Time Dilation
- Thread starter chinglu
- Start date
So, you can prove time dilation for light pulse emitted from origins when they are at same place.
Prove time dilation for positive x-axis points and negative x-axis points of light pulse for any x and -x.
Do not forget, time dilation is an absolute truth.
chinglu:
I note that you have still pointed out no errors in posts #117 and #118.
I did that in post #117, which has no errors.
Coordinates are points. They have no length. Only spatial intervals can expand or contract.
You still haven't explained exactly what you want me to prove, or how it is different from what I showed you in posts #117 and #118.
I can't start until you explain what you want me to do.
You still haven't thanked me for teaching you. Were you not taught to say thankyou when somebody helps you?
I don't know what you mean by this. Please explain in more detail.
You'll need to explain what you mean in more detail. I can't understand you.
No. Time dilation is a derived consequence of the postulates of special relativity. If the postulates are correct, then time dilation is a mathematical consequence.
Which of the two postulates of special relativity do you think is wrong?
I note that you have still pointed out no errors in posts #117 and #118.
I did that in post #117, which has no errors.
Coordinates are points. They have no length. Only spatial intervals can expand or contract.
You still haven't explained exactly what you want me to prove, or how it is different from what I showed you in posts #117 and #118.
I can't start until you explain what you want me to do.
You still haven't thanked me for teaching you. Were you not taught to say thankyou when somebody helps you?
I don't know what you mean by this. Please explain in more detail.
You'll need to explain what you mean in more detail. I can't understand you.
No. Time dilation is a derived consequence of the postulates of special relativity. If the postulates are correct, then time dilation is a mathematical consequence.
Which of the two postulates of special relativity do you think is wrong?
You do not understand his proof.
Set x = vt and plug this into Lorentz transform for t'.
You will notice he does not use any of your false ideas. Check his 1905 paper and understand how he derives time dilation before you claim expert opinion.
Coordinates are used to make intervals.
First, read Einstein's proof for time dilation. Then write your statement.
Maybe we can find your error in thinking with answer simple question.
Given (t,x,y,z) mapped by Lorentz Transform to (t',x',y,z) what exactly is this telling you. Say everything you know about this mapping.
chinglu:
Please quote the wrong part of post #117 or #118 (I don't know which post you're talking about) and remind me what is wrong with it.
Ok. I had a look at Einstein's 1905 paper.
He agrees with me.
If you dispute this, please show where my results in posts #117 and #118 differ from Einstein's.
It's strange that you're quoting Einstein as an authority, though. I thought you said Einstein was wrong.
Do you think Einstein was right or wrong? It's not clear. Are you now agreeing that Einstein was right?
Ok. I've done that.
My statements are all still correct and in agreement with Einstein.
What point are you making?
Einstein and I both know where you went wrong, and I explained it to you in posts #117 and #118.
Oh, it would take me years to tell you everything I know about this mapping.
I'll write down what the mapping is for you, ok? That's a good starting point.
$$t = \gamma (t' + \frac{vx'}{c^2})$$
$$x = \gamma (x' + vt')$$
$$y = y'$$
$$z = z'$$
This is assuming motion of the primed frame as speed v in the positive x direction.
Do you think there's an error in this? If so, please point it out.
Also, don't forget to keep searching for any error you can find in posts #117 and #118. Let's not get side-tracked from examining your mistakes, which I pointed out in those posts.
Do you see where you went wrong yet?
chinglu said:
Please quote the wrong part of post #117 or #118 (I don't know which post you're talking about) and remind me what is wrong with it.
chinglu said:
Ok. I had a look at Einstein's 1905 paper.
He agrees with me.
If you dispute this, please show where my results in posts #117 and #118 differ from Einstein's.
It's strange that you're quoting Einstein as an authority, though. I thought you said Einstein was wrong.
Do you think Einstein was right or wrong? It's not clear. Are you now agreeing that Einstein was right?
chinglu said:
Ok. I've done that.
My statements are all still correct and in agreement with Einstein.
What point are you making?
Einstein and I both know where you went wrong, and I explained it to you in posts #117 and #118.
Oh, it would take me years to tell you everything I know about this mapping.
I'll write down what the mapping is for you, ok? That's a good starting point.
$$t = \gamma (t' + \frac{vx'}{c^2})$$
$$x = \gamma (x' + vt')$$
$$y = y'$$
$$z = z'$$
This is assuming motion of the primed frame as speed v in the positive x direction.
Do you think there's an error in this? If so, please point it out.
Also, don't forget to keep searching for any error you can find in posts #117 and #118. Let's not get side-tracked from examining your mistakes, which I pointed out in those posts.
Do you see where you went wrong yet?
Take the primed frame as stationary.
The moving unprimed origin moves to x'. In the primed frame, how long does this take?
It is simple and consistent with Einstein's time dilation proof.
x' = vt' or x'/v = t'.
Your calculations calculate a different t' and so you contradict simply physics in the frame.
You calculate t' in your posts and come up with a different value. That mean you are in error.
It takes the time elapsed in the primed coordinate between the start time at some event and the end time when x'=-k and x=0 are in the same place.
The end event (place and time) has primed coordinates
x' = -k
t' = k/v
But unless observers in the primed and unprimed coordinate systems agree on what point in space time the start is, they are going disagree on which is moving and how far it moved.
A reasonable start for the "moving unprimed origin" as seen by the primed frame is x = 0 and t' = 0. [1] A very little math later, and we see that is:
x' = 0
t' = 0
So the interval (Δx', Δt') = end - start = (-k, k/v) - (0, 0) = (-k, k/v).
This is the interval that describes a particle moving at a speed of w' = Δx'/Δt' = -v.
So in the primed coordinate system, the start and end describe the "moving unprimed origin" is moving at speed -v, and to answer your question, the time this takes is Δt' = k/v.
Since the question is about time dilation, let us ask about the same two events in the unprimed coordinate system.
For any event the relationship:
x' = γ(x - vt)
t' = γ(t - vx/c²)
means
x = γ(x' + vt')
t = γ(t' + vx'/c²) [2]
So in the order presented:
End event (place and time) in unprimed coordinates:
x = γ(-k + vk/v) = 0
t = γ(k/v - vk/c²) = (γk/v)(1 - v²/c²) = k/(γv).
Start event:
x = γ(0 + v0) = 0
t = γ(0 + v0/c²) = 0
(Δx, Δt) = (0, k/(γv)).
Δt = k/(γv)
w = Δx/Δt = 0
So the conclusion is that the primed coordinate system measures k/v between the events which represent something moving at speed w' = -v, and the unprimed coordinate system, like a clock which is not moving, w=0, on the origin of place for the unprimed coordinate system measures a smaller number: k/(γv).
We can get the same conclusion in a number of ways [3]. For two events on the world line of a uniformly moving clock, the elapsed time is shortest in the inertial frame where the clock is not moving. [4]
Appendix [1]
Given:
x = 0
t' = 0
x' = γ(x - vt)
t' = γ(t - vx/c²)
What is x' ?
From t' = 0 and x = 0 and t' = γ(t - vx/c²) and γ≥1 we conclude t = 0.
From x = 0 and t=0 and x' = γ(x - vt) we conclude x' = 0.
Appendix [2]
Given:
x' = γ(x - vt)
t' = γ(t - vx/c²)
x'' = γ(x' + vt')
t'' = γ(t' + vx'/c²)
1 = γ²(1 - v²/c²)
what is x'' and t'' in terms of x and t?
x'' = γ(x' + vt') = γ(γ(x - vt) + vγ(t - vx/c²)) = γ²(x - vt + vt - v²x/c²) = γ²(x - v²x/c²) = xγ²(1 - v²/c²) = x.
t'' = γ(t' + vx'/c²) = γ(γ(t - vx/c²) + vγ(x - vt)/c²) = γ²(t - vx/c² + vx/c² - v²t/c²) = γ²(t - v²t/c²) = tγ²(1 - v²/c²) = t
So
x' = γ(x - vt)
t' = γ(t - vx/c²)
means
x = γ(x' + vt')
t = γ(t' + vx'/c²)
Appendix [3]
Given two events (ℓ,n) and (o,p) and defining the interval (Δx, Δt) as (ℓ-o,n-p), does the difference of the Lorentz transforms of the events equal the Lorentz transform of the interval?
Does (Δx', Δt') = (ℓ'-o',n'-p') ?
Well, by the Lorentz transform we have:
ℓ' = γ(ℓ - vn)
n' = γ(n - vℓ/c²)
o' = γ(o - vp)
p' = γ(p - vo/c²)
And so:
Δx' = γ(Δx - vΔt) = γ((ℓ-o) - v(n-p)) = γ((ℓ-vn)-(o - vp)) = γ(ℓ-vn)-γ(o - vp) = ℓ' - o'.
Δt' = γ(Δt - vΔx/c²) = γ((n-p) - v(ℓ-o)/c²) = γ((n - vℓ/c²) - (p - vo/c²)) = γ(n - vℓ/c²) - γ(p - vo/c²) = n' - p'
So if the interval is the space-time difference of the events, then the Lorentz transform of the coordinate differences is the same as the space-time difference of the Lorentz transformed coordinates.
Appendix [4]
Given a particle which is moving at speed w for a certain time Δt, what is the Lorentz transform of that time in a frame where the particle is not moving.
The speed of the particle in the unprimed coordinates is w = Δx/Δt so Δx = wΔt.
if we set v = w, then
Δx' = γ(Δx - vΔt) = γ(wΔt - wΔt) = 0
Δt' = γ(Δt - vΔx/c²) = γ(Δt - v²Δt/c²) = Δt/γ
So in the special frame where the particle is not moving (w' =Δx'/Δt' = 0), we see that the measured time is shorter by a factor of √(1 - w²/c²) than in the frame where it moves with speed w.
I stop your lack of understanding above.
The standard configuration mandates both origins are the same for the start of the event. This is not up for discussion.
So, you setting x'=0 is complete false because that is not the situation and indicates you do not understand what is going on with relativity.
x' is some negative number for the unprimed origin to reach.
Are you not able to follow Einsten's proof of time dilation? He set x = vt. You are so willing to accept this reasoning.
I am setting x' = -vt', which is exactly the same thing and you and James R are totally sliced by this.
We have x' = -vt' for some negative x'. We plug the result into the equation t = ( t' + vx'/c² )γ.
The result is t = t'/γ. There is nothing you can do except refute Einstein's proof of time dilation.
Since the LT matrix is invertible, then from the unprimed frame at rest, tγ = t' or the primed moving clock coming toward the unprimed origin beats time expanded.
There is no choice.
Assuming the origins were co-located at t'=0, this takes time t'=x'/v in the primed frame, where v is the velocity of the unprimed frame.
No. My calculations agree with yours in this instance.
So, we agree now. That was easy.
Are we finished?
You still haven't pointed out any mistakes in posts #117 and #118.
If you think I've made an error, you'll have to go through those two posts and show where the error is. Do it line by line.
All you've managed to do since I posted those posts is to try to divert the topic onto something else, avoid the issue, deny your mistakes and generally be evasive.
And you still haven't thanked me for helping you see your mistakes.
Since your calculations agree with mine, then you understand the unprimed frame claims the moving clock at x' is time expanded and thus time dilation is false.
I guess we are done then. You have learned a lot.
Chinglu wrote this:
Obviously, the "place" where x=0 is seen as moving by primed frame -- the x' of this "place" is a non-constant function of t'.
$$x(t) = 0 \quad \Rightarrow \quad x'(t') = -vt'$$
Both are uniform motions and this is true in both Galilean and Special Relativity. In Special Relativity, we can write:
$$x'(t') = -vt' \quad \Rightarrow \quad x(t') = \gamma \left( x'(t') + vt' \right) = 0 \\ x'(t') = -vt' \quad \Rightarrow \quad t(t') = \gamma \left( t' + vx'(t')/c^2 \right) = \frac{1}{\gamma} t' \\ x'(t') = -vt' \quad \Rightarrow \quad x(t) = 0$$
This is the core content of the principle of relativity. What is a "place" for one frame can "move" in another and both descriptions are equally valid.
What I have done is showing assuming we are talking about the moving "place" which is the spatial origin of the unprimed frame x=0, the choice of t' = 0 implies that the event we are talking about is the common space-time origin of both coordinate systems. This is also (more obviously) true if you pick x=0 and t=0. But my proof that t'=0 means t=0 depends on x=0, and is not true if x ≠ 0.
So we are agreed that (x,t) = (0,0) is the same point (event) in space-time as (x',t') = (0,0) and is the start of movement of the "unprimed origin [place]" that we wish to follow.
Since we are talking about the start of the journey, where the "place" picked out by the world-line of x=0 has not yet met the "place" picked by the "not-moving" world-line of x'=-k, the expectation is at the start x' ≠ -k, otherwise there has been no movement. That's why the math of Appendix [1] demonstrates that for the world-line of x = 0, the choice of t' = 0 uniquely picks the start event that forces x' = 0. More than 100 posts ago, I would be even more explicit by writing $$x_O = 0 \, \textrm{and} \, t'_O = 0 \quad \Rightarrow \quad (x_O,t_O) = (0,0) \, \textrm{and} \, (x'_O,t'_O) = (0,0)$$, or even $$\left. x = 0 \, \textrm{and} \, t' = 0 \quad \Rightarrow \quad (x,t) = (0,0) \, \textrm{and} \, (x',t') = (0,0) \right| _{Start}$$.
You can invert the Lorentz transform all you want, but you have lost the thread of your physical-geometrical reasoning. (tγ = t' is just another way of saying t = t'/γ, and is only valid for describing events on the world-line that is moving relative to the the primed spatial origin and stationary in the unprimed coordinates.) But if you want to talk now about (like you did in post #1) events on a different world line that is moving relative to the unprimed spatial origin and stationary in the primed coordinates, then the correct expression is Δt' = Δt/γ. If you flip-flop which coordinates see the object as stationary, then you have to flip-flop which coordinate measure the shortest elapsed time. See Appendix [4] where I already demonstrated this concept in anticipation of your continued (at least four forums now) failure to undertand place, time, space-time, event, and coordinate, which are prerequisites to having a working understanding of Special Relativity.
You take the "place" defined in the primed frame x'=-k as stationary and ask "how long from now" the moving "place" defined in the moving unprimed frame as x=0. So you need to know what you mean by the time of "now" and the time of "when" the "moving unprimed origin moves to x'=-k".
Obviously, the "place" where x=0 is seen as moving by primed frame -- the x' of this "place" is a non-constant function of t'.
$$x(t) = 0 \quad \Rightarrow \quad x'(t') = -vt'$$
Both are uniform motions and this is true in both Galilean and Special Relativity. In Special Relativity, we can write:
$$x'(t') = -vt' \quad \Rightarrow \quad x(t') = \gamma \left( x'(t') + vt' \right) = 0 \\ x'(t') = -vt' \quad \Rightarrow \quad t(t') = \gamma \left( t' + vx'(t')/c^2 \right) = \frac{1}{\gamma} t' \\ x'(t') = -vt' \quad \Rightarrow \quad x(t) = 0$$
This is the core content of the principle of relativity. What is a "place" for one frame can "move" in another and both descriptions are equally valid.
Note that there is math connecting the claim that t'=0 means t=0 when x=0. See below for why that math is important.
The only thing I don't understand is why you think you understand this material better than Einstein and all who followed in the 105 years.
The standard configuration says when the space-time origin of the unprimed frame (x,t) = (0,0) is the same point (event) in space-time as the space-time origin of the (x',t') = (0,0). Since you started a new question unrelated to the questions of post #1, it was not clearly communicated that you assumed that this event was meant as the start, that you know that every event has both a place and time, or that you have a thesis statement in mind.
What I have done is showing assuming we are talking about the moving "place" which is the spatial origin of the unprimed frame x=0, the choice of t' = 0 implies that the event we are talking about is the common space-time origin of both coordinate systems. This is also (more obviously) true if you pick x=0 and t=0. But my proof that t'=0 means t=0 depends on x=0, and is not true if x ≠ 0.
Please demonstrate superiority before asserting authority; you will still be wrong but at least you will have a reasoned argument for your authoritarian claim.
So we are agreed that (x,t) = (0,0) is the same point (event) in space-time as (x',t') = (0,0) and is the start of movement of the "unprimed origin [place]" that we wish to follow.
You claim this, but have not demonstrated this. Just because you use horribly confusing notation like x' to mean the spatial coordinate in the primed coordinate frame of the end of the journey of the "unprimed origin [place]" is equal to some constant, does not forbid me from using the much more standard notation x' = -k to say the same thing. More than 100 posts ago, I would be even more explicit by writing $$x'_Q = -k$$, or even $$\left. x' = -k \right| _{End}$$.
Since we are talking about the start of the journey, where the "place" picked out by the world-line of x=0 has not yet met the "place" picked by the "not-moving" world-line of x'=-k, the expectation is at the start x' ≠ -k, otherwise there has been no movement. That's why the math of Appendix [1] demonstrates that for the world-line of x = 0, the choice of t' = 0 uniquely picks the start event that forces x' = 0. More than 100 posts ago, I would be even more explicit by writing $$x_O = 0 \, \textrm{and} \, t'_O = 0 \quad \Rightarrow \quad (x_O,t_O) = (0,0) \, \textrm{and} \, (x'_O,t'_O) = (0,0)$$, or even $$\left. x = 0 \, \textrm{and} \, t' = 0 \quad \Rightarrow \quad (x,t) = (0,0) \, \textrm{and} \, (x',t') = (0,0) \right| _{Start}$$.
Exactly why I said that was unnecessarily confusing. x' and x and t' and t are values which vary throughout space-time depending on what event we are talking about. Since you are talking about something moving to a "place" in the primed frame, at the start the place of that something cannot be the same place. So it is better to use some other symbol to represent that place, which is the world-line which consists of all events where the primed spatial coordinate is equal to that negative constant number. x' = -k is a valid way to describe that world-line in the language of algebra, just like x = 0 is a valid way to describe the world-line of the "unprimed origin [of space]".
Well I also have exposure to the notation of Cartesian coordinates and the 300+ year history of physics to know that we are talking about the algebraic geometry of straight lines in space-time. In which case the most general expression of that concept is $$\left. \vec{x}_A(t) = \vec{x}_{0,A} + \vec{v}_{A} t \right|_S$$.where A selects the world-line and S selects the frame. The Lorentz transformation converts this expression into an expression of the same form for a different frame, say $$S'$$. What is at issue here is not Einstein's thesis, but what your thesis is and why you see unable to communicate it or a reason to trust it.
I doubt we are "sliced." But here you are confusing apples and oranges, since you are criticizing me for talking about the start of the movement and here you are talking about the end of the movement. In my notation, your "x' = -v t'" would mean that for all points on the world-line of the "origin" are described by (x',t') = (-vt',t'), but in your notation you are just talking about the end of the journey, which in my notion is x' = -vt' = -k.
It would be more useful to name this as x' = -k like James R did or x' = -5/6 light seconds as your post #1 did.
Here, based a situation such that t measures the elasped time in the frame where there is no movement of the relevant object Chinglu, Einstein, James R, and myself conclude that t = t'/γ < t' .
None of the above refutes Einstein. The frame that sees the object as moving says it takes more time (t') to get from O to Q than the frame that sees the same object as not moving (t).
ERROR.
You can invert the Lorentz transform all you want, but you have lost the thread of your physical-geometrical reasoning. (tγ = t' is just another way of saying t = t'/γ, and is only valid for describing events on the world-line that is moving relative to the the primed spatial origin and stationary in the unprimed coordinates.) But if you want to talk now about (like you did in post #1) events on a different world line that is moving relative to the unprimed spatial origin and stationary in the primed coordinates, then the correct expression is Δt' = Δt/γ. If you flip-flop which coordinates see the object as stationary, then you have to flip-flop which coordinate measure the shortest elapsed time. See Appendix [4] where I already demonstrated this concept in anticipation of your continued (at least four forums now) failure to undertand place, time, space-time, event, and coordinate, which are prerequisites to having a working understanding of Special Relativity.
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No. I have only agreed with one particular point - your primed frame calculation that I quoted above.
I note once again that you have still not addressed the content of posts #117 and #118 in any way.
I must assume at this point that you have no reply to those posts and are therefore avoiding responding to them because you can't admit you were wrong.
Is that right?
Just to be repeat myself: your error is that you have wrongly assumed that the time interval you measured in the unprimed frame is a proper time. In your example, it is the primed frame that measures the proper time between the two events you have used. Therefore, we expect the time interval for the same two events to be LONGER in the unprimed frame than it is in the primed frame, and this is exactly what we see in post #117.
Your problem is that you think you are calculating the time interval between events 1 and 3 in post #117, when in fact you have calculated the interval between events 2 and 3.
But all this is already clear from posts #117 and #118.
You have not shown any error in either of those posts and you're avoiding the issue because you know there is no error. At this point you're just stringing us along like a troll. Your credibility has sunk to nothing.
I thought I did address #117.
1. (x,t) = (0,0)
2. $$(x,t)=(-\gamma k, -\frac{\gamma k v}{c^2})$$
3. $$(x,t)=(0,\frac{k}{\gamma v})$$
1 and 3: $$\Delta t = \frac{k}{\gamma v}$$
2 and 3: $$\Delta t = \frac{\gamma k}{v}$$
In the context of the unprimed frame, if it happened to be the case that the observer had been watching som object moving to the origin based on its own coordinate system that began at -k when the origins were the same, then we would calculate
$$\Delta t = \frac{k}{v}$$
I am not sure it you agree.
Now, if on the other hand, this coordinate happened to be in the coordinates of the moving frame, then when the origins are the same, its distance to travel to the origin is $$\Delta t = \frac{k}{\gamma v}$$ instead because of length contraction. I am not considering the clock at -k at all. The only thing being considered is the view of the unprimed origin observer.
Do we agree on this?
ERROR.
You can invert the Lorentz transform all you want, but you have lost the thread of your physical-geometrical reasoning. (tγ = t' is just another way of saying t = t'/γ, and is only valid for describing events on the world-line that is moving relative to the the primed spatial origin and stationary in the unprimed coordinates.) But if you want to talk now about (like you did in post #1) events on a different world line that is moving relative to the unprimed spatial origin and stationary in the primed coordinates, then the correct expression is Δt' = Δt/γ. If you flip-flop which coordinates see the object as stationary, then you have to flip-flop which coordinate measure the shortest elapsed time. See Appendix [4] where I already demonstrated this concept in anticipation of your continued (at least four forums now) failure to undertand place, time, space-time, event, and coordinate, which are prerequisites to having a working understanding of Special Relativity.
This is not the case. In the view of the primed frame, all can agree in the view of the primed frame, the moving clock at the unprimed origin beats time dilated, that is the part all agree, the origin clock. So, in the primed frame, it calculates Δt' = γΔt or that the moving unprimed origin clock is time dilated. So, the origin moves to the primed coordinate -k a distance k at v or Δt' = k/v.
Using substitution, Δt' = γΔt = k/v or Δt = k/(γ v).
In fact, when unprimed is take stationary, based on length contraction, this is exactly waht the unprimed concludes for the primed coordinate -k to move to it origin such aht both frame agree on the start and end events.
This also is maintains the invertibility of the Lorentz Transform matrix and so this method is consistent with that.
I assume you conclude from the view of the unprimed frame, it must conclude Δt = γΔt' for this same start event of the origins same and primed -k and unprimed origin same.
Since, Δt = k/(γ v) = γΔt'. Then, in your calculation Δt' = k/(γ² v) for the calculation of the primed -k clock to reach the unrpimed origin.
This violates the calculation in the primed frame for the same events and also violates the invertibility of Lorentz transform matrix.
My calculation violates nothing.
Just advertise it on ebay like thousands of others do with unwanted presents. . . (make sure it's still in the original wrapper, though).rpenner said:
chinglu:
If the object began at x=-k and moved to x=0 at speed v, I agree the time taken would be as you say.
Well, that's a complicated statement, isn't it? The problem is the part where you say "when the origins are the same". The origins are assumed to be co-located at $$t=t'=0$$. But events that occur at $$x=-k/\gamma$$ or $$x'=-k$$, for example, are not located at $$x=x'=0$$, and so cannot be simultaneous with events at both $$x=0$$ and $$x'=0$$ at $$t=t'=0$$.
So, you need to carefully check when, in EACH frame, the event at, say $$x=-k/\gamma$$ occurs.
This was the source of your original error, and I quite clearly explained the problem to you in posts #117 and #118. It seems to me that you are still making the same mistake.
The solution to your problem is this: two spatially-separated events that are simultaneous in one reference frame cannot be simultaneous in any other reference frame.
If the object began at x=-k and moved to x=0 at speed v, I agree the time taken would be as you say.
Well, that's a complicated statement, isn't it? The problem is the part where you say "when the origins are the same". The origins are assumed to be co-located at $$t=t'=0$$. But events that occur at $$x=-k/\gamma$$ or $$x'=-k$$, for example, are not located at $$x=x'=0$$, and so cannot be simultaneous with events at both $$x=0$$ and $$x'=0$$ at $$t=t'=0$$.
So, you need to carefully check when, in EACH frame, the event at, say $$x=-k/\gamma$$ occurs.
This was the source of your original error, and I quite clearly explained the problem to you in posts #117 and #118. It seems to me that you are still making the same mistake.
The solution to your problem is this: two spatially-separated events that are simultaneous in one reference frame cannot be simultaneous in any other reference frame.
This part is resolved.
We are not writing about simultaneity as I do not care what the clock at -k in the moving system says. If I cared about its time, you have point and our debate would move on different tract.
So, I am only calculating how long that coordinate in moving system takes to reach origin and nothing more.
I assume you believe in length contraction. So when origins same, that coordinate is a distance k/γ from the unprimed origin.
Again I do not care what the time is on that clock and do not even care if clocks are synchronized in that frame. I only care how long the unprimed frame claims it takes to move to origin and nothing more.
So if length contraction true, then when origins same the distance to the unprimed origin is k/γ. Do you agree or disagree.
You are less than honest.
In your forum you wrote
Physics Forums: Backed into a corner
Let's be honest. I posed this question in your forum. You then banned me and I moved to
http://www.physicsforums.com/showthread.php?t=455613&page=2
to intellectually discuss the issue further. As an internet stalker, you follow me there. Here you droned on and on with false mathematics and even DaleSpam a respected moderator in that forum supported and agreed with my conclusions as anyone can see below.
http://www.physicsforums.com/showpost.php?p=3032436&postcount=20
I demonstrated you falsified my post in your forum here. You are less than honest.
http://www.physicsforums.com/showpost.php?p=3032669&postcount=26
Since that thread concluded with my conclusions as validated, I came here. Again, as an internet stalker, here you came again posting your failed math that I easily refute. I never chased you.
Then, I went back to your forum in which your bullies on that forum attacked me. Then, as all can see here,
http://www.physforum.com/index.php?showtopic=28676&st=120
I corrected the bully with pure math and back him down based only on math.
So, your bullies can dish it out but can't take it. Then, you banned me because it was impossible for your home bullies to push me around.
Again, DaleSpam validates my math and refutes yours.
http://www.physicsforums.com/showpost.php?p=3032436&postcount=20