Time in Special Relativity

OK, I do not want to talk anymore.

 

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So, you've finally come to the conclusion that you're wrong? Good for you!

 

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Perhaps the table is not as clear as it could be. Feel free to ask for clarification.
Look at the row for t=rγ/c (obviously?):

\(\begin{array}{c|cc|cc|cc|cc} & O(x,\ \ t) & O(x',\ \ t') & O'(x,\ \ t) & O'(x',\ \ t') & LL(x,\ \ t) & LL(x',\ \ t') & LR(x,\ \ t) & LR(x',\ \ t') \\ \hline t=\frac{r}{\gamma c} & (0,\ \ \frac{r}{\gamma c}) & (\frac{vr}{c},\ \ \frac{r}{c}) & (\frac{vr}{\gamma c},\ \ \frac{r}{\gamma c}) & (0,\ \ \frac{r}{\gamma^2c}) & (\frac{-r}{\gamma},\ \ \frac{r}{\gamma c}) & (-r(1-\frac{v}{c}),\ \ \frac{r}{c}(1-\frac{v}{c})) & (\frac{r}{\gamma},\ \ \frac{r}{\gamma c}) & (r(1+\frac{v}{c}),\ \ \frac{r}{c}(1 + \frac{v}{c}) \\ \hline \end{array}\)

The times in the moving frame are the t' coordinates:
At the location of O, t'=r/c.
At the location of O', t'=r/γγc.
At the location of LL, t'=r(1-v/c)/c.
At the location of LR, t'=r(1+v/c)/c.

The answer is in the row for t'=r/c (again, I thought that would be obvious?):

\(\begin{array}{c|cc|cc|cc|cc} & O(x,\ \ t) & O(x',\ \ t') & O'(x,\ \ t) & O'(x',\ \ t') & LL(x,\ \ t) & LL(x',\ \ t') & LR(x,\ \ t) & LR(x',\ \ t') \\ \hline t'=\frac{r}{c} & (0,\ \ \frac{r}{\gamma c}) & (\frac{-vr}{c},\ \ \frac{r}{c}) & (\frac{\gamma vr}{c},\ \ \frac{\gamma r}{c}) & (0,\ \ \frac{r}{c}) & (-\gamma r(1-\frac{v}{c}),\ \ \frac{\gamma r}{c}(1-\frac{v}{c})) & (-r,\ \ \frac{r}{c}) & (\gamma r(1+\frac{v}{c}),\ \ \frac{\gamma r}{c}(1+\frac{v}{c})) & (r,\ \ \frac{r}{c}) \\ \hline \end{array}\)

Look at the light flash events at that time: LL is at (x',t')=(-r,r/c), LR is at (x',t')=(r,r/c)
So, O' is at x'=0, the Left Light flash is at x'=-r, the Right Light flash is at x'=r.

 

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Very good, you accept what I said in my post?
That you were wrong about several facts regarding time dilation in your previous post?
That in the moving frame when the clock at O elapses rγ/c, the clock of O' elapses r/c?
That in the stationary frame when the clock at O elapses r/c, the clock of O' elapses rγ/c?

 

We agree that a moving clock at x'=-r reads t'=r/c at t=r/(γ(c+v))

Wrong.
You've done a single reading on a clock at a single time.
To get an elapsed time and a rate, you need a second reading, such as t'=0.

At what time in the stationary frame does the moving clock at x'=-r read t'=0?
t = -γrv/c²

So in the stationary frame, how long does it takes the moving clock at x'=-r to elapse r/c?
It reads zero at t = -γrv/c²
It reads r/c at t = r/(γ(c+v))
The difference is r/(γ(c+v)) + γrv/c² = γr/c

So, the moving clock at x'=-r takes γr/c to elapse r/c.

 

I don't think he'll be able to leave it alone. But maybe he was just playing all along, and now his fun is done.

 

You demonstrate SR is not viable.

But, you did not proceed with time dialtion and then the light sphere.

Do you need help?

 

When the clock at O' is r/c, why doesn't your table have light r in all directions?

 

Non sequitors will get you nowhere.
I wrote a post, you responded "Very good". Why? Because you agreed with the content of that post?

Do you, or do you not, agree that according to SR:
In the moving frame when the clock at O elapses rγ/c, the clock at O' elapses r/c?
In the stationary frame when the clock at O elapses r/c, the clock at O' elapses rγ/c?

 

It does, Jack. Just like I spelled out in the post you quoted. Did you not read it?

Here it is again:

Look at the light flash events at that time: LL is at (x',t')=(-r,r/c), LR is at (x',t')=(r,r/c)
So, O' is at x'=0, the Left Light flash is at x'=-r, the Right Light flash is at x'=r.

 

We have a winner!

 

This is reciprocol time dilation.

So what. You did not address the problem and you cannot escape using reciprocol time dilation.

If you try, I will put you into an infinite sequence that will converge to t=0 for any event.

 

Nope, we are operating from the stationary frame.

Are you contending the relativity postulate is false and that the rules of physics are different frame to frame?

This means you are contending the stationary frame cannot solve a problem and the moving frame is required.

That places the moving frame as preferred.

 

do you do anything?

 

Did you know that Reciprocol Harum is composed entirely of twins to Procol Harum? They skipped the light speed fandango and are now by "reciprocol" time dilation entirely immune to reason.

 

Jack, you said earlier:

Are you satisfied that those questions are answered?
Have you learned anything from those answers?

Yes, Jack, it's reciprocal time dilation.
So, do you now understand that your statement: "When the clock of O elapses rγ/c, the clock of O' elapses r/c" is not true in the moving frame?

You've tried the infinite sequence thing before, Jack. You made mistakes (don't feel bad - others have done the same), which were pointed out to you.
Any event that is transformed from the stationary frame to the moving frame will be transformed back to the original event.
Try it. Take any event (x, t). Transform it to (x', t'). Transform it again to (x, t).
You'll always return to the same event.

You made a similar mistake when you contended that the Lorentz transform says that a moving clock at x'=-r beats faster in the stationary frame. You never responded to the correction in post 85 - does that mean you accept you made a mistake, and that you now understand that the lorentz transform does predict time dilation?

Perhaps you should state your question more clearly.
You said:
"When the clock at O' is r/c, why doesn't your table have light r in all directions?"
I assumed you meant:
"In the moving frame, when the clock at O' is r/c, why doesn't your table have light r in all directions from O'?"

I can't see any other way that your question would make sense.
In the stationary frame, the clock at O' reaches r/c at t=rγ/c.
At that time, the light is rγ in all directions from O.
It isn't r in all directions from anything, and nor should it be.

No. How does that follow from my post?

 

This means time dilation is false. You have disproven SR. Good job, but you failed.

You do not understand what this means.
Assume -r and calculate t in the stationary frame.

Then calculate it in the moving frame.

You will find t' > t.

Do not debate this with me until you have the math.

 

I see you've ignored several direct questions in my post, Jack.
Why? Are you getting lost?

Let me remind you of a few things. I expect actual answers to all questions, even if it's "No, I do not agree". Explanations of your reasoning are encouraged.

Do you see that the table does in fact answer those questions?
What did you learn from those answers?

Take any event (x, t). Transform it to (x', t'). Transform it again to (x, t).
Have you tried this exercise?
Do you see that there is no infinite regress implied in the Lorentz transformations?

I assumed you meant:
"In the moving frame, when the clock at O' is r/c, why doesn't your table have light r in all directions from O'?"
Is that what you meant? If so, do you see that the table does have light r in all directions from O' in the moving frame?
If it's not what you meant, then what did you mean?

Do you see that time dilation and light spheres are a necessary consequence of the Lorentz transforms? (See the maths in posts 54 and 55.)

What is your reasoning?
Do you see that in the moving frame, time dilation says that when the clock of O' elapses rγ/c, the clock of O elapses r/c, not the other way around?

The math is in post 85, Jack.
The moving clock at x'=-r reads t'=0 at t = -γrv/c², and t'=r/c at t = r/(γ(c+v)).
So it takes a time of γr/c for the moving clock to elapse r/c.

Do you see that the Lorentz transform says that the moving clock at x=-r runs slowly in the stationary frame?

 

There's a mistake in this last line of the previous post. It should read:

Do you see that the Lorentz transform says that the moving clock at x'=-r runs slowly in the stationary frame?

 

We assume x'=-r.

t' = r/c

Stationary frame at that same point.

t = ( t' + vx'/c²)γ

t = r/(γ(c+v))

So, from the view of the stationary frame, the light is a distance r in the negative direction from the oriigin of the moving frame when the clock at that point reads t = r/γ(c+v).

The clock in the moving frame reads r/c.

So, whern light is a distance "-r" from O', O belives the clock at the point reads t = r/γ(c+v) and the clock in the moving frame at the at that point reads r/c.

What am I missing?

The moving clock is faster at that point.