# Res Ipsa Loquitor-- Disproved:The Impossiblity of absolute motion detection.

Discussion in 'Physics & Math' started by geistkiesel, Oct 23, 2004.

1. ### RawThinkTankBannedBanned

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What about the fact that a speeding objects clock slows down compared to the stationary one. U all psycophisists have been hiding like a mouse from this fact now for too long , shame on U. But I am not gona rest before ruining U people.

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3. ### geistkieselValued Senior Member

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Your statement "messed up the math somewhere" is very enlightening. It tells me you understood the post you are scrutinizing here in the exact sense it was written. Yes you uinderstand it competely. I use my own math conventions

The direction of relative motion of the respective frames determine the sign.

The following expression [post=717586]was copied from here.[/post]

"dA’/dt = d (Va’a)/dt = dk/dt a constant > 0 for t < 1 and k = constant for t = 1 and dk/dt = 0 we take dk/dt in the –x direction."

It appears yiou miscopied the expression and forgot the derivative for the Va' term. This is probably just an oversight on your part, its so big deal is it?

What does MacM have to do with this he isn't a direct party to these discussions as far as I remember. I don't nderstand your reply. Va' is the velocity of the A' and initially is equal to A, until it begins motion in the oopposite direction, in the -X direction. MacM would never say anyhing like your misprint, this I am confident.

See below.
What you are really saying is that You, Persol can't determine absolute velocity from relative velocity. However, I a able to do this as you have discovered with your incisive and acutely appropriate scrutiny.

Maybe you disclaimer is just a reflection of your inability to follow the logic, correct?

I select my own math conventions, Va' = -Va holds for Va' = 0 = Vb.
The "35000" was a typo it shouldf have been 3500 as above. It should be obvious.
"35000" is a typo it shoud be 3500.
What do you mean by "your starting position"? I assume you mean A' starting position with respet to Vb. Va' = Va at t= 0. Therefore, the relative velocity of Vab = 10000 = Va'b at t = 0.

Vb is determined by the A' frame measuring the changing relative velocity Va' - Vb until the is value is zero. This is the same relative velocity that would be added to Va to maintain the constant Va + Vb = 10000. Now we didn't have a real test here so the answer "3500" was a an example. We knw that Vb could not have been absolutely zero as testing for the Va' + Vb would have resulted in an increasing relative velocity when tested in both direction along X ,

If Vb were actually in the +X direction then the measure of relative velocity of the Va' + Vb would have increased in the first measurement and the A' would simple reverse directions come to rest with respect to A and accelerate int the +X direction, therefore, for this example the B frame was moving actually, absolutely, in the -X direction wrt A which is moving in the +X direction wrt B.
At t = 0 Va' + Vb = 10000, the instant A' begins to accelerate in the -X direction the same direction as the B frame is moving. This relative velocity is monitored at the same time the relative velocity of the A' and A frame are measured. Va'b is decreasing at the same absolute rate as the Va'a relative velocity is increasing. Va'a goes from 0 - > max, the current relative velocity Va'a at t = 1, which we assume to be arbitrarily 3500 unis of measured relative velocity. This is the same as the decrese in Va'b during the same time frame. Thus, the change in relative velocity of Va'b = -Va'a. I use the minus sign to indicate that Va'a is an increase in relative velocity. So Va', which was assuming the role of an accelerated Vb means simply that Vb = 3500.
Using The A and B frame alone

There is not an absolute need for the A' relative velocity probe as the two frames A and B can get the same results, but to explain this the clearest lets assume the relative velocity of A and B is 10000 units as before and B decides to determine his own absolute velocity.

We start with B changin direction and moves in teh same direction as A moving in the same direction (aftr B has turned around) and B accelerates at a constant rate and notices how long (say dt = 1 unit) it requires to reach a conditon that Va = Vb or Vab = 0. We designate this the first phase of measuements designate this relative velocity term as Vab(1). The navigator on Vb has a cup o coffee accelerates in the negative X direction at the same rate and for a dt = 1 and notes the relative velocity say it is Vab(2) = 3500 units (as an example).This is the second phase of measuements. This is the same amount of relative velocity the B, Vb frame added to his own absolute velocity to achieve a relative velocity wrt A during dt = 1 resulting in Vab = 0 when accelerating at the same rate and for the same duration as the B frame accelerated in the -X direction. Therefore, as before, the Vab(1) = - Vab(2), but Vab(2) is just Vb as the mesuement started from the position ath Vab = 0 absolute velocity. Therefoe Vb was 3500 at t = 0 the begininng of the measurements. As Va + Vb = 10000, Va = 10000 - Vb = 10000 - 3500 = 6500.

.

The problem does not need A' to get the same answer.
QED
Geistkiesel

.

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5. ### PersolI am the great and mighty Zo.Registered Senior Member

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No. What I quoted was from your first post in this thread. Don't expect me to go searching through other threads of use to determine your notation.
Ok.

Do the test in the reverse direction, what is your result?

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7. ### geistkieselValued Senior Member

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As an exercise gfor the readers let us go through some useless egotisms. The expression above was characterized as so far emoved from real that Persol thought he could insult both myself and MacM by attributing the "Va' = dVa/dt " expression as being far removed from physical reality.

Let us go to my response:

From above Persol says no, he was quoting from a previous post in a companion thread.So I went and looked at what I had wrirttten And I came up with[post=717586] the link to the whole post[/post] with the exception of an edit where I changed a number from "35000" to "3500" (this an obvious typo.) after being aprised of my error by Persol.

Then Persol tries to throw the quesation aside
[post=717586]
No. What I quoted was from your first post in this thread. Don't expect me to go searching through other threads of use to determine your notation.Ok"[/post]

The theory of relativity isn't about the truth, or seeking errors or correcting, is it Mr. Persol? This is the man, folks, that goes around slandering those that are in disagreement with whom ever is paying his salary. MacM is a man who works very hard, and from what I can gather from reading his posts, if MacM's theory or anti-theory happens to prove contrary to what he has been urging he would probably just shrug his shoulders (albeit reluctantly) and go on about his business.. Now, Persol sees it as a botrher some nettle if he must be, " determining your notation", which is a a notch or two down from the horror he described earlier.

James R is this the kind of rhetoric scientists are encouraged to offer on this thread in this forum? I have a certain warm personal feelings about this thread though I have learned to shrug my shoulders too, but in the meantime, I say we try charge Persol withr scientific misconduct, try him, and then burn him at the stake, after the trial of course, well as long as we get all three of the tasks completed we will have satisfied the minimum requirement for "honest effort in reaching justice", which is good, enough for me.

When I see exactly what you mean then I will tell you the result of what "doing the test in reverse means".

Geistkiesel.

Do the test in the reverse direction, what is your result?[/QUOTE]

8. ### PersolI am the great and mighty Zo.Registered Senior Member

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No, I didn't say that. What I quoted was from this thread. I said I WASN'T going to go searching through your other thread.
From B to A.

9. ### geistkieselValued Senior Member

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By what you said "from B to A" I 'm not clear. I could discard A' , sunbstitute B' and repeat the experiment starting with the B'B at rest but we get nothing of value. If we repeat the last setup and assume the same measured values we get the same answer.

I will let you describe what you mean, but you recall Vb + Va = Vab constant.

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11. ### geistkieselValued Senior Member

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The expression started from the one immediately below. This was from the "Original thread" quoting James R in the4 title regarding this subject.In differenctiating the terms I substituted dA'/dt for d(Va')/dt then the expression highlighted below would read properly and should read:

dVa'/dt = dV(Va'a)/dt;

Va’ + Va = Va’a = k for t >= 0 and Va’ ¹ Va.
Va’ – Vb = Va’b = l for t >= 0.

At t = 0, the A’ inertial reference frame accelerates in the -x direction as,

dA’/dt = d (Va’a)/dt = dk/dt a constant > 0 for t < 1 and k = constant for t = 1 and dk/dt = 0 we take dk/dt in the –x direction

Geistkiesel

12. ### PersolI am the great and mighty Zo.Registered Senior Member

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But nowhere in there does Va’ = Va.

13. ### James RJust this guy, you know?Staff Member

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I have merged the two threads on this topic.

14. ### geistkieselValued Senior Member

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Challenge to James R holding that absolute motion cannot be detected or m,easured.

Va' = Va at t < 0, at which time the A' frame began moving in the-X direction.

All this copied from a my post of a yesteday.

1. Va’ + Va = Va’a = k for t >= 0 and Va’ ¹ Va.
2. Va’ – Vb = Va’b = l for t >= 0.

3.At t = 0, the A’ inertial reference frame accelerates in the -x direction as,

4. dA’/dt = d (Va’a)/dt = dk/dt a constant > 0 for t < 1 and k = constant for t = 1 and dk/dt = 0 we take dk/dt in the –x direction

I numbered the expressions for convenience. 1) says the Relative velocities of the A' and A frames are Va'a for t greater than zero when the A' frame began its motion in the -X direction. Likewise the 2nd statement describes the relative velocity of the A' and B frames moving parallel in the -X direction where A' eventually "catches up" with the B frame welocity catches up, when the relative velocity becomes zero wrt Va' and Vb. Remember Va + Vb = Vab is a constant throughout and it is only their intrinsic motion that continues.

The A'B relative motion is separated from that of the A frame when A' acelerates away from the both A and B. \$, should read d(Va') /dt = d(Va'a)/dt t, and [not d(A')/dt' = -d(Va'a)/dt] which says the acceleration of A' is just the negatve of the acceleration of the A'A relative velocity.

Geiskiesel

Last edited: Nov 22, 2004
15. ### geistkieselValued Senior Member

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Do not you think it would be more approproiate to have all the summary numbers reflect the truth of the threads. There were over 1800 visitors of the first thread. James R put the correct numbers whre they are supposed to go, OK?

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17. ### geistkieselValued Senior Member

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James R had no options regarding the accumulated number of visitors. The preceding post was permature and unnecessary.
Geistkiesel.

18. ### geistkieselValued Senior Member

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This is probably why you can't understand the post. It is clear by inspection that Va' = Va at t < 0. This is where the probe starts out.

What it is Persol, you understand all too well, the thread disproving the postulate of James R that it is impossible to measure or detect unaccelerated translatory motion of systems in free space.

You are confused about "relativity", not "measuring absolute velocity". First you are unclear abut what "using elativity" means, in the context of measureing velocity. You aren't referring to using 'gammas and the like are you?

Secondly, I never claimed to "use relativity " to determine absolute velocity, I have dirscarded relativity as anything but a relativisticaly unreal contrived contortion of anything concievably deserving inclusion ion the scientific language as a model of physical laws, remember?

I suppose then that the source of your inability to understand is directly caused by your lack of mathematical fluency; that you are unable to follow trivial algebraic notation in the posts, such as the following which answers the following question in the affirmative: Is the James R postulate claim denying the possibility of detecting absolute motion false?
________________________________________________________________________________
At t < 0,
Va = Va'

where the terms are the absolute velocity terms of the A' and A inertial system where the A frame carried, and subsequently launched ,the A' inertial framee used to measure the relative velocioties, Va'a and Va'b, where Va + Ab = Vab, the relative velocity expression for the motion of the B and A inertial frames. Therefore, Va' + Vb = Vab at t < 0.
________________________________________________________________________________
At t = 0 ,
the A' frame (measuring Va'a and Va'b) accelerates opposite to the A frame and parallel to the B frame motion and therefore, Va' + Va = Va'a > 0, and Va' - Vb = Va'b < [Va' + Va = Vab](t < 0).
________________________________________________________________________________
At t = 1,
Va' - Vb = 0 or , Va' = Vb, and measurement of relative velocity Va'a
ceases, therefore at t > 1:
Therefore as Va + Vb = Vab, a constant, and as Va' = Vb ,
Va + Va' = Vab or,

Va = Vab - Va'. (1)
_______________________________

Hence, the measurement of the absolute motion of Vb has been accomplished and incorporated into the measurement of the Va absolute velocity (wrt B inertial fame). Hence, the postulate claiming the impossibility of detecting absolute velocity of unaccelerated syatems has been shown to be false.

Similarly, d(Va) + d(Vb) = d(Vab) = 0 or,

d(Va) = - d(Vb) is a statement expressing the conservation of relative motion of inertial systems.

______________________________________________________________________________
QED

Persol, you can respond to this, or whatever you choose, Personally, I have you observed as a dishonest coward.

Geistkiesel.

19. ### geistkieselValued Senior Member

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First What clock are you referrmg to?
Second, what are you talking about? Which speeding object?

Here is how it goes Premonitioner: If there be shown but one contradictory fact to a theory the total uselessness of that theory has been verified, This paper in this thread had alreaady neutraized any SRT, so Premonitioner, you are misnamed. Your handle should be Postmonitioner. Hiding from what? Hey, Postmonitioner, publish my name.

I am not playing anyone's time dilation game, nor do I diddle daddle in the laughable frame contracton of mass jus' cause' nudder guy sed a
dew edd.Time dilation and frame contraction are not in my physicics book. Hey, Poszmon I am already making reservations for a brief stay in Stockholm in the summer, What a beautiful place, Sweden in the summertime. Do you weep Posnitioner? I mean do you sob and shake like a drunken Peruvian llama, with screeching moans of disppointment catching in your throat and rattling your cage? What a beautifukl song you sing your the panic of the age.

Poezmon, go have yourselves a drinky poo and start bragging how you helped Geistkiesel solve the qwazhunn. Oh Poezmon Poezmon dry the tears that doth bring the storming rain down rain upon your watchful eyes I have already been cried over,a dozen time over by the Princess of Paradise. Is that a twitch you have there wads jerkin' at your lips and sending messages to all over town boud the dirdi sumbitch?
Few gona pruuv sumpm bedar ged onwidid,

Geistkiesel

20. ### VernRegistered Senior Member

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I've spent several hours trying to understand this thread. I would like very much for there to be an AFR detector. So far I haven't seen such; all I see is an exercise to calculate relative movement between objects.

Maybe we're not clear as to what is the AFR. To me it is the state of motion that represents the "at rest" property of the universe. If this thread is about something else; then that's why I'm confused

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Last edited: Feb 21, 2005
21. ### superluminalI am MalcomRValued Senior Member

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Vern, threads like this give us neither knowelege or truth, and unfortunately your long hours trying to understand this thread are largely wasted and lead to the only outcome possible - complete confusion.

All actual scientists and experimenters agree that there is no absolute frame of reference. SRT is proven beyond a doubt. This does not mean that another theory will not be proven better, in the future, at explaining reality, but it will not happen in Geist's or MacM's garage. Tune up your crackpot detection skills here:

http://www.autobahn.mb.ca/~het/detector.html

22. ### VernRegistered Senior Member

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They are facinating because of the confusion. Yet; I don't think I've seen a thread like this where the confusion was finally worked out.

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But there's always hope.

23. ### superluminalI am MalcomRValued Senior Member

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Go for it. Have fun!