Measuring distance, speed and clock

In summary, an expert summarizer of content would say that an individual can measure their speed by using the Doppler effect, measure distance by using clocks, and measure other times by using the speed and the Lorentz formula. However, given proper time, an individual can also measure distance to an object at rest relative to them, as well as the time dilation caused by relative motion.
  • #1
Stephanus
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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.
 
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  • #2
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
 
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  • #3
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
Mentz114 said:
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.
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
Stephanus said:
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.

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
A few issues:
Stephanus said:
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.
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.
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.
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.
Clocks are synchronized and they start sending signals.
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.
[..]
So basically, A can determine its speed through something like Doppler effect. Is this true?
Yes.
[..] 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.
[..]
Same problem with "synchronized" as before: according to which reference system...
 
  • #7
Mentz114 said:
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).
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
harrylin said:
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.
A and B is at rest. Their distance at rest is 900ls.
harrylin said:
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.
harrylin said:
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.
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.
harrylin said:
Synchronized according to S, or according to S' ??
According to S, they haven't move yet. A will move at t=900
harrylin said:
[..]And according to which clock(s) you send the signals for 6s.
Their respective clock. A sends signal every 6s by his clock. B sends signal every 6s according to B clock.
harrylin said:
Same problem with "synchronized" as before: according to which reference system...
According to B frame. B1 and B2 moves toward A from B frame.
 
  • #9
Stephanus said:
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.

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
Mentz114 said:
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.
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
Stephanus said:
Yes, by using Space Time diagram. But I calculate the space time diagram by Cartesian, not by Lorentz. It gives the same result tough
I don't know what to say to this. Speechless.

But I have to know "why" Lorentz transformation is thus, although just boost in x direction only.
You will not answer the 'why' question. No one can tell you 'why' things are the way they are.
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'm glad you remember that. Even if you forget who told you.

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.

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
harrylin said:
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.
Mentz114 said:
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.
Thank's Mentz114 for your answer. If you, I still remember, that you are THE ONE who taught me Space Time diagram. :smile:. 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.
Signal01.JPG
But, judging from the answer, I think this is what it means.
Signal02.JPG

The length of the signal is 6 seconds. That's why
If the signal is a pulse at 1 Hz frequency then there will be 6 signal pulses between A and B
but this is not what I meant.
Thanks Harrylin for your effort.
 
  • #13
Stephanus said:
Thank's Mentz114 for your answer. If you, I still remember, that you are THE ONE who taught me Space Time diagram. :smile:. It's not that I'm ignoring any answers. .

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
OK then, retake with the lacking info added:
Stephanus said:
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. Their distance at rest is 900ls.
And supposed A and B keeps exchanging signal for 6 seconds interval. what I mean is every 6 seconds, A emits a signal.
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. A sends signal every 6s by his clock. B sends signal every 6s according to B clock.
Now at T900, then the first signal from A reaches B (and the first B signal reaches A)
A will move at t=900. Now, A moves toward B...
OK the case now well defined.
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.
You mean probably that B clock reads 2400s and A clock reads 2100s. Looks right to me.
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.

That is, for classical case, assuming medium in rest and you are moving.
WAIT. This is not the case, I study SR, I know Lorentz contraction, so I have to calculate it again. [..]

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. :wink:
 
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  • #15
harrylin said:
Stephanus said:
B clock will shows 900 + 1500 = 2500 seconds and B clock shows 900 + 1200 = 2100 seconds.
You mean probably that B clock reads 2400s and A clock reads 2100s. Looks right to me.
A gross typo or calculation :smile:
harrylin said:
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. :wink:
I still don't understand it. I'll contemplate your answer. Thanks Harrylin
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..
 
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  • #16
Stephanus said:
A gross typo or calculation :smile:
I still don't understand it. I'll contemplate your answer. Thanks 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
 
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  • #17
Stephanus said:
Dear PF Forum,
I'd like to know how we measure speed, distance clock in space?

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 received 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.
 
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  • #18
Thanks Harrylin, thanks pervect, thanks Mentz114 for your answer.
I'll contemplate your lengthy answer first.
But before I go further. I'd like to ask about this particular subject first.
harrylin said:
- the clock synchronization that was done on the assumption that clocks A and B were first in rest, is wrong
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.
 
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  • #19
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.
SpaceTimeSample.JPG
 
  • #20
Stephanus said:
Thanks Harrylin, thanks pervect, thanks Mentz114 for your answer.
I'll contemplate your lengthy answer first.
But before I go further. I'd like to ask about this particular subject first.
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"
A. If at first A and B is at rest, separated by 900ls away.
B. They synchronized their clock
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.

C. And some 900 ls later, A moves toward B.
So what is wrong with this? [..]
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!
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? [..]
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.
 
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  • #21
harrylin said:
[commenting on "WAIT. This is not the case, I study SR, I know Lorentz contraction, so I have to calculate it again. ":]

[..] As long as you consider that A [starts] 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. [..]
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?
 
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  • #22
harrylin said:
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?
:oldlaugh: Of course not !
Thanks for the response, though.
I just want to 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
harrylin said:
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?
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 :smile:.
Okayy, duck for a while.
 
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  • #24
Stephanus said:
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 :smile:.
[..].
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). :wink:

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).
 
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  • #25
harrylin said:
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?

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.
 
  • #26
1977ub said:
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.
The particle's change of motion cannot affect the distance between Geneva and Lyon, and Stephanus understood that very well. As I next explained, one (you or the particle or your friend) may freely choose any inertial reference system for doing the physics.
 
  • #27
Because distance in SR is observer dependent, though, the particle, co-located with Geneva but moving at a different velocity than Geneva is, can find that the distance to Lyon in its own (moving) frame is less than the distance between Geneva and Lyon in the Geneva-Lyon rest frame.
 
  • #28
pervect said:
Because distance in SR is observer dependent, though, the particle, co-located with Geneva but moving at a different velocity than Geneva is, can find that the distance to Lyon in its own (moving) frame is less than the distance between Geneva and Lyon in the Geneva-Lyon rest frame.
Exactly - I elaborated the same in post #24
 
  • #29
harrylin said:
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?

Perhaps the best answer is that there is no such thing.
 
  • #30
1977ub said:
Perhaps the best answer is that there is no such thing.

Rather than say that distance "doesn't exist", I find it more expedient to say that it's a confusing concept because it's observer dependent. I spend a fair amount of time trying to explain to non-physicists why, in the context of special relativity, the observer independent Lorentz interval is actually simpler than the observer-dependent notion of distance, with rather limited success.
 
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Likes 1977ub
  • #31
Thank you mentors and readers for your responses.
I'm sorry, if I haven't got the chance to study all of them. Someone in this forum gives me a very good software to draw space time diagram.
So, I can draw ST less than 5 minutes compared to using Excel in 1 hour. So I can draw ST anytime I want.
I still study the software.
I don't even have the chance to study
HarryLyn - #147
HarryLyn - #16
Pervect - #17
And surely the next posts.
But thanks anyway.
I will response later.
 
  • #32
pervect said:
Rather than say that distance "doesn't exist", I find it more expedient to say that it's a confusing concept because it's observer dependent. I spend a fair amount of time trying to explain to non-physicists why, in the context of special relativity, the observer independent Lorentz interval is actually simpler than the observer-dependent notion of distance, with rather limited success.
I don't think it's confusing at all. Nugatory gave a very simple explanation in the very first response on this thread:
Nugatory said:
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.
And in fact, that's exactly the same process that could be used for a space-like Lorentz interval where an inertial observer passes through one of the two events in question and the "moment" of the reflection is midway between the sending and receiving of the light signals.

But in reality, the Lorentz interval is not the same kind of a thing as distance because usually when we are talking about distance, we don't mean the distance between two events but rather the distance between two objects at a particular time, which of course, will be different events depending on the selected reference frame. In Nugatory's definition, the assumed reference frame is the rest frame of the observer (who is also one of the objects) but unless we are willing to teach, and have those non-physicists learn, that distance is frame-dependent (not necessarily observer-dependent), then we haven't taught, and they haven't learned, Special Relativity.
 
  • #33
Mentz114 said:
[..]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.
YES!. That absolutely makes sense! Even if the object has moved several cm or distance away, all we KNOW from the radar information is the time when we emit the signal and the time WHEN we RECEIVE the signal. Not the position of the object 'now'.
 
  • #34
Mentz114 said:
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.
Sorry, I tought you were referring this

ghwellsjr said:
https://www.physicsforums.com/threa...n-but-velocity-invariant.819113/#post-5142973
Good question. I think some spacetime diagrams may help to answer your question.

Let's take a scenario where an observer sends out a light pulse to a reflector 6 feet away and measures with his clock how long it takes for the reflection to get back to him. Since light travels at 1 foot per nanosecond, it will take 12 nsecs for him to see the reflection and he will validate that the light is traveling at 1 foot per nsec for the 12 feet of the round trip that the light takes. If he is following the precepts of Special Relativity, he will define the time at which the reflection took place at 6 nsecs according to his clock and I have made that dot black:
This is what you actually referred, Nugatory's answer.
Nugatory said:
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.
 
  • #35
I know, it's been a week since this post. But I'd like to ask anyway. I'm new in SR, so there are many symbols which I don't recognize. What is the meaning of these symbols?
pervect said:
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.
What does ##t_r\text{ time of the reception}## mean?
Which one is correct?
A: tr = January 1st 2015, 18:00; te = January 20th 2015, 19:30, or something like
B: tr = 20 seconds, te = 30 seconds
It's likely B, but if it's B then, what is t? The reverse of frequency? Needs confirmation here.
Thanks a lot.
 
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