Can someone help?

[Jack_]

False dichotomy. You should have learnt about such logical fallacies on that imaginary course on logic you want people to believe you took.

Once again you quoted my entire post, including a direct simple question and you ignored it all. Are you so afraid to discuss things for fear of exposing more of your ignorance that you'll repeatedly be blatantly dishonest and hypocritical?

I have repeatedly asked you to say whether or not you understand a LT always maps an expanding light sphere to an expanding light sphere. If you answer 'no' to this then you admit you can't apply LT properly, which completely contradicts your claims about being able to put SR into a contradiction. Is this the reason you won't answer my question, you know you'll contradict yourself?

How did you do with the twins thread?

You failed.

Otherwise, show everyone here how you can refute my math.

 

[Jack_]

False dichotomy. You should have learnt about such logical fallacies on that imaginary course on logic you want people to believe you took.

Once again you quoted my entire post, including a direct simple question and you ignored it all. Are you so afraid to discuss things for fear of exposing more of your ignorance that you'll repeatedly be blatantly dishonest and hypocritical?

I have repeatedly asked you to say whether or not you understand a LT always maps an expanding light sphere to an expanding light sphere. If you answer 'no' to this then you admit you can't apply LT properly, which completely contradicts your claims about being able to put SR into a contradiction. Is this the reason you won't answer my question, you know you'll contradict yourself?

Is this what you mean?

$$ \frac{(x - vr/c)^2}{r^2\gamma^2} + \frac{y^2}{r^2} + \frac{z^2}{r^2} = 1 $$

Every point on this object when mapped with LT produces τ=r/c and

ξ² + η² + ς² = c² τ² = r²

[EDIT]

Here is the proof.

http://www.proofofabsolutemotion.com/AN.pdf

 

[Pete]

This is true.

However, there exists an infinite number of times in the rest frame such that t'=r/c and x' = r.

Not right, Jack.
The only time and place in the rest frame at which t'=r/c at x'=r is
when $$t=\frac{\gamma r}{c}(1+\frac{v}{c})$$
at $$x=\gamma r(1+\frac{v}{c})$$
When will you learn to apply a Lorentz transform?

This is really old stuff, by the way. Remember this?
Look at the t'= r/c row, in the LR column.
$$\begin{array}{c|cc|cc|cc|cc}
& O(x,\ \ t) & O(x',\ \ t') & O'(x,\ \ t) & O'(x',\ \ t') & LL(x,\ \ t) & LL(x',\ \ t') & LR(x,\ \ t) & LR(x',\ \ t') \\
\hline
t=t'=0 & (0,\ \ 0) & (0,\ \ 0) & (0,\ \ 0) & (0,\ \ 0) & (0,\ \ 0) & (0,\ \ 0) & (0,\ \ 0) & (0,\ \ 0) \\
\hline

t=\frac{r}{\gamma c} &
(0,\ \ \frac{r}{\gamma c}) &
(\frac{vr}{c},\ \ \frac{r}{c}) &

(\frac{vr}{\gamma c},\ \ \frac{r}{\gamma c}) &
(0,\ \ \frac{r}{\gamma^2c}) &

(\frac{-r}{\gamma},\ \ \frac{r}{\gamma c}) &
(-r(1-\frac{v}{c}),\ \ \frac{r}{c}(1-\frac{v}{c})) &

(\frac{r}{\gamma},\ \ \frac{r}{\gamma c}) &
(r(1+\frac{v}{c}),\ \ \frac{r}{c}(1 + \frac{v}{c}) \\

t=\frac{r}{c} &
(0,\ \ \frac{r}{c}) &
(\frac{-\gamma vr}{c},\ \ \frac{\gamma r}{c}) &

(\frac{vr}{c},\ \ \frac{r}{c}) &
(0,\ \ \frac{r}{\gamma c}) &

(-r,\ \ \frac{r}{c}) &
(-\gamma r(1+\frac{v}{c}),\ \ \frac{\gamma r}{c}(1+\frac{v}{c})) &

(r,\ \ \frac{r}{c}) &
(\gamma r(1-\frac{v}{c}),\ \ \frac{\gamma r}{c}(1-\frac{v}{c})) \\

t=\frac{\gamma r}{c} &
(0,\ \ \frac{\gamma r}{c}) &
(\frac{-\gamma^2vr}{c},\ \ \frac{\gamma^2r}{c}) &

(\frac{\gamma vr}{c},\ \ \frac{\gamma r}{c}) &
(0,\ \ \frac{r}{c}) &

(-\gamma r,\ \ \frac{\gamma r}{c}) &
(\frac{-r}{1-\frac{v}{c}},\ \ \frac{r}{c-v}) &

(\gamma r,\ \ \frac{\gamma r}{c}) &
(\frac{r}{1+\frac{v}{c}},\ \ \frac{r}{c+v}) \\

\hline

t'=\frac{r}{\gamma c} &
(0,\ \ \frac{r}{\gamma^2c}) &
(\frac{-vr}{\gamma c},\ \ \frac{r}{\gamma c}) &

(\frac{vr}{c},\ \ \frac{r}{c}) &
(0,\ \ \frac{r}{\gamma c}) &

(-r(1+\frac{v}{c}),\ \ \frac{r}{c}(1 + \frac{v}{c}) &
(\frac{-r}{\gamma},\ \ \frac{r}{\gamma c}) &

(r(1-\frac{v}{c}),\ \ \frac{r}{c}(1-\frac{v}{c})) &
(\frac{r}{\gamma},\ \ \frac{r}{\gamma c}) \\

t'=\frac{r}{c} &
(0,\ \ \frac{r}{\gamma c}) &
(\frac{-vr}{c},\ \ \frac{r}{c}) &

(\frac{\gamma vr}{c},\ \ \frac{\gamma r}{c}) &
(0,\ \ \frac{r}{c}) &

(-\gamma r(1-\frac{v}{c}),\ \ \frac{\gamma r}{c}(1-\frac{v}{c})) &
(-r,\ \ \frac{r}{c}) &

(\gamma r(1+\frac{v}{c}),\ \ \frac{\gamma r}{c}(1+\frac{v}{c})) &
(r,\ \ \frac{r}{c}) \\

t'=\frac{\gamma r}{c} &
(0,\ \ \frac{r}{c}) &
(\frac{-\gamma vr}{c},\ \ \frac{\gamma r}{c}) &

(\frac{\gamma^2vr}{c},\ \ \frac{\gamma^2r}{c}) &
(0,\ \ \frac{\gamma r}{c}) &

(\frac{-r}{1+\frac{v}{c}},\ \ \frac{r}{c+v}) &
(-\gamma r,\ \ \frac{\gamma r}{c}) &

(\frac{r}{1-\frac{v}{c}},\ \ \frac{r}{c-v}) &
(\gamma r,\ \ \frac{\gamma r}{c}) \\

\hline
\end{array}$$

How did you do with the twins thread?
You failed.

:roflmao:
Have you read that thread, Jack? Your argument was comprehensively demolished in the first page or two. You spent the rest of the thread with your fingers in your ears saying "La la la I can't hear you!"

 

[Jack_]

Not right, Jack.
The only time and place in the rest frame at which t'=r/c at x'=r is
when $$t=\frac{\gamma r}{c}(1+\frac{v}{c})$$
at $$x=\gamma r(1+\frac{v}{c})$$
When will you learn to apply a Lorentz transform?

This is really old stuff, by the way. Remember this?
Look at the t'= r/c row, in the LR column.
$$\begin{array}{c|cc|cc|cc|cc}
& O(x,\ \ t) & O(x',\ \ t') & O'(x,\ \ t) & O'(x',\ \ t') & LL(x,\ \ t) & LL(x',\ \ t') & LR(x,\ \ t) & LR(x',\ \ t') \\
\hline
t=t'=0 & (0,\ \ 0) & (0,\ \ 0) & (0,\ \ 0) & (0,\ \ 0) & (0,\ \ 0) & (0,\ \ 0) & (0,\ \ 0) & (0,\ \ 0) \\
\hline

t=\frac{r}{\gamma c} &
(0,\ \ \frac{r}{\gamma c}) &
(\frac{vr}{c},\ \ \frac{r}{c}) &

(\frac{vr}{\gamma c},\ \ \frac{r}{\gamma c}) &
(0,\ \ \frac{r}{\gamma^2c}) &

(\frac{-r}{\gamma},\ \ \frac{r}{\gamma c}) &
(-r(1-\frac{v}{c}),\ \ \frac{r}{c}(1-\frac{v}{c})) &

(\frac{r}{\gamma},\ \ \frac{r}{\gamma c}) &
(r(1+\frac{v}{c}),\ \ \frac{r}{c}(1 + \frac{v}{c}) \\

t=\frac{r}{c} &
(0,\ \ \frac{r}{c}) &
(\frac{-\gamma vr}{c},\ \ \frac{\gamma r}{c}) &

(\frac{vr}{c},\ \ \frac{r}{c}) &
(0,\ \ \frac{r}{\gamma c}) &

(-r,\ \ \frac{r}{c}) &
(-\gamma r(1+\frac{v}{c}),\ \ \frac{\gamma r}{c}(1+\frac{v}{c})) &

(r,\ \ \frac{r}{c}) &
(\gamma r(1-\frac{v}{c}),\ \ \frac{\gamma r}{c}(1-\frac{v}{c})) \\

t=\frac{\gamma r}{c} &
(0,\ \ \frac{\gamma r}{c}) &
(\frac{-\gamma^2vr}{c},\ \ \frac{\gamma^2r}{c}) &

(\frac{\gamma vr}{c},\ \ \frac{\gamma r}{c}) &
(0,\ \ \frac{r}{c}) &

(-\gamma r,\ \ \frac{\gamma r}{c}) &
(\frac{-r}{1-\frac{v}{c}},\ \ \frac{r}{c-v}) &

(\gamma r,\ \ \frac{\gamma r}{c}) &
(\frac{r}{1+\frac{v}{c}},\ \ \frac{r}{c+v}) \\

\hline

t'=\frac{r}{\gamma c} &
(0,\ \ \frac{r}{\gamma^2c}) &
(\frac{-vr}{\gamma c},\ \ \frac{r}{\gamma c}) &

(\frac{vr}{c},\ \ \frac{r}{c}) &
(0,\ \ \frac{r}{\gamma c}) &

(-r(1+\frac{v}{c}),\ \ \frac{r}{c}(1 + \frac{v}{c}) &
(\frac{-r}{\gamma},\ \ \frac{r}{\gamma c}) &

(r(1-\frac{v}{c}),\ \ \frac{r}{c}(1-\frac{v}{c})) &
(\frac{r}{\gamma},\ \ \frac{r}{\gamma c}) \\

t'=\frac{r}{c} &
(0,\ \ \frac{r}{\gamma c}) &
(\frac{-vr}{c},\ \ \frac{r}{c}) &

(\frac{\gamma vr}{c},\ \ \frac{\gamma r}{c}) &
(0,\ \ \frac{r}{c}) &

(-\gamma r(1-\frac{v}{c}),\ \ \frac{\gamma r}{c}(1-\frac{v}{c})) &
(-r,\ \ \frac{r}{c}) &

(\gamma r(1+\frac{v}{c}),\ \ \frac{\gamma r}{c}(1+\frac{v}{c})) &
(r,\ \ \frac{r}{c}) \\

t'=\frac{\gamma r}{c} &
(0,\ \ \frac{r}{c}) &
(\frac{-\gamma vr}{c},\ \ \frac{\gamma r}{c}) &

(\frac{\gamma^2vr}{c},\ \ \frac{\gamma^2r}{c}) &
(0,\ \ \frac{\gamma r}{c}) &

(\frac{-r}{1+\frac{v}{c}},\ \ \frac{r}{c+v}) &
(-\gamma r,\ \ \frac{\gamma r}{c}) &

(\frac{r}{1-\frac{v}{c}},\ \ \frac{r}{c-v}) &
(\gamma r,\ \ \frac{\gamma r}{c}) \\

\hline
\end{array}$$


:roflmao:
Have you read that thread, Jack? Your argument was comprehensively demolished in the first page or two. You spent the rest of the thread with your fingers in your ears saying "La la la I can't hear you!"

You did not refute anything in the twins thread or you would post the refutation.

As to your claim there is only one point in the rest frame with t'=r/c, we will examine (-r,0,0), (0,r,0) and (r,0,0) in the moving frame except we will use our brain.

According to the light postulate, t'=r/c for the points above, do you understand this?

Now, you apply LT as an instructional aid and you will find simultaneity in the moving frame occurs at an interval in the rest frame. You will see different x coords in the rest frame for these points with t'=r/c as per the light postulate in the moving frame.

 

[James R]

Moderator note:

Jack_ has been permanently banned from sciforums for repeated trolling in the Physics & Math forum.

He has accumulated enough infraction points for a permanent ban.

 

[Hercules Rockefeller]

Oh no, the international scientific orthodoxy manages to suppress yet another free-thinker! It’s so tragic that these professional educated scientists are locked in their established ways. If only they could break free of their conformist theories, think outside the box and discard the scientific method they would see the way forward and could recognise the genius of the internet’s community of unpublished self-taught science forum physicists.

 

[AlphaNumeric]

I know Jack's been banned but I'll comment anyway.

How did you do with the twins thread?

You failed.

Otherwise, show everyone here how you can refute my math.

Its in the threads. Your denial of reality doesn't make your position true.

Is this what you mean?

Jack proves his lack of understanding of SR.

Here is the proof.

Its always funny when a hack, who professes absolute belief in their own abilities, has time to write up an argument against someone on a forum and yet doesn't have the time to email said argument to a journal.

Anyway, he's banned now. SR, QT, logic, evolution. All subjects Jack not only doesn't understand but deliberately searches out people who do and then argues with them!! Talk about picking your battles poorly!