Measuring distance, speed and clock

[Mentz114]

"Don't believe anything the plotter is doing??"" Oh, I'm so sorry Mentz114. It's not a bug.
I edited the text file!
This software is very good, but what troubles me is the resolution. So, I worked out the coordinate in Microsoft Excel and I edit the text file, and I load the text in the plotter.
For x coordinate it is 300 + x*20
For t coordinate it is 300 - t*20
Then I get the picture, then I save it to bitmap.
It's completely my fault. I'm afraid it's the resolution.

Yes, I worked out that you hacked the file. You swapped the start and end points - but the software always rearranges them when you create the wl. The program got the velocity reversed because the endpoints were swapped after creation.

Please don't hack the saved files. You can delete a worldline if you right-click on it.

I hoped the animation would help you to see that each horizontal line is a time-slice, and the objects on slanted lines will change position in each slice.

 

[Stephanus]

Please don't hack the saved files. You can delete a worldline if you right-click on it.

not "hack", please. But "editing". And I think there's a bug in deleting an event. Suppose we have 5 events,
ev1, ev2, ev3, ev4 and ev5. If we delete for example ev4 then ev1 is deleted, not ev4, and if delete ev4 for the second time, ev2 is deleted. That's why I deleted the text manually.

I hoped the animation would help you to see that each horizontal line is a time-slice, and the objects on slanted lines will change position in each slice.

I haven't observed transformation just watched. I just use it to draw wl and light ray and event. It's a very good software, it helps me very much. Thanks.
[EDIT] For now, but later, I think I have to study at least acceleration wl

 

[Dale]

If we have two observer (B) Blue and (G) Green.
- They are at rest,
- Clocks are synchronized (is this related to our discussion? I don't think so, but I'll write anyway)
- Then at a preagreement time, G moves toward B at 0.6c γ=1.25\gamma = 1.25
What is the correct way to draw the ST diagram?

Picture 2 is obviously wrong. In it the green line is discontinuous. That means that green disappears from one location and reappears at another location. You are saying that turning on a rocket engine causes not only acceleration, but also teleportation.

 

[Stephanus]

Picture 2 is obviously wrong. In it the green line is discontinuous. That means that green disappears from one location and reappears at another location. You are saying that turning on a rocket engine causes not only acceleration, but also teleportation.

Teleportation , Now that's funnier than

A little elaboration: suppose that at CERN in Geneva a particle is accelerated to 0.99999c in the direction of Lyon. Does that contract the distance between Geneva and Lyon?

Thanks DaleSpam, Mentz114 has already answered me. #64

 

[Stephanus]

However, you can work out this formula for yourself following Bondi's apporach, simply from knowing that "k" exists. Suppose observer A is approaching observer b. At some time 1 second before they meet, A sends out a signal which B receives at a time k seconds (k<1) before they meet. B retransmits the signal, which is received by A, at a time k^2 seconds before they meet.
Example - if k=1/4, then A sends a signal 1 second before meeting, B receives this signal 1/4 = .25 seconds before they meet, and retransmits it. A receives the return signal 1/16 of a second before they meet.
From this exchange of signals, plus the knowledge that the speed of light is constant in his frame, A can compute the velocity relative to B as follows:[..]

I don't quite sure, but I think "A receives the return signal 1/16 of a second before they meet" is wrong.
I tried to work it out:
https://www.physicsforums.com/attachments/85844
Okay,...

The picture is not in scale to avoid measuring the distance to find the answer.

F1: "Suppose observer A is approaching observer b. At some time 1 second before they meet, A sends out a signal which B receives at a time k seconds (k<1) before they meet. "
At E0 A emits a signal to B 1 second before they meet. So L1 distance is V_A * 1 sec

F2: "Example - if k=1/4, then A sends a signal 1 second before meeting, B receives this signal 1/4 = .25 seconds before they meet..."
So, B is receiving the signal at $1 sec - 0.25 sec = 0.75 sec$ after A sends the signal.
So L1 distance is 0.75 ls.
Combining F1 and F2, so V_a is 0.75c, because it takes 1 sec for A to travel to B0.75c
So I want to work out, what is E1 distance from E0.
Supposed: E1 is A position when the light reaches B, see F2. So it takes 0.75 sec for A which travels at 0.75 c to reach E1.
So E1 is 0.5625 or $\frac{9}{16} ls or \frac{63}{112}$

F3: "B receives this signal 1/4 = .25 seconds before they meet, and retransmits it. A receives the return signal 1/16 of a second before they meet."
So, when A is at E1, B retransmit the signal.
Supposed A meets the signal at E2. To answer that we have to calculate L3 and L4
We know that $L2+L3+L4 = 0.75 ls = \frac{12}{16} or \frac{84}{112}$
$L2 = \frac{63}{112}$
So $L3+L4 = \frac{21}{112}$
A travels at 0.75c and light travels at c.
So combining both,
$t = d/(v+c)$
$t = \frac{21}{112} / 1.75 = \frac{3}{28}$
I think A will not receive the signal 1/16 before they met, but at $\frac{3}{16}-\frac{3}{28} = \frac{9}{112}$ seconds.
Or perhaps my calculations are wrong?
Or my understanding is wrong?

 

[Stephanus]

Oh, no. The attachment failed. Ok, I ask again my question.

However, you can work out this formula for yourself following Bondi's apporach, simply from knowing that "k" exists. Suppose observer A is approaching observer b. At some time 1 second before they meet, A sends out a signal which B receives at a time k seconds (k<1) before they meet. B retransmits the signal, which is received by A, at a time k^2 seconds before they meet.

Example - if k=1/4, then A sends a signal 1 second before meeting, B receives this signal 1/4 = .25 seconds before they meet, and retransmits it. A receives the return signal 1/16 of a second before they meet.

From this exchange of signals, plus the knowledge that the speed of light is constant in his frame, A can compute the velocity relative to B as follows:[..]

This is my picture. Again, not to scale because at first we (I) didn't know what their positions, so we have to calculate every point. This picture just makes us easier to do the calculation.
Okay,...
"Example - if k=1/4, then ..."
F1: $k = 0.25$
F2: "At some time 1 second before they meet, A sends out a signal" A sends the signal at $E0$, $L1 = V_A * 1 sec= V_A$, for simplicity the second $V_A$ is not a velocity, but a distance.
F3: "B receives this signal 1/4 = .25 seconds before they meet", supposed B received the signal when A is at $E1$, so $L3+L4 = \frac{1}{4}V_A$
F4: "B receives the signal and retransmits it. A receives the return signal 1/16 of a second before they meet." Supposed A received the signal when A is at $E2$
------------------------------------------------------------------------------
F3: B receives the signal at $0.25 seconds$ before they meet, or $1-0.25 = 0.75seconds$ after A sends the signal.
So length $L1 = 0.75c$
So $V_A = 0.75c$
F3: $L3+L4 = \frac{1}{4}V_A = \frac{3}{16}$light seconds
So A travels at 0.75c and c travels at -c. The will meet at E2

Again, not to scale.
$L3+L4 = \frac{3}{16}$
The time it takes for A to reach E2 (or C reaches E2) is
$\frac{L3+L4}{V_A-C} = \frac{3}{16}/1.75 = \frac{3}{28} seconds$
The time for A from E2 to reach E3 is $\frac{7}{28}-\frac{3}{28} = \frac{1}{7}$ seconds.
Did I do it right?
Do I understand the problem correctly?

 

[Stephanus]

Okay, today I learned that
https://www.physicsforums.com/threads/measuring-distance-speed-and-clock.821115/#post-5155592
This

and this,

are actually the same. And we can use the same formula to calculate them.
$k=\sqrt{\frac{1-\beta}{1+\beta}}$ or $k=\sqrt{\frac{1+\beta}{1-\beta}}$

 

[Mentz114]

Okay, today I learned that
'''
'''
are actually the same. And we can use the same formula to calculate them.
$k=\sqrt{\frac{1-\beta}{1+\beta}}$ or $k=\sqrt{\frac{1+\beta}{1-\beta}}$

I don't understand the diagrams.

This is a clip I made some time ago that might help to understand the Doppler effect. The moving object emits light pulses at a regular frequency and the two stationary observers detect the frequency to be different. The script is attached and you should get the same result if you do an 'animate' of this.

 

[Stephanus]

I don't understand the diagrams.

This is a clip I made some time ago that might help to understand the Doppler effect. The moving object emits light pulses at a regular frequency and the two stationary observers detect the frequency to be different. The script is attached and you should get the same result if you do an 'animate' of this.

Wow, thanks Mentz114. I have had this file a month ago. At least the movie not the text file. You sent me. Of course I didn't get the text file, just the representation graph and the video. How can STPlotter make the video out of it?
Btw, what I mean about my picture is this.
I kept imagining that B sends a signal for every, say, 10 seconds. And if B moves toward A, A will receive the signal less then 10 seconds.
And the good advisors/mentors/members kept saying about sending light and blue shifted and red shifted.
What I didn't realize was, both were the same.
Sending signals for every 10 seconds is the same as, for example, sending a green light.
But in green light case, you'll send signal for every 540 pico seconds.
https://en.wikipedia.org/wiki/Color#Spectral_colors
That's what keeps holding me understanding their answers.
But thanks anyway for your responses. I'll read the previous post' again.

 

[Mentz114]

Wow, thanks Mentz114. I have had this file a month ago. At least the movie not the text file. You sent me. Of course I didn't get the text file, just the representation graph and the video. How can STPlotter make the video out of it?

I forgot that I'd sent this earlier. The frames can be saved and a video making program used to join them into a clip.

Btw, what I mean about my picture is this.
I kept imagining that B sends a signal for every, say, 10 seconds. And if B moves toward A, A will receive the signal less then 10 seconds.

Yes and no. The receiver detects the time gap between signals. The transmission time is not important. And a detector moving away sees the gap as more than 10 secs.

And the good advisors/mentors/members kept saying about sending light and blue shifted and red shifted.
What I didn't realize was, both were the same.

Blue shifted means the frequency is increased and the wavelength is decreased. Red shifted means the reverse.

Sending signals for every 10 seconds is the same as, for example, sending a green light.
But in green light case, you'll send signal for every 540 pico seconds.

Yes. The colour we perceive depends on the frequency of the light wave hence 'red' and 'blue' shifted.
It is similar for gaps between pulses. That is how ground-based radar can detect the speed of a vehicle.

https://en.wikipedia.org/wiki/Color#Spectral_colors
That's what keeps holding me understanding their answers.
But thanks anyway for your responses. I'll read the previous post' again.

I hope it is clearer.

 

[Stephanus]

Blue shifted means the frequency is increased and the wavelength is decreased. Red shifted means the reverse.

Just like in Franhover lines? (or Frauhover, I forget the exact name)
Thanks for the respond Mentz114.
Perhaps you can understand my confusion.
At first I tought that:
1: B burst a signal every 10 seconds.
While all the answers sugest that
2: B sends colored light, for example Green. Or B sends signal very but not at 540 pico seconds as I mistakenly wrote above. 540THz is the frequency, so B actually sends signals every $\frac{1}{540E12} = 0.0018 pico seconds$
Conclusion:
1: B burst a signal every 10 seconds (any color)
2: B burst an EM wave signal every 0.0018 pico seconds (Green light)
Conclusion: The logic is the same.
And if B moves toward A, the signal is blue shifted. Perhaps the frequency can rise to 500 THz at the orange range?. Forgive the pun.

 

[pervect]

Suppose B sends a radio signal, like WWV, that has a 10 Mhz frequency, but is also amplitude modulated at 1khz. Then we can say that both the carrier frequency and the modulation frequency are red/blue shifted by the same amount due to doppler effects. This follows from the fact that at the transmitter, the ratio of the modulating frequency to the carrier frequency is 10,000:1, and the fact that this ratio is expected to be constant even if the carrier is red or blue shifted.

It doesn't really matter what the details of the modulation are. We can generalie from the above example with a sinusoidal carrier and a sinusoidal modulation signal to "bursty" modulation once every 10 seconds (.1 hz). The point is that the carrier signal and the modulating signal will both be red/blue shifted by the same amount, this is necessary for logical consistency.

[add]
It might be easier to explain this way. Suppose we send out "bursts" each of which contain 1000 carrier pulses. Then everyone will agree that there are 1000 pulses in a burst, even if they don't agree about the carrier frequency due to the doppler effect.

 

[Stephanus]

Finally, pervect...

Suppose B sends a radio signal, like WWV, that has a 10 Mhz frequency, but is also amplitude modulated at 1khz. Then we can say that both the carrier frequency and the modulation frequency are red/blue shifted by the same amount due to doppler effects.

Yes, I understand.

This follows from the fact that at the transmitter, the ratio of the modulating frequency to the carrier frequency is 10,000:1, and the fact that this ratio is expected to be constant even if the carrier is red or blue shifted.

Yes

It doesn't really matter what the details of the modulation are. [..]The point is that the carrier signal and the modulating signal will both be red/blue shifted by the same amount, this is necessary for logical consistency.

Yes

[..]Suppose we send out "bursts" each of which contain 1000 carrier pulses. Then everyone will agree that there are 1000 pulses in a burst, even if they don't agree about the carrier frequency due to the doppler effect.

Yes

But, pervect, can I ask you about your previous post? About sending a signal where k = 1/4. I think there's something not match there. Or the mistake is mine. It is in https://www.physicsforums.com/threads/measuring-distance-speed-and-clock.821115/page-4#post-5167834

 

[pervect]

Finally, pervect...
Yes, I understand.YesYesYes

But, pervect, can I ask you about your previous post? About sending a signal where k = 1/4. I think there's something not match there. Or the mistake is mine. It is in https://www.physicsforums.com/threads/measuring-distance-speed-and-clock.821115/page-4#post-5167834

Yes, there's a problem, which I'd describe as you not taking into account the fact that simultaneity is relative and the effects of time dilation. A's watch and B's watch keep different time. This means that A's watch and B's watch read different times, we have to be careful not to conflate them.

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. A receives a return signal at -.0625 (-1/16 a second) according to A's watch. The returned signal from B is conveniently timestamped with B's time reading. B's clock reads -.25 seconds. But A knows not to confuse the reading on B's clocks with his own readings.

A considers himself at rest in his own frame, with B moving towards him. 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. 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.

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.

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

Now, for the velocity. A meets B at 0 seconds, 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.

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.

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).

 

[Stephanus]

Thanks pervect for your answer.

A receives a return signal at -.0625 (-1/16 a second) according to A's watch. The returned signal from B is conveniently timestamped with B's time reading. B's clock reads -.25 seconds (add: according to B's watch).

Ahh, that makes a big different. Okay, I'll start again.

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

It takes 1 second for A to travel L1 distance. Supposed A speed is V, then $L1 = V$, A receives the return signal at 1/16 seconds, $L4 = \frac{1}{16}V$ Light travels at $V+\frac{1}{16}V = \frac{17}{16}V distance$
A travels at $\frac{15}{16}V$ distance. So A speed is $V = \frac{15}{17}c$ Okay,... Conform your calculation

[..]thus B's velocity was 15/17 of the speed of light.

Motion is relative, right. It's A who's at rest and B travels.
$\gamma = \frac{1}{\sqrt{1-V^2}} = \frac{1}{\sqrt{1-\frac{15^2}{17^2}}} = \frac{1}{\frac{8}{17}} = \frac{17}{8} = 2.125$
The signal will reach B according to A clock at $\frac{2}{17} seconds$ before impact. Because B travels at $\frac{15}{17}c$, so, B clock according to B is
$\frac{2}{17}*\frac{17}{8} = \frac{2}{8}$ Gosh, the number match!
But I still can grasp it intuitively. I'll study it again. Thanks pervect.
Now, go back to Post #17

 

[Mentz114]

Thanks pervect for your answer.
Ahh, that makes a big different. Okay, I'll start again.
View attachment 86067

It takes 1 second for A to travel L1 distance. Supposed A speed is V, then $L1 = V$, A receives the return signal at 1/16 seconds, $L4 = \frac{1}{16}V$ Light travels at $V+\frac{1}{16}V = \frac{17}{16}V distance$
A travels at $\frac{15}{16}V$ distance. So A speed is $V = \frac{15}{17}c$ Okay,... Conform your calculation
Motion is relative, right. It's A who's at rest and B travels.
$\gamma = \frac{1}{\sqrt{1-V^2}} = \frac{1}{\sqrt{1-\frac{15^2}{17^2}}} = \frac{1}{\frac{8}{17}} = \frac{17}{8} = 2.125$
The signal will reach B according to A clock at $\frac{2}{17} seconds$ before impact. Because B travels at $\frac{15}{17}c$, so, B clock according to B is
$\frac{2}{17}*\frac{17}{8} = \frac{2}{8}$ Gosh, the number match!
But I still can grasp it intuitively. I'll study it again. Thanks pervect.
Now, go back to Post #17

Here is a diagram of the above ( nearly). Between A (blue) and B(green) we have $\gamma=2.13$. The grid scale is (As coords) 1sec = 20 grid units.

The light is sent 1 sec before they meet ( interval AR ) and the light returns at t=0.63 s before they meet ( interval RD).

The time on Bs clock when the light arrives can be read off as about 4.85/20 secs assuming synchronization at t=0 (A time).

This seems to agree close enough with Pervects (and your) numbers.

 

[Stephanus]

Here is a diagram of the above ( nearly). Between A (blue) and B(green) we have $\gamma=2.13$. The grid scale is (As coords) 1sec = 20 grid units.

The light is sent 1 sec before they meet ( interval AR ) and the light returns at t=0.63 s before they meet ( interval RD).

The time on Bs clock when the light arrives can be read off as about 4.85/20 secs assuming synchronization at t=0 (A time).

This seems to agree close enough with Pervects (and your) numbers.

What?? You beat me by one hour

 

[Mentz114]

What?? You beat me by one hour

What do you mean ?

[edit]
I get it. Sorry.

Your diagram gives the same results but with a bit more error because it is less than half the scale of mine.

Well done. You now have the relevant times calculated by two methods. I prefer the diagram way.

 

[Stephanus]

What do you mean ?

[edit]
I get it. Sorry.

Your diagram gives the same results but with a bit more error because it is less than half the scale of mine.

Well done. You now have the relevant times calculated by two methods. I prefer the diagram way.

Well, if you know me. I calculate the speed $\frac{15}{17}$
The easiest way to do it is to put some point at 15 at the x-Axis and 17 at the t-axis. Calculate all those by spread sheet. I have the formula now, for world line, light and event. Open load project at STPlotter, see if I made a "bug". What is holding me, is the coordinate for light differs than world line. X in light is T in world line and vice versa. Depends on the speed 1 or -1. So does the coordinate for event is t,x not x,t as in wordline. That's why I didn't calculate it from 20 units as you do. This is my data:

[B]
Name  X          T        X          T
W0    0.0000 -3000.0000   0.0000  3000.0000; A world line
W1 1500.0000     0.0000   0.0000  1700.0000; B world line
E1    0.0000     0.0000          
E2 1500.0000     0.0000             
E3  796.8750   796.8750               
E4    0.0000  1593.7500               
E5    0.0000  1700.0000              
L0    0.0000     0.0000 796.8750  796.8750
L1  796.8750   796.8750   0.0000 1593.7500[/B]

Scale it down by 200, convert it to screen coordinates.

Name    X             T                 X           T
W0    300.000000    600.000000        300.000000    0.000000
W1    450.000000    300.000000        300.000000  130.000000
E1    300.000000    300.000000            
E2    300.000000    450.000000            
E3    220.312500    379.687500            
E4    140.625000    300.000000            
E5    130.000000    300.000000            
L0    300.000000    300.000000        379.687500    220.312500
L1    220.312500    379.687500        300.000000    140.625000

Add the necessary code for STPlotter and save it to ST-01.txt. Actually I like the combination of both. Calculate the numbers on spread sheet, convert it to world line, light or event code and save it to text file and load it to STPlotter.

 

[Mentz114]

Well, if you know me. I calculate the speed $\frac{15}{17}$
The easiest way to do it is to put some point at 15 at the x-Axis and 17 at the t-axis. Calculate all those by spread sheet. I have the formula now, for world line, light and event. Open load project at STPlotter, see if I made a "bug". What is holding me, is the coordinate for light differs than world line. X in light is T in world line and vice versa.

Add the necessary code for STPlotter and save it to ST-01.txt. Actually I like the combination of both. Calculate the numbers on spread sheet, convert it to world line, light or event code and save it to text file and load it to STPlotter.

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

 

[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.