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

[Reff]

Hi CJames
I do feel I am fighting a losing battle but still see the clarity in my observation and geometry. I thought about the "slow photon" several years ago and eventualy decided to put a little maths to it and crudely reached the required figures eventualy using c as 1 and transposing a 3 4 5 triangle to reach the rounded figures. It works with any speed. I started with GR initialy and still reached the "slow photon" geometry but the maths for that is way above me and the geometry still tricky. I would find it hard to concede having worked out SR calculations for myself.

Sure I feel a photon has a heading-- not found yet but the right experiment perhaps. If a photon can hit an electron over a distance-- sure it has a heading.

Absolute rest observable predictons not found yet but yes some smarty will devise prediction experiments and if found it gives a new meaning to relativity in general for instance as I have just said, radial construction of tranverse waves and the acceptance of an aether.


Quantum mechanics being incompatable with GR is interesting, With incompatability comes the chance of reconciling the two. So who is wrong. I would put my money on GR.
If space was classed as a liquid with transverse waves being formed by continuous intersection of the lateral frame and not a curved solid perhaps then perhaps we can class gravity as a pure flow of space and again the transverse photon moves slower than c.
Perhaps that reconciliation is closer than we think.

 

[Reff]

Hi Bessie 11
I am not a research person. I have an interest in anything geometrical. For many years I have puzzled over GR and SR and now feel I have a reasonable view of the phenominum.
I am sort of retired and on occasion make weird things! sort of inventions. If you wish to see a couple try utube search under "VAWT with rudder". There are a couple of bits there and perhaps more eventualy.
As for SR calculations they are easy to work out geometricaly but not easy to convince others how I arrived at the answers. For the moment the geometry of relativity is not going too far. GR geometry was a little more accepted.

 

[Dale]

If Absolute rest has been shown to be incorrect then I would be incorrect, but I don't believe that to be so because was it conveniently voted out not by observation but by consensus.

This is incorrect. It is more than a century of the most sophisticated observation possible: http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html

At this point, a belief in an absolute rest frame is on par with a belief in the Tooth Fairy. It cannot be ruled out in principle, but it doesn't appear to actually do anything since all available data can (more) easily be explained without it.

 

[Dale]

For the moment the geometry of relativity is not going too far.

Complete BS. Modern relativity is all about geometry. You should learn about the following:
Spacetime diagram
Minkowski norm
Spacetime interval
Four-vectors
Tensors
Riemannian geometry

 

[Reff]

Hi DaleSpam
You could be right.

 

[CJames]

Hi CJames
Sure I feel a photon has a heading-- not found yet but the right experiment perhaps. If a photon can hit an electron over a distance-- sure it has a heading.

But not an absolute heading. In one reference frame that heading is across the surface of the table. In another it is from a point where the table used to be. (I also feel the need to say that when you're talking about a photon, instead of "a pulse of light," you run into quantum mechanical issues, where the photon doesn't have a precise location or momentum.)

Absolute rest observable predictons not found yet but yes some smarty will devise prediction experiments and if found it gives a new meaning to relativity in general for instance as I have just said, radial construction of tranverse waves and the acceptance of an aether.

Lorentz ether theory makes the same predictions as SR, except that it arbitrarily chooses one reference frame as the reference frame of the ether, which only serves to complicate matters without adding any additional observable predictions.

Quantum mechanics being incompatable with GR is interesting, With incompatability comes the chance of reconciling the two. So who is wrong. I would put my money on GR.
If space was classed as a liquid with transverse waves being formed by continuous intersection of the lateral frame and not a curved solid perhaps then perhaps we can class gravity as a pure flow of space and again the transverse photon moves slower than c.
Perhaps that reconciliation is closer than we think.

GR doesn't postulate that space is either a liquid or a solid. Not sure what you're getting at here?

 

[Reff]

Hi CJames.
I feel shortly I will be consigned to the Crackpot drawer but for now
Quoting you
But not an absolute heading. In one reference frame that heading is across the surface of the table. In another it is from a point where the table used to be. (I also feel the need to say that when you're talking about a photon, instead of "a pulse of light," you run into quantum mechanical issues, where the photon doesn't have a precise location or momentum.)
This is interesting and in fact as I was about to have a browse into QM, your post came up.
Your geometry would seem to be strong and you have understood most of my intent even if you disagree re the reference frame of the crossing photon. I believe that it is --for the want of better words-- the frame of the photon which regulates everything. It is the geometry of that to me which is the king frame. A photon particle in free flight has its own datum speed. what effect has a frame on c after the photon has left. For sure it has an effect on a beam of light which must eminate from the frame but not the particle.
From an event just look at two photons leaving in opposite directions -- the moving frame has gone-- they are synchronised together and move at 2c relative to each other. They each can impart energy to a distant electron and so have a heading and a path reversal of that is back to a meeting point which has remained inertial throughout. Any frame which has to accellerate back in the opposite direction to meet that point is moving in my books. You may not agree but consider the path of the photon as being laser straight until it encounters gravity so in open gravity free space a single photon can hit a distant electron. It is also tidy to make that distant electron part of an absolute rest frame and all is precise. The heading of a photon towards a distant absolute frame, will hit that frame.
I once read a guys post re the impossibility of measuring the speed of light. He could be right but a reconciliation with QM would be interesting in that regard.
For sure I am frustrating but my geometry bas based on absolute rest for years and the maths agreeing sort of makes it hard to let go.
Quote
GR doesn't postulate that space is either a liquid or a solid. Not sure what you're getting at here?
At the risk of changing direction. I believe SR is an induced flow by frame speed and the lateral crossing photon is slow. With GR it is the flow of space where absolute rest is being taken into gravity so a frame on Earth has similar geometry to SR and the crossing photon is still slow due to its absolute rest frame having been taken in by the flow. My analogy is a launch in a river directly crossing the flow so its crossing speed is less than its hull speed.
I am sure you will be able to draw the geometry for that. I convinced a respected poster of that years ago who now uses a canoe as his analogy.
Quote
GR doesn't postulate that space is either a liquid or a solid. Not sure what you're getting at here?
The man himself said If space is a liquid then transverse waves are not allowed. If we have absolute rest and transverse waves are formed from the continuous emission of photons from a continuous movement of the photon at rest frame, then transverse waves can be formed from the process.
Thats geometry for a headache. Hard to follow if I have no credibility.
Lorentz has some interesting stuff I have not touched on for a while

 

[Dale]

I believe that it is --for the want of better words-- the frame of the photon which regulates everything.

Again, there is no such thing. https://www.physicsforums.com/showthread.php?t=511170 This is not even something that is open to the "tooth fairy" wiggle room of the idea of an absolute frame. Even if there is an absolute rest frame it is not the frame of a photon because such a concept is not even logically self consistent.

 

[CJames]

I believe that it is --for the want of better words-- the frame of the photon which regulates everything. It is the geometry of that to me which is the king frame. A photon particle in free flight has its own datum speed. what effect has a frame on c after the photon has left. For sure it has an effect on a beam of light which must eminate from the frame but not the particle.

As DaleSpam has said, and you seem to partially acknowledge, there is no "frame of the photon." I'm sure you're referring to a hypothetical electromagnetic field that is "stationary."

From an event just look at two photons leaving in opposite directions -- the moving frame has gone-- they are synchronised together and move at 2c relative to each other. They each can impart energy to a distant electron and so have a heading and a path reversal of that is back to a meeting point which has remained inertial throughout. Any frame which has to accellerate back in the opposite direction to meet that point is moving in my books.

I'm going to insist on talking about pulses of light and mirrors instead of photons and electrons, because of the quantum mechanical effects that would severely mangle this thought experiment.

So imagine two pulses of light travel outward from an origin point, hit two mirrors, and return to the origin point.

Now suppose that this setup was on the table in your previous thought experiment. The experiment would play out exactly the same way, and it would be the external reference frame that would need to "accelerate back in the opposite direction to meet that point."

For sure I am frustrating but my geometry bas based on absolute rest for years and the maths agreeing sort of makes it hard to let go.

I can understand that, but insisting on an absolute reference that has no observable predictions only leads to confusion in my opinion.

As for the rest of your post, it appears more philosophical than scientific. Whether you imagine space as a liquid or a solid, all we can really say is that it is something we can measure using speed and time.

 

[Reff]

Hi Dalespam and CJames
Thanks for your replys.
Could I just go back to what CJames calls a pulse of light. I presume it is reasonable to suggest a pulse of a very short duration say its width 1cm and its length 1cm and is also a parrallel- non convergant or divergant light. Now from the event which creates the pulse we watch two pulses of heading in opposite directions. Are you both saying they do not have a heading. Are you also saying their heading cannot be reflected directly back to meet again
If the pulse was from a .8c frame are you saying the reflected pulse will go back to the pulse generator on that frame or somewhere else because that frame has moved on.
That point between two pulses of light can be used to create more pulses in sequence which will remain centered on all the others irrespective of the speed and directions of the frames they leave. What possible effect can frame speed and direction have on a light pulse which has left it. It is free to move by its own frames laws the moment it leaves
Two pulses-- and I prefer particles close on the same trajectory from different frames, What does it matter where their frames have moved to. One photon cannot overtake another.
I have suggested before, that I have geometry to my formula whilst your formula does not re DaleSpam. How about showing me how you would do your SR calculations using geometry including vectors.
How can c be anything more than a reference to itself, photons going in the opposite direction for instance, Absolute rest What has c to do with any moving frame a pulse of light leaves other than a complicated surreal abberation.
We all know about the slowest clock, but where is the fastest clock.

 

[ghwellsjr]

Hi Dalespam and CJames
Thanks for your replys.
Could I just go back to what CJames calls a pulse of light. I presume it is reasonable to suggest a pulse of a very short duration say its width 1cm and its length 1cm and is also a parrallel- non convergant or divergant light. Now from the event which creates the pulse we watch two pulses of heading in opposite directions. Are you both saying they do not have a heading. Are you also saying their heading cannot be reflected directly back to meet again
If the pulse was from a .8c frame are you saying the reflected pulse will go back to the pulse generator on that frame or somewhere else because that frame has moved on.
That point between two pulses of light can be used to create more pulses in sequence which will remain centered on all the others irrespective of the speed and directions of the frames they leave. What possible effect can frame speed and direction have on a light pulse which has left it. It is free to move by its own frames laws the moment it leaves
Two pulses-- and I prefer particles close on the same trajectory from different frames, What does it matter where their frames have moved to. One photon cannot overtake another.
I have suggested before, that I have geometry to my formula whilst your formula does not re DaleSpam. How about showing me how you would do your SR calculations using geometry including vectors.
How can c be anything more than a reference to itself, photons going in the opposite direction for instance, Absolute rest What has c to do with any moving frame a pulse of light leaves other than a complicated surreal abberation.
We all know about the slowest clock, but where is the fastest clock.

Reff, if you would relate what you are saying to my animations, I think it could help us understand what you are trying to say, since you have not produced your own diagrams.

My first animation seems to correspond to your scenario concerning two pulses of light that head in opposite directions and are reflected back to their point of origin, except that I have a whole bunch of pulses of light going in every possible direction, but you could draw a line through the head of the green stick man and that would correspond to the headings of your two pulses. They go out and hit a portion of the mirrror on opposite sides, simultaneously and reflect back and collapse on the man at the same time. Doesn't that fit with what you are saying for the first part?

Then for the next part, you talk about a .8c frame but I used a .5c "frame" except I hate to call it a frame because we're still looking at it from the same frame as the first situation. What we have now is another observer who is traveling at .5c and when he gets to the point where the first man was, the same flash of light is emitted in all directions. It's the same flash as in the first case but this time the moving man carries along with him his own set of mirrors and when the light hits his mirror, it reflects back differently than it did for the first man who was stationary in the frame. And when it reflects back it hits the moving man at a new location.

You can do the same thing I suggested for the first part, draw a line through the moving man's head at the point where the flash is emitted and pretend like there are just two pulses of light going in opposite directions, but they travel relative to the stationary frame, not relative to the moving man. They eventually hit the moving man's mirror and you have to draw a new line from the point of contact to the point where the man will be when he finally gets to the point where all the pulses of light collapse simultaneously on his head.

I drew the black dashed line so that you would recognize it as an ellipse with the two foci corresponding to the starting point of the flash and the collapsing point of the flash and I'm sure you're aware that every line going from the first focus point to a point on the ellipse and back to the second focus point has the same total distance as any other line. This is a requirement of the light traveling at a constant speed in the stationary frame, not in the moving frame.

I cannot understand what you mean here:

That point between two pulses of light can be used to create more pulses in sequence which will remain centered on all the others irrespective of the speed and directions of the frames they leave. What possible effect can frame speed and direction have on a light pulse which has left it. It is free to move by its own frames laws the moment it leaves​

And the reason this doesn't make any sense is because you keep talking about light pulses leaving frames. What does that mean? Whenever you're dealing with a scenario like this, you must use just one frame for everything. All the light pulses travel at c in that one frame. It doesn't matter how they got emitted, by a source stationary in the frame or a source moving at a high speed in the frame. Light pulses don't leave frames.

And I can't make any sense out of the rest of your post.

So could you please relate the two situations of an observer at rest in a frame to my first animation and an observer moving with respect to a frame in my second animation and see if you agree with what they are depicting. If not, what do you find objectionable with them?

 

[Dale]

Are you both saying they do not have a heading. Are you also saying their heading cannot be reflected directly back to meet again

They have a heading in any reference frame. The heading is frame-dependent. When a pulse of light hits a reflective surface it obeys the laws of reflection in all frames.

I have suggested before, that I have geometry to my formula whilst your formula does not re DaleSpam. How about showing me how you would do your SR calculations using geometry including vectors.

Certainly. For reference see the following links which describe the geometric elements used here:
http://en.wikipedia.org/wiki/Four-vector
http://en.wikipedia.org/wiki/Minkowski_space

I will use a convention where capital letters are for four-vectors, and $X \cdot Y$ denotes the Minkowski inner product of the four-vectors X and Y.

A sphere of light emitted at t=0 from the origin is given by all events X such that:
$$X \cdot X = 0$$

If $\eta$ is a Lorentz transform then it preserves the Minkowski inner product such that in another reference frame we have
$$X' = \eta X$$
$$X' \cdot X' = X \cdot X = 0$$

So in any other frame the events also form a sphere of light emitted at t=0 from the origin.

How can c be anything more than a reference to itself, photons going in the opposite direction for instance, Absolute rest What has c to do with any moving frame a pulse of light leaves other than a complicated surreal abberation.
We all know about the slowest clock, but where is the fastest clock.

Reff, there is no physical significance to the concept of absolute rest to the best experimental accuracy possible to date. This has been demonstrated over and over and over again during the course of the last 100+ years of increasingly sophisticated and accurate experimental tests. See: http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html . The fact that you are unaware of the experimental evidence against your position does not invalidate that evidence.

Your geometrical reasoning is also incorrect. This was demonstrated by me in post 7. The fact that you are unable to follow the algebra does not invalidate the math.

You need to spend less effort trying to justify your mistake and more effort learning the relevant scientific and mathematical concepts that you are missing. Once you have done so then you will either understand why you are wrong or you will have the tools to defend your position rationally.

 

[CJames]

Now from the event which creates the pulse we watch two pulses of heading in opposite directions. Are you both saying they do not have a heading. Are you also saying their heading cannot be reflected directly back to meet again

As Dalespam said, they have a heading but that heading is different in each reference frame. Yes, they can be reflected directly back to meet again. All I'm saying is that you could place the two mirrors on a table that is "moving" at .8c, or you could place the two mirrors on the floor which is "at rest," and in both scenarios the light would be reflected back to the origin.

What looks like two pulses of light moving away from the origin are returning to it in one reference frame, looks like two pulses of light moving away from one another, and meeting again at some other point in space.

If the pulse was from a .8c frame are you saying the reflected pulse will go back to the pulse generator on that frame or somewhere else because that frame has moved on.

Within any reference frame, if there are two mirrors and a source of light bisecting them, two pulses of light will reflect off of the mirrors and return to the source of light simultaneously. It doesn't matter if that frame is "at rest," because all frames are at rest wrt themselves.

What possible effect can frame speed and direction have on a light pulse which has left it.

Redshift/blueshift.

It is free to move by its own frames laws the moment it leaves
Two pulses-- and I prefer particles close on the same trajectory from different frames, What does it matter where their frames have moved to. One photon cannot overtake another.

Nobody is saying one photon can overtake another. Light pulses don't have their own frame.

How can c be anything more than a reference to itself, photons going in the opposite direction for instance, Absolute rest What has c to do with any moving frame a pulse of light leaves other than a complicated surreal abberation.
We all know about the slowest clock, but where is the fastest clock.

It is not traveling at c wrt itself, it's traveling at c wrt everything else. The fastest clock is in any inertial reference frame, measured from within that reference frame.

 

[Reff]

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.

 

[Reff]

Hi CJames and DaleSpam
You have both given me plenty of information to get on with. Thanks for the effort. I would have to be more than a little obstinate to try and convince you guys otherwise. Blind and cannot see. perhaps I will have a look at QM,there would appear to be conflict of theory there. For sure I believe I am right but I understand to go on will mean contravening the agreement to post plus I now fancy a beer. If this remains open I may just answer your posts on a nicer day.

 

[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.