Measuring distance, speed and clock

[Stephanus]

That is interesting but it shouldn't be necessary. I can use the mouse interface to draw any worldlines in a few seconds.

Yep, that's right. I'm sorry, if you still remember what I've done. I'm still cartesianing ST diagram.
Consider this.

This graph clearly shows that Green moves at 0.6c
Green distance is 900 units. I like 900 when c is 0.6
$\gamma = 1.25$, and $\frac{900}{1.25} = 720$
So, where will the light that comes from E2 will cross Green world line?
E2 is 300 seconds.
Y = X + 300
Y = (X - 900)*3/5,
Sorry,
F1: $X = t - 300$ Light ray from E2
F2: $X = vt + 900$, Green world line
Eliminates F1 by F2, you'll have...
$0 = t(1-v)-1200; t = 750; x = 450$, then I put those numbers in spread sheet. I have the formula to convert world line, light ray and event to screen coordinate.
W,W0, 0, 300, 400, 300, 150, 8388608
W,W1, -0.6, 390, 300, 300, 150, 32768
W,W2, 0, 390, 350, 390, 300, 32768
E,E2, 270, 300, 0
E,E3, 225, 345, 0
E,E1, 300, 390, 0
E,E4, 150, 300, 0
L,L0, 0, 270, 300,0, 0, !none, 345, 225, 1
and load it to STPlotter. This is much simpler for me.
The world line color is in RGB mode,
R*1 + G * 100h + B * 10000h

 

[Mentz114]

Yep, that's right. I'm sorry, if you still remember what I've done. I'm still cartesianing ST diagram.
Consider this.
View attachment 86082
This graph clearly shows that Green moves at 0.6c
Green distance is 900 units. I like 900 when c is 0.6
$\gamma = 1.25$, and $\frac{900}{1.25} = 720$
So, where will the light that comes from E2 will cross Green world line?
E2 is 300 seconds.
Y = X + 300
Y = (X - 900)*3/5,
Sorry,
F1: $X = t - 300$ Light ray from E2
F2: $X = vt + 900$, Green world line
Eliminates F1 by F2, you'll have...
$0 = t(1-v)-1200; t = 750; x = 450$, then I put those numbers in spread sheet. I have the formula to convert world line, light ray and event to screen coordinate.
W,W0, 0, 300, 400, 300, 150, 8388608
W,W1, -0.6, 390, 300, 300, 150, 32768
W,W2, 0, 390, 350, 390, 300, 32768
E,E2, 270, 300, 0
E,E3, 225, 345, 0
E,E1, 300, 390, 0
E,E4, 150, 300, 0
L,L0, 0, 270, 300,0, 0, !none, 345, 225, 1
and load it to STPlotter. This is much simpler for me.
The world line color is in RGB mode,
R*1 + G * 100h + B * 10000h

That is weird.

If I want to draw a similar diagram

1. click on 'Draw worldline'
2. Put cursor on the start point
3. Hold down left mouse button and move to end point. I can see $\gamma$ and $\beta$ in the status bar.
4. Release cursor.

Repeat for B.

To draw the light pulse path
1. select 'Draw light ray'
2. Put cursor on start point
3. Hold down left button and move to end point.
4. Release button.

There was no need to calculate the intersection of the light and WL B. The plotter forces the correct point.

So what do you use the plotter for ? Can you read off the lengths of intervals for instance ?

 

[Stephanus]

Can you read off the lengths of intervals for instance ?

But, I can't. I wouldn't know the precise distance. And if I save the project to text file, the number just isn't rounded.
And I have one problem. I like 0.6c, because gamma is round. And I like 900, because 900/1.25 is 720, and if I want to draw an event which is at 60 interval, than 900/60 is 15 round and 720/60 is 12 round and 15/12 is gamma, again round.
Again for 1-V or 1.6, so 720/1.6 is 45 and 900/1.6 is 56.25, hmmhh, still round nicely. At least not some number like 1.333333333
And I'm having trouble to draw the precise coordinate at 900, the plotter only draw in 20 units pixel for its square.
For -0.6c, supposed if I draw a wordline at (15,0) to (0,25), the plotter can't draw beyond (0,15).
I like 15 because 15/gamma is 12. The world line is supposed to cross (12,0) if I "match speed" it.
So I calculate all the numbers, based on 900 units. Then divide them by 100 scale so it can be uploaded to ST plotter limit: 20 squares, and convert it to text. Have to do it in spread sheet to speed up calculations.

So what do you use the plotter for ?

After I calculate all the numbers, than I load project them to the plotter and I upload it in the forum, if I have a question regarding SR, so I can explain my problem clearly. Too bad the plotter does not have line for simultaneity of event. Only world line (<45⁰) and light ray (45⁰ or -45⁰). But it's still a very, very good software.

 

[Stephanus]

Sorry, I have to cartesian ST diagram, because I have known cartesian since junior high school. I only knew Hendrik Lorentz 3 months ago . I have to do cartesia it over and over until I can lorentz them intuitively.

 

[Mentz114]

Sorry, I have to cartesian ST diagram, because I have known cartesian since junior high school. I only knew Hendrik Lorentz 3 months ago . I have to do cartesia it over and over until I can lorentz them intuitively.

You don't have to apologise for calculating the intersection of two lines.

But, I can't. I wouldn't know the precise distance. And if I save the project to text file, the number just isn't rounded.
And I have one problem. I like 0.6c, because gamma is round. And I like 900, because 900/1.25 is 720, and if I want to draw an event which is at 60 interval, than 900/60 is 15 round and 720/60 is 12 round and 15/12 is gamma, again round.
Again for 1-V or 1.6, so 720/1.6 is 45 and 900/1.6 is 56.25, hmmhh, still round nicely. At least not some number like 1.333333333
And I'm having trouble to draw the precise coordinate at 900, the plotter only draw in 20 units pixel for its square.
For -0.6c, supposed if I draw a wordline at (15,0) to (0,25), the plotter can't draw beyond (0,15).
I like 15 because 15/gamma is 12. The world line is supposed to cross (12,0) if I "match speed" it.
So I calculate all the numbers, based on 900 units. Then divide them by 100 scale so it can be uploaded to ST plotter limit: 20 squares, and convert it to text. Have to do it in spread sheet to speed up calculations.
After I calculate all the numbers, than I load project them to the plotter and I upload it in the forum, if I have a question regarding SR, so I can explain my problem clearly. Too bad the plotter does not have line for simultaneity of event. Only world line (<45⁰) and light ray (45⁰ or -45⁰). But it's still a very, very good software.

I don't understand this but it shows you have no idea what the plotter is for. Why do you say that it is 'good software' when it won't do the stuff you mention above ?

You find ways to wriggle out of anything that might be relativistic.

Look at this diagram. Can you say what the time is on Greens clock at the event when the first light pulse intersects with the green WL ?

 

[Stephanus]

You don't have to apologise for calculating the intersection of two lines.

Because you once said something like "Don't say cartesianing ST diagram, this is nonsense."

Look at this diagram. Can you say what the time is on Greens clock at the event when the first light pulse intersects with the green WL ?

What?? Where did you get this picture? Upload ST-03.txt and you'll see. Open "Original Source for ST-03.txt" in notepad. Don't open it in ST plotter. This is my original number. You'll see that I use 1500 and 1700 coordinate to match $v = \frac{15}{17}c$ I haven't upload it yet at PF Forum. I still have a question regarding pervect post. But your post came before I get the chance to finish my question.
Okay where it intersect Green Line? At (9.5,9.5) [(796.875, 796.875) according to my numbers].
But I have to count each square carefuly. I tought it was 10.5, not 9.5. Second calculation gives 9.5.
It's still a good software. If I "match speed" it, it will show the Lorentz transformation for one. And it helps me much than drawing ST diagram in Microsoft Excel or in Microsoft Paint.

 

[Stephanus]

It may help to draw a space-time diagram of this.

The sequence of events from A's point of view goes like this:

A sends a signal at -1 second (1 second before impact) according to A's clock -> E1
A receives a return signal at -.0625 (-1/16 a second) according to A's watch -> E4
The returned signal from B is conveniently timestamped with B's time reading. B's clock reads -.25 seconds -> E3

But A knows not to confuse the reading on B's clocks with his own readings.

PRECISELY! There's not ASSURANCE that their clock are synchronized first. Is this true?
They can only rely that each other is using an accurate atomic clock. Is this true?
And there's NO WAY for B to know WHEN they will meet. So we can rely to -0.25 number. -0.25 seconds is just an arbitrary number, is this true?
Of course at E4 A WILL KNOW when they will meet, is this true?

A considers himself at rest in his own frame, with B moving towards him.
- Yes
A knows that the time it takes for the signal to reach B is equal to the time that the signal takes to return from B.
- Yes
Interpreting his radar results, A concludes that the round-trip time for the light/radar signal was (-1/16 - (-1)) seconds, i.e. the round trip time was 15/16 seconds, which implies that the one-way trip time was 15/32 of a second.
-Yes

A computes the time in A's frame that B received the signal as the start time (-1 second) plus the one-way travel time (15/32 of a second), for a result of -17/32 seconds, i.e. -.53125 seconds
This is obviously different from the time B assigned to the same event. If A is familiar with special relativity, he expects this - he expects B's clock to be running slow, and it is - just as much slower as predicted, reading only -.25 seconds rather than -.53125 seconds.
- But A just can't divide -0.53125 to $\frac{8}{17}$ before E5, is this true?

Now, for the velocity.
A meets B at 0 seconds -> E5, and A knows that at the time B received the signal (-.53125 seconds), that B's distance was c multipled by the one-way travel time, i.e 15/32 seconds * c, c being the speed of light. Recall that the speed of light in A's frame must be constant an equal to c in A's frame on both the outgoing and ingoing trip.
- Yes

So A computes B's velocity as the distance, (15/32) light seconds dived by the time it takes for A to reach B, (17/32) seconds, thus B's velocity was 15/17 of the speed of light.
- But A can deduce B velocity at E4, according to Doppler, right? A doesn't have to WAIT until E5 to determine B velocity. Is this true?

It might be easier to work out a closely related problem where all the times are integers - or to use algebraic variables. Using the former approach (integers), we can imagine A sending a signal at 16 seconds before impact (by A's watch), and receving the echo at 1 second before impact (by A's watch).

Now you tell me, after all the calculations above.

 

[Stephanus]

Dear PF forum,
Before I ask further, can someone explain to me about time dilation?

This problem is similar to Post - #84
Rather than starting at -1 second where we already know when Blue (B) and Green (G) will meet, I started the clock at 0, so we won't know at first when they will meet. "Nature can't be fooled."
Green (G) moves at 0.6c; $V = -0.6; \gamma = 1.25$
The distance before Green starts to move is 900 ls
From Blue (B) frame,
E1: -900
E3: 900
E4: 1500
From Green (G) frame
E2: 0
E4: 1200, again if E2 is 0
Clocks are NOT synchronized, is this important?
Okay,...

E1: B:-900
B sends a signal to G, containing B clock's: B:-900
E2: G:0
G receives the signal from B, reads the data B:-900, G sends the signal to B, containing G's clock, G:0
Is it relevant for G to compare B:-900 with G's clock? I think no, because their clocks are not synchronized. G can only write (B:-900;G:0) in G's notebook.
G knows that B is moving toward G their distance is receding by 0.6c because of Doppler effect, is this true?
E3: B:900
B receives the signal from G, reads the data G:0,
Is it relevant for B to compare G:0 with B's clock? Again, I think not. B will write G:0; B:900 in B's notebook
B knows that G is moving toward B their distance is receding by 0.6c.

E4:B:1500; G:1200
B and G meet:
B reads his notebook
E3: When B is 900s, G is 0
E4: When B is 1500s, G is 1200
$\Delta t_b = 600; \Delta t_g = 1200$ What is this? G's clock runs faster then B's?

G reads his notebook
E2: When B is -900s, G is 0
E4: When B is 1500s, G is 1200
$\Delta t_b = 2400; \Delta t_g = 1200$ B's clock runs faster than G's.

The situation from E3 and E2 is symmetrical both for B and G. $\frac{1}{2}$

But from G frame, it's B who moves, right? Motion is relative, and the clock for moving object runs SLOWER, not faster?
Is this how B should reconcile?
At E3, B receives the bounce signal that B has sent at -900 (E1),
$\frac{\Delta t}{2} = \frac{1800}{2} = 900$, so B knows that actually G received the signal 900 seconds ago E3. So the distance where G received the signal is 900 ls away when B clock reads $900-900=0$.
At E4, when they meet. B see G clock is 1200, and comparing to his notebook E3: G Clocks -> 0. B clocks read 1500 at E5
So $\Delta t_g = 1200$, while $\Delta t_b = 1500$. So this is actually G proper time where B receives G signal at E3. Is this true?
This clocks conform gamma factor. $\frac{1500}{1200}=\gamma = 1.25$

How B should reconcile? I can't find the solution here.
The situation should be symmetrical, right?
Thanks for any help.

 

[pervect]

For post #98, in blue's frame we can write:

$(E4 - E3) = k^2 (E4-E1)$

where k is the doppler shift factor, $\sqrt{\frac{1 - \beta}{1+\beta}} = \sqrt{.4 / 1.6} = 0.5$ in this example, since $\beta = v/c = .6$.

Here E1, E3, and E4 are time coordinates along the blue worldline, which are the reading of the blue clock at the specified events. It might be clearer to write timeof.E4.in.blue.frame instead of just E4, but that's too much work, so I just wrote down E4 , etc, and explained.

Note that E4-E1 is just the proper time along the blue worldline until it meets the green worldline. If we don't know when the two worldline's meet, we calculate that first (calculate E4), then use the above formula.So it's really not that much different, we use the same basic idea.

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

We can write down a few other things, too:

In the blue frame, $E2 = (E1 + E3)/2$, which is zero in your example. We also know that the distance from blue to green in the blue frame at time E2 is given by the relationship 2*distance = (E3-E1)

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.

The logical justification for the above formulae is just the fact that the relative velocity between blue and green is constant, implying that the doppler shift is constant, and that fact that for every transmitted signal, there is a unique time of reception, a 1:1 mapping.

 

[Stephanus]

Post #99

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

Post #17

Basically, if you have two observers, one of which is moving relative to the other, who synchronize their clocks such that they both read zero when they are colocated, you can write a very simple relationship between the proper time of emission for one observer, and the proper time of reception for the other:

$t_{r} = k t_{e}$

where $t_r$ is the time of reception, and $t_e$ is time time of transmission. If you insist on synchronizing your clocks differently , then you'd need to rewrite this equation as

=>$(t_r - c_r) = k \, (t_e - c_e)$<= 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.

Ah, I got it!

 

[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.