Relative Velocity Measurement – Frame and photon

Discussion in 'Pseudoscience Archive' started by geistkiesel, Jan 28, 2010.

  1. geistkiesel Valued Senior Member

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    Relative Velocity Measurement – Frame and photon

    note: this basic experiment was posted using a significantly different system which the current model has upgraded.
    1. An inertial frame A accelerates a transducer probe A’ in forward +x direction until some distance D reached and acceleration power cut with VA’ velocity set at VA = VA’. Now the clock tick rates of are A and A’ equal.
    2. A emits a series of pulse at t0 every 100 ticks toward A’ which then returns the signal with the embedded A’ arrival time t1 include in the signal.
    3. The return signal at A is recorded as t2.
    4. Frame A records the data as shown.
    Code:
    [U]t0 [/U]   [U]t1n[/U]   [U]t2[/U]
      0    t11  90
     190  t12  280
     380  t13  470
    etc.
    The delta t1n+1 – t1n = a constant, (60). The arrangement is shown.
    .
    Code:
    A t0  |_____________d1____________________|A’ 
          t2 |____________d2__________________|t1
          | D|
    
    
    D = d1 – d2 (where D is the distance A travels during t2 - t0); c = 1.
    D= (t1 - t2) – (t2 – t1) = Va (t2 – t0) = 2t1 – (t2 – t0)
    Va = 2t1/(t2 – t0)– 1

    QED
     
    Last edited: Jan 28, 2010
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  3. James R Just this guy, you know? Staff Member

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    The above is too unclear to make sense of.

    In particular, it is not clear where the signal is emitted from or where the return signal is received. geistkiesel says A is a reference frame, but then talks about emitting signals from A. As any reference frame covers the whole of space, it's impossible to tell where the signal is emitted from.

    geistkiesel claims that "delta t1n+1 - t1n" is constant. As far as I can tell, this is a time interval between a signal being emitted somewhere and received somewhere else. For that time to be constant, both the emitter and receiver must be moving. But geistkiesel hasn't told us how the initial emitter is moving. He only told us how A' is moving relative to the frame A.

    Also, despite the "QED" at the end, it is not clear what post #1 is supposed to be proving.

    In short, post #1 is meaningless, as far as I can tell.
     
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  5. geistkiesel Valued Senior Member

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    Here is the schematic of the arrangement of the experiment.
    Code:
    A t0  |_____________d1____________________|A’ 
          t2 |____________d2__________________|t1
    
    The light is emitted from A at t0 after A' has adjusted its velocity to Va'=Va;
    The light is received at t1 by A' moving at the same velocity as A.
    Then the light is reflected from A' back to A at t1 (using transponder technology).
    The light is then received by A at t2.
    The initial emitter is inertial frame A; the transponder-receiver is inertial frame A'. A and A' have the same velocity.
    Good catch James R. The QED refers to the thread title - Relative Velocity Measurement - Frame and photon. The expression Va = 2t1/(t2 – t0)– 1 is the 'frame' velocity. I left it to the reader to properly apply the requisite 'photon' velocity and to complete the picture of relative velocity of frame and photon.
     
    Last edited: Jan 29, 2010
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  7. James R Just this guy, you know? Staff Member

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    geistkiesel:

    Thanks for the clarification. When you wrote "An inertial frame A accelerates a transducer probe A’ ", I took that to mean that the probe was accelerated in some coordinate system A. But it appears that what you mean is that both A and A' can be considered as accelerating together.

    I assume you also mean that A and A' maintain a constant separation at all times in the frame of A. Thus, the distance d2 is the fixed distance between A and A', provided that d2 is measured in the frame of A and A'.

    The travel time of a light pulse from A to A' and back, is therefore 2 (d2)/c, as measured in the frame of A or A' (which are the same inertial frame). i.e.:

    t1 - to = (d2)/c
    t2 - t1 = (d2)/c

    Your claim that delta tn+1 - tn is constant is therefore correct.

    Now, you define:

    "D = d1 – d2 (where D is the distance A travels during t2 - t0)"

    Which reference frame was D measured in? And which frame was d1 measured in? Which was d2 measured in? It is problematic to measure D in the reference frame of A or A', since A and A' both accelerated prior to time to.

    We can perhaps avoid this problem by using a third reference frame - one that sees A and A' accelerate from rest to speed Va. Call that frame, frame X.

    Let us suppose that the various distances D, d1 and d2 are measured in frame X, then.

    In frame X, it is still true that

    t1 - to = (d2)/c
    t2 - t1 = (d2)/c

    (where we now understand the times to be measured using frame X's clocks, and not A or A' 's clocks).

    Now you say:

    "D = d1 – d2 (where D is the distance A travels during t2 - t0)"

    As I see it D = d1 - d2 where D is the distance A travels BEFORE to. The distance A travels in time t2 - t0 is: vA (t2 - to), but this is not equal to D.

    You next say:

    "D= (t1 - t2) – (t2 – t1) = Va (t2 – t0) = 2t1 – (t2 – t0)
    Va = 2t1/(t2 – t0)– 1"

    This is still a muddle, as far as I can tell. Can you please explain what you're saying here?
     
  8. AlphaNumeric Fully ionized Registered Senior Member

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    6,702
    Geist is a prime demonstration that the best way to undermine a theory is to actually learn it. He refuses to actually learn how to do special relativity and, as James comments in his first post, is unable to formulate even coherent special relativity systems, never mind ones which demonstrate inconsistency in the theory. Its my experience that the more I learn about an area of physics the more I see the unresolved issues, the unanswered questions and in some cases the glaring mistakes. Cranks seem to have a problem with this, as if learning is something to be avoided.

    Unfortunately Geist is plagued by an even deeper problem, he's trying to demonstrate the mathematical inconsistency of special relativity. This is entirely equivalent to demonstrating the the geometry of Minkowski space-time is inconsistent. But then that would filter through to ALL of geometry. And all group theory. The mathematical consistency of special relativity is as true as the mathematical consistency of group theory (specifically the SO(3,1) group) and Riemannian geometry in general. It doesn't matter how convoluted a physical system you construct, ultimately you are asking the same question, "Can I form an inconsistency in geometry?". The answer is no.

    The only way special relativity can be put in the bin by physicists is if someone shows that, despite its mathematical consistency, it is not the correct model of nature. Newtonian physics is mathematically consistent (relating to the SO(3) group) but we found it doesn't predict the right results for experiment. Mathematically consistent, physically inaccurate.

    You aren't going to come up with a thought experiment which shows special relativity is inconsistent. The only way to make physicists look at other things is to provide an actual experiment which special relativity doesn't work with. Geist, every single one of your threads is about a thought experiment and thus not going to achieve what you want them to. If you truely do think special relativity is not the way physicists should model the world then so be it, I see nothing wrong with that in principle, but you're going about it in the wrong way.

    /edit

    http://en.wikipedia.org/wiki/Euclidean_geometry#Logical_basis

    Godel's incompleteness theorem doesn't apply to geometry, it is provably consistent.
     
  9. geistkiesel Valued Senior Member

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    2,471
    No. A’ accelerates from A in order to get a sufficient distance apart such that the measurements may be made. This is the only reason for having the A’ probe available in the first place.
    It all sounds good to me.
    Look at the schematic. A and A’ are continually moving in one direction the distance from A’ back to A is constantly shrinking and expanding proportional to the relative directions of frame and photon. As the pulse leaves A and heads toward A’, A and A’ always move the +X direction, the light direction change in a regular fashion. If you take Figure 1 and repeat the light pulse sequence, each round trip will be the same duration.
    Your statement here made me sit up and take notice.

    Lets start with a simple model and set ct = d1 + VAt. This says the light in leaving A must travel a distance d1, plus the distance VAt, the distance the frame moves during the time the light has moved a distance d1 (see the schematic – Figure 1).
    All measurements in the experiment are in the AA' frame when Va = Va'.
    But now both have the same velocity and therefore the same clock tick rates. Check the schematic below James R. D results from the measurement of t0, t1 and t2, where at least two pulses are used in order to confidently generate a known t1. As all math is performed on the A frame d1 = t1 - t0 and d2 = t2 - t1. Therefore, d1 - d2 = D, the distance the frame travels during t2 - t0. Getting ahead slightly, D = Va(t2 - t0) = (t1 – t0) - (t2 - t1) where the first and second expressions on the right are d1 and d2 respectively.
    The following was stated by James R in the previous 5 lines above.
    “Let us suppose that the various distances D, d1 and d2 are measured in frame X, then.”
    [I didn’t see it at first, but the second expression above should read, t2 - t1 = (d1)/c, you have “d2”, clearly a typo(?)]

    What you are suggesting would add unnecessary complications and, without more than an instinct, I do not see how an inertial frame X, albeit in the same frame of reference as A, could make the same measurements with a close approximation to the resolution (accuracy) as produced on the A frame, taking into account the necessity of the light motion being intimately tied to the frame motion, or utilizing an added reference frame that I see as “problematic. Perhaps you see something I missed or that I was unable to detect in your post.

    It is critical that all measurements be conducted on the A frame. This frame emits the pulse along the axis of motion and on the pulse is reflected back to the A frame. The outward pulse is a measure of d1, that is, d1 = t1 – t0. The in bound reflected pulse is a measure of d2 = t2 – t1. Here it is important to understand that the measurement of t2 completes the measurement protocol.

    Similarly, I am a loss to see how the previous motion of the A frame has any rationally arrived at effect on the results and am not quite following your claim that

    “It is problematic to measure D in the reference frame of A or A', since A and A' both accelerated prior to time t0.”
    You must explain this to me.

    Up until the A’ was launched, the motion histories of A and A’ are identical as A’ began the journey as cargo on the A frame.

    I wanted to relate this thread as occurring exclusively in "deep space" where A and its cargo, including the A' transponder probe, exist as a solitary unit. Remember the measurement is of unaccelerated translatory motion, which to my 1st edition of “Handbook of Astronautical Engineering”, Chapter 11 ‘Relativistic Rocket Mechanics”, the SRT postulate is stated as, “It is impossible to measure or detect unaccelerated translatory motion in deep space”. I intended to maintain the thread with a minimum of actors, physical, parameters if you will, which to my thinking would require restricting the experiment to one inertial frame measuring its own motion. Your reasoning for including a third frame I find difficult to follow, perhaps this is because my post was not understood as intended.

    VA(t2 – t0), is most certainly equal to D. Check the schematic as D, as described there, only appears after t0.

    D cannot be the distance A travels before t0, because no measurements of any kind were made beforet0; no pulses occurred before t0; Your modification using the X frame drastically alters the experimental arrangement, and together with the statement that D was established before t0, make no sense, to me. This experiment is like a mini-Big Bang that is nothing happens before t0.

    Slow down James R. As stated d1 and d2 are the measurements that begin with t0, include t1 and finally end with t2. d1 = t1 - t0 and d2 = t2 - t1.

    There is nothing of interest here prior to t0, nor after t2. I considered clock perturbations due to acceleration of the A' frame, but up to the time A' was launched A and A' shared an identical history. Now, once A' was launched I considered some perturbation to A' tick rate (acceleration effects, some might refer this to SRT time dilation). However, I discarded the possibility that the A' suffered some permanent tick rate changes and presumed both frame clocks tick the same, they are, after all the same frame of reference. But then perhaps the time-of-day indicated on both frame clocks are not the same after A’ surged ahead. There is no unambiguous way of asking A’ what the time of day (tod) is indicated on his clocks as the spatial separation precludes any confident 'time hacking' of the A and A’ clocks. Mulling over this problem brought me to the solution of using the delta-tn as indicated which brought a soft wind of evaporation to the enigma as t1 is not a simple ad hoc substitution for observed data.
    The first term in your post above after the “=” is incorrect. Either I made a typo or you did. The expression should read (see the underlined t0), D = (t1 – t0) – (t2 – t1) = Va (t2 – t0) = 2t1 – (t2 – t0)

    Va (t2 – t0) = 2t1 – (t2 – t0)
    And dividing through by the t2 – t0 term we arrive at,
    Va = 2t1/(t2 – t0) – 1
    [See below for an interesting and different method of calculation.]
    From t0 on and until t1, the time of flight of the out going pulse, the A frame moved a distance Va(t1 - t0) during the time the light moved a distance d1. The light moves a bunch, the A frame but a relatively small distance. But we do not know what t1 is until the pulse is returned to A and to the measurement of t2. The return signal from A' includes t1 embedded in the signal. Now we have a t0, and an ambiguous t1 and we are not confident
    that t1 is usable until a second identical pulse is emitted with the identical protocol as the previous pulse.

    Now we useful data are available for some arithmetic. The frame is moving in the direction of the emitted pulse (and toward A' and always along the X axis). When the light traveled from A to A' the A frame traveled a distance Va(t1 - t0), but the value of VA is not known from this alone, obviously. Remember, velocity, V, is defined as, in general,
    V = (Xn+1 - Xn)/(Tn+1 - Tn)
    At this point d1 may not be calculated before d2 is measured, as this then and only then, is the t1 data available as t1 is embedded in the signal received by A at t2. Simpler said no calculations are permitted until the completion of the round trip travels of the pulse. When t2 is measured the d1 calculation is performed using appropriate data, the only data available, d1 = t1 - t0 for c = 1. Nothing is known of d1 until the return of the light from A' to A, which is received by A at t2 with the additional information that the time t1 is embedded in the signal ala a transponder signal (So I repeat myself occasionally).

    Here is the crux. As the light moves a known distance during the time of flight from A to A' (which is available when the signal returns at t2) the A frame moves some distance, unknown up to here. After reflection from A', the light continues its travels to the A frame where/when the light and the A frame meet. This received pulse signals the end of the total distance the light travels during which time the frame is moving steadily forward. The light motion began at t0, where now the values of (X0,t0) are known, or are they? Actually X0 is not known because we did not measure an actual X0, but this is of no concern as we are not measuring X0 and X1 individually, directly, or otherwise. Rather we are measuring the difference of X1 - X0, the distance the frame travels during t2 - t0.

    The schematic may be of some use here.

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    1. As the pulse moves a distance d1, the frame moves a distance Va(t1 - t0), Va being unknown at this point.
    2. The pulse then travels the distance d2, (A' back to A) during t2 - t1, while the frame continues its constant velocity. The A frame will continue moving, hypothetically speaking, until the light motion ceases.
    3. When the pulse is received at A at t2 the measurement of the continuing frame motion ceases, or has ceased. There are three parameters observed in the experiment’s data base, t0, t1, and t2 with a minimum of six data points, two for each of the three times measured.
    The actual direction of the pulse motion is irrelevant, the crucial parameter here is the time-of-flight of the pulse along d1 and d2 as all distances can be determined, for example, by reflections between mirrors ala Michelson-Morley. Any sufficient number of arbitrary reflections will suffice as we are measuring non-vectored distances which should be only be sufficiently large to insure an acceptable experimental error in measurement, which can be as fine tuned by extending the time of flight until reaching some distance where gains in accuracy can no longer be determined – the point where the return on investment in accuracy has reached a point of diminished returns that approach zero. Why subtract d1 - d2? We use the difference to determine the ΔX for our upcoming velocity calculation. Using Figure 1 we see that if the light simply continued along an un-reflected path that D would not change, but in this case we would not be able to determine its value. When the motion of the frame and reflected pulse meet (t2 is the indicated point in Figure 1), we then have confidence in the calculated distance D, which is the delta-X in the velocity expression, as opposed to being the numbers describing the individual X0 and X1. D is the distance the frame moves during t2 – t0.

    To now answer your specific question, we begin by equating D = (d1 - d2), when expressed mathematically (for c = 1) becomes,
    D = (t1 – t0) – (t2 – t1) = VA(t2 – t0).
    The first () term is d1, the second, d2, and D expressed as VA(t2 – t0) where here the delta-t covers the entire time of flight of the pulse.

    VA = (t1 – t0) – (t2 – t1)/(t2 – t0).
    = (t1 – t0 – t2 + t1)/(t2 – t0)
    = [2t1 – (t2 – t0)]/(t2 − t0)
    = (2t1)/(t2 - t0) – (t2 − t0)/(t2 − t0)
    = (2t1)/(t2 - t0) – 1
    There are a lot of interesting twists and turns up to now and I trust some of your ‘muddle’ has been clarified. I am looking forward to further comments.
     
  10. James R Just this guy, you know? Staff Member

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    39,397
    geistkiesel:

    Ok. It looks like I misinterpreted what this distance D is supposed to be, but for good reason. You said:

    Now, in the frame AA', the distance between A and A' is constant at all relevant times from to to t2. But your diagram appears to show A moving a distance D in that period. In the AA', A NEVER moves at all! A only moves in some other reference frame (like the one I called X above).

    Now, if we work in the X frame, then your diagram starts to make some sense. Let's split the problem up a bit. For the pulse moving from the location of A at to to A', we have (setting the initial location of A to x=0):

    x(A) = vA t
    x(A') = D + vA t
    x(pulse) = c t

    Here, the vA is measured in the X frame - it is the velocity of frame AA' relative to X.

    The pulse reaches A' at a time given by:

    D + vA t1 = c t1

    which means

    t1 = D / (c - vA)

    At this time, A' is at location:

    x1 = D (1 + vA / (c - vA))
    x(A) = vA D / (c - vA)

    The pulse is immediately reflected and received by A at time t2. The locations of A and the pulse between times t1 and t2 are:

    x(A) = vA D / (c - vA) + vA (t - t1)
    x(pulse) = x1 - c (t - t1) = D (1 + vA / (c - vA)) - c (t - D / (c - vA))

    The pulse arrives back at A at time t = t2, given by:

    vA D / (c - va) + vA (t2 - t1) = D + vA D / (c - vA) - c t2 + c D / (c - vA)

    Simplifying, we find:

    vA (t2 - t1) = D - c t2 + c D / (c - vA)

    or

    t2 (vA + c) = D + vA t1 + c D / (c - vA)

    t2 (vA + c) = D + vA D / (c - vA) + c D / (c - vA)

    t2 (vA + c) = D + D (vA + c) / (c - vA)

    t2 (vA + c) = [D (c - vA) + D (c + vA)] / (c - vA)

    t2 (vA + c) = 2 c D / (c - vA)

    t2 = 2 c D / (c - vA)(c + vA)

    t2 = 2 c D / (c^2 - vA^2)

    Now, from this expression, if we know D and c, and we measure t2, we can work out vA.

    But, what does this mean? Have we worked out the absolute velocity of frame AA'? Of course not. The distance D and the time t2 are both measured in frame X. So, the value of vA that we calculate is the velocity of frame AA' relative to frame X.

    We can't get anything about the absolute speed of frame AA' from this thought experiment.
     
  11. geistkiesel Valued Senior Member

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    2,471
    .
    James R, When the A' to A trajectory is heading to an oncoming A frame, d2 distance must be less that d1.

    No. This a departure from the constraints establishged in opening post oif this thread. You are making a direct observation of VA from frame X. You are measuring a differenjt VA than the one I am detrermining from the measured times t0, t1, and t2. Your VA is not measured in A frame, mine is. You must defeat the thread as posted.
    Stop here. If you are using my D, that D is nogt determined until t2.
    No. You originally began you argument by stating as follows:

    James R stated above that
    "x(A) = vA t
    x(A') = D + vA t
    x(pulse) = c t

    Here, the vA is measured in the X frame - it is the velocity of frame AA' relative to X."

    .
    I said nothing about "absolute speed", not that I am trying hide some fond affection for the concept. The reason you cannot get anything about the absolute speed of frame AA' is that you mjade no attempt to measure the distance D, for just one example.

    You are using SRT to determine your result, when this thread has made a measurement of unaccelerated translatory motion in free space. This being a fact, there isn't an SRT from which the analysis based, any more, at least. The D I clearly referred to is determined from d1 - d2 = D and where d1 and d2 are determined from t0, t1 and t2, and then ultimately with the use of c = 1, the unit velocity of light. You don't measure t0, t1, and t2 from frame X anyway, do you??

    Under what concept or analytical methods do you employ in an attempt to reach a preconceived notion this thread is flawed when my measurements, acquired from the deck of frame A, produce data sufficient to conclude that VA = (2t1)/(t2 - t0) - 1, a result different from your result for the simple reason that your experiment, as you set the the relevant parameters as you saw fit, was not the same as my experiment, even when you use some of my data, but I acquired the three time measurements from the deck of the frame for which the frame velocity VA is determined.

    It is to this point that I direction your attention and request you explain any the flaw in my described experiment with a reference to my experiment. You aren't asserting that my result is baseless for the reason that you and I disagree. This is why I pointed out the necessity of determining a flaw in the protocol I described as a viable VA, as I determined, means your method would need adjustment. If my methodology includes flaws it should be a trivial matter to point to the warts.

    James R, you are using a model based extensively of determining the t2, t1 and t0 that I measure directly.
     
  12. James R Just this guy, you know? Staff Member

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    geistkiesel:

    In the frame AA', the distances d1 and d2 are the same, and there is no distance D. In that frame, A and A' remain a constant distance apart at all times, for the simple reason that in that frame neither A nor A' is moving at any time. A does not move towards A', and nor does A' move away from A. Your diagram therefore cannot apply to frame AA', but only to a frame such as X, defined in my previous post - a frame that sees A and A' moving.

    In the frame AA', t1 = d1 / c and t2 = 2 d2/c = 2 d1/c, and D=0. There is no way to determine the velocity of frame AA' from these measurements.
     
  13. Montec Registered Senior Member

    Messages:
    248
    In an attempt to clarify I give the following:

    Let t1 equal the time it takes light to go from A to A'
    Let t2 equal the time it takes light to go from A' to A
    Let s1 be the distance light travels from A to A'
    Let s2 be the distance light travels from A' to A
    Let L be the measured distance between A and A'
    Let V be the velocity of a frame with respect to the emission points of light.
    c is the speed of light

    Then
    s1 = L + Vt1 = ct1 Distance between A and A' plus the distance A' moved during t1
    s2 = L - Vt2 = ct2 Distance between A' and A minus the distance A moved during t2

    Solving for L we get
    L = t1(c-V) and L = t2(c+V) and the ratio

    t1/t2 = (c+V)/(c-V) Which tells us that if V=0 then t1=t2. It also tells us that t1 cannot equal t2 if a frame has any velocity with respect to the emission points of light. (And think that said points don't move.)

    All of the above calculation and statements are based on the idea that the speed light does not depend on the speed of the observer or light source. Notice that the equation works with any "V". (Which is generally defined to be from zero up to but not including "c").

    Note: It also just happens that when you compare the total distance the light traveled (s1 + s2) with the frame distance of 2L you get 2L/(1-v^2/c^2) = s1+s2.

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  14. Montec Registered Senior Member

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    As an addendum to my previous post;

    Using t1/t2=(c+V)/(c-V) and solving for V you get

    V=c(t1-t2)/(t1+t2)

    Which is the difference between the one-way (to and fro) times divided by the total time multiplied by the speed of light to give a velocity value.

    An absolute velocity in this case. To bad that no one has figured out how to measure the one-way (not sure if this is even possible) speed of light or even measure the difference between the one-way times.

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  15. Uno Hoo Registered Senior Member

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    You will most likely not give my following comment sufficient serious study to have any chance to retort cogently. Here's trying, anyway:

    You seem to equate mathematical consistency with logical consistency. I would say that there is often a tenuous relationship. Never a rigorous one-to-one relationship.

    In Special Relativity, do you say that there is always a rigorous and direct connection of logical consistency with mathematical consistency?
     
  16. Uno Hoo Registered Senior Member

    Messages:
    383
    Montec has mis-spoken. What Montec really meant to say is that Montec has not figured out any way to test the one way speed of light.
     
  17. AlphaNumeric Fully ionized Registered Senior Member

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    6,702
    Mathematics is a form of logic and sometimes vice versa. Can you point me to a section of mathematics which is not based on logic?

    SR is basically the statement that space-time has associated G-bundle with group SO(3,1). Its consistency is entirely equivalent to the consistency of group theory.

    The fact you might think the statements of special relativity are counter intuitive doesn't mean they are illogical or mathematically inconsistent. Too often people think their intuition is equivalent to logical consistency. They are wrong. We've already seen your understanding of how science works to be flawed, given you think those who learn science are just 'calculators' for the people with ideas. Proof you fail to understand how science is developed so I hardly think you're the person to ask about the notion of logic within mathematical sciences.

    So, can you provide me with an example of mathematics which is not built from logical methods? How do you prove something if you aren't using logic, since proof requires logical reasoning.
     
  18. Montec Registered Senior Member

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    Hello Uno Hoo
    Actually, I have a couple ideas that may work but why should I let you in on them?

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  19. Uno Hoo Registered Senior Member

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    In this post, I see that you say that no one has figured it out.
     
  20. Uno Hoo Registered Senior Member

    Messages:
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    Now, in this post, you say that some one has figured it out. A little less confusion, please!

    Telling them to me is irrelevant. Don't worry much about it. It is doubtful that you have figured out any possible one way light speed tests which I have not already examined.
     
  21. Uno Hoo Registered Senior Member

    Messages:
    383
    I guessed right. You obviously did not consider my comment carefully enough to make a cogent response. Well, for whatever reason, your response was not of much scientific value.

    My serious logical study of Special Relativity has covered more decades than you have even been physically alive. My reliance upon intuition to evaluate any science matter is minimal. It is somewhat amusing to read that you imagine that you somehow can read my mind and clairvoyantly know how I use my mind.

    You have completely missed the point of my post, which I predicted.

    Special Relativity contains mathematics which seems to me to be impeccably internally consistent. The math is not a problem for me. The problem(s) is that Special Relativity was presented containing a quantity of terrible logical inconsistencies.

    The Einstein presentation of The Relativity Of Simultaneity is perfectly logically inconsistent with his Postulate that the speed of light is always observed to be the same value. It is as obvious as that full grown pink elephant in your office.
     
  22. geistkiesel Valued Senior Member

    Messages:
    2,471
    All this is very inteesting but you have just manipulated another experiment not the lone described in my thread. The thread is facially correct. Being so, you have no relativity theory from which you may applyt as my post describesw a technique to measure and detect unaccelerated translatory motion in free space.

    You are either thoroughly confused or you are just being cheeky and attempting to confuse any readers.

    You, while responding to my post also moved it away from the main physics forum without the slightest attemp to justify your belief that the the thread wasn't deserving a normal scrutiny. Why do you do these things?

    James R, I can only make guesses because you haven't been completely open in this very simple measuring system. You ask from which frame of reference is the velocity measured. The t0 pulse records the instantaneous initial X0. The t2 measurement defines X1. By definition V(t2 - t0) = (X1 - X0)/(t2 -t0).

    These Xn are all instantaneous values determined when measured. Velocity determination needs no fixed physical frame to be correct. The AA' frames velocityu are Va = Va' for all relevant times here. There is no physical necessity to determine a physical entity to establish a physical frame of reference - the reason Jameds R is that velocity V = delta_Xn/delta_tn.elocity is a relative concept V. It is the difference in the Xn divided by the difference in the pertinent tn.

    Give us all a break and provide us a source for your persitent nagging question - "velocity with respecrt to what". Prove such a question has its own legs and can walk.

    You are exposing yourself as either incompetent or purposefully trying to muddy the waters or the purpose of killing the thread , or perhapos you jjust don't see it. I know AN hasn't a clue about how to defet tghe thesis her by falsifying the thread from the four corners of the opening post and following discussion. Jame R it should be obvious that AN is maneuvering for a slot opening for a mentor's position so he can be an official and can chew up anyone's post as an authority, meaning he wouldn't require him to demonstrate his lack of analytical ability. He's a power mongering control freak that will never be able to recognize any of that snot dripping from his nose.

    Do the mootors here have any sense of ethical restraints, of politeness or professionalism? Give him the badge James R and you can both play good cop bad cop.
     
  23. Neddy Bate Valued Senior Member

    Messages:
    2,548
    LOL, I will let everyone in on my idea, for free.

    This is a method to test the one-way speed of light. It is also a fool-proof method to synchronise clocks.

    Put two clocks far away from each other in space. Put an honest observer next to each clock. Provide the observers with very powerful telescopes so they can view the other guy's clock.

    Each observer must keep a log of what they see on both clocks. So one observer might have a log that looks like this:

    Mine .. His
    0.00 .. 5.00
    1.00 .. 6.00
    2.00 .. 7.00

    And the other observer might have a log like this:

    Mine .. Hers
    0.55 .. 0.00
    1.55 .. 1.00
    2.55 .. 2.00

    From this data, I can tell that the speed of light is 1.8c in one direction, and 0.2c in the other direction.

    It's like a magic trick. See if you can tell how I did it.
     

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