It appears from your remark that (T' = T) means .........
No, you yourself said T=T', see below
My calculations, where γ = (1-v^2/c^2)^-1/2;You keep posting things like this without defining them. What you said tells us nothing about what those quantities are/mean.
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 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 mis-assumptions 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 (One-Way Events)
WRT Einsteinís 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:
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.
First off I've asked you several times what "There are two parallel linear events (A and B) with uniform velocity along the positive x-axis." 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 +x-direction and ends up at event A. That means that it's a line which is 45 degrees with respect to the +x-axis (and of course its also a line which is 45 degrees with the ct-axis).
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 +x-axis.
This means that it is a line which is 45 degrees with respect to the +x-axis (and of course itís also a line which is 45 degrees with the ct-axis).
Therefore it follows that these two worldlines are not parallel. So what in the world do you mean by ďparallel eventsĒ?
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 non-rt 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 TK'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
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 uK'Bsay.
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.