# The equivalence of observed separation and relativistic velocity.

The final conversatation between Doc Al and Russ_Watters in post 78 in the Speed of light thread went as:

Doc Al said:
Right. Don't confuse this "separation velocity" with a relative velocity.

As another example: Someone (observer A) observes two rockets (B and C) traveling in opposite directions, each with speed 0.9c. The "separation velocity" of the two rockets, as observed by A, is 1.8c. But the speed of C as observed by B (the velocity of C with respect to B) is only 0.994c.
Reading this last post led me to question whether the conversation between Doc Al and Russ_Watters ought not be left unanalyzed. I had finished reading the entire thead earlier and began to look backwards for something interesting to scrutinize.

There is a way for B and C to observe their relative separation velocity to be equal to the separation velocity measured by A who measure 1.8c, which is twice the velocity of each ship moving at .9c in opposite directions. I use c=1 the unit speed of light.

We start the measurement when B and C are 1/9 either side of the zero point A. At this instant B moving left and C moving right a photon is emitted parallel to each of the oppositely moving ships from A.The wave lengths L of each photon moving toward their respective ships is, L(B) not equal to L(C).

As B and C expand in distance from A the photons catch up with B and C after 1 second, measured in the stationary frame. Here each photon is reflected back to the origin arriving there at t = 2. At this time B and C are each 1.8 for c = 1, unit speed of light, on either side of A.

The reflected photon from B, or L(B) is allowed to proceed to C and the photon L(C) is allowed to proceed to B. From symmetry considerations each photon will arrive at their respective target ships when t= 4 8/10 or t = 48/10 seconds, if I haven't miscounted. B amd C having timed the emission of the photons can accurately determine when the alien photon wave length length arrives at the domestic detector, hence observers on each ship deduce that the expanding velocity of each ship is identically 1.8c, unambiguously.

Knowing their intrinsic speed, the B and C clocks on the ships can be calibrated to run identically with the stationary clocks. Here the expansion/relativity velocty are calculated in stationary and moving frame times.

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geistkiesel said:
From symmetry considerations each photon will arrive at their respective target ships when t= 4 8/10 or t = 48/10 seconds, if I haven't miscounted.
This will be the time as measured by the observer A. How do you know that B and C will make the same measurement?

B amd C having timed the emission of the photons can accurately determine when the alien photon wave length length arrives at the domestic detector, hence observers on each ship deduce that the expanding velocity of each ship is identically 1.8c, unambiguously.
But to get a velocity from a time you must have a distance. What value for the distance is each observer using?

The point is that if observer A deduces that the relative velocity between B and C is greater than c then this is no problem. However, B anc C will not measure their relative velocity to be greater than c. You have only calculated what happens in A's frame, but have not told us what B and C will measure.

Matt

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baffledMatt said:
This will be the time as measured by the observer A. How do you know that B and C will make the same measurement?

But to get a velocity from a time you must have a distance. What value for the distance is each observer using?

The point is that if observer A deduces that the relative velocity between B and C is greater than c then this is no problem. However, B anc C will not measure their relative velocity to be greater than c. You have only calculated what happens in A's frame, but have not told us what B and C will measure.

Matt
Maybe they discussed he experiment before they performed it/ I dunno, it wan't my hypo. Ask DocAl and uss_wattes, it was the one they wee discussing. They have all the answers to you questions.

B and C are in the stationary frame. confirm with Doc Al and russ_watters.
Distance , yes, how do you find the distance? Let me look in my Feynman "lectures on physics" Text, here it is: distance = rate times time. Every body has a rate, every body has a time, need I actually grind this sucker out? Hey you're the mentor. Aren't you supposed to check my work? Give ehlpful suiggestions, steer me to my natural directed end based on what I think the answer is? Nope. Nope, you're supposed to bleed it to death, aren't you..
How do we know B and C will make the same measurement? Symmetry. my good man, symmetry. Like I said they can calculate using any theory, even using dilated time, its a free hypo, so to speak. I don't care what they scheme they use. The post was published to show a non SR model works, is reasonable and uses physics. What does SR use?
They are using the same light scheme. B and C can use dilated time, stationary time or whatever, but since the hypo said the were moving at .9c they prolly used the stationary frame as a reference. I wasn't there, were you?

baffledMatt said:
This will be the time as measured by the observer A. How do you know that B and C will make the same measurement?

But to get a velocity from a time you must have a distance. What value for the distance is each observer using?

The point is that if observer A deduces that the relative velocity between B and C is greater than c then this is no problem. However, B anc C will not measure their relative velocity to be greater than c. You have only calculated what happens in A's frame, but have not told us what B and C will measure.

Matt
Wrong I told you B and C will measure their relative velociy as 1.8c. It is a given they are weach moving at .9c.

Don't screw up the problem by intejecting what A will do to B and C calculations. It is irelevant what A does, or doesn't do. B and C can calculate in their moving and/or the stationary frame. they have all the assets to do this. If you want to kill the thread, do it. Ask me to make an SR calculation. and when I refuse you can say, "see"!

geistkiesel said:
B and C are in the stationary frame.
Hang on, you just said they were each moving at 0.9c relative to A didn't you?

Distance , yes, how do you find the distance? Let me look in my Feynman "lectures on physics" Text, here it is: distance = rate times time. Every body has a rate, every body has a time, need I actually grind this sucker out?
In relativity it is not quite as simple as that. First of all, is everyone going to agree on the time? you seem to be dismissing the whole issue of measurement as if it were trivial but I'm afraid this only reinforces our belief in your lack of understanding of SR. Anybody will tell you that SR is all about measurement.

Aren't you supposed to check my work? Give ehlpful suiggestions, steer me to my natural directed end based on what I think the answer is? Nope. Nope, you're supposed to bleed it to death, aren't you..
Check your work? yes. Do your work for you? No. I'm afraid that there is no evidence you have even attempted to understand the SR point of view, let alone tried doing any calculations.

How do we know B and C will make the same measurement? Symmetry. my good man, symmetry.
This may be, B and C will agree with each other. But they will not agree with A. You have still only provided us with the measurements that A will make.

The post was published to show a non SR model works, is reasonable and uses physics. What does SR use?
But don't you see? The non-SR model doesn't work. There is no way for the non-SR model to work and for the principle of relativity to hold. So, you can give me any explanation you like but if it isn't SR then we will be able to find a way to distinguish between inertial frames in your theory - something which you cannot do in reality.

They are using the same light scheme. B and C can use dilated time, stationary time or whatever, but since the hypo said the were moving at .9c they prolly used the stationary frame as a reference. I wasn't there, were you?
Well, it was a thought experiment so I would hope we were both there.

Wrong I told you B and C will measure their relative velociy as 1.8c. It is a given they are weach moving at .9c.
I know what you told me, but it's nonsense.

Don't screw up the problem by intejecting what A will do to B and C calculations. It is irelevant what A does, or doesn't do.
I'm not saying that what A does is affecting the measurement of B and C. I'm saying that all you have provided is the measurement A will make. You still need to show whether or not B and C will measure the same thing.

B and C can calculate in their moving and/or the stationary frame.
No. B and C are in their respective frames and that is where they measure things. You cannot say "B is moving relative to A, but he makes measurements as if he were in A's frame".

Matt

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russ_watters
Mentor
geistkiesel said:
Knowing their intrinsic speed, the B and C clocks on the ships can be calibrated to run identically with the stationary clocks. Here the expansion/relativity velocty are calculated in stationary and moving frame times.
The clocks on GPS satellites are calibrated to run slow when on the ground so that their clocks remain in sync once in orbit. Is that what you mean?

Hurkyl
Staff Emeritus
Gold Member
Geistkeisel, you seem to be saying:

If B and C compute how things would be measured in the A frame, then they can compute their seperation velocity in the A frame.

This seems an awfully trivial point to make; am I missing something?

russ_watters said:
The clocks on GPS satellites are calibrated to run slow when on the ground so that their clocks remain in sync once in orbit. Is that what you mean?
no, I mean that the moving observers, knowing their relative velocity can calibrate clocks and other machinery and makes ccalculations as if in another frame,

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baffledMatt said:
Hang on, you just said they were each moving at 0.9c relative to A didn't you?
Actually Doc Al and Russ_wattes said it.

baffledMatt said:
In relativity it is not quite as simple as that. First of all, is everyone going to agree on the time? you seem to be dismissing the whole issue of measurement as if it were trivial but I'm afraid this only reinforces our belief in your lack of understanding of SR. Anybody will tell you that SR is all about measurement.
Who is "our" any way?

What significance does the phrase "our belief" have to do with any scientific assertion? I am not clear just what level of confidence or value I can place on your statement as having anything useful, meaningful, or wortwhile.

Actually I am disimissing the whole issue of special relativity. Whyen i discuss the subject with claimed experts I get dogmatic habitual confusion.

Anybody can tell me that in SR measurement is erything. Do I have to make an appointment for this? Did somebody tell you this?

The scrutiny of the specifics of oflight motion and intrinsic processes are prohibited by theoretical fiat Objects moving in a straight line do not present any problems of measuement of the kind envisoned by special relativity theory.

baffledmatt said:
Check your work? yes. Do your work for you? No. I'm afraid that there is no evidence you have even attempted to understand the SR point of view, let alone tried doing any calculations.
I'm not missing anything that I desire to asimilate as knowledge or information. I am prepared to dismiss SR as casually as I woud wave of a pesty fly bizzing aorund my ear. Special relativity is the highest achievenment of scenific blindness and ignorance in the history of humanity.

[quoet=baffledMAtt]This may be, B and C will agree with each other. But they will not agree with A. You have still only provided us with the measurements that A wille make.[/quote]
B and C determine their resective velocities based on roundtrip travel time of light photyns. B and C agree exactly with each other., see the post. They each agree they are travelling at .9c and ar which is in ageemnet with the emesurements A would make. It is a trivial execise you might try making the effort to make the calculation.
I didn't find the "formula" in a book. I had to make it up as I went along.

baffledmtt said:
But don't you see? The non-SR model doesn't work. There is no way for the non-SR model to work and for the principle of relativity to hold. So, you can give me any explanation you like but if it isn't SR then we will be able to find a way to distinguish between inertial frames in your theory - something which you cannot do in reality.
Yiou can look for this if you like. I have akleady done what you said I couldn 't do. I hope you aren't too upset.

If the straight forward method iscontrdity t SR then my advice isd to scrap SR before you get to far along in your educaitonal career and the SR concepts
calcify to rock solid dogma.

baffledMatt said:
Well, it was a thought experiment so I would hope we were both there.

I know what you told me, but it's nonsense.
but from your satement you know how I feel about the matter.

baffledMatt said:
I'm not saying that what A does is affecting the measurement of B and C. I'm saying that all you have provided is the measurement A will make. You still need to show whether or not B and C will measure the same thing.

What are you talking about? The post was a reflection of B and c making measurements regading ther respectibve velocity anmd position.

It is statements like these that get met thinking in terms of "why are these people taking something I have done and completely reverse my express meaning and intentions? Are they trying to consciously obliterate my effort dfronm scrutiny and perusal by others? whatsove??

No. B and C are in their respective frames and that is where they measure things. You cannot say "B is moving relative to A, but he makes measurements as if he were in A's frame".

Matt[/QUOTE]

russ_watters
Mentor
geistkiesel said:
no, I mean that the moving observers, knowing their relative velocity can calibrate clocks and other machinery and makes ccalculations as if in another frame,
Absolutely. That's the whole point of Lorenz transformations and it is pretty much the same as what is done with GPS satellites (just with GPS satellites, their behavior is first predicted instead of measured. GPS satellites travel in one frame but report the time passing in another frame. Either way, it validates the theory.
Actually I am disimissing the whole issue of special relativity. Whyen i discuss the subject with claimed experts I get dogmatic habitual confusion.
If you launch a satellite with a clock in it and compare it to a clock on earth every time your orbit that point (doesn't matter how as long as you are consistent), you can "tune" the clock on the satellite to run at the same rate as the one on earth, without using Relativity. However, you can also calculate ahead of time what rate to run the clock at so you don't have to tune it. That's the purpose of Relativity - and indeed, physics itself. Understanding the actual realationships between physical phenomena means you can make predictions about how similar phenomena will act. Otherwise everything is just guessing and trial/error.
Anybody can tell me that in SR measurement is erything. Do I have to make an appointment for this? Did somebody tell you this?
??? That has nothing at all to do with SR. That's the SCIENTIFIC METHOD.
Yiou can look for this if you like. I have akleady done what you said I couldn 't do. I hope you aren't too upset.
As said before, if you have done this, you haven't shown us: it requires math and lots of it.

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geistkiesel said:
Wrong I told you B and C will measure their relative velociy as 1.8c. It is a given they are weach moving at .9c.

Don't screw up the problem by intejecting what A will do to B and C calculations. It is irelevant what A does, or doesn't do. B and C can calculate in their moving and/or the stationary frame. they have all the assets to do this. If you want to kill the thread, do it. Ask me to make an SR calculation. and when I refuse you can say, "see"!

??

Tom Mattson
Staff Emeritus
Gold Member
geistkiesel said:
OK I just read the reference.
No, you didn't. You couldn't possibly have read and understood that reference in less than an hour.

We both assume it "true". What is your point?
His point is that the paper correctly analyzes the problem that you are describing.

Tom Mattson said:
No, you didn't. You couldn't possibly have read and understood that reference in less than an hour.

His point is that the paper correctly analyzes the problem that you are describing.
Thanks Tom, I do wish that some try to develop a little understanding to their posts, but I suppose every-one has an opinion, but some have more defined methods than others.

Its a really good paper, had I not seen this thread (and argumentative other postings) over the last number of weeks, I would have posted the Paper in it's own deserved thread .

Its there in Black and White