Naked Science Forum
On the Lighter Side => New Theories => Topic started by: Waste of Time on 16/02/2015 22:54:22

The following scenario utilizes events that are applicable to Einstein’s Special Relativity Theory. What follows requires further analysis regarding application of the first postulate WRT SRT. For information with images, see http://gsjournal.net/ScienceJournals/Research%20PapersRelativity%20Theory/Download/5916.
 Definitions; (Beta) β = v/c, (Gamma) γ = (1v^2/c^2)^1/2.
 Electromagnetic relations (second postulate); x = cΔτ, x’ = cΔτ’.
 There are two parallel linear events (A and B) with uniform velocity along the positive xaxis.
 Event A is an electromagnetic event (velocity = c) following the path from point x = 0 to x.
 Event B is an inertial body event (velocity = u) following an identical and parallel path from point x = 0 to x.
EVENTS OCCURRING IN REST FRAME (v = 0)
 These events now occur within a reference frame considered to be at rest (rest frame).
 The event time for A is; Δτ = x/c.
 The event time for B is; Δt = x/u.
 The ratio of these event times gives us the relationship between their event times;
Δτ/Δt = (x/c)/(x/u) = u/c.
EVENTS OCCURRING IN INERTIAL FRAME (v)
 These events now occur within a reference frame considered to be in uniform motion (inertial frame).
 The SRT event time for A is; Δτ = (Δτ’+vx’/c^2)γ. Applying the second postulate;
Δτ = (Δτ’+vx’/c^2)γ, x’ = cΔτ’
Δτ = (Δτ’+v(cΔτ’)/c^2)γ
Δτ = (Δτ’+vΔτ’/c)γ
Δτ = Δτ’(1+ β)γ
 The SRT event time for B is; Δt = (Δt’+vx’/c^2)γ.
 The ratio of these event times gives us their the relationship between their event times;
Δτ/Δt = [Δτ’(1+ β)γ]/[(Δt’+vx’/c^2)γ]
Δτ/Δt = [Δτ’(1+ β)]/[(Δt’+vx’/c^2)]
Δτ/Δt = [Δτ’(1+ β)]/[Δt’(1+vx’/(c^2)Δt’)]
Δτ/Δt = [Δτ’(1+ β)]/[Δt’(1+ βx’/cΔt’)]
In order to satisfy the first postulate, the relationship between their event times must remain unchanged in order to prevent one from ascertaining the motion of the inertial frame utilizing this scenario. As we can clearly see, the relationship between their event times can only be regained by allowing an electromagnetic relation (x’ = cΔt’) to be utilized to reduce the Δt relation to (1+ β) in order to regain equality for these events.
Δτ/Δt = [Δτ’(1+ β)]/[Δt’(1+ βx’/cΔt’)], x’ = cΔt’
Δτ/Δt = [Δτ’(1+ β)]/[Δt’(1+ β)]
Δτ/Δt = Δτ/Δt
If we were to allow this substitution, then we would have to admit that in order for the first postulate to remain valid, then the Lorentz transformations (in their current format) could only apply to electromagnetic events. It appears that the only recourse would be to reformat the Lorentz temporal transformation to;
***Δτ = Δτ’(1+ β)γ

http://en.wikipedia.org/wiki/Principle_of_relativity
Principle of relativity
"In physics, the principle of relativity is the requirement that the equations describing the laws of physics have the same form in all admissible frames of reference.
For example, in the framework of special relativity the Maxwell equations have the same form in all inertial frames of reference. In the framework of general relativity the Maxwell equations or the Einstein field equations have the same form in arbitrary frames of reference.
Several principles of relativity have been successfully applied throughout science, whether implicitly (as in Newtonian mechanics) or explicitly (as in Albert Einstein's special relativity and general relativity)."
Now I haven't even had the time to read carefully through your post but are you saying this is not correct? Are you saying it needs to be modified?

So what you are saying is that light is intrinsically linked to gravitation.

All that I show is that the current format of the Lorentz temporal transformation complies with the first postulate for roundtrip events, but not oneway events. We can either choose to accept this as a limitation of STR, or choose a format for the temporal transformation that does comply with the first postulate for oneway events.

All that I show is that the current format of the Lorentz temporal transformation complies with the first postulate for roundtrip events, but not oneway events. We can either choose to accept this as a limitation of STR, or choose a format for the temporal transformation that does comply with the first postulate for oneway events.
Hi Robert
Hadn't come across this argument before, and as this is not my subject area I, like Jeffrey, will need to read though a few times. But a couple of initial observations:
I can't see why what you've written applies to one way events and not to round trip events.
Some of the reasoning in your gsjournal paper is a little obscure and it's easy to lose track of what you are saying.
In your uniform motion frame you appear to apply dilation to the time of travel of the EM wave, have I misunderstood?
In your paper you say
"Since the principle of relativity requires that the laws of physics remain valid for all
reference frames, then the resulting measurements of the two events must be identical in a state of motion as they were observed at rest."
At first sight it appears to me that this could be a confusion between who is observing and from which frame, can you make this clearer please?. The introduction of your 'unity' as a physical law could well be compounding this confusion.
As I say these are only my first thoughts and I need to go away and think over. Greater minds than mine may well see instantly what you are saying.
Just a comment, some people here refuse to read nonpeer reviewed papers and self published theories, so you might find limited responses to your post.

I will be coming back to this.

So what you are saying is that light is intrinsically linked to gravitation.
I thought variability of light was only seen in accelerating frames eg influence of gravity, not inertial frames.
Did I miss something in the arguments?

 There are two parallel linear events
What is "parallel linear events" supposed to mean?
 Event A is an electromagnetic event (velocity = c)...
There is no such thing as an electromagnetic event. What do you mean when you use that phrase?
 Event B is an inertial body event (velocity = u) ...
There is no such thing as an inertial body event . What do you mean when you use that phrase?
EVENTS OCCURRING IN REST FRAME (v = 0)
 These events now occur within a reference frame considered to be at rest (rest frame).
 The event time for A is; Δτ = x/c.
 The event time for B is; Δt = x/u.
What is "x"?
What is the difference between τ and t? Or are those terms defined by those relations?
 The ratio of these event times gives us the relationship between their event times;
Δτ/Δt = (x/c)/(x/u) = u/c.
EVENTS OCCURRING IN INERTIAL FRAME (v)
 These events now occur within a reference frame considered to be in uniform motion (inertial frame).
 The SRT event time for A is; Δτ = (Δτ’+vx’/c^2)γ. Applying the second postulate;
Δτ = (Δτ’+vx’/c^2)γ, x’ = cΔτ’
Δτ = (Δτ’+v(cΔτ’)/c^2)γ
Δτ = (Δτ’+vΔτ’/c)γ
Δτ = Δτ’(1+ β)γ
 The SRT event time for B is; Δt = (Δt’+vx’/c^2)γ.
 The ratio of these event times gives us their the relationship between their event times;
Δτ/Δt = [Δτ’(1+ β)γ]/[(Δt’+vx’/c^2)γ]
Δτ/Δt = [Δτ’(1+ β)]/[(Δt’+vx’/c^2)]
Δτ/Δt = [Δτ’(1+ β)]/[Δt’(1+vx’/(c^2)Δt’)]
Δτ/Δt = [Δτ’(1+ β)]/[Δt’(1+ βx’/cΔt’)]
It's not enough to merely write down formulas. One has to explain exactly what those formulas mean and what they're describing. I can't see that from what you've written down so far.
In order to satisfy the first postulate, the relationship between their event ..
I don't understand. Who are "they" when you say "their event"?
times must remain unchanged in order to prevent one from ascertaining the motion of the inertial frame utilizing this scenario.
That's not at all what the first postulate means. All it means is that you can't determine an inertial frames absolute motion through space. It doesn't mean that you can't determine its motion relative to another inertial frame.

So what you are saying is that light is intrinsically linked to gravitation.
I thought variability of light was only seen in accelerating frames eg influence of gravity, not inertial frames.
Did I miss something in the arguments?
No you didn't miss anything. It just gave me a thought which I am following up on. I followed his train of thought with difficulty but saw something interesting. It might be garbage but I am following it up.

This is what HeyBert's time dilation profile for one way trips looks like.
Here is the standard plot of time dilation.
http://commons.wikimedia.org/wiki/File:Time_dilation.svg
EDIT: reuploaded the graph as I forgot the square root in the first one.

So what you are saying is that light is intrinsically linked to gravitation.
I thought variability of light was only seen in accelerating frames eg influence of gravity, not inertial frames.
Did I miss something in the arguments?
No you didn't miss anything. It just gave me a thought which I am following up on. I followed his train of thought with difficulty but saw something interesting. It might be garbage but I am following it up.
Understood, sometimes good ideas come from thinking laterally about garbage.
However, I've made my mind up about this one. My gut feeling was right:
At first sight it appears to me that this could be a confusion between who is observing and from which frame, can you make this clearer please?. The introduction of your 'unity' as a physical law could well be compounding this.
Confirmed by PmbPhy  thanks PmbPhy
In order to satisfy the first postulate, the relationship between their event ..
I don't understand. Who are "they" when you say "their event"?
OK HeyBert, I see what's happened. I will refer to the frames you call 'at rest' and 'moving'.
If we take an observer who was 'at rest' and is now moving, call him RM, at rest he will measure v, u, x, t, T and when moving v', u', x', t', T' and find that v=v', u=u' etc in other words everything appears as it was when at rest. As implied in the first postulate, RM cannot determine from physical laws whether he is at rest or not.
However, you have applied transforms to the moving frame and that is only valid for an observer who has remained, ie still is, at rest (call him RR).
You now appear to try to apply the equality* valid only for RM, to the observations of RR. These two sets of observations cannot be mixed, hence you create a false conclusion. (* what you state as "the relationship between their event times must remain unchanged")
I think you have misled yourself by using lots of ratios using Δ. If you had stuck to plain v, u, t etc for each observer you would have spotted the problem earlier. To be fair, you were rather set up by the paper you quote which makes the same errors.
GIGO
I wish you well in your studies, but would suggest you check out the excellent resources on the net describing relativity, correctly. Your teacher should be able to point out some relevant sites or texts.

Perhaps we can look at it this way as my original choice of words may have not been understood clearly by all...(τ and t indicate time spans, or time intervals measured by a clock).
In a stationary laboratory frame (considered to be at rest), a photon and electron race each other along the positive xaxis for an identical distance (they both start the race simultaneously at x = 0). Although the photon and electron travel an identical distance along the xaxis, their velocities and time span for traversing this distance will be different. Utilizing the Lorentz transformations with respect to the stationary laboratory frame, calculate the ratio of the time span it takes for the photon to traverse this distance to the time span it takes for the electron to traverse this distance (τ/t).
Now we introduce another inertial frame at uniform velocity (v), relative to the stationary frame, whose x'axis moves along the stationary frames' positive xaxis. We reproduce the identical photon vs. electron race in the inertial frame at velocity (v), yet we start this race when the coordinate system origins of the stationary frame and inertial frame at velocity (v) coincide. Utilizing the Lorentz transformations with respect to the stationary laboratory frame, calculate the ratio of the time span it takes for the photon to traverse this distance to the time span it takes for the electron to traverse this distance (τ/t).
Does the ratio of these two scenarios mathematically equal each other (same race happening in two different inertial frames)? If so, then how? If not, then what would stop us from determining the velocity of the inertial frame at velocity (v) by conducting such a "race" experiment and measuring how the ratio changes with increasing velocity of the inertial frame?
Fascinating discussions BTW.

My calculations, where γ = (1v^2/c^2)^1/2;
When the race occurs within the stationary laboratory frame (v = 0), no calculation is needed (inverse Lorentz transformation reduces to Galilean format) and the ratio of the photon time span to the electron time span WRT the stationary laboratory frame is simply;
Photon: τ = τ'
Electron: t = t'
Ratio #1: τ/t = τ'/t'
When the race occurs within the inertial frame at velocity (v), the calculation of the ratio of the photon time span to the electron time span WRT the stationary laboratory frame IAW the inverse Lorentz transformation is;
Photon: τ=(τ'+vx'/c^2)γ
Since it is a photon, the second postulate gives x' = cτ'
τ=(τ'+v(cτ')/c^2)γ
τ=(τ'+v(τ')/c)γ
τ=τ'(1+v/c)γ
Electron: t=(t'+vx'/c^2)γ
Ratio #2: τ/t = [τ'(1+v/c)γ]/[(t'+vx'/c^2)γ]
τ/t = τ'(1+v/c)/(t'+vx'/c^2)
τ/t = τ'(1+v/c)/(t'+vx't'/t'c^2)
τ/t = τ'(1+v/c)/t'(1+vx'/t'c^2)
τ/t = τ'(1+v/c)/t'(1+(v/c)(x'/ct'))
Clearly these ratios differ. Why?

What about a roundtrip race? Well, let's walk the dog on that one shall we...
We will utilize the same photon/electron race scenario as before, only now we add the requirement for the photon and electron to be reflected (after traveling an identical distance along the positive xaxis) and travel an identical distance back along the negative xaxis. The new roundtrip times will be reflected as rτ for the photon and rt for the electron.
When the race occurs within the stationary laboratory frame (v = 0), no special calculation is needed (Lorentz transformations reduce to Galilean format) and the ratio of the photon time span to the electron time span WRT the stationary laboratory frame is (1/2 the trip is in the positive xaxis direction, the other 1/2 of the trip is in the negative xaxis direction, where both photon and electron end up returning to their perspective points of origin);
Photon: rτ = τ'+τ'
rτ = 2τ'
Electron: rt = t'+t'
rt = 2t'
Ratio #1: rτ/rt = 2τ'/2t'
rτ/rt = τ'/t'
When the race occurs within the inertial frame at velocity (v), the calculation of the ratio of the photon time span to the electron time span WRT the stationary laboratory frame IAW the Lorentz transformations is;
Photon: rτ = (τ'+vx'/c^2)γ + (τ'+v(x')/c^2)γ
rτ = (τ'+vx'/c^2)γ + (τ'vx'/c^2)γ
Since it is a photon, the second postulate gives x' = cτ'
rτ = (τ'+v(cτ')/c^2)γ + (τ'v(cτ')/c^2)γ
rτ = (τ'+v(τ')/c)γ + (τ'+v(τ')/c)γ
rτ = τ'(1+v/c)γ + τ'(1v/c)γ
rτ = τ'γ+τ'(v/c)γ+τ'γτ'(v/c)γ
rτ = 2τ'γ
*NOTE: (2τ'γ) was expected as this is the same roundtrip time that would occur for the photon if the race occurred along the yaxis (reference "http://en.wikipedia.org/wiki/Michelson–Morley_experiment", 'Length contraction and Lorentz transformation' section). See also "http://www.people.fas.harvard.edu/~djmorin/chap11.pdf".
Electron: rt = (t'+vx'/c^2)γ + (t'+v(x')/c^2)γ
rt = (t'+vx'/c^2)γ + (t'vx'/c^2)γ
rt = t'γ+t'(vx'/c^2)γ+t'γt'(vx'/c^2)γ
rt = 2t'γ
*NOTE: (2t'γ) was expected as the same relativistic delay (γ factor) encountered by the photon would also be expected to affect the electron in order for the measured race results to appear the same to the inertial frame at velocity (v) as when the race occurred within and was measured by the stationary laboratory frame.
Ratio #2: rτ/rt = (2τ'γ)/(2t'γ)
rτ/rt = τ'/t'
The ratios are identical as expected since "The laws of physics are the same in all inertial frames of reference." This means that the race results measured by the stationary laboratory frame when the race occurs within the stationary laboratory frame as well as the race results measured by the inertial frame at velocity (v) when the race occurs within the inertial frame at velocity (v) will be identical. This is Einstein's first postulate in action!
Why are the ratios equal for the roundtrip races, yet unequal for the oneway races?

OK, things are staring to look clearer, but before we confuse ourselves with maths let's be really clear that we are talking about the same things.
Perhaps we can look at it this way as my original choice of words may have not been understood clearly by all...(τ and t indicate time spans, or time intervals measured by a clock).
Which clock? Sorry to labour this but there are 2 clocks, one in the rest frame and one in the moving frame.
For the observer in the moving frame the moving clock measures the same time intervals as when he made the same measurements in the rest frame so T/t is the same. So by doing a physics experiment he sees no difference between the frames. From his point of view he is stationary (because he is moving with the frame) and no Lorentz transforms are required.
For the observer in the rest frame observing the moving frame, he sees the clock in the moving frame measuring time more slowly than the one he has next to him in the rest frame. So the T/t is not the same and this difference is calculated using the Lorentz transforms.
Does the ratio of these two scenarios mathematically equal each other (same race happening in two different inertial frames)? If so, then how? If not, then what would stop us from determining the velocity of the inertial frame at velocity (v) by conducting such a "race" experiment and measuring how the ratio changes with increasing velocity of the inertial frame.
As we can see, it depends who is doing the observing.
For the observer who was stationary and is now moving there is no difference and this is what the postulate is all about. The observer cannot tell by the race experiment whether he is moving or not. He can however, look out of the window and see he is moving away from the rest frame, so would perceive himself as moving.
For the observer who is stationary observing the moving frame, no the ratios are not the same. That's what relativity is all about, frames moving relative to one another.
Again sorry to labour this, but it think there is a confusion of observers.
I hate to say it but I don't think you are comparing light with light! [:)]

My calculations, where γ = (1v^2/c^2)^1/2;
When the race occurs within the stationary laboratory frame (v = 0), no calculation is needed (inverse Lorentz transformation reduces to Galilean format) and the ratio of the photon time span to the electron time span WRT the stationary laboratory frame is simply;
Photon: τ = τ'
Electron: t = t'
You keep posting things like this without defining them. What you said tells us nothing about what those quantities are/mean.

Which clock? Sorry to labour this but there are 2 clocks, one in the rest frame and one in the moving frame.
For the observer in the moving frame the moving clock measures the same time intervals as when he made the same measurements in the rest frame so T/t is the same. So by doing a physics experiment he sees no difference between the frames. From his point of view he is stationary (because he is moving with the frame) and no Lorentz transforms are required.
I use the same definitions for the variable "types" as Einstein within his book "Relativity: The Special and General Theory", i.e. the primed variable is with respect to the coordinate system K' and the unprimed variable is with respect to the coordinate system K. I use the Lorentz transformations in the same manner as well.

My calculations, where γ = (1v^2/c^2)^1/2;
When the race occurs within the stationary laboratory frame (v = 0), no calculation is needed (inverse Lorentz transformation reduces to Galilean format) and the ratio of the photon time span to the electron time span WRT the stationary laboratory frame is simply;
Photon: τ = τ'
Electron: t = t'
You keep posting things like this without defining them. What you said tells us nothing about what those quantities are/mean.
"Utilizing the Lorentz transformations with respect to the stationary laboratory frame, calculate the ratio of the time span it takes for the photon to traverse this distance to the time span it takes for the electron to traverse this distance (τ/t)." Already defined both time variables here.
"When the race occurs within the inertial frame at velocity (v)." Already defined what this velocity is here.
For all the primed variables, the clarification is as follows. IAW Einstein's book "Relativity: The Special and General Theory", the primed variable is with respect to the coordinate system K' and the unprimed variable is with respect to the coordinate system K.

I am definitely learning a lot about the specific vernacular used to communicate these physics ideas to others on this forum...thanks for the feedback.

OK, things are staring to look clearer, but before we confuse ourselves with maths let's be really clear that we are talking about the same things.
Perhaps we can look at it this way as my original choice of words may have not been understood clearly by all...(τ and t indicate time spans, or time intervals measured by a clock).
Which clock? Sorry to labour this but there are 2 clocks, one in the rest frame and one in the moving frame.
For the observer in the moving frame the moving clock measures the same time intervals as when he made the same measurements in the rest frame so T/t is the same. So by doing a physics experiment he sees no difference between the frames. From his point of view he is stationary (because he is moving with the frame) and no Lorentz transforms are required.
For the observer in the rest frame observing the moving frame, he sees the clock in the moving frame measuring time more slowly than the one he has next to him in the rest frame. So the T/t is not the same and this difference is calculated using the Lorentz transforms.
Does the ratio of these two scenarios mathematically equal each other (same race happening in two different inertial frames)? If so, then how? If not, then what would stop us from determining the velocity of the inertial frame at velocity (v) by conducting such a "race" experiment and measuring how the ratio changes with increasing velocity of the inertial frame.
As we can see, it depends who is doing the observing.
For the observer who was stationary and is now moving there is no difference and this is what the postulate is all about. The observer cannot tell by the race experiment whether he is moving or not. He can however, look out of the window and see he is moving away from the rest frame, so would perceive himself as moving.
For the observer who is stationary observing the moving frame, no the ratios are not the same. That's what relativity is all about, frames moving relative to one another.
Again sorry to labour this, but it think there is a confusion of observers.
I hate to say it but I don't think you are comparing light with light! [:)]
I think we can come to a common interpretation by "walking the dog" if you're up to it...wadda ya say?

I am definitely learning a lot about the specific vernacular used to communicate these physics ideas to others on this forum...thanks for the feedback.
I'm glad to hear that, but how about learning to answer the question? [:)]
I use the same definitions for the variable "types" as Einstein within his book "Relativity: The Special and General Theory", i.e. the primed variable is with respect to the coordinate system K' and the unprimed variable is with respect to the coordinate system K. I use the Lorentz transformations in the same manner as well.
This doesn't tell us how you are interpreting what you are working with.
Nor does it answer the question 'which clock'
Let me take you back to your original post
In order to satisfy the first postulate, the relationship between their event times must remain unchanged in order to prevent one from ascertaining the motion of the inertial frame utilizing this scenario.
When you write this I have to assume you are working from the viewpoint of an observer in K', with a clock in K', moving at the same speed as K'. In this case that observer will say that c=c' and u=u'. So the race is the same as observed in K.
When you apply transforms to K' I have to assume you are working from the viewpoint of an observer in K, with a clock in K, observing events in K'. For this observer c=c', but u≠u' the value of u' has to be calculated using transforms. Hence the race is not the same as the race run in K.
We have to be careful how we use physical laws in relativity.
If we define that atomic clocks always show the 'correct' time, this only valid in inertial frames, not ones moving relative to one another.
The same is true if we define that a 1m platinum rod at temp x is always the same length. Relativity say only in inertial frames, not between frames moving relative to one another.
I think we can come to a common interpretation by "walking the dog" if you're up to it...wadda ya say?
Only if you can confirm which clock, which observer, and that you understand what I have written in this and my last post.

EVENTS OCCURRING IN REST FRAME (v = 0)
 These events now occur within a reference frame considered to be at rest (rest frame).
 The event time for A is; Δτ = x/c.
 The event time for B is; Δt = x/u.
 The ratio of these event times gives us the relationship between their event times;
Δτ/Δt = (x/c)/(x/u) = u/c.
No wonder I had a problem understanding what it was you were doing. It was as if you actually went out of your way to take a very simple thing and say it in a very complex way. You should have simply said the following: Let Δx be the distance traveled by a photon and a particle. Let u be the speed of the particle (thus u < c). Let Δt be the time it takes the particle travel the distance Δx and ΔT the time it takes a photon to travel the same distance. Then Δx = uΔt = cΔT. Therefore ΔT/Δt = u/c.
See how simple that was? [:)]

EVENTS OCCURRING IN REST FRAME (v = 0)
 These events now occur within a reference frame considered to be at rest (rest frame).
 The event time for A is; Δτ = x/c.
 The event time for B is; Δt = x/u.
 The ratio of these event times gives us the relationship between their event times;
Δτ/Δt = (x/c)/(x/u) = u/c.
No wonder I had a problem understanding what it was you were doing. It was as if you actually went out of your way to take a very simple thing and say it in a very complex way. You should have simply said the following: Let x be the distance traveled by a photon and a particle. Let v be the speed of a particle where v < c. Let t be the time it takes the particle travel the distance x and T the time it takes a photon to travel the same distance. Then x = vt = cT. Therefore T/t = v/c.
See how simple that was?
I managed to get it. Probably because I'm still picking things up. I bet if I had a degree in physics it wouldn't have made any sense. That is why I try to read so much. So at least I may be able to explain things in terms that other members may readily understand. I am not always successful.

No wonder I had a problem understanding what it was you were doing. It was as if you actually went out of your way to take a very simple thing and say it in a very complex way. You should have simply said the following: Let x be the distance traveled by a photon and a particle. Let v be the speed of a particle where v < c. Let t be the time it takes the particle travel the distance x and T the time it takes a photon to travel the same distance. Then x = vt = cT. Therefore T/t = v/c.
See how simple that was?
Yes, but he makes it even more complicated in the following section where he considers the moving frame. He uses the same T and t for the values seen by the observer in the rest frame when viewing the moving frame, but T and t are what the observer within the moving frame sees and are seen as reduced by the observer in the rest frame. He needs new symbols for these transformed values as he is making the mistake of equating unequal values.
I find the entire analysis over complicated and it is very unclear. To be honest I haven't even looked to see whether the maths is correct as there are too many false assumptions before you even start on the maths. No, no, I'll be really honest, it's because I'm not a mathematician and I don't really enjoy the maths bit [;)]

No wonder I had a problem understanding what it was you were doing. It was as if you actually went out of your way to take a very simple thing and say it in a very complex way. You should have simply said the following: Let x be the distance traveled by a photon and a particle. Let v be the speed of a particle where v < c. Let t be the time it takes the particle travel the distance x and T the time it takes a photon to travel the same distance. Then x = vt = cT. Therefore T/t = v/c.
See how simple that was?
If that is a format that you understand better, then yes...go with it. The original format makes perfect sense to me, but then again I wrote it which makes me biased to understanding.

Only if you can confirm which clock, which observer, and that you understand what I have written in this and my last post.
Just as Einstein defines his primed variables (such as t'), so I define mine. With regards to the Lorentz transformation t' = (t  vx/c^2)γ, which clock and observer does Einstein use for t'?

When I asked "I think we can come to a common interpretation by "walking the dog" if you're up to it...wadda ya say?", I meant to start at the beginning and step through the progression (changing as necessary in small pieces) of the scenario ensuring that you can follow my logic (as you best understand) until we get to the conclusion. My original writing style was how I think and write, which is obviously not best for a general audience that is used to another style.

If that is a format that you understand better, then yes...go with it. The original format makes perfect sense to me, but then again I wrote it which makes me biased to understanding.
I'm sorry to see that you don't understand what we're trying to get through to you yet. Here you made the mistake of interpreting my response as a complaint of format whereas it was actually a complaint against your overly complicated and very unclear description of the problem. The best way to communicate something to someone else is to make it as clear and as simple as you can. You didn't do that. You should have first stated the problem very clearly in words. After that's done, and only after it's done, do you translate it into the language of spacetime physics. Do you understand this now? I'll try to do this with your opening post.
First off I've asked you several times what "There are two parallel linear events (A and B) with uniform velocity along the positive xaxis." means and I've yet to get a response. You didn't even state what the worldlines are which are supposed to be parallel. I'm going to assume that you're referring to the following worldlines;
Worldline A: Worldline connecting origin with event A
Worldline B: Worldline connecting origin with event B
Worldline A is the worldline of a photon which is emitted from the origin and moves in the +xdirection and ends up at event A. That means that it's a line which is 45 degrees with respect to the +xaxis (and of course its also a line which is 45 degrees with the ctaxis).
Worldline B is the worldline of a particle which moves at a speed less than the speed of light and ends up at event B. That means that it's a line which is greater than 45 degrees with respect to the +xaxis.
This means that it is a line which is 45 degrees with respect to the +xaxis (and of course it’s also a line which is 45 degrees with the ctaxis).
Therefore it follows that these two worldlines are not parallel. So what in the world do you mean by “parallel events”?

In order to satisfy the first postulate, the relationship between their event times must remain unchanged in order to prevent one from ascertaining the motion of the inertial frame utilizing this scenario.
Okay. Now I see where you're going with all of this. This statement is not true. The first postulate states that the laws of physics are the same in all frames of reference. That means that no experiment can be done which would tell the observer that he's in motion relative to an absolute frame of reference. It does not mean that the relationship you established is invariant. This is the error which led you to the error in the paper you posted in post #1. You forget, those relationships are well tested so if you thought that you found a mistake in them then you can be certain that you're wrong.

First off I've asked you several times what "There are two parallel linear events (A and B) with uniform velocity along the positive xaxis." means and I've yet to get a response. You didn't even state what the worldlines are which are supposed to be parallel. I'm going to assume that you're referring to the following worldlines;
Worldline A: Worldline connecting origin with event A
Worldline B: Worldline connecting origin with event B
Worldline A is the worldline of a photon which is emitted from the origin and moves in the +xdirection and ends up at event A. That means that it's a line which is 45 degrees with respect to the +xaxis (and of course its also a line which is 45 degrees with the ctaxis).
Worldline B is the worldline of a particle which moves at a speed less than the speed of light and ends up at event B. That means that it's a line which is greater than 45 degrees with respect to the +xaxis.
This means that it is a line which is 45 degrees with respect to the +xaxis (and of course it’s also a line which is 45 degrees with the ctaxis).
Therefore it follows that these two worldlines are not parallel. So what in the world do you mean by “parallel events”?
I have no idea where you are getting this information or interpretation. When I say parallel, I mean parallel. A photon travels parallel to the xasix, and an electron travels parallel to the xaxis. Simple geometry, like your car travels parallel to the surface of the road. No need for world lines or tilting through any degrees. You don't understand my original scenario...I get it. No need to keep stating the same thing...I get it.

Paraphrasing is the cause for much of this confusion (all my mistakes), so let's return to Einstein's definitions as the source instead of my failed attempts.
In accordance with Einstein's "Relativity: The Special and General Theory", the Lorentz transformations will transform an event (event as referenced within "Relativity: The Special and General Relativity) that occurs within K (x,y,z,t) to a system K' (x',y',z',t') or vice versa. Refer to Part 1, Chapter 11 of his book.
Within this coordinate system scenario given by Einstein, we define an experiment based upon two events;
(A) As seen by an observer in K, a photon propagates (c) from (x = 0) to (x) in a time interval (τ).
(B) As seen by an observer in K, an electron travels in uniform motion (u<c) from (x = 0) to (x), identical length along the xaxis as the photon, in a time interval (t).
Does this clarify the beginning of this scenario?

I have no idea where you are getting this information or interpretation.
Well, I've asked you directly what that meant in post #7 but you never answered me. Therefore I took a guess.
In post #7 I asked
What is "parallel linear events" supposed to mean?
When I say parallel, I mean parallel.
That's the problem. You might think that such a phrase makes sense but in reality it doesn't. You spoke of parallel events and since events are but mere points in spacetime its meaningless to speak of two points being parallel. Also when you're speaking of events its natural to speak in terms of spacetime diagrams and worldlines. That was the motivation for my response.
A photon travels parallel to the xasix, and an electron travels parallel to the xaxis. Simple geometry, like your car travels parallel to the surface of the road. No need for world lines or tilting through any degrees. You don't understand my original scenario...I get it. No need to keep stating the same thing...I get it.
Then you made an error when you spoke of two parallel linear events (A and B) and that caused a problem since I was unable to understand what you meant by it. I now see the problem. What you said was meaningless. Do you understand why? In post number 1 you wrote about "two parallel linear events" (why you called them 'linear' is beyond me) when in fact there was nothing parallel about them.
By the description that you just posted you weren't talking about parallel events. You were talking about parallel lines. One line terminated on event A and the other line terminated on event B. Each line originated at the origin.
Do you now understand why you can't say that two events are parallel?
In any case your conclusion is incorrect.

The following scenario utilizes events that are applicable to Einstein’s Special Relativity Theory. What follows requires further analysis regarding application of the first postulate WRT SRT. For information with images, see http://gsjournal.net/ScienceJournals/Research%20PapersRelativity%20Theory/Download/5916.
Okay. Given your responses which clarify what you were trying to do I went over this carefully and now see that your conclusion is wrong. The Lorentz transformation (LT) works for any two events whatsoever so long as they can be observed in reality. After all, that's exactly what the LT is for. The problem is your erroneous belief that
In order to satisfy the first postulate, the relationship between their event times must remain unchanged in order to prevent one from ascertaining the motion of the inertial frame utilizing this scenario.
This assertion is quite wrong and based on a misunderstanding of the first postulate of special relativity. Here's why: The result of the LT ends up being
Δτ/Δt = Δτ’(1+ β)/Δt’(1+ βx’/cΔt’), x’ = uΔt’
Δτ/Δt = Δτ’(1+ β)/Δt’(1+ β(u/c))
This means that the ratio as measured in frame K has a different value than that measured in K'. The transformation of the ratio is a function of the relative speed of the two frames. It's okay for observers in one frame to measure his velocity relative to another inertial frame. Therefore your assumption that the first postulate is wrong is incorrect.

Okay. Given your responses which clarify what you were trying to do I went over this carefully and now see that your conclusion is wrong.
While you were composing yours, I was composing mine, you beat me to it by moments. I approach it from a different direction, but we reach the same conclusions. Perhaps you can confirm my logic as I see HeyBert creating an invalid conclusion by failing to perform an audit trail of observers and frames.
Perhaps between the two responses he will be able to understand how he should be viewing his scenarios.
My reply to HeyBert
In accordance with Einstein's "Relativity: The Special and General Theory", the Lorentz transformations will transform an event (event as referenced within "Relativity: The Special and General Relativity) that occurs within K (x,y,z,t) to a system K' (x',y',z',t') or vice versa. Refer to Part 1, Chapter 11 of his book.
The fact that you are starting to repeat your previous analysis even after we  PmbPhy and I  have posted replies, means you have either not read them or not understood them.
Just as Einstein defines his primed variables (such as t'), so I define mine. With regards to the Lorentz transformation t' = (t  vx/c^2)γ, which clock and observer does Einstein use for t'?
This confirms that you have not thought this through, nor have you understood the replies we have posted which answered this. Copying the formulae without understanding which observer, which clock, which frame means you are creating logical inconsistencies that result in wrong answers.
In order to cut this short:
When the race occurs within the stationary laboratory frame (v = 0), no calculation is needed (inverse Lorentz transformation reduces to Galilean format) and the ratio of the photon time span to the electron time span WRT the stationary laboratory frame is simply;
Photon: τ = τ'
Electron: t = t'
Ratio #1: τ/t = τ'/t'
Wrong
This experiment is performed in the stationary laboratory frame K and for the observer and clocks in this frame, let's call him A, the experiment delivers 2 results T and t. It tells us nothing about K', T' or t'. Just so we are really clear about the observer and frame let's call these results T_{KA}  this means T as measured in frame K by A. Similarly t_{KA}
The way we find out about T' and t' is to allow A to travel in frame K' which is moving at v relative to K, and use the clocks in this frame. He performs the same experiment as in K but now gets results T' and t'. However, when he compares T and T' he finds they are the same, T=T', similarly t=t'. This means that by performing this experiment he is unable to determine whether he is moving relative to K, he also confirms that the laws of physics are the same in these 2 frames of reference.
When the race occurs within the inertial frame at velocity (v), the calculation of the ratio of the photon time span to the electron time span WRT the stationary laboratory frame IAW the inverse Lorentz transformation is;
Photon: τ=(τ'+vx'/c^2)γ
Since it is a photon, the second postulate gives x' = cτ'
τ=(τ'+v(cτ')/c^2)γ
τ=(τ'+v(τ')/c)γ
τ=τ'(1+v/c)γ
τ=(τ'+vx'/c^2)γ wrong
Since it is a photon, the second postulate gives x' = cτ' correct
τ=(τ'+v(cτ')/c^2)γ wrong
τ=(τ'+v(τ')/c)γ wrong
τ=τ'(1+v/c)γ wrong
At this point I am not interested in whether the transform formulae are correct or not, I am still looking at observers, frames and clocks, and the logic of their relationships. Until we understand this, the formulae are of little use.
Let us look at your final formula:
τ=τ'(1+v/c)γ
We know from the experiments performed by A in K and K' that T=T' so:
T=T(1+v/c)γ
Really? Do you believe that? 1=(1+v/c)γ, where γ = (1v^2/c^2)^1/2
All this comes about because you are copying formulae without considering observers, frames or clocks.
Let's look at the observation of K' from K.
A second observer, let's call him B, sits in K and observes A conducting his experiment in K'. Although he can perform the same experiment as A performed in K, and similarly derive T and t, what does he see when he observes the experiment in K'.
B sees that K' is moving at v and knows that because of this the clocks in K' appear to be running slow and any time T' or t' he observes has to be adjusted by the Lorentz transforms.
So we can call the value of T' that B observes in K' as T_{K'B}.
Let's not worry about the exact maths but say that:
T_{K'B}=ƒT' (where ƒ is the transform and ƒ is not 0 or 1)
But we know that T'=T so
T_{K'B}=ƒT
Hence T_{K'B}≠T
In other words
T_{K'B}≠T_{KA}≠T
which is why your formula τ=(τ'+vx'/c^2)γ is wrong, you assumed the wrong value for T, you should have been deriving T_{K'B} not T_{KA}.
This is why your ratios don't work. You have not considered in which frames and for which observers and clocks your defined terms are valid.
To be fair you were misled by the paper you posted.
I don't intend to go through the detail of the calculations or your round trip scenario, but leave you to work through the correct versions.

While you were composing yours, I was composing mine, you beat me to it by moments. I approach it from a different direction, but we reach the same conclusions. Perhaps you can confirm my logic as I see HeyBert creating an invalid conclusion by failing to perform an audit trail of observers and frames.
Perhaps between the two responses he will be able to understand how he should be viewing his scenarios.
Sure thing. I'll do my best. However it will have to wait since I have to go to the store and get an ink cartridge for my printer. :)

It appears from your remark that (T' = T) means that you are assuming I am discussing what K measures in their own frame vs what K' measures in their own frame. This was not at all what I was discussing. Of course they will not measure anything different, that is merely a result of the first postulate. Since all the following remarks are based upon this view, I can see how you came to your conclusion.

It appears from your remark that (T' = T) ...
To whom are you speaking to? I don't see Colin making any such comment.
...means that you are assuming I am discussing what K measures in their own frame vs what K' measures in their own frame. This was not at all what I was discussing.
That is exactly what it appears that you're discussing. If it isn't then your posts are extremely deceptive. When someone places a prime on a variable it means that the quantity that has the prime on it is measured in the primed frame and vice versa. I.e. A' is measured in frame K', B is measured in frame K, etc.
You also speak of measuring observables in the "rest frame (v = 0)" where variables have no prime on them and in the "inertial frame (v)" where variables have a prime on them. By the way, this is a bad name for a frame since all frames used in special relativity are inertial frames. Then you apply the LT which means you're relating variables between inertial frames. That's why your comment
...means that you are assuming I am discussing what K measures in their own frame vs what K' measures in their own frame. This was not at all what I was discussing."
is quite misleading if its at all true. In fact you yourself defined the primed and unprimed variables in post #17 where you wrote
For all the primed variables, the clarification is as follows. IAW Einstein's book "Relativity: The Special and General Theory", the primed variable is with respect to the coordinate system K' and the unprimed variable is with respect to the coordinate system K.
Of course they will not measure anything different, that is merely a result of the first postulate.
The ratio is not invariant as you claim that it is. Are you unable to understand that? If not then you have a serious problem understanding special relativity. I explained why you're wrong above. Did you not read it?

It appears from your remark that (T' = T) ...
To whom are you speaking to? I don't see Colin making any such comment.
I was referring to his comment about "Really? Do you believe that? 1=(1+v/c)γ, where γ = (1v^2/c^2)^1/2"
Can you see it?

It is obvious that my original presentations suffer many flaws due to the wording and the related mathematical results as a result of this wording. What if I strip out all the extraneous referencing to observers, rest frames, etc.?

PART 1 (OneWay Events)
WRT Einstein’s book (Relativity: The Special and General Theory), Part 1, Chapter 11 gives the Lorentz transformation (LT) as;
t’ = (tvx/c^2)γ, where γ = (1v^2/c^2)^1/2
Referencing Einstein’s book (Part 1, Chapter 11), we merely apply the transformation to two events originating in K that travel from (x = 0) to (x) in unequal times (τ) and (t).
WRT K’:
τ’ = (τvx/c^2)γ
t’ = (tvx/c^2)γ
Evaluating the ratio of these times at (v) gives the results WRT K’ as;
* τ’/t’ = (τvx/c^2)/(tvx/c^2)
Evaluating this ratio at (v = 0) gives the results WRT K as;
τ’/t’ = (τ0*x/c^2)/(t0*x/c^2)
* τ’/t’ = τ/t
PART 2 (RoundTrip Events)
Referencing Einstein’s book (Part 1, Chapter 11), we merely apply the transformation to two events originating in K that travel from (x = 0) to (x) in unequal times (ϖ) and (T), then are reflected back such that they return to their points of origin (x = 0) in this same time (ϖ) and (T).
WRT K’:
2ϖ’ = (ϖvx/c^2)γ + (ϖ+vx/c^2)γ
2ϖ’ = ϖγ(vx/c^2)γ+ϖγ+(vx/c^2)γ
* 2ϖ’ = 2ϖγ
2T’ = (Tvx/c^2)γ + (T+vx/c^2)γ
2T’ = Tγ(vx/c^2)γ+Tγ+(vx/c^2)γ
* 2T’ = 2Tγ
Evaluating the ratio of these times at either (v or v = 0) gives the results WRT to K’ or K as;
* ϖ’/T' = ϖ/T

It appears from your remark that (T' = T) means .........
No, you yourself said T=T', see below
My calculations, where γ = (1v^2/c^2)^1/2;
When the race occurs within the stationary laboratory frame (v = 0), no calculation is needed (inverse Lorentz transformation reduces to Galilean format) and the ratio of the photon time span to the electron time span WRT the stationary laboratory frame is simply;
Photon: τ = τ'
Electron: t = t'
You keep posting things like this without defining them. What you said tells us nothing about what those quantities are/mean.
You then make an incorrect assumption about what I wrote which makes me think you didn't read thoroughly what I did write:
It appears from your remark that (T' = T) means that you are assuming I am discussing what K measures in their own frame vs what K' measures in their own frame.
No I didn't make that assumption I saw you were comparing K and K' as above when you said:
Photon: τ = τ'
Electron: t = t'
I merely expanded on what you said in order to make clearer the relationship between K and K'
My assumption is that you are trying to relate what happens in K' back to K using LT. In fact I wrote:
When you apply transforms to K' I have to assume you are working from the viewpoint of an observer in K, with a clock in K, observing events in K'.
However, these misassumptions you are making are not the only problem
It is obvious that my original presentations suffer many flaws due to the wording and the related mathematical results as a result of this wording.
No, not just the wording, you are making mistakes of logic which I pointed out in my last post  did you really read it properly?
...... What if I strip out all the extraneous referencing to observers, rest frames, etc.?
Why?
This is relativity. The clue is in the name, relativity = observers, frames, clocks all moving relative to one another. If you take out those references you will be even more confused,
PART 1 (OneWay Events)
WRT Einstein’s book (Relativity: The Special and General Theory), Part 1, Chapter 11 gives the Lorentz transformation (LT) as;
NO, NO, NO.
Don't start on another scenario until you understand the errors you are making on the first.
PmbPhy would explain this is terms of worldlines and spacetime geometry, which is by far the best way as you would understand more easily where you are going wrong. However, your dismissive response to his post:
First off I've asked you several times what "There are two parallel linear events (A and B) with uniform velocity along the positive xaxis." means and I've yet to get a response. You didn't even state what the worldlines are which are supposed to be parallel. I'm going to assume that you're referring to the following worldlines;
Worldline A: Worldline connecting origin with event A
Worldline B: Worldline connecting origin with event B
Worldline A is the worldline of a photon which is emitted from the origin and moves in the +xdirection and ends up at event A. That means that it's a line which is 45 degrees with respect to the +xaxis (and of course its also a line which is 45 degrees with the ctaxis).
Worldline B is the worldline of a particle which moves at a speed less than the speed of light and ends up at event B. That means that it's a line which is greater than 45 degrees with respect to the +xaxis.
This means that it is a line which is 45 degrees with respect to the +xaxis (and of course it’s also a line which is 45 degrees with the ctaxis).
Therefore it follows that these two worldlines are not parallel. So what in the world do you mean by “parallel events”?
Your response indicates that you do not understand spacetime geometry, if you did it would be easier to explain where you are going wrong. So I will try one last time using the terms you are familiar with K, K', t, t', etc.
Fortunately LT gives us a way of interpreting the geometry in the same way that Pythagoras allows us to interpret rt angled triangles. However, as with Pythagoras we must be careful not to assume that because a value is given a symbol in a formula the value is the same for another symbol of the same name. That is equivalent to saying that in Pythagoras a, b and c have the same value for all triangles, also we must be careful not to apply Pythagoras to nonrt angled triangles. Yet this is what you have been doing with LT!
Your formula τ=(τ'+vx'/c^2)γ transforms a value of T from T'. That is transforming from K' to K.
But earlier you derived a value for T in K, when you considered the rest frame v=0. It is important to realise that although they have the same symbol T, these two quantities are NOT THE SAME. That is why when I went through the logic of this in my last post, I assigned T_{K'B} to the LT derived T.
Do you understand. If not then your understanding of STR is very flawed.
I hate to labour this, but as this will be my last post I will go through it with a simpler version.
We will use K and K'
Also you have previously agreed that T/t=u/c. So if we consider u and c we will also be calculating what happens to T and t.
Because you seem to be having problems with LT we will consider a case where v is small. That mean we can ignore any relativistic effects that are due to v being a significant proportion of c, and our scenario looks closer to Newtonian rules:
Again we start in K (your original rest frame) and calculate u and c. Because K' is also an inertial frame then u=u' and c=c'.
So if we are in K looking at what is happening in K' we will see that c being constant for all observers is still c.
But u clearly = v+u'
But wait a minute, we already decided that u'=u so substituting we get
u=v+u
Clearly a logical impossibility. The reason being that the u in u=v+u' is not the same as the u for K in the rest frame, it needs a different symbol u_{K'B}say.
I hope it is easier to see in this example where the circular augment was in your calculations. Go back to my previous post and work it through.
If you cannot see it then we cannot help you. You will need some face to face sessions with your teacher who should be able to explain it. I would also suggest you stop reading papers such as the one you posted, very misleading.
As I say, my last post. Other things to do.

Calm down...it's just a math problem. No need to huff and puff in front of the world. Did your teacher act this way when you made mistakes learning in class? I agreed that there were mistakes within my original scenarios, so stop pretending I didn't.
It is obvious that my original presentations suffer many flaws due to the wording and the related mathematical results as a result of this wording. What if I strip out all the extraneous referencing to observers, rest frames, etc.?
Just because you disagree with how someone answers your question does not mean they didn't answer your question. I tried changing the wording of my scenario based upon the wonderful feedback I received from posts like yours. Continually going backwards in the posts (after I make changes based upon replies) is like going to jail over and over for the same crime. Get over it! Do me a favor, don't ever become a teacher...I don't think you have the patience for it.
And what law states I have to answer your questions anyways?

My latest edit based upon great feedback. Not interested in arguing older scenarios as I already admitted they were flawed. That was the point of posting in the first place.
PART 1 (OneWay Events)
WRT Einstein’s book (Relativity: The Special and General Theory), Part 1, Chapter 11 gives the Lorentz transformation (LT) as;
t’ = (tvx/c^2)γ, where γ = (1v^2/c^2)^1/2
Referencing Einstein’s book (Part 1, Chapter 11), we merely apply the transformation to two events (i.e. moving electrons) in K from (x = 0) to (x) in unequal times (τ) and (t).
WRT K’:
τ’ = (τvx/c^2)γ
t’ = (tvx/c^2)γ
Evaluating the ratio of these times at (v) gives the results WRT K’ as;
* τ’/t’ = (τvx/c^2)/(tvx/c^2)
Evaluating this ratio at (v = 0) gives the results WRT K as;
τ’/t’ = (τ0*x/c^2)/(t0*x/c^2)
* τ’/t’ = τ/t
PART 2 (RoundTrip Events)
Referencing Einstein’s book (Part 1, Chapter 11), we merely apply the transformation to two events (i.e. moving electrons) in K from (x = 0) to (x) in unequal times (ϖ) and (T), then from (x) to (x = 0) in these same times (ϖ) and (T).
WRT K’:
2ϖ’ = (ϖvx/c^2)γ + (ϖ+vx/c^2)γ
2ϖ’ = ϖγ(vx/c^2)γ+ϖγ+(vx/c^2)γ
* 2ϖ’ = 2ϖγ
2T’ = (Tvx/c^2)γ + (T+vx/c^2)γ
2T’ = Tγ(vx/c^2)γ+Tγ+(vx/c^2)γ
* 2T’ = 2Tγ
Evaluating the ratio of these times at either (v or v = 0) gives the results WRT to K’ or K as;
* ϖ’/T' = ϖ/T

And what law states I have to answer your questions anyways?
Nothing
But if you don't want my answers, that's also ok.
I'm sorry you feel you have been to jail over and over, but if you look back you will see that you have rebuffed my replies a number of times including my post where I take the trouble to write it down in some detail.
It appears from your remark that (T' = T) ...
To whom are you speaking to? I don't see Colin making any such comment.
...means that you are assuming I am discussing what K measures in their own frame vs what K' measures in their own frame. This was not at all what I was discussing.
That is exactly what it appears that you're discussing. If it isn't then your posts are extremely deceptive. When someone places a prime on a variable it means that the quantity that has the prime on it is measured in the primed frame and vice versa. I.e. A' is measured in frame K', B is measured in frame K, etc.
It is quite irritating to be rebuffed so offhand and I suspect that irritation came through in my final post, where I felt you were ignoring the points I made. Despite that rebuff I had the patience to write it out with even more examples, and then to be told you don't want to go over it again. So it looks like I wasted my time!
That's ok you don't have to read it, but it takes time to take the trouble to write out these replies.
You tell me I no longer have the patience for this, so as the Dragons say "I'm out"
PS  470 views (including repeats) is hardly the world watching.

Thanks for stopping by.

HeyBert = I've grown tired of following those derivations everyday. There are only two points to be stated and they are
1) If you got anything but Δτ/Δt = Δτ’(1+ β)/Δt’(1+ β(u/c)) then you made a mistake. Since this is not what you got then you did make another mistake. It's now up to you to find it. If you can't then please let me know and I'll show you one last time.
2) The first postulate is correct and your interpretation of it that you started this thread with is wrong.

Working things out on another forum right now. Learned to frame the problem in a "relativity friendly" format from another user and have good positive communication going. Thanks for the offer though, and I'll let you know If I come back to this or identify where my logic went astray.

Working things out on another forum right now. Learned to frame the problem in a "relativity friendly" format from another user and have good positive communication going. Thanks for the offer though, and I'll let you know If I come back to this or identify where my logic went astray.
Yeah. I know. I saw that and I saw that you're making mistakes there too. I've already shown you how to do this right so what's your objection to it?

Yeah. I know. I saw that and I saw that you're making mistakes there too. I've already shown you how to do this right so what's your objection to it?
5:1 against him getting it on his own, he so much wants others to do his thinking for him. I would give you better odds, but as the rate of exchange is beers I would be worried about your health. Catch is I'm unlikely to come over your way so you would have to come here to collect.
Who is HeyBert, what is he?
When he 1st came on I assumed one of the self publishers you seem to get on this site checking out their pet theory. That's why I called him Robert. I was a little ratty because I don't like people trying to advertise their products on sites like this.
As the posts went on I was surprised at the low level of knowledge of basic relativity. Particularly for someone who has self published a book on the subject (shudder to think what mis guidance it gives). I was also surprised at the lack of eagerness to learn. If I didn't know about worldlines your comment on non parallel events would have me asking you to explain, so I could learn, but he wasn't interested and seemed quite ratty about it. I decided to start really goading to get a response. The one I got made me think student with a deadline on a homework task or forum and wanting to be 1st to get the answer, but the number of ownership phrases brought me back to my original guess and he is still trying to get others to do his thinking for him. Looking at his paper I see he has modified it at the same time he reposted here. I was about to start looking for the other forum, but I see you are ahead of me.
If he comes back, I'll offer him some advice
You should have seen this as soon as you saw the solution to the first problem, don't move on too quickly, spend time to consolidate and learn. When tackling problems like this it is best to try as many tools as possible, visualisation, bounding cases, worldlines, sanity checks and audit trails. Use these 'maths problems' to develop good verbal reasoning and logic. Look at it from as many angles as possible and you won't make these mistakes in the future.
Sorry if you thought my goading was sending you to jail, but I did want to check you out, and also I was irritated that you dismissed a lot of time spent answering your questions, without really wanting to learn holistically.
I hope Naked Scientist and the other forum gets mentioned in the credits for your papers.
PS PmbPhy, the beers are here anytime you are over this way [:o)]

His problem is that he wants his idea to be correct, not get the physics correct.

His problem is that he wants his idea to be correct, not get the physics correct.
Ah, so there is an agenda, I just couldn't put my finger on it.
He calls it a maths problem, but to me it seems like bad methodology.
I'll put the beer on ice.

His problem is that he wants his idea to be correct, not get the physics correct.
Ah, so there is an agenda, I just couldn't put my finger on it.
He calls it a maths problem, but to me it seems like bad methodology.
I'll put the beer on ice.
Colin  You never answered me. I sent you a PM asking you a question. What is your response?

Colin  You never answered me. I sent you a PM asking you a question. What is your response?
Sorry, hadn't noticed it, reply sent.
Been busy repairing a microscope and haven't logged on to email either for a while.

Colin2B + PmbPhy,
It's a hobby...It's a math problem...even if I'm wrong, it's fun to me...stop being so offensive and judgmental! You're both acting like a couple drama queens on here.

Colin2B + PmbPhy,
It's a hobby...It's a math problem...even if I'm wrong, it's fun to me...stop being so offensive and judgmental! You're both acting like a couple drama queens on here.
There's no reason to insult us. That's very rude and extremely uncalled for. Also there's nothing that you claimed here that is correct. I haven't insulted you or judged you whatsoever. And I know that's true because only **I** know what I'm thinking. You don't.
You've been operating under false assumptions, perhaps because you don't understand how you've been coming across to people.
Perhaps what you've perceived as insults and judgments has been attempts at constructive criticism. Point out some of these socalled "insults" and "judgments" and I'll explain why you're wrong.
I will admit that I've been frustrated with you because I've asked you so many questions and you haven't answered them. You've been appearing to ignore them or trying to talk around them. It's come across to both myself and Colin that you've simply been unwilling to admit that you made a mistake or your skills are so bad that you've been unable to recognize them.
It's also possible that we've said that you don't understand something or something of that nature. That is **not** an insult. That is merely an observation of a fact and nothing else.
Colin and I were unaware that you considered this a hobby. When visitors come to this part of the forum its almost always because they think that they've proved SR or QM to be wrong and they want to prove it to us, sometimes they provide what appears to us to be some sort of attempt at a scientific paper. That's how you've come across to us. And that's not our fault, that's your fault.
Your insults will be reported to the forum moderator.

Posts #48#49. And don't bother "explaining", not interested.

Posts #48#49. And don't bother "explaining", not interested.
Well, HeyBert. I really don't care whether you're interested or not. I'm defending myself so that bystanders aren't misled by your claims.
Before I go on I'd like to point out that if this is the way you react to criticism then in a real meeting with real physicists criticizing your claim that SR is wrong then you'd walk out of that meeting crying your eyes out because physicists can sometimes be brutal in their criticism. They don't shy away from telling it like it is just because you insult them by calling them drama queens. In fact you'd probably be banned from all other meetings with such a terrible attitude.
As for myself I said His problem is that he wants his idea to be correct, not get the physics correct. which is exactly how you're coming across to all of us. However that's certainly a far cry from all the bogus accusations that you leveled against us, that's certainly for sure!
As for Colin, he was both describing how you're coming across to all of us and also criticizing your style of attempting to work this problem. It was certainly constructive criticism. It's clear now that you're seeing constructive criticize and claiming that its insults merely because your mistakes are being pointed out to you.
You can't go to a forum and post claims that SR is wrong and then assume that everyone is going to admire you for it. Countless people go to physics forums with their own pet theories with the belief that they've also proven SR to be wrong. All of them fail and they all believe that we're just too stupid or too closed minded to see that they're right when in fact 99% of them have never studied physics.

In the 'New Theories' section of "Got a new theory on something? Post your hypotheses here..."
This is where you choose defend the pillars of science? Your actions do nothing but make people afraid to freely discuss their ideas on this forum. Great job Ace!

In the 'New Theories' section of "Got a new theory on something? Post your hypotheses here..."
This is where you choose defend the pillars of science? Your actions do nothing but make people afraid to freely discuss their ideas on this forum. Great job Ace!
And you actually thought that means that nobody is going to show him the mistakes that he's making which are causing the false claims that a theory in physics is wrong? You're really deluded "Ace".
His (and perhaps your) problem is that he's as sensitive as a teenage girl who can't stand people showing them what they did wrong. If he didn't want people to show him his mistakes then he simply should say so in his first post, e.g. say something like this
Here is my new theory which shows that relativity is wrong. If it's wrong then keep it to yourself because I don't care. If what I'm writing is confusing then please don't ask me to clarify. etc.
That will get you what he wanted.
Now please stop being such a jerk. People like you are not welcome in this forum.

Posts #48#49. And don't bother "explaining", not interested.
I suggest that if you're "not interested" then we'll have to assume that you're also unprepared to learn or share with us. If such is the case, I suggest we all ignore you as well!

Posts #48#49. And don't bother "explaining", not interested.
I suggest that if you're "not interested" then we'll have to assume that you're also unprepared to learn or share with us. If such is the case, I suggest we all ignore you as well!
Comments like this are why Ethos_ is so well respected in this forum. :)

Colin2B + PmbPhy,
It's a hobby...It's a math problem...even if I'm wrong, it's fun to me...stop being so offensive and judgmental! You're both acting like a couple drama queens on here.
Why are you posting using two names? That's against the forum rules. You sure are turning out to be a scumbag.