q-hat^-1 ?
in [Relativity]
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From: Ken S. Tucker on 28 Feb 2010 03:44 On Feb 27, 6:08 am, "Jay R. Yablon" <jyab...(a)nycap.rr.com> wrote: > My query here is in reference to the position operator q-hat, which > commutes with the momentum operator p-hat along any single space > dimension according to the canonical commutation relationship: > > [q-hat,p-hat]=i h-bar (1) > > Query: Is there a known *inverse* position operator q-hat^-1? If so, > what is it and can you point out a reference(s) in which it is derived > and discussed. Jay, when are you going to post an easy problem? Ok, I'll give it a shot... Allow me to simplify the rotation to, m*[V(y)*x - V(x)*y] = h where V(y) is the mass m Velocity in direction y. Allow me to unitized (for ascii sake) m and q to follow, then, sub a positional potential O(x) = (q/r^2) * x for the inverted q-hat, (I'll use "O" because positional potentials are measured w.r.t an Origin), but that - with q (charge) unitized is dimensionally 1/L , O(x) = x/r^2 . Putting that together yields, V(y) O(x) - V(x) O(y) = h/r^2, (h=action). Interpretation's can be found from r=ct. Let me know if we should go there. Regards Ken S. Tucker
From: Igor on 28 Feb 2010 10:24 On Feb 28, 3:44 am, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote: > On Feb 27, 6:08 am, "Jay R. Yablon" <jyab...(a)nycap.rr.com> wrote: > > > My query here is in reference to the position operator q-hat, which > > commutes with the momentum operator p-hat along any single space > > dimension according to the canonical commutation relationship: > > > [q-hat,p-hat]=i h-bar (1) > > > Query: Is there a known *inverse* position operator q-hat^-1? If so, > > what is it and can you point out a reference(s) in which it is derived > > and discussed. > > Jay, when are you going to post an easy problem? > Ok, I'll give it a shot... > Allow me to simplify the rotation to, > > m*[V(y)*x - V(x)*y] = h > > where V(y) is the mass m Velocity in direction y. > > Allow me to unitized (for ascii sake) m and q to follow, then, > sub a positional potential O(x) = (q/r^2) * x for the inverted q-hat, > (I'll use "O" because positional potentials are measured w.r.t an > Origin), > but that - with q (charge) unitized is dimensionally 1/L , > > O(x) = x/r^2 . > > Putting that together yields, > > V(y) O(x) - V(x) O(y) = h/r^2, (h=action). > > Interpretation's can be found from r=ct. > Let me know if we should go there. > Regards > Ken S. Tucker Good job, Ken. The OP asked about A=B and you naturally showed something related to Q=V.
From: Ken S. Tucker on 28 Feb 2010 14:18 On Feb 28, 7:24 am, Igor <thoov...(a)excite.com> wrote: > On Feb 28, 3:44 am, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote: > > > > > On Feb 27, 6:08 am, "Jay R. Yablon" <jyab...(a)nycap.rr.com> wrote: > > > > My query here is in reference to the position operator q-hat, which > > > commutes with the momentum operator p-hat along any single space > > > dimension according to the canonical commutation relationship: > > > > [q-hat,p-hat]=i h-bar (1) > > > > Query: Is there a known *inverse* position operator q-hat^-1? If so, > > > what is it and can you point out a reference(s) in which it is derived > > > and discussed. > > > Jay, when are you going to post an easy problem? > > Ok, I'll give it a shot... > > Allow me to simplify the rotation to, > > > m*[V(y)*x - V(x)*y] = h > > > where V(y) is the mass m Velocity in direction y. > > > Allow me to unitized (for ascii sake) m and q to follow, then, > > sub a positional potential O(x) = (q/r^2) * x for the inverted q-hat, > > (I'll use "O" because positional potentials are measured w.r.t an > > Origin), > > but that - with q (charge) unitized is dimensionally 1/L , > > > O(x) = x/r^2 . > > > Putting that together yields, > > > V(y) O(x) - V(x) O(y) = h/r^2, (h=action). > > > Interpretation's can be found from r=ct. > > Let me know if we should go there. > > Regards > > Ken S. Tucker > > Good job, Ken. The OP asked about A=B and you naturally showed > something related to Q=V. Igor, if there was something immediate Jay the genious would have got it. What I demo'd was an inversion of the Length dimension, in an algebraic way, from an antisymmetrical spin, to arrive at the q-hat^-1 spec. Two ways to intrepretation are as follows, 1) E=hf = h/t = h/ct = m, h/r^2 == m/r (gravitational potential). 2) E/t = Power A quick glance at this UFT article explains the fusion, of (1) and (2), http://physics.trak4.com/MST_UFT.pdf wherein potential and power are related by, dg_00 = - dW_00. Power is something that's not dicussed much, it's like a hidden alien lurking in physics. Regards Ken S. Tucker
From: eric gisse on 28 Feb 2010 17:27 Ken S. Tucker wrote: > On Feb 27, 6:08 am, "Jay R. Yablon" <jyab...(a)nycap.rr.com> wrote: >> My query here is in reference to the position operator q-hat, which >> commutes with the momentum operator p-hat along any single space >> dimension according to the canonical commutation relationship: >> >> [q-hat,p-hat]=i h-bar (1) >> >> Query: Is there a known *inverse* position operator q-hat^-1? If so, >> what is it and can you point out a reference(s) in which it is derived >> and discussed. [snip irrelevancies] Ken, did you notice how he asked specifically about obtaining an inverse position operator? Notice how what you said is completely irrelevant to what he said? Furthermore, have you noticed that sci.physics.relativity is completely irrelevant to your followup? http://en.wikipedia.org/wiki/Dunning?Kruger_effect
From: eric gisse on 28 Feb 2010 17:27 Ken S. Tucker wrote: > On Feb 28, 7:24 am, Igor <thoov...(a)excite.com> wrote: >> On Feb 28, 3:44 am, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote: >> >> >> >> > On Feb 27, 6:08 am, "Jay R. Yablon" <jyab...(a)nycap.rr.com> wrote: >> >> > > My query here is in reference to the position operator q-hat, which >> > > commutes with the momentum operator p-hat along any single space >> > > dimension according to the canonical commutation relationship: >> >> > > [q-hat,p-hat]=i h-bar (1) >> >> > > Query: Is there a known *inverse* position operator q-hat^-1? If so, >> > > what is it and can you point out a reference(s) in which it is >> > > derived and discussed. >> >> > Jay, when are you going to post an easy problem? >> > Ok, I'll give it a shot... >> > Allow me to simplify the rotation to, >> >> > m*[V(y)*x - V(x)*y] = h >> >> > where V(y) is the mass m Velocity in direction y. >> >> > Allow me to unitized (for ascii sake) m and q to follow, then, >> > sub a positional potential O(x) = (q/r^2) * x for the inverted q-hat, >> > (I'll use "O" because positional potentials are measured w.r.t an >> > Origin), >> > but that - with q (charge) unitized is dimensionally 1/L , >> >> > O(x) = x/r^2 . >> >> > Putting that together yields, >> >> > V(y) O(x) - V(x) O(y) = h/r^2, (h=action). >> >> > Interpretation's can be found from r=ct. >> > Let me know if we should go there. >> > Regards >> > Ken S. Tucker >> >> Good job, Ken. The OP asked about A=B and you naturally showed >> something related to Q=V. > > Igor, if there was something immediate Jay the genious would > have got it. > What I demo'd was an inversion of the Length dimension, > in an algebraic way, from an antisymmetrical spin, to > arrive at the q-hat^-1 spec. > > Two ways to intrepretation are as follows, > > 1) E=hf = h/t = h/ct = m, h/r^2 == m/r > (gravitational potential). > > 2) E/t = Power > > A quick glance at this UFT article explains the fusion, > of (1) and (2), > > http://physics.trak4.com/MST_UFT.pdf > > wherein potential and power are related by, > dg_00 = - dW_00. > > Power is something that's not dicussed much, it's like a > hidden alien lurking in physics. > Regards > Ken S. Tucker http://en.wikipedia.org/wiki/Dunning?Kruger_effect
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