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

[Dale]

Reff, this site is for mainstream physics education, not for the promotion or discussion of personal theories. It is clear that you are not interested in the former, so you should find another forum that encourages speculation.

Blind and cannot see.

That is pretty hypocritical and rude for someone who cannot even follow a few lines of algebra and is completely ignorant of Minkowski geometry and its relevance to SR.

 

[CJames]

Reff,

I have understood and responded to your geometry, and responded with answers stating why a pulse of light does not have a definite heading. I can't understand why you would call me blind in that context.

 

[ghwellsjr]

Hi ghwellsjr
thanks for replying again.
Yes your animations are really very good although I don't necesserily agree. Now Re the first video yes I totaly agree with this one provided that it represents an inertial frame centered on the propagation. Now I did explain my geometry of SR earlier on for a frame moving at .8c and this was further clarified by CJames I believe. Now if you draw my geometry and I presume you agree it gives the correct answer for .8c frame and any other frame speed you care to name. Draw a 200mmdia circle and draw a vertical line right through the middle of it now this is your first animation. The circle is as far as propagation has reached. Your inertial frame is at the center. Photons are heading in all directions but we are only interested in one photon or a small pulse of light which is heading along the surface of the tabletop. When it hits a reflector at the end of the tabletop, it returns straight back on the surface of the table to its point of origin. that pulse of light or particle or photon is moving at c and back from the source.
Now do my .8c diagram and note the frame speed compered with the vector of the photon the edge of the table is following-- it is marked off simply at .8c. That is where the left hand edge of the tabletop is .8c up from the event. Now draw a photon vector from the event to where the tabletop intersects the sphere. There is a photon on the edge of the sphere and it is on the surface of the tabletop. That very same photon has been on the surface of the tabletop right from the event up to where it is now. It has moved .6 of the propagation we are looking down on so without any maths at all we know the moving frame is .6time dilated because it is simply a transposed 345 triangle. I can do this for all frame speeds and it works fine. The .6 time dilation is simply happening because it is actualy moving at c like absolutely any other photon in the radius off the sphere, they are all there to see. BUT what is the table crossing cspeed .6 it is a little tricky to understand because it is not crossing the table top it is constantly intersecting the tabletop. Big difference.
Let's go back to your first diagram. If we put a tabletop in your diagram the left hand side of this is where your little animated man is- as I have said-- its light progresses at c and returns at c Is there a vector to the crossing photon---no-- is there any time dilation--- no--because it crosses at c. has the left hand side of the tabletop moved up the vertical line No. his is a clock that in my books reads absolute time. There is no faster clock than this.
If you do my geometry as described and overlay yours as I have described then you can perhaps understand how I derive my figures purely geometricaly.
Your animation would be interesting to show how a moving frames tabletop crossing photon crosses in a state of constant intersection and the intersection speed of an inertial moving frame is always less than c. The point of origin of the photon and the constant intersection could be hilighted.
Its tricky geometry. Just ask yourself, if a clock stops at c, there must be a clock that runs faster and faster with a decreases in frame speed. The fastest is centered on your first diagram.

Is the purpose of your diagram to graphically determine the time dilation factor as a function of speed? Would this diagram, where I called the time dilation factor "age" work? At a speed of .8c the time dilation is .6:

 

[Reff]

Hi
Thanks for your persistence. I apologise to all I have pd off but I would like to point out a fundamental error in your understanding of the geometry I have described.
(Re the last posting of ghwellsjr your neat little chart predicts time dilation for sure and there are any number of similar charts that could do the same and that is not what I am on about).
From the start I have been on about the release of a Photon-particle from the event. I am not on about a beam of light or a pulse of light. Let me elaborate. If you believe I am on about a beam or pulse of light then indeed in my diagram of light expanding to a sphere of 200mm, light will still be being transmitted at the frame transmitting point as it reaches the edge of the sphere. The source will still be the frame transmitter. If we consider one millionth of a second builds a sphere of 600 meters, then indeed my geometry is not clear. If it is a beam or pulse then indeed it will be red or blue shifted viewed by another frame. This is not how I described the sphere initialy. I did make a regretful mistake in part of my posting by conceeding to a pulse with qualifications as to the length of the pulse but this would have confused people more. Photons or particles on the edge of a sphere cannot be red or blue shifted, there is nothing behind them to "wave" an advancing wall of light beam propagation moves at c, red or blue shifted. I say this believing a photon can exhibit two qualities

This is how I said it in my first posting.
"Use three photons on the sphere to create a target. 1. Any one photon 2. A photon heading out on the reciprocal of the first photon. 3. A photon heading out at 90 degrees to the line formed by the first two photons".

How do you guys get a beam or pulse of light from that. The geometry I have described is to scale- it is not a chart that many can produce and has a simple geometric logic that some people find hard to understand.
If anyone is still game and now can understand the use of a single particle from an event I can use a variation to the diagram you may understand.
Draw a 200mm sphere with a line from the compas point up the page and now a line from the same point at right angles to the right.
There is a photon-particle on the sphere intersection of each line.
Imagine or draw a thin tube in line with the tragectory of both lines so the photon particles pass inside the tubes without touching the inside of the tubes. Now fix the tubes together and mount them to a moving frame at .8c moving in the direction of the photon moving up the page.
Re create the same event at the entrance point of the two tubes and watch the photons and the tubes move. The tube up the page will have a photon in it moving through without touching the inside of the tube. Now-- the tube out to the right is a different story-- the right angle photon cannot pass through the tube without touching the inside of the tube. Identify a photon that can move through the tube. It is NOT the right angle photon. Even an observer on the moving frame could confirm that.This is why we have time dilation and it is how it can be calculated using a ruler. and yes pythagarus is more acurate. The propagation symetry cannot be moved by the moving frame because it was a zero time generated event and as such will not be influenced by frame speed. I say zero time generated because I consider the time it takes for a photon to leave a point on any frame is close to zero. A beam of light is another story as it is being constantly generated on the moving frame as in a constant event but at anyone moment a sphere of photons is being generated from the moving event generator with all photons moving radialy from it.
Go back to the right angle tubes with the right angle trajectory photon passing through without touching the sides. How fast is that tube frames clock moving compared to any other moving frame. When you say absolute rest is not required are you saying it cannot exist.

I believed this group would understand (Not believe) the geometry. Some struggle.
At your desk in a GR situation, does the right angle photon cross the tabletop. Would QM agree?, I really don't know for the moment, but perhaps not.

 

[Dale]

From the start I have been on about the release of a Photon-particle from the event. I am not on about a beam of light or a pulse of light.

I am fine with photons, but if you don't have the mathematical background to follow the simple algebra I posted above then you are essentially guaranteed to make mistakes naively using photons. For example:

Photons or particles on the edge of a sphere cannot be red or blue shifted

This is incorrect. It is an experimentally verified fact that photons red and blue shift according to the predictions of SR. The seminal experiment was by Ives and Stilwell in 1938, and has been followed by increasingly accurate experiments all confirming the same fact.

Imagine or draw a thin tube in line with the tragectory of both lines so the photon particles pass inside the tubes without touching the inside of the tubes. Now fix the tubes together and mount them to a moving frame at .8c moving in the direction of the photon moving up the page.
Re create the same event at the entrance point of the two tubes and watch the photons and the tubes move. The tube up the page will have a photon in it moving through without touching the inside of the tube. Now-- the tube out to the right is a different story-- the right angle photon cannot pass through the tube without touching the inside of the tube. Identify a photon that can move through the tube. It is NOT the right angle photon. Even an observer on the moving frame could confirm that.

Actually, an observer on the moving frame would disagree. There is a photon that would pass through the moving tube without touching the inside of the tube and in the moving frame that photon IS the right angle photon.

The headings are frame variant, as I said back in post 48. For every possible frame if you constructed the apparatus there would be a photon which goes through the tube to the right, and that photon would be the right angle photon in that frame.

There is nothing in this setup (nor any other possible set up) to distinguish one frame from another. The most you can do is measure relative velocities.

 

[Reff]

We may be getting somewhere now DaleSpam
re
Originally Posted by Reff
Photons or particles on the edge of a sphere cannot be red or blue shifted

This is incorrect. It is an experimentally verified fact that photons red and blue shift according to the predictions of SR. The seminal experiment was by Ives and Stilwell in 1938, and has been followed by increasingly accurate experiments all confirming the same fact.

I made a careful effort to point out the duality of a photon in the last post.
Quote you
There is a photon that would pass through the moving tube without touching the inside of the tube and in the moving frame that photon IS the right angle photon

For me.
Now this is precisely where I am in conflict DaleSpam

Going back to the first two tube geometry, Identify the two photons and mark their arrival at the end of the tubes.. Passing through the tubes not having touched the sides I believe they can be identified specificaly I believe they can transfer energy on their arrival.
Now do the same with the tubes moving at .8c. The right angle photon must touch the inside of the right angle tube.
Look carefuly at the geometry and there is a photon that has passed perfectly through the right angle tube and has not touched the sides. It is not the same photon. You can identify it. Its energy tranferring trajectory is not the same as the identified right angle photon. This photon is right on the point where the sphere crosses the centerline of the tube. A .8c frame observer would see the tube conflict with the right angle photon but not with the identified other photon which will pass through. Is it possible that he and others believe it is the right angle photon even when exiting the tube it is found to not be the identified right angle test photon.
Just as a slight variation to this, place a tabletop flat face away from the direction of travel.
180 degrees turned over from my original example. and do the same thing. Where is the right angle photon now. There will be another photon doing the crossing, under the table as it were. It will not be the right angle photon. Look where the photon is after both the table has moved and sphere has expanded.

During the whole time this experiment has been carried out, all frames have not touched the perfection of the right angle photon except to highlight an error, so how can a now
moving frame drag the right angle photon to suit your statement. Both observers frames can identify internal tube conflict with the rightangle photon.

 

[Dale]

I made a careful effort to point out the duality of a photon in the last post.

I don't know how that in any way justifies your incorrect assertion that a photon cannot be redshifted.

Going back to the first two tube geometry, Identify the two photons and mark their arrival at the end of the tubes.. Passing through the tubes not having touched the sides I believe they can be identified specificaly I believe they can transfer energy on their arrival.
Now do the same with the tubes moving at .8c. The right angle photon must touch the inside of the right angle tube.

The stationary frame right angle photon does touch the inside of the moving frame right angle tube, because it is not the right angle photon in the moving frame. Similarly the right angle photon in the moving frame does touch the inside of the stationary frame right angle tube. There is no difference between the two frames in this respect.

Suppose we have a photon going along the y axis, then its worldline is given by:
r=(ct,0,ct,0)
which has a heading of atan(ct/0)=90º

In a frame moving at .8c its worldline is given by:
r'=(1.66ct,-1.33ct,ct,0)
which has a heading of atan(-ct/1.33ct)=143º

Do you understand that?

I have said this three times now, but you seem to not be getting it. The direction that something is traveling depends on the reference frame. Do you understand this point? Please respond to this question.

 

[Reff]

You said
The stationary frame right angle photon does touch the inside of the moving frame right angle tube, because it is not the right angle photon in the moving frame. Similarly the right angle photon in the moving frame does touch the inside of the stationary frame right angle tube. There is no difference between the two frames in this respect.

Now look at photons moving from sequential zero time events which occur at the same point regardless of the speed and direction of any sphere generating frame you wish. They will all propagate in perfect symetry. I place my right angle tubes over any two photons moving at right angles to each other. They are the right angle photons. Now move your right angle frame. The right angle photon that can cross in a moving frame is no longer a right angle photon within the sphere. frame speed has nothing to do with propagation but everything to do with time dilation because of the right angle photon does not cross, it constantly intersects and that very same photon is moving at c from its own event just like any other photon in the sphere. As the photon exits the tube, the event point is no longer at the start of the tube. The crossing is at right angles in a moving frame but the photons trajectory is not.

No I don't understand the maths, the geometry will do me for now.
You need an answer to this
I have said this three times now, but you seem to not be getting it. The direction that something is traveling depends on the reference frame. Do you understand this point? Please respond to this question.


Yes I do
There are at least two reference frames for me and that is remaining inertial at the center of a propagating sphere of photons or using right angle tubes allowing two photons moving at right angles to each other from a zero time event to pass through the tubes without touching and that is absolute rest to me When the directional photon matches the right angle photon in a moving frame, then the frame is moving at c, all referenced to a propagating sphere.
Indicate the right angle crossing photon in your moving frame and you can calculate your time dilation by its trajectory to your directional photon--all referenced to the stationary or dare I say absolute rest frame. Your right angle photon in a moving frame does not cross, it intersects. The tube and the photon point in different directions in a moving frame unlike the same tube and photon at rest. If the right angle photon and tube are pointing at a distant at rest frame, then the photon will hit it unlike in a moving frame. Big difference. (to me of course)

 

[Dale]

Yes I do

You say you understand, but then you make statements like this one:

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

In which frame? If you really understood the point then you would have said: "I place my right angle tubes over any two photons moving at right angles to each other in the frame of the tubes." The direction of travel is a frame-variant quantity so it is meaningless to talk about the direction of travel without specifying the reference frame.

As the photon exits the tube, the event point is no longer at the start of the tube.

Yes, it is, in every frame.

There are at least two reference frames for me and that is remaining inertial at the center of a propagating sphere of photons or using right angle tubes allowing two photons moving at right angles to each other from a zero time event to pass through the tubes without touching and that is absolute rest to me When the directional photon matches the right angle photon in a moving frame, then the frame is moving at c, all referenced to a propagating sphere.
Indicate the right angle crossing photon in your moving frame and you can calculate your time dilation by its trajectory to your directional photon--all referenced to the stationary or dare I say absolute rest frame. Your right angle photon in a moving frame does not cross, it intersects. The tube and the photon point in different directions in a moving frame unlike the same tube and photon at rest. If the right angle photon and tube are pointing at a distant at rest frame, then the photon will hit it unlike in a moving frame. Big difference. (to me of course)

No difference at all. You can construct the same apparatus in the "moving" frame, perform the same experiment, and get the same result. There is no distinction between the frames this way.

The laws of physics are Lorentz invariant, therefore there is no absolute rest frame. It is as simple as that.

 

[Reff]

Hi DaleSpam
Thanks
Yes I did say this and stand by it
I place my right angle tubes over any two photons moving at right angles to each other.

I am saying That I observe an event just beginning and go to it and as it progresses I observe two photons beginning to move at right angles to each other. It can be an event from any frame, the photons are immediately up to their own governing speed laws. I place my tubes over these two photons and the photons pass through without touching.
This is absolutely the only frame one can do this with.
Let me elaborate on my concept of direction of travel.
Take anyone photon from an event and follow it. Now if you were able to survive the exercise I am saying you would be in an inertial frame, in other words the photon has its own specific heading and is not accellerating byturning in any way. Reverse the directions of all the photons on the sphere and they will all return to meet again and even be able to tranfer some energy to create the datum point I have always talked about which is in my books is now the marked event point.
In any moving frame I am saying a tube passing photon is not a 90 degree photon relative to the event and the moving frame tubes have moved on from the marked event.
All photons must have speed laws and not be able to overtake another regardless of the frame they leave.
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.
Create two identical frames one light year apart. Propagate and select two right angle photons in each frame and place the tubes over the propagation so that the photons move through without touching the inside of the tubes. Will these tubes ever meet.
I presently believe they will remain inertial and will never meet. They cannot meet because they are governed by the direction of each tube photon. The tubes must remain aligned with the photons however distant. All photon "directions" in this example will return to two detectable points of origin at the two respective Stationary- absolute rest frames.

Take any moving frame and if they are moving towards each other or on a conflicting course yes they can meet. If they have right angle tubes then the photons used in the tubes are not sphere right angle photons. Also a moving frame can move to absolute rest frames but two absolute rest frames can never meet as I have previously explained.
.

 

[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.

 

[Reff]

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.

Yes I absolutely, agree, that we can see the red man is NOT in the center of the blue sphere and as you say--"but he cannot see that" That is what it is all about, I use this assymetry for direct calculation of his frames time dilation. Who cares what green and red men think is happening when they are ignoring the big picture and experiencing an abberation

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.

Absolutely, I think you are beginning to understand the geometry, and yes the photon is taking longer to reach a mirror and yes he will measure c because he has no option but to do so. Every movement in red mans frame is directly in proportion to the crossing---- diagonal photon. No photon can cross faster because it would violate our only constant c. Try crossing one. Every single photon in red mans and green mans frames-- is moving radialy at c from the center of the blue sphere so who cares where any frame of any direction is within the sphere unless you wish to calculate its time dilation.

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.

Smart and neat.

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.

Yes, look I have a confusing ramble and use incorrect terminology on occasion which many quite rightly find confusing but on my side I have a raw unrefined view of relativity, not too bad at geometry and I worked it out myself and it works out fine. I was convinced on geometry alone and further so when a little maths also worked and even further so when it could be drawn precisely to scale.
I talk about frames as being a specific speed of an individual observer ie red mans and green mans as frames. and yes your green man is at absolute rest.
While I think about it-- green mans tabletop photon crosses at c and he can use any photon within his sphere to make a direct crossing of his table------ NO diagonal----NO intersection means no time dilation and thus an absolute time clock which will measure a non intersecting -- straight trajectory photon at c but at-- absolute rest . He will experience and believe the same as any other frame trying measuring c.

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?

Now there is a question to ponder but perhaps sometime in the future it may be done with clocks or interference patterns or sagnac or light shift --etc etc
I did read a while ago about "the impossibility of measuring the speed of light" mmmmmm
may be a message in that.

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

Yes I do believe that as long as it is said that they "believe" that too.

 

[Dale]

I am thinking right now that your maths is impressive but I am not to sure we are understanding each other.

I am pretty sure that you are not understanding me, but after your posting the drawings I am confident that I am understanding you. Even if you cannot understand the math you can at least look at the equations in the different frames and see that they are essentially the same.

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

In his frame he IS at the center and green man is not.

So what makes green man's frame correct and red man's frame incorrect in your opinion? Simply because we chose to draw green man's frame? If so, then any frame is the absolute rest frame as long as we choose to draw it. Is that what you really want?

I think that the problem is that you don't even understand what "absolute rest" means. There are millions of ways to determine whether or not two things are moving relative to each other, and so far that is all any of your geometry has demonstrated. In order to experimentally determine if something is at absolute rest you need to perform an experiment where the identical experiment performed in different frames gives different results. That is simply not possible under the known laws of physics.

 

[ghwellsjr]

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?

Now there is a question to ponder but perhaps sometime in the future it may be done with clocks or interference patterns or sagnac or light shift --etc etc
I did read a while ago about "the impossibility of measuring the speed of light" mmmmmm
may be a message in that.

Reff--this is the future.

I'm sure your read that it's impossible to measure the one-way speed of light. That just means that it's impossible to measure a state of absolute ether rest. Or to put it another way, any measurement that either man makes will lead him to believe that he is the one at rest in the absolute ether and the other one is moving through it. It's only when each man realizes that the other man has the same experience that they begin to realize that neither one of them can make an exclusive claim about an absolute ether rest state.

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?

Yes I do believe that as long as it is said that they "believe" that too.

Assuming that they do, which do you believe you are, the green man or the red man?

 

[Reff]

=ghwellsjr;3446671]Reff--this is the future.

Are you serious

 

[ghwellsjr]

Reff, this is at least the second time you have mentioned the Sagnac effect as something that might identify an absolute ether rest state and those experiments were done a centure ago. The Sagnac effect is used in inertial guidance systems to measure rotational acceleration. Accelerations are absolute, not relative. This has nothing to do with identifying an absolute rest state. Both Special and General Relativity have passed every test that is thrown at them and they always pass with flying colors. How many more centuries do you think it will take before some experiment will prove them wrong?

But that's beside the point. You have been taking a different position in this thread, which is that right now, you know a way to prove that an absolute rest state exists, even if you cannot locate it. Your position has been that if photons travel at c in one frame of reference, those same photons cannot travel at c in another frame of reference moving with respect to the first one, correct?

 

[JDoolin]

Is the purpose of your diagram to graphically determine the time dilation factor as a function of speed? Would this diagram, where I called the time dilation factor "age" work? At a speed of .8c the time dilation is .6:

I never noticed that the equation for time dilation factor was so... circular.

 

[JDoolin]

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.

Oh, it is both ways, exactly. The sphere is not frame dependent, but the angles ARE frame dependent. The angles are frame dependent even in Galilean relativity. If I throw an object off a truck, it will appear to be going at a different angle than you would see it if you were standing still.

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.

I have a flash demo that let's you set the speed up to .99c if that helps.

http://www.wiu.edu/users/jdd109/stuff/relativity/Circle.swf

and another one which sort of demonstrates the angle question:

http://www.wiu.edu/users/jdd109/stuff/relativity/gardner.swf

I just want to say to gh, though: I didn't realize that you had made those animations before you saw mine! (I didn't see that you had made similar animations until this afternoon.) It is a good thing when two people independently come up with the same results.

 

[JDoolin]

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.

On further thought, what are you saying here? I'm not entirely certain what the LET idea entails. Does LET assume an absolute ether rest state, or an observer dependent rest state? Or, regardless, don't your animations suggest observer dependence, rather than any absolute nature of the spacetime?

Or are you just saying that there is no detectable difference between the two?

 

[JDoolin]

It's only when each man realizes that the other man has the same experience that they begin to realize that neither one of them can make an exclusive claim about an absolute ether rest state.

Assuming that they do, which do you believe you are, the green man or the red man?

Ahh, I see. You came back to it. How very dialectical of you!

 

[Reff]

Both Special and General Relativity have passed every test that is thrown at them and they always pass with flying colors. How many more centuries do you think it will take before some experiment will prove them wrong?

Sorry I have been slow
Any theory should be able to withstand any amount of challenge and retain its integrity, no matter how long the theory is generaly acceptable, rather like the flat Earth society time v integrity, or the Earth being the center of all things, time v integrity or us being the only galaxy in the universe.
Are you suggesting time gives a theory integrity then I absolutely disagree.
I believe you are ignoring some of the facts here including some of the evolution of SR.
If we go back to the 20s when the formation of Lorentz was proposed, do we not find many pysicists hard at work trying to find a purely mechanical interpretation of SR laws
and in that process had many problems. Now I say that if a mechanical theory were introduced at the same time as Lorentz and the man himself ( Einstein) were to make an assessment of both--
which one would Einstein himself have chosen
During his consideration of two theories would he have debunked the mechanical by using elements of the Lorentz theory.--- I don't think so
Would Einstein have been impressed by the Lorentz theory formulated on " substracting from the aether its mechanical and from matter its electromagnetic qualities" as having any form of validity. ----I don't think so
Lorentz formulated a theory that cannot be challenged if we use the time v validity formula
Without using the Lorentz rabbit can you prove that absolute rest does not exist.
How about using the sort of logical thinking that Einstein would have used against a mechanical SR theory, to say one will never exist closes an open mind.