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

[Dale]

we find he is measuring a slow photon.

No such thing.

every single photon will be emitted from its own at rest frame.

Also, no such thing.

observed by an absolute rest observer

If there is such a thing nobody knows, including the lucky absolute rest observer. It is a physically useless concept, but feel free to use it if it makes you feel better. Lorentz aether theory is experimentally indistinguishable from SR, and basically consists of doing normal SR and occasionally throwing in a transformation to a frame labeled "aether" which is moving with an unknown velocity.

Why I say this is simply because two photons heading in opposite directions have a start point and if they return from their heading after each has completed a specific distance they
will arrive at the same time. This I believe is the frame of the photon and that point would have to be absolute rest.

Then according to this all frames are absolute rest.

 

[ghwellsjr]

Hi ghwellsjr- Quoting
I'm not sure if you answered my third question. I asked about a single sphere of light and you answered about spheres of light. Are you thinking that the light is behaving differently for each observer, that is, there is one expanding sphere of light for the stationary observer and a different expanding sphere of light for the moving observer?

I believe both observers will see an expanding sphere of light centered on them but if we stay centered on all the the headings or reciprocals--world lines and observe the two frames, then the moving observer I believe is mistaken. A marking event from this observer will show up the aberation.
We should be able to center ourselves on any "event" and consider a photon on the edge of the sphere at each radius and a measurement of each radius should be identical. Whilst still observing the sphere, a frame which created it moving at say .9c would no longer be centered after any propagation and frame movement. The frame would need to reverse to be hit by all the photons if the photons reversed direction. I see light moving at c absolutely independently of the speed of the frame it leaves. That part of the frame it leaves must be within the sphere even if the frame is fast moving and close to the edge. The only no aberation frame to view this from is centered on the sphere and the center of it marks absolute rest-- yes this is purely my view for the present. My geometry works fine on with this

OK, in this and other posts, you have indicated that you don't have a correct understanding of what's going on with an expanding sphere of light. I have created a series of animations to illustrate and hopefully help you understand what's happening.

First of all, you have indicated that you think that an absolute rest state, one that has been described as an ether, is the only one in which light travels at c. That's fine, we'll start with that idea and later I will show that it is identical to selecting any arbitrary inertial frame of reference as defined in Einstein's Theory of Special Relativity. So let's pretend that there exists an absolute ether rest state in which light travels at c in all directions and only in that state does light travel at c in any direction.

Now imagine an observer at rest in this ether and he sets off a flash of light. Wouldn't everyone agree that this will form an expanding sphere of light, enlarging at the speed of light, with the observer in the center? (We won't concern ourselves with the individual photons, we'll assume that the light is so bright that it creates a wavefront of light traveling away from our observer.) Now we have to ask ourselves the question: how does the observer decide if he is in the center of this expanding sphere of light? He cannot see the light that is traveling away from him, can he? It's gone away from him.

Well, the easiest way for him to figure out if he is in the center is to place a bunch of mirrors equally spaced from himself and wait for the sphere of light to hit the mirrors and reflect back to him. If he sees the reflections from all the mirrors arrive back to him at the same time, he can validly deduce that he was in the expanding sphere of light. In our diagrams and animations, we get to watch the wavefront of the light both while it is expanding and after the reflection, while it is collapsing, but we have to understand that the observer in the center of all this activity only knows of the initial emission of the light and its final reception, both of which happen at his location.

Now an animation to illustrate an expanding sphere of light in three dimensions would be very difficult to implement, but we can easily demonstrate the salient points by limiting it to a plane with an expanding circle of light, a circle of mirrors, and a contracting circle of reflected light. In the video that follows, the observer is shown as a green stick man with the light emitted as a blue circle from the center of his round head. He has placed a full circle of yellow mirrors all around him at an equal distance from his round head. The expanding circle of blue light reflects off the circular mirror as a collapsing circle of green light. Watch this:



Pretty simple, isn't it?

Next we want to consider another observer who is traveling at one-half the speed of light and see what happens to him. He also will place what he thinks is a circle of mirrors some equal distance around him. But it turns out that due to length contraction along the direction of his motion, his mirrors actually form an ellipse as shown in this animation:



Here, the moving observer is shown as a contracted red stick man and the reflection of the circle of light is also in red. Do you see how the light always travels at c no matter whether it is traveling away from him or reflected back towards him? Do you also see how the elliptical shape of the mirrors actually transplants the center of the expanding circle of light to a new location where the collapsing circle of light ends up?

The thin black dashed line shows the locus of points where the two partial circles of light contact the mirror and can be used to illustrate the reflections that occur in a light clock, although, usually a light clock is shown with just the horizontal sections of mirror and a light flash traveling up and down between them.

Finally, in order to show both stick men at the same time, they need to make their mirrors be partial, otherwise one of them will collect all of the reflected light leaving none for the other one:



Notice how the reflected light first collapses on the green stationary man and then later on the red moving man. This illustrates the time dilation experienced by the moving man.

Now it turns out that as far as the men can determine, they each are having exactly the same experience. The moving man not only concludes that he is in the center of the expanding circle of light, he also believes, and has every reason to believe, that the light struck every part of his mirrors at the same time and that his mirrors formed a perfect circle. Furthermore, he believes that the speed of light is a constant c in all directions relative to him because that is what he measures and even though he cannot see the progress of the light, it behaves exactly as if he were stationary in an absolute ether rest state.

It also turns out that any Frame of Reference that is defined according to the requirements of the Theory of Special Relativity will look exactly like these same animations--you cannot tell any difference between them and the absolute ether rest frame of the Lorentz Ether Theory.

 

[CJames]

Just consider this
Look closer at this beam of light and look at the photon make up. I believe we need to consider the release of each photon particle being radialy emitted from the at rest target and as the moving frame progresses another particle is emitted from a new at rest target point. This is a continuous process and the statement I have in mind is yes you can have a lateral beam of light which is not at rest orientated but every single photon will be emitted from its own at rest frame. So I believe all light waves from a moving frame have a lateral wave but all particles within it are -radialy- emmitted from their own event observed by an absolute rest observer. I presently happy with that I believe, so is my geometry.

As I have already said, you can arbitrarily choose the external reference frame to be at "absolute rest." It works. It just doesn't have any physical meaning in any known way.

In both reference frames, the beam of light started at the left side of the table. In both reference frames, the beam of light is now located 6cm across the surface of the table. In one reference frame, the table was moving, and the light has traveled 10 cm. In another reference frame, the table was at rest, and the light has traveled 6 cm.

There is no known scientific way to choose which reference frame was at rest. It is completely arbitrary.

If you insist on saying that one reference frame is at rest, you can still get the right answers. It looks like this is what you have done. Maybe it is simply easier for you to visualize it that way. I just hope you aren't insisting that there has to be a reference frame at absolute rest, because nothing you've portrayed here indicates that there must be.

 

[Reff]

Hi ghwellsjr and other posters
Well thanks for your last post. The videos are brilliant. I need a little time here to digest them, as you may have noted I am more of a procrastinator. I had a realisation overnight re my geometry of the .8c frame dilation which was explained in clearer terms by CJames, that readers are believing that I am on about a beam of light and not a particles, I will explain. If we go back to the .8c geometry and I am not discounting yours ghwellsjr. If we use a beam of light and allow it to form a sphere of light 200mm dia then it is obvious to most that the beam is still being generated at the scource in any frame.The trajectory is straight across the table which would seem to make me incorrect. If we take frames at any speed we have the same effect. All frames experience the same as you have impressively demonstrated.
With my geometry, it is a single particle on a constant state of intersecting the tabletop and not a beam of light. If that photon returns straight back on its heading, then the frame must reverse to meet it. If the frame was at absolute rest then the photon will return to that frame and that frame remains inertial. The beam of light is a different scenario. If the beam has moved 100mm, it is still being generated at the left hand side of the tabletop which logicaly one would take as a point of origin but, consider a constant stream of photons within the beam as being generated from a constant new point of absolute rest.
There is no length contraction for my geometry as the tabletop lies in the lateral plane and the maths work fine. There is no pythagarus required as the 200mm dia circle and the associated relativity are precisely to scale.
Quoting Drakkith re absolute rest.
That is what happens when things are shown to be incorrect. They become obsolete and aren't used anymore. This has happened countless times in the past, why is absolute rest any different than anything else before it?

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.

CJames had a good understanding of the geometry although he had problems with it.
I would ask that consideration be given to a single particle crossing a tabletop and its point of origin, because it is not a beam of light which is still being transmitted during propagation and which would return to the light generator, now the particle has an origin at the confluence of all the particle headings now allow the frame to progress a further cm and create the next event, which now becomes another photon in a state of constant intersection with the tabletop. Now constanly release particles while the moving frame continues on and the constructed beam would appear to be crossing smoothly across the table with the beam starting at the event generator, but every single particles heading will go back to absolute rest.
Was it Einstein who said the " Aether must be the nature of a solid body, because transverse waves are not possible in a fluid, but only in a solid"
What I believe is All photons are radialy emitted from absolute rest and form transverse waves appearing to come from any moving frame.
Does QM have some incompatability issues with SR and perhaps GR.

 

[CJames]

If that photon returns straight back on its heading, then the frame must reverse to meet it. If the frame was at absolute rest then the photon will return to that frame and that frame remains inertial.

How do you determine whether or not a photon is returning straight back on its heading? There is no known experiment you can perform to prove that this is what is happening.

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.

The point is, the idea of absolute rest ceased to have any observable predictions. What scientific use is a theory that doesn't make predictions?

Does QM have some incompatability issues with SR and perhaps GR.

Quantum field theory reconciles quantum mechanics and SR. Quantum mechanics is currently incompatible with GR. It is every theoretical physicist's dream to figure out how to reconcile the two.

 

[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 totally 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 totally 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.
.