate more clearly.

perhaps I will have a go at posting a drawing

The rest of your post mmm

I believe the true heading of a photon and not the apparent heading is the key
your moving r angle tube is not sighted at the target before it is hit unlike the directional tube. Both the at rest tubes are sighted on targets which will be hit. I am still happy with that.

 

[Dale]

Hope this works.

That worked reasonably well, thanks. It makes your posts much more readable.

I think this is another step in the right direction.
And I believe I can answer this constructively. I believe we have to determine what is the true direction-heading of a photon in both frames.
Take the moving frame tube pointing in the direction of travel--The photon within it is on a true heading. Where the tube is pointing, the photon will go, confirmed by the photons collision with a distant object and the tube still sighting the photon up to the collision.
True heading works fine for me.

Excellent description. However, it applies for the right-angle moving tubes in the moving frame also. I.e. for the right-angle tube in the moving frame "Where the tube is pointing, the photon will go, confirmed by the photons collision with a distant object and the tube still sighting the photon up to the collision.". Therefore, in the moving frame the right-angle photon's "true direction-heading" is 90º.

Just as the direction of travel is frame variant, so is where the tube is pointing. This is known as relativistic abberation, and has been experimentally observed.

Now we look at the moving frame with your right angle tube and the photon moving within it.
If we consider the True heading of your photon I am saying it does not perform a 180 degree reversal from a mirror at the end of the tube.

This is not correct. In the moving frame it does perform a "true direction-heading" 180º reversal. It is only in other frames that it does not. Again, this is confirmed by the Michelson Morely experiment more than 100 years ago.

I believe the true heading of a photon and not the apparent heading is the key
your moving r angle tube is not sighted at the target before it is hit unlike the directional tube. Both the at rest tubes are sighted on targets which will be hit. I am still happy with that.

Sure, but "true direction-heading" is frame variant. Simply sticking the word "true" in front of something does not make it frame invariant.

 

[Reff]

Excellent description.(((Thank you ))) However, it applies for the right-angle moving tubes in the moving frame also. I.e. for the right-angle tube in the moving frame "Where the tube is pointing, the photon will go, confirmed by the photons collision with a distant object and the tube still sighting the photon up to the collision.". Therefore, in the moving frame the right-angle photon's "true direction-heading" is 90º.

Ok yes I absolutely agree with your above description because the photon is sighted within the tube right up to the collision. The big difference here is that as the photon in the moving frame exits the r angle tube and we watch it on its journey the background it is heading for is moving due to frame speed. Look through the directionaly aligned tube and the photon is steady to the background. Now consider my rest frame and the background-- or should I say --distant target-- is not moving relative to the photon moving away from me in both tubes. Can you see my logic there.
I further elaborate in this way.
A simple sphere is generated with zero time emission and propagates.
The moment of creation of the event, the sphere is absolutely independent of the frame it leaves and it propagates to the laws of c . Perhaps you would agree that absolutely no velocity change to any of the photons is given by the frame direction or frame velocity.
As a sphere, every single photon has moved the same amount from the event irrespective of where the event generator frame is at that time. . Mark the sphere as it propagates to identify each photon. Allow the photons to travel on till they individualy colide with anything at all. Set a tight line 180 degrees instantly from their collision points till the tight lines begin to cross at the original event point. That point is my at rest point. All photons will go back to that point and all move at c. All the headings are true headings

Just as the direction of travel is frame variant, so is where the tube is pointing. This is known as relativistic abberation, and has been experimentally observed

.

Your background as I have said is moving relative to the photon on first looking down your right angle tube right after your photon has left-- what you see is not what you are going to hit (Barring of course moving an item to specificaly colide)

This is not correct. In the moving frame it does perform a "true direction-heading" 180º reversal. It is only in other frames that it does not. Again, this is confirmed by the Michelson Morely experiment more than 100 years ago.

Sure, but "true direction-heading" is frame variant. Simply sticking the word "true" in front of something does not make it frame invariant

I have a couple of diagrams of my geometry where I use a couple of examples from David Darlings website. He has used the correct formula and I have used plain geometry and taken the moving frames time dilation figure directly off the drawing with a ruler. The time dilation figure is the intersecting speed of the photon moving over the tabletop. The photon like all the others in the sphere moves at c and its true heading is the hypotenuse. At no time all the photons done anything different than all the others. The true heading of the hypotenuse photon will go on to hit the same target as the moving frame r angle photon. The big difference is the background of the hypotenuse target is not moving. The moving frame observer is measuring the intersection speed-- not the photon speed and he has no option but to find it is moving at c because his clock must run at its subsequent time dilated speed as per the rules of c.
To further see how the intersection occurs .8c frame movement is easy. create the event it is at the start of the tube. Expand the event to 1 cm. The hypotenuse has formed to one photon on the tabletop or in the tube because frame movement and propagation have both moved now go to two cm-- the same photon is still at the tube-table top-- keep on expanding the sphere and move the tabletop-- the very same photon is at the tabletop. At one frame speed, it is always the same photon. At no time has the photon touched the tabletop right from its own event-- That photon is obeying its own laws and is moving precisely the same as all the other photons in the sphere apart from all their true headings.
To further confirm that photons true heading-- let it colide with anything and with the benefit of instantaneous communication-- reverse the true headings of all the photons and they all meet together again. The background against the hypotenuse photon does not move---so its a true heading and not an intersecting heading. I did not simply stick the word true in front of it.

Just go back a little to a moving frame with a mirror and consider how I qualified a true heading as being the one on the hypotenuse. Then the mirror on the end of the r angle tube.
will reflect the photon precisely back on another intersecting heading with the inside of the tube, which on a .8c frame is not a 180 degree reversal. Yes of course the observer will swear it is the photons true course as it returns back down his moving r angle tube.
I will figure out how to send my drawings soon.

 

[Dale]

Can you see my logic there.

Yes, I fully understood your logic. What you continue to not understand is that there is no difference between frames in any of your logic. It applies equally to any frame.

The big difference is the background of the hypotenuse target is not moving.

All you are doing is determining your velocity wrt the target. Pick a different target, moving wrt the first, and do this experiment and then the moving tube has a 90º "true direction-heading" and the stationary tube does not. If this is what you consider a "big difference" then that is fine, but it is not what anyone else is talking about when they say "absolute frame".

 

[Reff]

DaleSpam;3434839]Yes, I fully understood your logic. What you continue to not understand is that there is no difference between frames in any of your logic.

It applies equally to any frameAll you are doing is determining your velocity wrt the target

.

I said this in my post
Your background as I have said is moving relative to the photon on first looking down your right angle tube right after your photon has left-- what you see is not what you are going to hit (Barring of course moving an item to specificaly colide) You are moving the background target for convenience.



For sure I am clear in my mind that the intersecting photon moves through the tube and it will not go to the aimed point of the right angle moving tube. The background targetis moving and is not in the sights of the tube till the photon hits.
I am saying that every single photon in the universe is radialy emitted and does not move in a combined double vector just one single radial vector. You are looking at a constant intersection which does not cross at c.

Pick a different target, moving wrt the first, and do this experiment and then the moving tube has a 90º "true direction-heading" and the stationary tube does not. If this is what you consider a "big difference" then that is fine, but it is not what anyone else is talking about when they say "absolute frame

".[/QUOTE]

For a start, I am talking of an absolute rest frame, I made another regretable concession when I talked about an absolute frame. I should stick to my own wording. Both absolute rest tubes are aimed at a distant object-- (barring being moved for convenience)
The photons will hit those targets.
One of the moving frames will stay aimed at its target which it hits and no, the right angle moving tube does not point to its target till the photon hits it. You can move the target if you wish but you would be missing the point.

 

[Dale]

You are moving the background target for convenience.

It isn't a question of convenience, it is a question of knowledge. How do you know if a given target is stationary or moving. If there are two distant targets which are moving wrt each other how do you know which one (if any) to pick?

For a start, I am talking of an absolute rest frame

Not unless you can think of some way of experimentally identifying a target which is at rest in the absolute rest frame. Unless you can do that you are merely talking of the target's rest frame, which is no more privileged than any other frame.

 

[Reff]

Hi DaleSpam

It isn't a question of convenience, it is a question of knowledge. How do you know if a given target is stationary or moving. If there are two distant targets which are moving wrt each other how do you know which one (if any) to pick?

The Earth is used as far as I know as a stationary example so at .8c using it as a target it would be classed as background moving with respect to the aim point. If it was a frame of how I explain absolute rest then that would be background moving whilst sighting down the tube. I believe we are near a yes it is, no it isn't scenario but perhaps I can do better with another example.

Not unless you can think of some way of experimentally identifying a target which is at rest in the absolute rest frame. Unless you can do that you are merely talking of the target's rest frame, which is no more privileged than any other frame.

Well that is an interesting thought but it is also worth considering that over recent histry re Einstein, there were a series of experiments after his claims which proved him correct.
Perhaps the first step is to prove experimentally that it does not exist. I would be interested in steps in that direction but physics in all frames being the same (virtualy) ie Michelson
Morley?(spelling) With later experiments-- isn't there a minor consistent confirmation of variation. Sagnac perhaps is trying to tell us something- How close to a zero time photon emission are we-- Particle colider impact emissions. Interference paterns.

How about another example.
You are at .999c and you create a sphere of photons and your right angle tube is about 10ft long. As your tube photon exits you create another event at the start of the tube.
Now I understand you to point to the start of the tube when I ask you where did the photon start from. If that is so I would ask you to step outside ond view the complete picture.
We see the first event sphere having expanded in perfect symetry and we locate the center of the sphere,( where all the photon reciprocals cross.) Now we look at your second event or marker event where you tell me the first event started from. The second sphere is is not centered on the first. The tube start point is now way out close to the edge of the sphere at .999c.
If you are talking about a beam of light, that would be another story.
All photons on the sphere are either there or not. They have the ability to prove their existence by an exchange of energy. They have all traveled the same distance--- irrespective of the frame they have left. They all have their own heading--direction. They cannot be in two places at the same time. They cannot have scribed two headings in our view of the sphere. Dont you agree that on absolutely any frame speed and direction of a frame passing through the center of a sphere and creating an event at the center of the sphere, that every subsequent sphere is centered on the first because it is zero time generated.
Now what would you call the inertial marker point of the expanding spheres. I would call it absolute rest. There is absolutely no time dilation at that point. Absolute time would work for me.
All photons within all the spheres move at c from their respective events so consider the geometry which keeps the tube photon in a state of constant intersection at less than c which can only be measured at c by the frames clock which is absolutely speed variable and has no option but to measure c in its frame.
I still don't see any laws I have broken to date.
Il try to post the geometry.

 

[Reff]

Hi Dalespam
I hope this works

The hypotenuse photon has moved the same distance as any other photon in the sphere in both examples. The table intersection speed is less than c and so its clock must be dilated to always make that speed c. A slow photon must always be measured by a the slow clock at c. Invarience is fine for me.
All photons move at c but not the intersection speed
[PLAIN]http://http://i1190.photobucket.com/albums/z460/Tapapakanga/photon2.jpg
[PLAIN]http://http://i1190.photobucket.com/albums/z460/Tapapakanga/Photon.jpg
Drawn absolutely to scale and time dilation measured with a ruler to show it is so. Pythagarus is more precise of course.

 

[Dale]

The Earth is used as far as I know as a stationary example so at .8c using it as a target it would be classed as background moving with respect to the aim point. If it was a frame of how I explain absolute rest then that would be background moving whilst sighting down the tube.

Sorry, your wording is confusing. Are you saying that you are picking the Earth as your stationary target? If so then you are merely measuring speed relative to the earth. The idea of the Earth being an absolute rest frame has been dismissed since the days of Copernicus.

perhaps I can do better with another example.

There is no reason to go to another example when you have not addressed this key problem of how to determine if a target is stationary in the absolute rest frame.

 

[Dale]

The hypotenuse photon has moved the same distance as any other photon in the sphere in both examples. The table intersection speed is less than c and so its clock must be dilated to always make that speed c. A slow photon must always be measured by a the slow clock at c. Invarience is fine for me.
All photons move at c but not the intersection speed

Drawn absolutely to scale and time dilation measured with a ruler to show it is so. Pythagarus is more precise of course.

Thanks for the drawing. This is exactly the geometry I had understood, so we can be sure that there is no misunderstanding of the geometry now.

So my previous objections all hold.
1) a sphere of photons centered on the origin in one frame is a sphere of photons centered on the origin for all frames.
2) the heading of each individual depends on the frame chosen, so the photon going down the moving right-angle tube is a 90º photon in the moving frame.
3) determining if a photon hits a distant target and drawing a line back from that distant target only determines your speed relative to that target.
4) there is no experimental way to determine if your target is stationary in the absolute rest frame.

 

[Reff]

Firstly, I think you may have corrected a mistake re the posting of my geometry DaleSpam.
If so , thank you.

Sorry, your wording is confusing. Are you saying that you are picking the Earth as your stationary target? If so then you are merely measuring speed relative to the earth. The idea of the Earth being an absolute rest frame has been dismissed since the days of Copernicus.

No I am not but I am using the Earth as the best of a set of bad examples re an unchallenged quote from Drakkith post 28

There is no absolute frame that we can say is at rest. To keep things simple we usually refer to the frame of the Earth as being at rest, but it is not.

There is no reason to go to another example when you have not addressed this key problem of how to determine if a target is stationary in the absolute rest frame

.

That is a little unfair ---But---no problem.
As I have said before. I observe the center of a zero time event and many other frames that pass through it also creating an event concentricaly to the first sphere. They all expand concentricaly to the first irrespective of the speed or direction of their frames, because they are all zero time events. This is the geometry I am interested in. This is the point of absolute rest. I now fabricate a frame of tubes as many as you wish- all pointing radialy away from the center of the frame. I place this frame over the events concentric center. It will be inertialialy located at that point. Observing the departing photons I see at least one in each tube move into the distance-- they will all remain in sight of their tubes. Now consider the background view. is it moving relative to the path of the photon- then it is moving relative to the absolute rest frame. If the background is not moving we could have one of three scenarios, One- it is moving towards us. Two it is moving away from us or three it is also at absolute rest. Time would resolve the first two eventualy. Do the same exercise with two initial events ten miles apart just for an example. A whole set of concentric expanding spheres at one end and a whole set of concentric spheres ten miles away. Mark each concentric point. The marks will stay 10 miles apart. They will never meet, because they are at absolute rest. Absolutely any frame able to cross the gap between the two, will be time dilated. At rest clocks must run faster than any other able to move to it. The key in all this, is to remain at the origin point of every single photon in all the spheres. All their photon headings return to that point.
In either of these two frames you can ask me where did all the photons originate from and to prove it, a fresh definitive event would also be proven to expand concentricaly with whichever example you chose.
I suppose it is fair you address this

There is no reason to go to another example when you have not addressed this key problem of how to determine if a target is stationary in the absolute rest frame.

What I am saying here is you have not addressed this key problem of how to determine the origin of the right angle photon in a moving frame. addressed by my next example. Are you really saying that a second event, created to determine the point of origin will be concentricly located on the first event. I believe a rabbit and a hat would be useful here.

You are at .999c and you create a sphere of photons and your right angle tube is about 10ft long. As your tube photon exits you create another event at the start of the tube.
Now I understand you to point to the start of the tube when I ask you where did the photon start from. If that is so I would ask you to step outside ond view the complete picture.
We see the first event sphere having expanded in perfect symetry and we locate the center of the sphere,( where all the photon reciprocals cross.) Now we look at your second event or marker event where you tell me the first event started from. The second sphere is is not centered on the first. The tube start point is now way out close to the edge of the sphere at .999c.
If you are talking about a beam of light, that would be another story, but we are not.
All photons on the sphere are either there or not. They have the ability to prove their existence by an exchange of energy. They have all traveled the same distance--- irrespective of the frame they have left. They all have their own heading--direction. They cannot be in two places at the same time. They cannot have scribed two headings in our view of the sphere. Dont you agree that on absolutely any frame speed and direction of a frame passing through the center of a sphere and creating an event at the center of the sphere, that every subsequent sphere is centered on the first because it is zero time generated.
Now what would you call the inertial marker point of the expanding spheres. I would call it absolute rest. There is absolutely no time dilation at that point. Absolute time and rest would work for me.
All photons within all the spheres move at c from their respective events so consider the geometry which keeps the tube photon in a state of constant "intersection" at less than c which can only be measured at c by the frames clock which is absolutely speed regulated and has no option but to measure c in its frame.
I still don't see any laws I have broken to date.

 

[Dale]

No I am not but I am using the Earth as the best of a set of bad examples re an unchallenged quote from Drakkith post 28

So the question remains. How do you experimentally determine if your target is at absolute rest. If you cannot do that then you are merely measuring velocity relative to the target.

I observe the center of a zero time event and many other frames that pass through it also creating an event concentricaly to the first sphere. They all expand concentricaly to the first irrespective of the speed or direction of their frames, because they are all zero time events. This is the geometry I am interested in. This is the point of absolute rest.

See post 7 where I proved that this geometry is the same in every frame, so by this geometry every frame is absolute rest.

 

[Reff]

H DaleSpam

So the question remains. How do you experimentally determine if your target is at absolute rest. If you cannot do that then you are merely measuring velocity relative to the target.

See post 7 where I proved that this geometry is the same in every frame, so by this geometry every frame is absolute rest.

I believe we do have a major difference here and that is at rest being concentricaly located to a sphere of photons from a zero time event .When you say every frame is absolute rest you have changed the frame dependence to suit. Let me explain a little more.

I believe we have established that a sphere of photons from a zero time event is---Not frame dependent---
Now when I ask you where the right angle photon originated from in a moving frame, you then make it---- frame dependent---.

I believe you cannot have it both ways to suit.

If the r angle photon in a moving frame is --- not frame dependent--- Then it must have originated from the center of the sphere like all the others from the same event, which makes the right angle photon in conflict with the right angle in that it is in a state of constant intersection with that right angle and therefore that moving intersection is -- not moving at c-- but like all the the photons in the sphere, it is moving at c from its non frame dependent event.

Staying concentricaly located with the event, there is no intersecting trajectory photon therefore we have absolutely --no-- time dilation for that frame. It is absolute rest.

I still invite you to indicate where the r angle photon started from in a .999c frame by using a marker event when we know the photon sphere is not frame dependent. It is only a beam or pulse of light which is frame dependent.
I believe a previous post described present belief as "wierd but that's how it is" A rabbit and a hat sounds more like it.

 

[Dale]

I believe we have established that a sphere of photons from a zero time event is---Not frame dependent---

It depends what you mean by this sentence. All frames agree that a sphere of photons centered on the origin in one frame is a sphere of photons centered on the origin in all frames. So that fact is not frame dependent.

However the spatial location of the origin at any t' \ne t \ne 0 is frame dependent. So, all frames agree that the sphere of photons centered on the origin is a sphere centered on the origin, but they disagree where the origin is except at t'=t=0, the zero-time event of the emission of the photons.

Now when I ask you where the right angle photon originated from in a moving frame, you then make it---- frame dependent---.

I believe you cannot have it both ways to suit.

Pay attention to what I actually said. I said that the direction something travels is frame dependent. The direction that something travels and the point that it travels from are two entirely different questions and may have two entirely different answers without changing the rules (the Lorentz transform).

Your complaint is like saying "you said 2+2=4 and now when I ask about 2+3 you say it is 5, you cannot change the rules to suit". Because you do not understand the math you see the answers as contradictory, when in fact they both follow directly from the math using the same set of rules.

I still invite you to indicate where the r angle photon started from in a .999c frame

And I still invite you to explain how to experimentally determine if the target is at absolute rest. You are clearly avoiding the question, so I think you realize now that there is no possible answer. There is no way to experimentally determine if the target is at absolute rest, so all you can do is measure your velocity relative to the target.

I am perfectly willing to address your question later, but, as I have already stated, I am not inclined to proceed on to the minor details of any other examples until we have resolved this central issue of the current example. Do you now agree that there is no way to experimentally determine if the target is at absolute rest?

 

[Reff]

It depends what you mean by this sentence. All frames agree that a sphere of photons centered on the origin in one frame is a sphere of photons centered on the origin in all frames. So that fact is not frame dependent.

Perhaps a thought concerning the only constant in the universe being c and if we locate ourselves at the beginning of the creation of a sphere and study that constant and indeed stay concentricaly centered to the sphere. When I say not frame dependent I mean it does not matter what speed and direction the frame was moving when the sphere was created. I have stayed concentricaly centered on the propagation from its very beginning. all subsequent events concentricaly centered on that point are not frame dependent, who cares where the frames are in relationship to the spheres that are created from these moving frames. the spheres are not frame dependent.
I ask each frame observer if they measure c-- yes, I ask are there photons in your tubes, they say yes and I ask where did they start from---- and prove it by creating an event at your start point the moment your right angle tube has a photon exiting. As each observer creates that event on his frame to say " this is where the photons originated" I say no you have missed to almost every observer and point out that almost every marker event has missed the absolute or constant sphere. Each moving observer is saying the sphere propagation is dependent on where the event happened on his frame, so I am saying almost all the observers are using their frame to Drag the perfect symetry of a photon sphere around with them. Perhaps Copernicus would say "you have done it again." Observers are incorrectly saying they are the center of a the perfection of a sphere created by a constant. They are saying the sphere is frame dependent. I have mentioned "almost all" because there was one frame which stayed central to all the spheres at that point I call absolute rest. When a photon exited absolutely any tube on his frame and he complied by creating an event at the start of the tube, then this new event and any number of events at that same point on the frame, will expand concentricaly to all the others.

Pay attention to what I actually said

.
We both seem to suffer from that eh!

And I still invite you to explain how to experimentally determine if the target is at absolute rest. You are clearly avoiding the question, so I think you realize now that there is no possible answer. There is no way to experimentally determine if the target is at absolute rest, so all you can do is measure your velocity relative to the target I am perfectly willing to address your question later, but, as I have already stated, I am not inclined to proceed on to the minor details of any other examples until we have resolved this central issue of the current example. Do you now agree that there is no way to experimentally determine if the target is at absolute rest?

no!
Conversly-- Do you say that there is no way to experimentally determine that absolute rest noes not exist.

Ok let's have a go. I will say now though that it is unfair to ask me to design an experiment to determine absolute rest when during the course of history plenty has been said which was much later verified by experiments (including michelson Morley now trying to tell us something)
But let's begin with a frame and observer remaining centered on an expanding sphere of photons from a zero time event. He is thus inertial. Let us take six frames observers and clocks. They are going to do a fly past at .9c .( any substantial common figure will do.) Each frame is moving at 90 degrees to the others-- ie all directions relative to the sphere located frame. As they pass they synchronise clocks with the sphere clock as an instantaneous exchange of information and proceed for a distance from the spheres frame. Let's use a distance of a long piece of string, say 300 000km. As each frame reaches the end of the piece of string, it exhanges instant clock information with messenger frames which return to the sphere located frame. If I am right then the period of time elapsed in each non concentric frame will be the same and it will be. If you are correct and there is variation I will purchase a rabbit.
I can perhaps do better than that with more than ten minutes to think about it.

 

[Dale]

I have stayed concentricaly centered on the propagation from its very beginning.

So have the observers at rest in every other frame.

As each observer creates that event on his frame to say " this is where the photons originated" I say no you have missed to almost every observer and point out that almost every marker event has missed the absolute or constant sphere.

And how do you determine which sphere is the absolute or constant sphere? Each observer in each frame has the same result from the same experiment. What experimentally distinguishes the absolute or constant sphere from every other sphere?

But let's begin with a frame and observer remaining centered on an expanding sphere of photons from a zero time event. He is thus inertial. Let us take six frames observers and clocks. They are going to do a fly past at .9c .( any substantial common figure will do.) Each frame is moving at 90 degrees to the others-- ie all directions relative to the sphere located frame. As they pass they synchronise clocks with the sphere clock as an instantaneous exchange of information and proceed for a distance from the spheres frame. Let's use a distance of a long piece of string, say 300 000km. As each frame reaches the end of the piece of string, it exhanges instant clock information with messenger frames which return to the sphere located frame. If I am right then the period of time elapsed in each non concentric frame will be the same and it will be.

If you perform this same experiment in any frame you will get the same result. This experiment does not distinguish one frame from another. You still have not described a way to experimentally determine if your target is at absolute rest.

 

[Dale]

Look Reff, the laws of physics are invariant under the Lorentz transform, and the Lorentz transform is derived from the principle of relativity. So it is simply not possible to make an experiment which will detect an absolute rest frame using the known laws of physics.

I believe that your problem is that you are unfamiliar with the Lorentz transform. You have not actually worked enough physics "homework" style problems involving the Lorentz transform. Because of that you don't have a good understanding about how physics will work in other frames besides the one that you draw.

You are chasing after ghosts. The laws of physics don't do what you want them to do. All that you have left is to hope that future discoveries will lead to new laws of physics that are not Lorentz symmetric.

 

[Reff]

Look Reff, the laws of physics are invariant under the Lorentz transform, and the Lorentz transform is derived from the principle of relativity. So it is simply not possible to make an experiment which will detect an absolute rest frame using the known laws of physics.

I believe that your problem is that you are unfamiliar with the Lorentz transform. You have not actually worked enough physics "homework" style problems involving the Lorentz transform. Because of that you don't have a good understanding about how physics will work in other frames besides the one that you draw.

You are chasing after ghosts. The laws of physics don't do what you want them to do. All that you have left is to hope that future discoveries will lead to new laws of physics that are not Lorentz symmetric.

Hi DaleSpam
There is only one constant in this universe and that is c
Using that constant and forming a perfect sphere of photons which can announce their presence by an exchange of energy. The sphere does exist.
All frames forming further spheres at a point concentricaly located on the first can be anywhere within their own spheres.
The propagation of the spheres are not dependent in any way on the speed and direction of the frames.
I have aknowledged Lorentz invarience and also frame contraction but as to the position of the frames within the spheres, who cares about Lorentz. It is purely frame related.
You have still not answered where a .999c frame is in its own circle of propagation of say 300 000km. I am sure you would say as that frames observer that you are at the center of the sphere. Staying with the bigger picture of the sphere, and anouncing your presence by a further event from your frame would prove that you are not anywhere near the center of the sphere and can no longer find your rabbit.
Every single moving frame observer within its own sphere will say they are at the center but just using our universal constant of c we know otherwise, so what has Lorentz to do with that.
How is Lorentz going to reposition your .999c frame to suit your position at the center. Copernicus would be turning in his grave.

 

[ghwellsjr]

Hi DaleSpam
There is only one constant in this universe and that is c
Using that constant and forming a perfect sphere of photons which can announce their presence by an exchange of energy. The sphere does exist.
All frames forming further spheres at a point concentricaly located on the first can be anywhere within their own spheres.
The propagation of the spheres are not dependent in any way on the speed and direction of the frames.
I have aknowledged Lorentz invarience and also frame contraction but as to the position of the frames within the spheres, who cares about Lorentz. It is purely frame related.
You have still not answered where a .999c frame is in its own circle of propagation of say 300 000km. I am sure you would say as that frames observer that you are at the center of the sphere. Staying with the bigger picture of the sphere, and anouncing your presence by a further event from your frame would prove that you are not anywhere near the center of the sphere and can no longer find your rabbit.
Every single moving frame observer within its own sphere will say they are at the center but just using our universal constant of c we know otherwise, so what has Lorentz to do with that.
How is Lorentz going to reposition your .999c frame to suit your position at the center. Copernicus would be turning in his grave.

Reff, I already answered your question about how two observers, one stationary in an absolute ether rest state and another one traveling at 0.5c, would both think that they are in the center of an expanding sphere of light. Please go back to post #32, watch the animations and understand what I'm presenting there. I don't talk about Lorentz Transformation--that should make you happy. And remember, this is all done from an absolute ether rest state--no frames at all. You previously indicated that you needed more time to digest them. Please take the time now. And then please ask questions if you don't understand or if you don't agree.

By the way, I could have done similar animations for 0.999c but you don't have a screen large enough and with enough resolution for the animation to make any sense. And you wouldn't have the patience to watch those animations, as they take a very long time.

 

[Dale]

Using that constant and forming a perfect sphere of photons which can announce their presence by an exchange of energy. The sphere does exist.

Certainly, that is not in dispute.

The propagation of the spheres are not dependent in any way on the speed and direction of the frames.

Nor is that.

I have aknowledged Lorentz invarience

If you actually understood Lorentz invariance then we wouldn't be having this conversation. Lorentz invariance is completely incompatible with an experimentally detectable absolute rest frame. For you to claim that you acknowledge Lorentz invariance is logically incompatible with the rest of your posts in this thread.

You have still not answered where a .999c frame is in its own circle of propagation of say 300 000km.

True, and although you still have not provided an experimental way to determine if your target is at absolute rest, you did at least make a sincere effort. So I will respond to that in a moment.

How is Lorentz going to reposition your .999c frame to suit your position at the center. Copernicus would be turning in his grave.

Actually, I think that Copernicus would be very comfortable with the principle of relativity and its logical implications.

 

[Reff]

Reff, I already answered your question about how two observers, one stationary in an absolute ether rest state and another one traveling at 0.5c, would both think that they are in the center of an expanding sphere of light. Please go back to post #32, watch the animations and understand what I'm presenting there. I don't talk about Lorentz Transformation--that should make you happy. And remember, this is all done from an absolute ether rest state--no frames at all. You previously indicated that you needed more time to digest them. Please take the time now. And then please ask questions if you don't understand or if you don't agree.

By the way, I could have done similar animations for 0.999c but you don't have a screen large enough and with enough resolution for the animation to make any sense. And you wouldn't have the patience to watch those animations, as they take a very long time.

Hi ghwellsjr and DaleSpam
Thanks for replys and patience. I have looked at your animations once ghwellsjr and partly absorbed them so I owe you some comment in that respect, so I will be having another look to see if I can fully understand.
Thanks DaleSpam I will be back.

 

[Dale]

You are at .999c and you create a sphere of photons and your right angle tube is about 10ft long. As your tube photon exits you create another event at the start of the tube.
Now I understand you to point to the start of the tube when I ask you where did the photon start from. If that is so I would ask you to step outside ond view the complete picture.
We see the first event sphere having expanded in perfect symetry and we locate the center of the sphere,( where all the photon reciprocals cross.) Now we look at your second event or marker event where you tell me the first event started from. The second sphere is is not centered on the first. The tube start point is now way out close to the edge of the sphere at .999c.

I will use units where c=1, and the standard Lorentz transform: http://en.wikipedia.org/wiki/Lorent...ormation_for_frames_in_standard_configuration

In the first frame
There is a moving tube (A) whose worldline is:
\left( t, -\frac{999}{1000}t, L, 0 \right)
and a stationary tube (B) whose worldline is:
\left( t, 0, L, 0 \right)
There is the sphere of light when the ends of the tubes are together
t^2=x^2+y^2+z^2
There is a sphere of light which emanates at t=10 from B
(t-10)^2=x^2+y^2+z^2
And there is a sphere of light which emanates at t=223.663 from A
\left(t-\frac{10000}{\sqrt{1999}} \right)^2= \left(x+\frac{9990}{\sqrt{1999}} \right)^2+y^2+z^2

Then by the Lorentz transform, in the second frame
B is a moving tube whose worldline is:
\left( t', \frac{999}{1000}t', L, 0 \right)
and A is a stationary tube whose worldline is:
\left( t', 0, L, 0 \right)
There is the sphere of light when the ends of the tubes are together
t'^2=x'^2+y'^2+z'^2
There is a sphere of light which emanates at t'=10 from A
(t'-10)^2=x'^2+y'^2+z'^2
And there is a sphere of light which emanates at t'=223.663 from B
\left(t'-\frac{10000}{\sqrt{1999}} \right)^2= \left(x'-\frac{9990}{\sqrt{1999}} \right)^2+y'^2+z'^2

So the results are the same in each frame. There is no experimental difference between the frame where one tube is stationary and the frame where the other is stationary.

 

[Reff]

=ghwellsjr;3442077]Reff, I already answered your question about how two observers, one stationary in an absolute ether rest state and another one traveling at 0.5c, would both think that they are in the center of an expanding sphere of light. Please go back to post #32, watch the animations and understand what I'm presenting there. I don't talk about Lorentz Transformation--that should make you happy. And remember, this is all done from an absolute ether rest state--no frames at all. You previously indicated that you needed more time to digest them. Please take the time now. And then please ask questions if you don't understand or if you don't agree.

Hi ghwellsjr
Yes neat little animations. Yes I understand how any moving frame observer would believe he was at the center of propagation by watching your animations. The third animation I found especialy inreresting. Look, for the moment could you just retain an open mind and consider a minor variation to to the third animation, purely using a logic which you would no doubt correct if it broke any laws.
Just for a moment could we go to your first animation. Would it be reasonable to mentaly mark the path of a few photons starting from the man and finishing at the mirrors, just simply the radius from the man to the sphere. I presume you are ok that they mark the path of the only constant in the universe--c
Now can we use the third animation and use both stick men and the blue sphere. I am interested in the propagation of the blue sphere.Both men were together at the moment of the event and propagation progressed independently of both men but the first--green man remained concentricaly centered on the sphere,
Start the animation and watch to the point where the red man receives the return reflection and determines he is the center of the blue sphere. If red man creates a marker event right then, can we see a red man assymetry to the blue sphere.
As a moving man, red man is progressing from left to right and is following a left to right track of a photon from his event which coincided with the green man. He is moving at .5c and so he will be half way along that c radius track.
He has a table, the surface of which is flat face to the direction of travel and we can simply indicate this by dropping a line from his head verticaly down and off the screen. This line representing the edge of the table moving with the red man.
Now go back to the original blue sphere event. It is adjacent to the edge of the table.
Now propagate the sphere slightly and move red man and his table. now scribe the path of a photon to what is now an intersecting line between the sphere and the table. Now propagate a little more and extend the radius line, it will intersect the sphere again. repeat the exercise in increments as many times as you wish. It is always the same photon on the end of the radius and on the surface of the table. During propagation and frame movement, that individual photon will complete a full table crossing of the moving frame table. Its speed is always the universal constant of c, like all the other photons on the spheres radius. Now consider the photon on the end of that one radius on the surface of the table is only in an intersecting state and while it moves at c on the radius just like any other radius in the sphere the intersection progress is less than c
I hope you can follow that.
Do you use an animation programme.
Your green man s tabletop crossing photon is not a table top intersecting photon so it crosses the table at c the absolute. Can you see there is no time dilation for him. Simply, a frame at c is fully time dilated-- there is no crossing photon. A table with your green man is fully at an absolute undilated time because his tabletop crossing photon must cross at c.
You might be able to relate this to my geometry.
Like your red man, any amount of frames creating a sphere from the green man will leave a concentric sphere based on the green man.
The green man must be inertial and in a time related state, so what would you call that state. I would call it absolute rest. Is there a faster clock than green mans anywhere.

 

[Dale]

Reff, how do you think things are in the red man's frame?

 

[ghwellsjr]

Reff, I already answered your question about how two observers, one stationary in an absolute ether rest state and another one traveling at 0.5c, would both think that they are in the center of an expanding sphere of light. Please go back to post #32, watch the animations and understand what I'm presenting there. I don't talk about Lorentz Transformation--that should make you happy. And remember, this is all done from an absolute ether rest state--no frames at all. You previously indicated that you needed more time to digest them. Please take the time now. And then please ask questions if you don't understand or if you don't agree.

By the way, I could have done similar animations for 0.999c but you don't have a screen large enough and with enough resolution for the animation to make any sense. And you wouldn't have the patience to watch those animations, as they take a very long time.

Hi ghwellsjr
Yes neat little animations. Yes I understand how any moving frame observer would believe he was at the center of propagation by watching your animations. The third animation I found especialy inreresting. Look, for the moment could you just retain an open mind and consider a minor variation to to the third animation, purely using a logic which you would no doubt correct if it broke any laws.
Just for a moment could we go to your first animation. Would it be reasonable to mentaly mark the path of a few photons starting from the man and finishing at the mirrors, just simply the radius from the man to the sphere. I presume you are ok that they mark the path of the only constant in the universe--c
Now can we use the third animation and use both stick men and the blue sphere. I am interested in the propagation of the blue sphere.Both men were together at the moment of the event and propagation progressed independently of both men but the first--green man remained concentricaly centered on the sphere,
Start the animation and watch to the point where the red man receives the return reflection and determines he is the center of the blue sphere. If red man creates a marker event right then, can we see a red man assymetry to the blue sphere.

Yes, we can see that the red man is not in the center of the blue sphere but he cannot see that. He has exactly the same experience as the green man. The only difference between them is that the green man sees the red man moving to his left and the red man sees the green man moving to his right.

As a moving man, red man is progressing from left to right and is following a left to right track of a photon from his event which coincided with the green man. He is moving at .5c and so he will be half way along that c radius track.
He has a table, the surface of which is flat face to the direction of travel and we can simply indicate this by dropping a line from his head verticaly down and off the screen. This line representing the edge of the table moving with the red man.
Now go back to the original blue sphere event. It is adjacent to the edge of the table.
Now propagate the sphere slightly and move red man and his table. now scribe the path of a photon to what is now an intersecting line between the sphere and the table. Now propagate a little more and extend the radius line, it will intersect the sphere again. repeat the exercise in increments as many times as you wish. It is always the same photon on the end of the radius and on the surface of the table. During propagation and frame movement, that individual photon will complete a full table crossing of the moving frame table. Its speed is always the universal constant of c, like all the other photons on the spheres radius. Now consider the photon on the end of that one radius on the surface of the table is only in an intersecting state and while it moves at c on the radius just like any other radius in the sphere the intersection progress is less than c
I hope you can follow that.

I think what you are saying is that as we watch the progress of a photon moving "downward" from the red man's moving position, it will appear to us that it is traveling at c along a diagonal, and we could say that it represents a legitimate photon in the green man's experience but from the red man's point of view we would have to say that it is traveling much slower than c because it is taking so much longer to get down to the mirror below him. But remember, time is going slower for the red man so from his point of view when he calculates the speed of the photon (if he could possibly know where it was), then he would believe that it was actually traveling at c.

Do you use an animation programme.

I use a general purpose program (LabVIEW) that is not specific to animation so it is a lot of work for me to produce these animations. I then use a screen capture utility (CamStudio) to make an avi that I can upload to YouTube.

Your green man s tabletop crossing photon is not a table top intersecting photon so it crosses the table at c the absolute. Can you see there is no time dilation for him. Simply, a frame at c is fully time dilated-- there is no crossing photon. A table with your green man is fully at an absolute undilated time because his tabletop crossing photon must cross at c.
You might be able to relate this to my geometry.
Like your red man, any amount of frames creating a sphere from the green man will leave a concentric sphere based on the green man.
The green man must be inertial and in a time related state, so what would you call that state. I would call it absolute rest. Is there a faster clock than green mans anywhere.

I have presented these animations from the point of view of LET which assumes an absolute ether rest state. I have not talked about frames at all. I don't know why you keep talking about frames when you believe in an absolute rest. What I'm trying to point out to you is that even from the viewpoint of LET and a single absolute ether rest state in which is c is constant and only in which c is constant, as long as you believe that an inertially moving observer will also measure the round-trip speed of light to be c (like in the real world), which you say you do because you agreed that my animations illustrate how both men will think they are in the center of the expanding sphere of light, then you can follow the interpretation of LET which assumes the actual real existence of an absolute ether rest state. In LET, the moving observer experiences time dilation and length contraction along the direction of motion through the ether. LET affirms that the green man is really stationary in the ether and the red man is really moving through the ether but it also affirms that the red man experiences everything the same way the green man does. In other words, the red man has every reason to believe that he is the one that is stationary in the ether and that it is the green man who is moving through the ether and exeriencing time dilation and length contraction and whose photons are bouncing off his mirrors at different times and whose photons are slowed down in some cases and speeded up in others.

So the question is: how can the green man prove that he really is stationary in the ether, even if he is. How can he tell? How can the red man prove that he is moving in the ether, even if he is? If the red man wants to believe that he is stationary in the ether, even if he isn't, how could the green man prove that he is wrong? How could you prove that he is wrong?

Remember, I said that the only difference between what the two men are experiencing is which direction the other one is traveling. Do you believe that?

 

[Reff]

Reff, how do you think things are in the red man's frame?

Hi DaleSpam
First thanks for the maths posting. I am a little confused re

There is a moving tube (A) whose worldline is:

and a stationary tube (B) whose worldline is:

There is the sphere of light when the ends of the tubes are together

I am thinking right now that your maths is impressive but I am not to sure we are understanding each other. The moving frame observer I am sure would believe he is the center of propagation. I am trying to suggest athough he believes this, he is not and using a little geometry I predict his time dilation within a scale drawing of a 20 cm sphere. If we use
ghwellsjr's neat animation-- the third one, green man and red man we see red mans blue sphere propagating around green man identical to green man and red man believing he is at the center of the sphere when the reflection returns. I am saying a marker event from red man is not centered on the green man. Now, am I to believe a sphere made up of the only constant in the universe c, ie any radius in the sphere, and thus its perfect symetry is superceded by red mans belief. Red man in 200mm at .999c is not at the center of the sphere All photons within the sphere move at c. He is .999 along one absolute--and constant radius
The one photon which can cross his tabletop is not crossing at c, in is intersecting at less than c. I hope you have seen that in my geometry.
The whole basis of the geometry is that the moving frame is not centered on propagation, which to me is a logical step to predict the crossing speed of the tabletop intersecting photon. The crossing speed is directly related to the frames time dilation and that works fine.
I don't need every frame to believe they are the center of propagation, to do so violates the only constant.
Re

Reff, how do you think things are in the red man's frame?

Y 09:47 AM
I have a feeling I would like to answer that but I am not quite sure what you want.