Experiment to test mutual time dilation
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From: Sam Wormley on 8 Aug 2010 15:45 On 8/8/10 7:38 AM, kenseto wrote: > On Aug 7, 2:32 pm, Sam Wormley<sworml...(a)gmail.com> wrote: >> On 8/7/10 12:51 PM, kenseto wrote: >> >> >> >> >> >>> On Aug 7, 9:47 am, Sam Wormley<sworml...(a)gmail.com> wrote: >>>> On 8/7/10 8:24 AM, kenseto wrote: >> >>>>> No he said that mutual time dilation doesn't apply to the GPS because >>>>> the GPS does not see the SR effect on the ground clock is ~7us/day >>>>> running slow. He tried to avoid the fact that the GPS sees the SR >>>>> effect on the ground clock as 7 us/day running fast so he said the GR >>>>> must be used for such calculations. >> >>>>> Ken Seto >> >>>> For whatever reason, Seto, you have a lot of trouble comprehending >>>> what knowledgeable people are trying to tell you. To calculate time >>>> dilation in satellite clocks requires general relativity, >> >>> Hey idiot we are talking about the SR effect only. You are so stupid. >> >> There is NO SR effect only... don't be so stooopid! > > Hey idiot...the SR effect uses the SR math to do calculations. The > gravitational effect uses separate equation to do calculations. No, Seto, When using general relativity, there is no need for special relativity separately. You don't understand relativity at all and should take the time to do some self education. AS far as GPS goes, I've tried to steer you to accessible literature but for whatever reason you won't even look at "Relativistic Effects on Satellite Clocks" http://relativity.livingreviews.org/open?pubNo=lrr-2003-1&page=node5.html Why is that Seto? Over your head?
From: Koobee Wublee on 9 Aug 2010 01:42 On Aug 6, 2:16 pm, "Paul B. Andersen" wrote: > On 06.08.2010 08:48, Koobee Wublee wrote: > > First, you are stating the following which stands on no ground. > > > ** w (t - x / c) = w' (t' - x' /c) > > It should be clear from the context that the event > with coordinates (t,x) in the unprimed frame > has the coordinates (t',x') in the primed frame. Believe me. I fully understood your ill-fated intention. <shrug> > The phase of the wave at a specific event is the same > in all frames of reference. > That is, phi(t,x) = phi'(t',x') > where (t,x) and (t',x') are the coordinates of the _same_ event. What properties of the Lorentz transform spells that one out again? > You do know that the LT transforms the coordinates > of an event, don't you? I think so. <shrug> > Or don't you? :-) Gee! Am I bluffing? Is it that obvious or what? > > Secondly, you can easily write the result as the following instead, > > > ** w = w' sqrt(1 + v / c) / sqrt(1 - v / c) > > Sure. > If the frequency of the wave in the primed frame is lower > than its frequency in the unprimed frame, then the frequency > of the wave in the unprimed frame is higher than its frequency > in the primed frame. I thought you used to work in the private industrial. There leaves no room for such ambiguous mathematical models as you have proposed. Is it why you are now working in the academics where your mathematical models don't affect anyone's lives? <shrug> > Do you find this obvious triviality remarkable? :-) Forgive me to be blunt. I do find your ambiguous mathematical model of the Doppler effect to be of fictional interest. <shrug> > > Where > > > ** v = Speed of w' relative to w > > > And proudly claim w-w' moving away (v > 0) as Doppler blue shift. > > Sure, what's wrong with that? When a mathematical model is able to predict just about anything under the sun, some less challenged individuals would hail that as the greatest thing after sliced bread, but true scholars of physics would call that what it is that is total bullshit. <shrug> > The wave is propagating in the positive x-direction! > > ------> wave > > |----> x' ->v > |----> x > > If the wave has the frequency w' in the primed frame, > an observer in the unprimed frame will measure the frequency > w = w' sqrt(1 + v / c) / sqrt(1 - v / c) > which is a blue shifted compared to w'. It is a shame that a professor working in the field of electromagnetism does not realize that an accurate mathematical model in Doppler effect must involve the dot product where it spells out exactly where the signal is going and received. Are you guys at Trondheim really that clumsy? Or is it just you who cannot take the pressure of the real world? > > Notice the classical Doppler effect for sound cannot be derived using > > your method. > > Why not? > "Classical Doppler" -> Galilean transform. According to the mathematical method come up with by the self-styled physics in the past 100 years, the Galilean transform should indicate no Doppler shift? Why is that? ** dt' = dt <shrug> > An acoustic wave propagating in the positive x direction > in the unprimed rest frame of the medium can be written: Then, show me the mathematical model built out of vectors. <shrug> > A cos(phi(t,x)) where phi(t,x) = wt - (w/c)x > where c is the speed of the wave in the medium. > > [...] Gee! The classical Doppler effect is simpler than that. It is merely an algebraic manipulation of vectors. I will leave the following as a homework assignment for our professor from the University of Trondheim. f2 = f1 sqrt[(c^2 - 2 [c] * [v2] + v2^2) / (c^2 - 2 [c] * [v1] + v1^2)] Where ** [v1], [v2] = Velocities relative to the stationary background of the medium ** * = Dot product of two vectors [] and [] Hint: ** f lambda = c ** Invariant lambda > > This should be an alarm ringing in your head on your > > part. Once again, your mistake is in the application of the Lorentz > > transform. > > There is no mistake. > You seem very confused about what Doppler shift is. Ahahahaha... This is the finest joke after coming back from spending an entire weekend in a mountain retreat. > Try again to find an error? :-) http://groups.google.com/group/sci.physics.relativity/msg/59b3d0dcb8deab2d?hl=en > > The time transformation can generally be written as follows. You must > > pay very close and crucial attention to the directions of these > > vectors. The self-styled physicists have been hand-waving and spoon- > > feeding this nonsense to their brain-washed pupils in the past 100 > > years. > > > ** dt' = (dt - [v] * d[s] / c) / sqrt(1 - v^2 / c^2) > > > Where > > > ** [v] = Velocity of dt' as observed by dt > > ** d[s]/dt = [c] = Velocity of light > > ** d[s]/dt * d[s]/dt = c^2 > > > Since you are making the same mistake again, I shall not count this > > one as a score of mine. Please study my posts below more carefully. > > It is not that hard.<shrug> > > >http://groups.google.com/group/sci.physics.relativity/msg/c06938d96ee... > > >http://groups.google.com/group/sci.physics.relativity/msg/fddbda5c1bb... > > > Since you have brought up the wave equations for light, solutions to > > Maxwell's equation in free space are: > > > SUM_n[[E_n] exp(a_n (w_n t + b_n [k_n] * [s]) + c_n)] > > > Where > > > ** a_n = +/- 1 or +/- sqrt(-1) > > ** b_n = +/- 1 > > ** [E_n] = Constant vector > > ** [k_n] = Direction vector > > ** [s] = Position vector > > ** k_n^2 = w_n^2 / c^2 > > ** * = Dot product of two vectors > > > Filtering out solutions that do not allow propagation of waves, what > > is left is the following representing one particular frequency of > > interest. > > > E cos(w t - [k] * [s] + theta) > > > Where > > > ** theta = Phase > > > You understand k^2 as 1 / wavelength^2.<shrug> > > No. > > k^2 = 4*pi^2/wavelength^2 > = w^2/c^2 for a non-dispersive wave Well, being a nick-picker when your professorship is under pressure, eh? > EM-waves in free space are non-dispersive. > And so are audible acoustic waves in air. > > <shrug> Hmmm... Trying to change the topics, eh? > So setting the arbitrary constant theta = 0, > and letting the wave vector be parallel > to the x-axis, we are back to my equation above. > > E cos(phi(t,x)) where phi(t,x) = wt - (w/c)x > or > E cos(wt - (w/c)x) > > <shrug> Yes, who gives a damn. It does not tell you anything. <shrug> > I took for granted that you knew that this is a solution > of Maxwell's wave equation. > (Or any wave equation, for that matter). > > <shrug> I take it that as a professor working closely related to electromagnetism, Maxwell's equations are not required to be fully understood at the University of Trondheim. <shrug> > > Should the Lorentz transform be applied to [k] and w? > > If you want to transform the wave to a different frame, > of course. > > <shrug> Please elaborate instead of shrugging it away. <shrug> > You can see above how k = w/c and w are transformed. > Note that k' = (w/c)sqrt((1-v/c)/(1+v/c)) > That means that lambda' = lambda ((1+v/c)/(1-v/c)) According to Maxwell's equations, k is a vector which in my terminology should be written as [k] in this text editor. It is more subtle in complexity than you think. <shrug> Try the following and holding c as invariant instead of lambda. <shrug> ** c = f lambda
From: Paul B. Andersen on 9 Aug 2010 09:36 On 09.08.2010 07:42, Koobee Wublee wrote: > On Aug 6, 2:16 pm, "Paul B. Andersen" wrote: >> On 06.08.2010 08:48, Koobee Wublee wrote: > >>> First, you are stating the following which stands on no ground. >> >>> ** w (t - x / c) = w' (t' - x' /c) >> >> It should be clear from the context that the event >> with coordinates (t,x) in the unprimed frame >> has the coordinates (t',x') in the primed frame. > > Believe me. I fully understood your ill-fated intention.<shrug> In that case, the following explanation of mine should be unnecessary. > >> The phase of the wave at a specific event is the same >> in all frames of reference. >> That is, phi(t,x) = phi'(t',x') >> where (t,x) and (t',x') are the coordinates of the _same_ event. > > What properties of the Lorentz transform spells that one out again? And you are serious? :-) >> You do know that the LT transforms the coordinates >> of an event, don't you? > > I think so.<shrug> So then you should be able to understand the calculation you have snipped. Koobee Wublee wrote: | The second post by Professor Roberts has a wrong conclusion on | relativistic Doppler effect. The Lorentz transform actually leads to | a reverse Doppler effect. The following derivaton has no other purpose than showing that the Lorentz transform don't lead to "a reverse Doppler effect". That the wave is propagating along the velocity of the primed frame in the unprimed frame is a special case chosen to keep the calculation simple. The challenge to Koobee Wublee is to point out an error in _this_ simple calculation. Try again Koobee? A wave propagating in the positive x direction in the unprimed frame can be written: E cos(phi(t,x)) where phi(t,x) = wt - (w/c)x Let the primed frame be moving at v along the positive x-axis. The same wave transformed to the primed frame can be written: E' cos(phi'(t',x')) where phi'(t',x') = w't' - (w'/c)x' Applying the LT transform: t = g(t' + vx'/c2) x = g(x' + vt') g = 1/sqrt(1-v2/c2) yields: wt - (w/c)x = wg((t' + vx'/c2) - (w/c)g(x' + vt') = wg(1-v/c)t' - (w/c)g(1-v/c)x' = w't' - (w'/c)x' Thus: w' = wg(1-v/c) = w sqrt((1-v/c)/(1+v/c)) The wave is red shifted in the primed frame. REPEAT: The challenge is to point out an error in the derivation. If you can't do that, your claim that the LT leads to "a reverse Doppler effect" is proven wrong. > >> Or don't you? :-) > > Gee! Am I bluffing? Is it that obvious or what? > >>> Secondly, you can easily write the result as the following instead, >> >>> ** w = w' sqrt(1 + v / c) / sqrt(1 - v / c) >> >> Sure. >> If the frequency of the wave in the primed frame is lower >> than its frequency in the unprimed frame, then the frequency >> of the wave in the unprimed frame is higher than its frequency >> in the primed frame. > > I thought you used to work in the private industrial. There leaves no > room for such ambiguous mathematical models as you have proposed. Is > it why you are now working in the academics where your mathematical > models don't affect anyone's lives?<shrug> > >> Do you find this obvious triviality remarkable? :-) > > Forgive me to be blunt. I do find your ambiguous mathematical model > of the Doppler effect to be of fictional interest.<shrug> Why these non sequiturs, Koobee? Have you realized that what you seemed to find remarkable indeed is a trivial obviousity? :-) > >>> Where >> >>> ** v = Speed of w' relative to w >> >>> And proudly claim w-w' moving away (v> 0) as Doppler blue shift. >> >> Sure, what's wrong with that? > > When a mathematical model is able to predict just about anything under > the sun, some less challenged individuals would hail that as the > greatest thing after sliced bread, but true scholars of physics would > call that what it is that is total bullshit.<shrug> > >> The wave is propagating in the positive x-direction! >> >> ------> wave >> >> |----> x' ->v >> |----> x >> >> If the wave has the frequency w' in the primed frame, >> an observer in the unprimed frame will measure the frequency >> w = w' sqrt(1 + v / c) / sqrt(1 - v / c) >> which is a blue shifted compared to w'. > > It is a shame that a professor working in the field of > electromagnetism does not realize that an accurate mathematical model > in Doppler effect must involve the dot product where it spells out > exactly where the signal is going and received. Are you guys at > Trondheim really that clumsy? Or is it just you who cannot take the > pressure of the real world? Why these non sequiturs, Koobee? Don't you understand that: If the wave has the frequency w' in the primed frame, an observer in the unprimed frame will measure the frequency w = w' sqrt(1 + v / c) / sqrt(1 - v / c) which is a blue shifted compared to w'. >>> Notice the classical Doppler effect for sound cannot be derived using >>> your method. >> >> Why not? >> "Classical Doppler" -> Galilean transform. > > According to the mathematical method come up with by the self-styled > physics in the past 100 years, the Galilean transform should indicate > no Doppler shift? Why is that? > > ** dt' = dt SIC!!!! :-) > <shrug> > >> An acoustic wave propagating in the positive x direction >> in the unprimed rest frame of the medium can be written: > > Then, show me the mathematical model built out of vectors.<shrug> > >> A cos(phi(t,x)) where phi(t,x) = wt - (w/c)x >> where c is the speed of the wave in the medium. >> >> [...] > > Gee! The classical Doppler effect is simpler than that. It is merely > an algebraic manipulation of vectors. I will leave the following as a > homework assignment for our professor from the University of > Trondheim. > > f2 = f1 sqrt[(c^2 - 2 [c] * [v2] + v2^2) / (c^2 - 2 [c] * [v1] + > v1^2)] > > Where > > ** [v1], [v2] = Velocities relative to the stationary background of > the medium > ** * = Dot product of two vectors [] and [] The derivation you snipped is still correct, so your claim: "Notice the classical Doppler effect for sound cannot be derived using your method." is proven wrong. | An acoustic wave propagating in the positive x direction | in the unprimed rest frame of the medium can be written: | A cos(phi(t,x)) where phi(t,x) = wt - (w/c)x | where c is the speed of the wave in the medium. | | Let the primed frame be moving at v along the positive x-axis. | | The same wave transformed to the primed frame can be written: | A' cos(phi'(t',x')) where phi'(t',x') = w't' - (w'/(c-v))x' | (The speed of sound in the prined frame is c-v) | | Applying the Galilean transform: | t = t' | x = x' + vt' | yields: | wt - (w/c)x = wt' - (w/c)(x' + vt') | = w(1-v/c)t' - (w/c)x' | = w't' - (w/c)x' | | Thus: | w' = w(1-v/c) | The wave is red shifted in the primed frame. | Note that k' = w/c = k | k' = k and thus lambda' = lambda | The wavelength is the same in all frames. And after having read this, you can snip it, and give me this hint. :-) > Hint: > > ** f lambda = c > ** Invariant lambda > >>> This should be an alarm ringing in your head on your >>> part. Once again, your mistake is in the application of the Lorentz >>> transform. >> >> There is no mistake. >> You seem very confused about what Doppler shift is. > > Ahahahaha... This is the finest joke after coming back from spending > an entire weekend in a mountain retreat. You have still failed to point out any mistake in my derivation. > >> Try again to find an error? :-) > > http://groups.google.com/group/sci.physics.relativity/msg/59b3d0dcb8deab2d?hl=en You have still failed to point out any mistake in my derivation. > >>> The time transformation can generally be written as follows. You must >>> pay very close and crucial attention to the directions of these >>> vectors. The self-styled physicists have been hand-waving and spoon- >>> feeding this nonsense to their brain-washed pupils in the past 100 >>> years. >> >>> ** dt' = (dt - [v] * d[s] / c) / sqrt(1 - v^2 / c^2) >> >>> Where >> >>> ** [v] = Velocity of dt' as observed by dt >>> ** d[s]/dt = [c] = Velocity of light >>> ** d[s]/dt * d[s]/dt = c^2 >> >>> Since you are making the same mistake again, I shall not count this >>> one as a score of mine. Please study my posts below more carefully. >>> It is not that hard.<shrug> >> >>> http://groups.google.com/group/sci.physics.relativity/msg/c06938d96ee... >> >>> http://groups.google.com/group/sci.physics.relativity/msg/fddbda5c1bb... >> >>> Since you have brought up the wave equations for light, solutions to >>> Maxwell's equation in free space are: >> >>> SUM_n[[E_n] exp(a_n (w_n t + b_n [k_n] * [s]) + c_n)] >> >>> Where >> >>> ** a_n = +/- 1 or +/- sqrt(-1) >>> ** b_n = +/- 1 >>> ** [E_n] = Constant vector >>> ** [k_n] = Direction vector >>> ** [s] = Position vector >>> ** k_n^2 = w_n^2 / c^2 >>> ** * = Dot product of two vectors >> >>> Filtering out solutions that do not allow propagation of waves, what >>> is left is the following representing one particular frequency of >>> interest. >> >>> E cos(w t - [k] * [s] + theta) >> >>> Where >> >>> ** theta = Phase >> >>> You understand k^2 as 1 / wavelength^2.<shrug> >> >> No. >> >> k^2 = 4*pi^2/wavelength^2 >> = w^2/c^2 for a non-dispersive wave > > Well, being a nick-picker when your professorship is under pressure, > eh? > >> EM-waves in free space are non-dispersive. >> And so are audible acoustic waves in air. >> >> <shrug> > > Hmmm... Trying to change the topics, eh? The topic is if you can find an error in my calculation. You have failed to do so. > >> So setting the arbitrary constant theta = 0, >> and letting the wave vector be parallel >> to the x-axis, we are back to my equation above. >> >> E cos(phi(t,x)) where phi(t,x) = wt - (w/c)x >> or >> E cos(wt - (w/c)x) >> >> <shrug> > > Yes, who gives a damn. It does not tell you anything.<shrug> It should tell you that E cos(wt - (w/c)x) is a solution of Maxwell's wave equation for EM-waves in free space. The challenge was to find an error in the transformation of this particlar wave to a moving frame of reference. You have failed to do so. > >> I took for granted that you knew that this is a solution >> of Maxwell's wave equation. >> (Or any wave equation, for that matter). >> >> <shrug> > > I take it that as a professor working closely related to > electromagnetism, Maxwell's equations are not required to be fully > understood at the University of Trondheim.<shrug> E cos(wt - (w/c)x) is a solution of Maxwell's wave equation for EM-waves in free space. The challenge was to find an error in the transformation of this particlar wave to a moving frame of reference. You have failed to do so. > >>> Should the Lorentz transform be applied to [k] and w? >> >> If you want to transform the wave to a different frame, >> of course. >> >> <shrug> > > Please elaborate instead of shrugging it away.<shrug> > >> You can see above how k = w/c and w are transformed. >> Note that k' = (w/c)sqrt((1-v/c)/(1+v/c)) >> That means that lambda' = lambda ((1+v/c)/(1-v/c)) > > According to Maxwell's equations, k is a vector which in my > terminology should be written as [k] in this text editor. It is more > subtle in complexity than you think.<shrug> > > Try the following and holding c as invariant instead of lambda. > <shrug> > > ** c = f lambda Well Koobee? Can you find en error in my calculation? :-) No? Thought so. There are none. -- Paul http://home.c2i.net/pb_andersen/
From: Koobee Wublee on 9 Aug 2010 12:52 As usual, you have challenged me to find your error, and I have done so. <shrug> See the link below. http://groups.google.com/group/sci.physics.relativity/msg/59b3d0dcb8deab2d?hl=en Unable to face your own short comings, you are once again becoming asinine with silly and childish denials. You need to calm down and stop babbling and repeating the same nonsense before I can continue to guide you out of mysticism. <shrug> Do you think you can behave like a responsible adult for a change?
From: Paul B. Andersen on 9 Aug 2010 14:37
On 09.08.2010 18:52, Koobee Wublee wrote: > As usual, you have challenged me to find your error, and I have done > so.<shrug> See the link below. > > http://groups.google.com/group/sci.physics.relativity/msg/59b3d0dcb8deab2d?hl=en > > Unable to face your own short comings, you are once again becoming > asinine with silly and childish denials. You need to calm down and > stop babbling and repeating the same nonsense before I can continue to > guide you out of mysticism.<shrug> > > Do you think you can behave like a responsible adult for a change? Repeating a posting where you failed to point out an error won't do. Here is the derivation again: A wave propagating in the positive x direction in the unprimed frame can be written: E cos(phi(t,x)) where phi(t,x) = wt - (w/c)x Let the primed frame be moving at v along the positive x-axis. The same wave transformed to the primed frame can be written: E' cos(phi'(t',x')) where phi'(t',x') = w't' - (w'/c)x' Applying the LT transform: t = g(t' + vx'/c2) x = g(x' + vt') g = 1/sqrt(1-v2/c2) yields: wt - (w/c)x = wg((t' + vx'/c2) - (w/c)g(x' + vt') = wg(1-v/c)t' - (w/c)g(1-v/c)x' = w't' - (w'/c)x' Thus: w' = wg(1-v/c) = w sqrt((1-v/c)/(1+v/c)) The wave is red shifted in the primed frame. Can you point out an error, Koobee? -- Paul http://home.c2i.net/pb_andersen/ |