# Measuring distance, speed and clock

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1. Jun 28, 2015

### Stephanus

Dear PF Forum,
I'd like to know how we measure speed, distance clock in space?
$v = 0.6c \gamma = 1.25$
Supposed the distance between A and B is 900 ls.
And supposed A and B keeps exchanging signal for 6 seconds interval.
Because the distance between A and B is 900 ls, there are 150 signals from A to B in that 900 ls space and 150 signals from B, too.
Clocks are synchronized and they start sending signals. Now at T900, then the first signal from A reaches B (and the first B signal reaches A) Now, A moves toward B...
When A reaches B, (I don't include the calculations here, I think this is very simple for the good mentors and advisors) B clock will shows 900 + 1500 = 2500 seconds and B clock shows 900 + 1200 = 2100 seconds.
A will receive (900/6 + 1500/6) 400 signals.
How does A reconcile with this?
"I'm receiving 400 signals for 1200 seconds, so 3 seconds for each signal. The frequency for I receive signal is twice what I should have, so I'm picking the signal and move at 1c, because I receive the signal at half what I should receive.
WAIT. This is not the case, I study SR, I know Lorentz contraction, so I have to calculate it again.
The distance of signal is NOT 6s . It's length contracted according to Lorentz Factor."

Remember A doesn't know the speed yet. So this is what A should have calculated.
The signal moves at c, A moves at V, the distance is contracted.
$V = \frac{F^2-1}{F^2+1}$ F is the frequency of receiving signal.
$V = \frac{2^2-1}{2^2+1} = 0.6$
I comes up with $V = \frac{F^2-1}{F^2+1}$ through a long way equations, so I don't include them.
So basically, A can determine its speed through something like Doppler effect. Is this true?
How can A determine distance? A will see the B clock is 900 sec late, so B is 900ls away. But clock can be deceiving right? What if B moves backward and forward and B light cone reaches A showing 900 ls late?
Second problem.
Consider this:
Three observer A, B1 and B2:
Distances:
A -> B1: 100 lys
B1 -> B2: 3 lys
Clocks are synchronized
B1 and B2 would move together toward A.
So here is the situation.
A stays
B1 and B2 move together toward A at say... 0.6c
100 years later A will see that B1 is moving toward A at 0.6c, distance is contracted. B1 would be at 80ly away. But A still see B2 at 103 lys away. Is this true? This statement seems contradicted length contraction. I'm not against the SR theory, much less disputing it, understanding it is very difficult.

_______________________________________________________________________________
So what I ask here is this:
With all the observers receive are signals from the other observers, all they see are lights/signals coming from the other observers...
1. How can we measure speed? Through Doppler effect?
2. How can we measure distance? Through clock? But clock can be deceiving right? Altough "nature can be fooled" (Richard Feynman)
3. How can we measure other time (dilation)? Through speed (Dopper) and therefore using Lorentz formula?
4. In problem 2. Concerning A, B1 and B2. How can we reconcile this? Drawing space time diagram would be cheating right. Because we 'already know' the problem.

2. Jun 28, 2015

### Staff: Mentor

The only time that we can directly measure is proper time, the time that will be measured by a clock that is at rest relative to us and colocated with us. We could count our heartbeats, or the ticks of our wristwatch, or any other time-dependent process that's happening where we are.

However, given proper time we can get all the other measurements. Distance to an object at rest relative to us is easy - bounce a light signal off it, divide the round-trip time by two, multiply by c, and we have the distance. If we get the same answer every time we know the object is at rest relative to us - if not, we know the distance to the object at the moment that the reflection happened.

Now we have several ways of measuring speed. We can bounce a signal of known frequency (say from a speed radar gun that is at rest relative to us so we can use our proper time to determine its frequency) and Doppler will tell us the speed. Or we can position two detectors at rest relative to us in its path at a known distance apart (we have distance measurements already), have each detector send a flash of light our way as the object passes, and then by subtracting out the light travel times we have the time difference and so can calculate the speed using $v=\Delta{x}/\Delta{t}$.

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

3. Jun 28, 2015

### Mentz114

Here is a diagram of the blue observer measuring the distance to two other objects. The distances are (B-A)/2 and (C-A)/2 where A,B, C refer to the proper times of the events.

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4. Jun 29, 2015

### Stephanus

Thanks, Mentz114. Judging from your Space Time diagram, the object (green) doesn't move. What if the object moves?
The left green will move closer to A when its light cone entering A world line, right. And then, the right green light cone entering A world line.
So, instead of length contraction, A will see that Green is length expanded? Or there's a simultaneity of event here. I'm at the office now, when I get home, I'll draw the sapce time diagram. See, if someone can correct it for me.

5. Jun 29, 2015

### Mentz114

Green does not have a length. This is a distance measurement i.e. the separation between the emission event and the reflect event. The result is always correct for this separation. However by the time Blue knows this Green has already moved away ( or closer if it was approaching).

6. Jun 29, 2015

### harrylin

A few issues:
What is missing here: distance AB according to which system at what time? There is no reason for the two systems to agree on that, so it should be specified.
Let's call system S the inertial rest system of A. Let's specify that according to S, distance AB is 900 ls at t=0.
As it takes 900 s to reach B, then in 6 s all signals that A emits will still be between A and B. If the signal is a pulse at 1 Hz frequency then there will be 6 signal pulses between A and B.
Synchronized according to S, or according to S' ??
What you can do, is synchronize them when they are at relative rest. And then agree on time t=0 that they start moving towards each other at a certain speed.
Now you have to decide when you want to send the 6s of signals: until they are moving, or from the time that they start moving. And according to which clock(s) you send the signals for 6s.
Yes.
Same problem with "synchronized" as before: according to which reference system...

7. Jun 30, 2015

### Stephanus

Of course. The vertical line, doesn't have length. It's not a distance, it's a time. What I meant is not "length", but the distance between left vertical green and right vertical green.

8. Jun 30, 2015

### Stephanus

A and B is at rest. Their distance at rest is 900ls.
No, what I mean is every 6 seconds, A emits a signal. The signal frequency? It's not in the equation here. The only frequency that they will receive, is the receiving frequency. B would receives signal from A every 6 seconds and vice versa. If they are at rest all the time.
According to S, they haven't move yet. A will move at t=900
Their respective clock. A sends signal every 6s by his clock. B sends signal every 6s according to B clock.
According to B frame. B1 and B2 moves toward A from B frame.

9. Jun 30, 2015

### Mentz114

The green line is a line representing the change in position of Green as time progesses.

Do you understand how to measure the distance to an object by bouncing a signal off it ? Nugatory has explained this above very clearly.

If the object is moving towards or away from you, you will stll measure the ditance between the emission and the reflection events.

10. Jun 30, 2015

### Stephanus

Yes, by using Space Time diagram. But I calculate the space time diagram by Cartesian, not by Lorentz. It gives the same result tough. But I have to know "why" Lorentz transformation is thus, although just boost in x direction only.
And one thing to remember about space time diagram. The hypotenuse length is not $\sqrt{x^2 + t^2}$, but it's $\sqrt{t^2-x^2}$
I forgot who throwed the problem before. But it's about the object, moving away from us, and bouncing a signal to its companion. The companion? At certain point, it moves toward us.

11. Jun 30, 2015

### Mentz114

I don't know what to say to this. Speechless.

You will not answer the 'why' question. No one can tell you 'why' things are the way they are.
I'm glad you remember that. Even if you forget who told you.

You will not stand still long enough to learn anything. You asked about measuring but you seem to ignore the answers and change the subject when people try to explain.

12. Jun 30, 2015

### Stephanus

Thank's Mentz114 for your answer. If you, I still remember, that you are THE ONE who taught me Space Time diagram. . It's not that I'm ignoring any answers. I'm afraid that Harrylin misunderstood my question.

This is what I mean. Every 6s, A send a signal. The time for A (or B) to send the signal? Well, perhaps for 1 second, perhaps for 1 ms, but The time that they send signal AGAIN is 6s. This is the picture that I meant.

But, judging from the answer, I think this is what it means.

The length of the signal is 6 seconds. That's why
but this is not what I meant.

13. Jun 30, 2015

### Mentz114

Please don't say that ! I tried to teach you the space-time diagram and failed. You still say completely wrong things like 'I use Cartesian and it gives the same results'.

14. Jun 30, 2015

### harrylin

OK then, retake with the lacking info added:
OK the case now well defined.
You mean probably that B clock reads 2400s and A clock reads 2100s. Looks right to me.

That is, for classical case, assuming medium in rest and you are moving.

NO that is wrong: you are moving relative to the assumed "rest system". You are using the calibration including synchronization of that system.
As long as you do that, you must assume that your clock is slowed down by the gamma factor, just as you already calculated.
Else you must first adjust the clock synchronization to a reference system according to which "moving clock A" is in rest. And it is that resync procedure that corrects for the error that you encountered here. It's probably the most occurring mistake that we see on the relativity forum, and I know as a fact that you even participated in a few threads in which this was rather well explained. You fell in that trap again, but this was a good example.

15. Jun 30, 2015

### Stephanus

A gross typo or calculation
No, I mean I have to draw the Space Time diagram first. I think if I see the other observer moves toward me from a distance, beside his/her length is contracted, his/her T0 is started earlier. Relative Simultaneity?
I have to draw two space diagram. A. For me in rest frame, B. For me in moving frame. Wait..

Last edited: Jun 30, 2015
16. Jun 30, 2015

### harrylin

It went already wrong with the classical calculation, as you apparently switched the physical interpretation without realising it. I have the impression that you try to understand SR by means of the effects such as time dilation and length contraction on top of classical physics, and that is certainly possible to do. If so, it may be useful to get that one right first.

The classical Doppler equation that you used, corresponds to clock A moving relative to space (the ether), and clock B remaining in rest.
But then you spoke about length contraction of the distance AB according to SR. That corresponds to a different physical perspective: the perspective that clock A and B were first moving together (so that the distance AB from that perspective is shorter), and that next clock A stopped moving.

But from that new, different perspective, both classically and in SR:

- the clock synchronization that was done on the assumption that clocks A and B were first in rest, is wrong
- the distance between signals in both directions differs
- you used the wrong Doppler equation - see https://en.wikipedia.org/wiki/Doppler_effect#General

17. Jun 30, 2015

### pervect

Staff Emeritus
I would recommend the method used by Bondi in "Relativity and common sense". I believe I've seen it online, but I'll let you google for it. to avoid possible issues with the links being less than legitimate.

You can also look for "Bondi k calculus", though there is no actual calculus involved, just algebra.

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.

Here k is the doppler shift factor, It's the same relativistic doppler shift factor that we discussed before. When the two observers are approaching, there will be a blueshift. This means that k, the ratio of transmission time to reception time, is less than 1, and the formula is $k = \sqrt{\frac{1 - \beta}{1+\beta}}$. When they are moving away from each other, $k = \sqrt{\frac{1 + \beta}{1-\beta}}$, there is a redshift, which means that the recep

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:

In A's frame, B is moving towards him at some velocity. The radar signal was sent at time -1 (1 second before meeting), and was recieved at a time of -k^2 (1/16 of a second before meeting). The round trip time is (1-k^2). The signal moves out at the speed of light and returns at the speed of light, so A concludes that the midpoint, -(1+k^2)/2 seconds, B was a distance of c * (1-k^2)/2 away. Taking the ratio, we can find the velocity of B relative to A in A's frame, which is just the ratio of distance / time, c * (1+k^2) / (1-k^2).

Lets go through the example with k = .25. Then the signal sent at 1 second before union arrives 1/16 second before union. A's conclusion from his radar measurement is that at a time of (1+1/16)/2 = 17/32 seconds before union, the distance was c * (1 - 1/16) / 2 = (15/32) light seconds, or a distance of c * (15/32 seconds). He concludes this because in his frame, the one-way trip time must be half the total trip time in his frame, using his frame's notion of what simultaneity means. Thus he knows that when k=.25, the approach speed must be 15/17 the speed of light.

To get k in terms of v, you need to do some algebra, the method gives v in terms of k, but it's reasonably straightforwards high school math to solve the equation.

After the meet, on the outbound trip, A sends a signal at 1 second after the union. The echo is received 16 seconds later, because now k=4 rather than 1/4. This means that at 17/2 seconds, the distance was 15/2 light seconds, the velocity again is 15/17 of the speed of light.

You can work out the time dilation factor from this too, if you assume that B timestamps his time signal when he replies to A.

If this explanation is too terse (likely), try tracking down Bondi's book and reading it.

18. Jul 1, 2015

### Stephanus

A. If at first A and B is at rest, separated by 900ls away.
B. They synchronized their clock
C. And some 900 ls later, A moves toward B.
So what is wrong with this?
1. A and B can't synchronized their clocks, because they are far apart away? If so what's the difference between 900ls away, or 900 miles, or even 9 mm? Colocated?
2. Even if A and B synchronized their clocks, it will be useless for A to read B clock (or vice versa), because the participants clocks are not synched anymore?
3. Is this how we should treat the clock in SR?
3a. My clock tick for 10 seconds, your clock tick for 12 seconds.
And this is useless in SR.
3b. My clock SHOWS July 1st 09:00 AM, yours shows July 1st 08:15 AM. We can't just substract the clock, before there's some relativity involved here.
So you mean by the clock synchronization that was done on asumption..., is wrong is because we can only measure 3a, not 3b?

Thanks for any clarification.

Last edited: Jul 1, 2015
19. Jul 1, 2015

### Stephanus

Duck for a while.
Every time I draw a space time diagram, it takes a lot of effort.
So I break for a while to make this simple software.
It helps me draw space time diagram.

20. Jul 1, 2015

### harrylin

Note that your quote truncates the sentence such that what remains is not what I wrote (and is not correct). I wrote:
"from that new, different perspective, both classically and in SR: the clock synchronization that was done on the assumption that clocks A and B were first in rest, is wrong"
Once more (I lost count): synchronization is "relative".

Classically (Maxwell-Lorentz theory): they synchronized their clocks according to the assumption that A and B are in rest in the light medium, so that the speed of light is assumed to be isotropic relative to the assumed "rest system".
SR: they synchronized their clocks according to the convention that A and B are in rest, so that the speed of light is pretended to be isotropic relative to the so defined "rest system".

Classically and in SR, that synchronization is invalid if you instead assume or define that A and B were moving at that time, because the speed of light was then not isotropic relative to that moving reference system.

As I already explained, there is nothing wrong with consistent measurements. As long as you consider that A is moving and B is in rest, all is OK just as you first calculated. If A starts moving and B remains in rest, then that cannot affect the distance between A and B at the instant that A starts moving. That would be weird physics!
No, A can use the observation to calculate its own movement, just as you assumed - before you erroneously introduced length contraction of a system that you assumed to be in rest.

And I see that pervect gives a detailed example but with two independent reference systems.

Last edited: Jul 1, 2015
21. Jul 2, 2015

### harrylin

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?

22. Jul 3, 2015

### Stephanus

Of course not !
Thanks for the response, though.
I just wanna duck a while. I'm writing a computer software that makes me easier to draw lines, for space time diagram. I drew lines in Microsoft Excel, and it's tidious.

23. Jul 3, 2015

### Stephanus

No, no. It's not funny. It's a good tought. Is that how I tought before? The universe must be contracted when something moves toward me at relativistic speed? I didn't say that, but it must have been implied in my calculations .
Okayy, duck for a while.

Last edited: Jul 3, 2015
24. Jul 3, 2015

### harrylin

You seemed to imply that with your choice of Doppler equation (=> particle A starts moving) and "WAIT. This is not the case, I study SR, I know Lorentz contraction [of Geneva Lyon], so I have to calculate it again" (emphasis mine).

See again #14 (and #20). If A was assumed to start moving from rest in the light medium, then you used the correct Doppler analysis (classical equation for moving receiver A and stationary source B). But then you should have written:
"WAIT. This is not the case, I study SR, I know time dilation, so I have to calculate it again".
Then you would have confirmed your earlier calculation about which I commented: "You mean probably that B clock reads 2400s and A clock reads 2100s. Looks right to me."

Alternatively, if you assumed A to stop moving (so that B is all the time moving), then you effectively switched reference systems before doing the Doppler calculation. In that case you used the wrong Doppler analysis to start with (you should have used equation for moving source B). And then you should consider matching clock synchronization as well as Lorentz contraction (yes, the distance Geneva-Lyon is Lorentz contracted according to a reference system relative to which Geneva-Lyon is moving).

Last edited: Jul 3, 2015
25. Jul 3, 2015

### 1977ub

In a way. In our IF we are comparing Geneva at a particular time with Lyon at "that same time" and we conclude that they are X miles apart. For the particle, the planes of simultaneity are different. Geneva at a particular time and Lyon "at that same time" are much closer together.