Measuring distance, speed and clock

[Stephanus]

Similarly, we can write that (E4-E2), measured along the GREEN worldline, is equal to k*(E4-E1), measured along the blue worldline.

Yes, we couldn't write $(E4-E2) = k(E4-E1)$ Because in $(E4-E2)$ E4 is as observerd by Green, which is different if observerd by Blue. That's why you wrote

$(t_r - c_r) = k \, (t_e - c_e)$ and set the values of $c_r$ and $c_e$ such that the receiving clock reads $c_r$ when the transmitting clock reads $c_e$ at the moment when the two clocks are colocated.

$(t_r - c_r) = k\,(t_e -c_e)$ Instead of $(t-c_r) = k\, (t-c_e)$

 

[Stephanus]

That's pretty much all we need to solve the problem, as far as I know. I'm not quite sure what you're puzzled about.

I'm puzzled about the symmetry of time dilation.
Motion is relative. A will see that B clock runs slower.
B will see that A clock runs slower.
But at E3, the first time A see that B is moving toward A their distance is receding, A read B clock's = 0. A clocks = 900.
At E4 when they meet: A = 1500, B = 1200. $\Delta t_A = 600; \Delta t_B = 1200$ A will see that B clock runs faster? Now, I realize A has to add his own clock according to AB distance by 900, do the calculation again.
600 + Distance = 1500. So it match Lorentz factor $1500 = 1200 * \gamma$
But I can't find the symmetry for B. Where or when does B see A's clock run slower? I'm almost close to the solution now

 

[Stephanus]

Yes, yes I got it!
The symmetry for time dilation I think is this.

$V = 0.6; \gamma = 1.25$
E2 = 900 ls from E6.
Two observer Blue (B) and Green (G).
At E4, Delta B clock will be 1500 and Delta G clock will be 1200 or simly B = 1500 and G = 1200.
Clocks do not have to be synchronized!. We'll calculate everything from E4.
At E3 B will see that G is moving. G sent G's clock read: G:0
At E4, B calculate it tooks 600 seconds (from E3 to E4) for G to reach B.
At that time G clock advanced 1200 seconds.
But B just can't divide 600/1200, B has to calculate his clock from E6 -> which is half way from B reading its bouncing signal. (E3-E1)/2 = 900 seconds
So $1500/1200 = 1.25 = \gamma$
What does G see?
G can't directly calculate from E2. G has to calculate everything from E5 where/when G receiving its bouncing signal. And calculate its distance/time from E7 -> which is half way from (E5-E2) = 450 seconds. Its time should be adjusted by gamma factor if using this diagram.
So G will see E4-E7 = 1200 - 450 = 750 seconds, everything is adjusted by gamma factor. Unless we use this diagram.

Using the same logic as B
Again B calculates its time ratio to G by $\frac{(E4-E3)+\frac{(E3-E1)}{2}}{E4-E2} = \frac{E4-\frac{E3+E1}{2}}{E4'-E2} = \frac{1500}{1200} = 1.25$
Manipulating those variables...
G calculates its time ratio to B by $\frac{(E4'-E5)+(E5-E2)/2}{E4-E3} = \frac{E4'-\frac{E5+E2}{2}}{E4-E3} = \frac{750}{600} = 1.25$

 

[Mentz114]

Okay where it intersect Green Line? At (9.5,9.5) [(796.875, 796.875) according to my numbers].

That is not what I asked for. The point of intersection is not the time on the green clock. That time is given by the marks on the green worldline.

So the elapsed time on the green clock betwen the start of the WL and the intersection with the light is about 4.9/20 which agrees with your k-calculation of 0.25.

I don't think you understand yet what proper time is.

 

[Stephanus]

Sorry, I hastily calculated.
About 4.8 or 4.9?
[Add:] And the close the angle to -45⁰ the knots will be separated farther and farther?
[Add:] Oh I see, with my spread sheet, it will be difficult to see the 4.8 and 4.9, I have to transform it manually according to Lorentz transformation boost in the x-direction.

 

[Mentz114]

Sorry, I hastily calculated.
About 4.8 or 4.9?
[Add:] And the close the angle to -45⁰ the knots will be separated farther and farther?
[Add:] Oh I see, with my spread sheet, it will be difficult to see the 4.8 and 4.9, I have to transform it manually according to Lorentz transformation boost in the x-direction.

The knots as you call them are the ticks of the green clock. If you count up it comes to about 4.9 and to get seconds we divide by 20 because we decided that 20 ticks is one second.
It is important to understand that every clock tells the proper length of its worldline. This is a big change from Gallilean relativity.

 

[Stephanus]

The knots as you call them are the ticks of the green clock. If you count up it comes to about 4.9 and to get seconds we divide by 20 because we decided that 20 ticks is one second.
It is important to understand that every clock tells the proper length of its worldline. This is a big change from Gallilean relativity.

Hmhhh, all this time every time I calculated time, I used the reverse of pythagoras. $t' = \sqrt{y^2-x^2)$ How can I didn't see the knots before?
Btw, this is my source numbers and the resulting text files. You might want to rename Source ST-03.txt to Source ST-03.rtf, because PF forum doesn't allowed uploading RTF files.

 

[Stephanus]

This is how I calculate ST diagram with the help of ST plotter and spread sheet.

Then, I calculate events that I already know the coordinates.

E6 (0,0), of course
E4 (1125,1875) -> 1500 * gamma = 1875, x = 0.6 * 1875 = 1125
E2 (1125,1275) -> E4 - 1200 = 1275
And I calculate all events by the help of algebra eliminations and the logical light rays.
Then, I connect the necessary events coordinate to draw light rays.
Wait, I'll a draw a Twins Paradox world lines

 

[Mentz114]

Hmhhh, all this time every time I calculated time, I used the reverse of pythagoras. $t' = \sqrt{y^2-x^2}$ How can I didn't see the knots before?
Btw, this is my source numbers and the resulting text files. You might want to rename Source ST-03.txt to Source ST-03.rtf, because PF forum doesn't allowed uploading RTF files.

I wondered why you didn't see the 'knots'.

You have made the connection with proper time, $\tau = \sqrt{t^2-x^2}$ which is good.

I'm not going to download any of the stuff you posted. I find your methods complicated and baroque but if it makes sense to you then that's fine.

 

[Stephanus]

Twins paradox

There is something very wrong here?
Those I area that I circle. Why there are so much lines? I've calculated my numbers twice (may be three times) But the lines at the red circle are crowded?? I'm afraid there's something in my text files that ST Plotter won't receive.

 

[jbriggs444]

Twins paradox
Those I area that I circle. Why there are so much lines?

If you looked at those south-east to north-west yellow lines as light signals sent from the green twin to the blue twin then that close spacing would be called a Doppler shift.

 

[Stephanus]

What about this? This looks right to me.

 

[Mentz114]

What about this? This looks right to me.
View attachment 86136

That looks OK. But, as usual, I'm wondering why you need all those light pulses ?

The time between parting and meeting on the blue clock is the number of blue knots between the events, and the time on the green clock is the total number of green knots.

You could also calculate the green clock times using $\tau=\sqrt{t^2-x^2}$ for the green triangles.

 

[Stephanus]

That looks OK. But, as usual, I'm wondering why you need all those light pulses ?

The time between parting and meeting on the blue clock is the number of blue knots between the events, and the time on the green clock is the total number of green knots.

You could also calculate the green clock times using $\tau=\sqrt{t^2-x^2}$ for the green triangles.

I don't know
I just try to draw ST diagram quickly. I saw an ST diagram for Twins Paradox, there were lights rays from every "knots". And I just copy paste the formula. Doesn't have to drag many lines. Okay, just don't explain to me about Twins Paradox now. Studying doppler first.

 

[Stephanus]

[..]Time dilation we cannot directly measure. This was discussed in another thread just a day or so ago: https://www.physicsforums.com/threa...ver-be-directly-observed.820770/#post-5152104

The link/thread you gave me is very useful. I just read it now. Thanks.