Electron’s puzzles.
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From: Ken S. Tucker on 8 May 2010 14:37 On May 8, 10:18 am, PD <thedraperfam...(a)gmail.com> wrote: > On May 8, 2:48 am, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote: > > > The spin for a given quantum state is usually represented by a vector, > > > where the components are the expectation values of the spin operator > > > -- not a tensor. > > Not quite, as far as I know, that has yet to be determined, >>Ken > I don't think so. > Yes, I know tensors, Ken, but spin is not a tensor. > > > I understand why that is, but you'll need to know tensors > > for me to explain it. > > > > You asked for a concept, not a mathematical representation. Some of > > > the references I gave you in another post in this thread introduce the > > > mathematical representation. > > > I guess you're ok with this, as you suggested,http://en.wikipedia.org/wiki/Spin_%28physics%29 > > Sub section,http://en.wikipedia.org/wiki/Spin_%28physics%29#Spin_and_rotations > > Paragraph, > > "Further, rotations preserve quantum mechanical inner product, > > and so should our transformation matrices:" > > You somewhat vaguely waved that ref to us, is that what you mean? > > Let's get some precion of thought, it's tough to work with mushy > > minded stuff. > > Do you need a reference to a book, Ken? Well Paul, I'm not sure we can simplify spin to your level of understanding, but if you're serious we can go to "spin connections" in Weinberg's QFT Vol.3, Eq.(31.5.17). The math is a bit detailed, but the concept is straight- forward, but my impression is you (Paul) enjoy science on a superficial level, so that's why I avoid math with you ... you post for some relaxation, fine with me. Regards Ken S. Tucker
From: PD on 8 May 2010 15:28 On May 8, 1:37 pm, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote: > On May 8, 10:18 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > On May 8, 2:48 am, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote: > > > > The spin for a given quantum state is usually represented by a vector, > > > > where the components are the expectation values of the spin operator > > > > -- not a tensor. > > > Not quite, as far as I know, that has yet to be determined, > >>Ken > > I don't think so. > > Yes, I know tensors, Ken, but spin is not a tensor. > > > > I understand why that is, but you'll need to know tensors > > > for me to explain it. > > > > > You asked for a concept, not a mathematical representation. Some of > > > > the references I gave you in another post in this thread introduce the > > > > mathematical representation. > > > > I guess you're ok with this, as you suggested,http://en.wikipedia.org/wiki/Spin_%28physics%29 > > > Sub section,http://en.wikipedia.org/wiki/Spin_%28physics%29#Spin_and_rotations > > > Paragraph, > > > "Further, rotations preserve quantum mechanical inner product, > > > and so should our transformation matrices:" > > > You somewhat vaguely waved that ref to us, is that what you mean? > > > Let's get some precion of thought, it's tough to work with mushy > > > minded stuff. > > > Do you need a reference to a book, Ken? > > Well Paul, I'm not sure we can simplify spin to your > level of understanding, but if you're serious we can > go to "spin connections" in Weinberg's QFT Vol.3, > Eq.(31.5.17). Indeed. A spin connection is not a spin vector. A spin connection is a connection in a spinor bundle. It helps to know what the words mean, Ken. > The math is a bit detailed, but the concept is straight- > forward, but my impression is you (Paul) enjoy science > on a superficial level, so that's why I avoid math with > you ... you post for some relaxation, fine with me. > Regards > Ken S. Tucker
From: glird on 9 May 2010 17:58 On May 8, 3:28 pm, PD <thedraperfam...(a)gmail.com> wrote: > On May 8, 1:37 pm, "Ken S. Tucker" wrote: > > On May 8, 10:18 am, PD wrote: > > > On May 8, 2:48 am, "Ken S. Tucker" wrote: ><<< The spin for a given quantum state is usually represented by a vector, where the components are the expectation values of the spin operator -- not a tensor. > > ><< Not quite, as far as I know, that has yet to be determined, Ken >< I don't think so. Yes, I know tensors, Ken, but spin is not a tensor.> snip >< Well Paul, I'm not sure we can simplify spin to your level of understanding, but if you're serious we can go to "spin connections" in Weinberg's QFT Vol.3, Eq.(31.5.17). > > > Indeed. A spin connection is not a spin vector. > A spin connection is a connection in a spinor bundle. > It helps to know what the words mean, Ken. > > > The math is a bit detailed, but the concept is straight-forward, but my impression is you (Paul) enjoy science on a superficial level, so that's why I avoid math with you ... you post for some relaxation, fine with me. Regards Ken S. Tucker > I find both of you reasonable and well informed. Both of you seem logically AND mathematically adept. Even so, PD is right in saying "It helps to know what the words mean". Until you both know exactly what the following words mean, you can't understand your own subjects: mass, energy, quantum state, renormalization, expectation value, probability, photon, point-sized, etc etc. Here is an example; In his 1905 SR paper Einstein wrote an equation, tau = a(), in which he said that a is a function of v; which he expressed as phi(v). A bit later he said that phi(v) = 1. He never did say what "a" dnotes all by itself. If we let a denote acceleration, then a = phi(v) = 1 denotes acceleration = d(dx/dt) = 1. That means that for any given change in the value of v, the value of _a_ changes by an identical amount. However, after writing eta = phi(v)y and finding that phi(v) = 1, instead of realizing that this means that the value of phi(v)y, thus of eta, CHANGES as v does; he DELETED the symbol from his equations. Although that did leave him with the LTE, in which eta and zeta are independent of v or any change in it, neither he nor anyone since seeems to have realized that the latter requires that phi(v) = 0; i.e. that there is NO change in the value of a = phi(v) as v changes. Regards to both of you, glird
From: waldofj on 9 May 2010 20:24 > Here is an example; In his 1905 SR paper Einstein wrote an equation, > tau = a(), actually tau a(t - vx'/(c^2 - v^2) > in which he said that a is a function of v; which he > expressed as phi(v). A bit later he said that phi(v) = 1. He never did > say what "a" dnotes all by itself. He never did because he showed it doesn't denote anything. It's just a constant function of v which always equals one. > If we let a denote acceleration, a is a function of v at present unknown. You can't just decide to give an arbitrary meaning and then start drawing conclusions. Well, you can but you'll just end up with nonsense. Garbage in garbage out. > then > a = phi(v) = 1 > denotes > acceleration = d(dx/dt) = 1. actually acceleration = (d/dt)(dx/dt) or (d^2/dt^2)x or, since we are talking physics here, dx/dt = v so acceleration = dv/dt of course I realize none of this means anything to you since you refuse to learn calculus. > That means that for any given change in the value of v, the value of > _a_ changes by an identical amount. now you don't know basic physics and you're contradicting yourself. You have set a = phi(v) = 1. One is a constant, constants don't change (didn't you learn that in algebra class?) Constant acceleration (one is a constant, right?) means velocity is always increasing at a constant rate. However that last point is moot since _a_ is not acceleration. > However, after writing > eta = phi(v)y > and finding that phi(v) = 1, instead of realizing that this means that > the value of phi(v)y, thus of eta, CHANGES > as v does; How can you say phi(v) is constant (one is a constant, right?) and then say it changes? > he DELETED the symbol from his equations. what's the point of writing one times something, you do realize multiplying by one doesn't change the expression, right? > Although that did leave him with the LTE, in which eta and zeta are > independent of v or any change in it, neither he nor anyone since > seeems to have realized that the latter requires that > phi(v) = 0; > i.e. that there is NO change in the value of a = phi(v) as v changes. Well, zero is also a constant but phi(v) is being MULTIPLIED in each expression. If you multiply by zero you'll just set everything to zero, not what you want. Or is it you don't know what phi(v) = 0 means?
From: Ken S. Tucker on 11 May 2010 19:00
On May 8, 12:28 pm, PD <thedraperfam...(a)gmail.com> wrote: > On May 8, 1:37 pm, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote: > > > > > On May 8, 10:18 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > On May 8, 2:48 am, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote: > > > > > The spin for a given quantum state is usually represented by a vector, > > > > > where the components are the expectation values of the spin operator > > > > > -- not a tensor. > > > > Not quite, as far as I know, that has yet to be determined, > > >>Ken > > > I don't think so. > > > Yes, I know tensors, Ken, but spin is not a tensor. > > > > > I understand why that is, but you'll need to know tensors > > > > for me to explain it. > > > > > > You asked for a concept, not a mathematical representation. Some of > > > > > the references I gave you in another post in this thread introduce the > > > > > mathematical representation. > > > > > I guess you're ok with this, as you suggested,http://en.wikipedia.org/wiki/Spin_%28physics%29 > > > > Sub section,http://en.wikipedia.org/wiki/Spin_%28physics%29#Spin_and_rotations > > > > Paragraph, > > > > "Further, rotations preserve quantum mechanical inner product, > > > > and so should our transformation matrices:" > > > > You somewhat vaguely waved that ref to us, is that what you mean? > > > > Let's get some precion of thought, it's tough to work with mushy > > > > minded stuff. > > > > Do you need a reference to a book, Ken? > > > Well Paul, I'm not sure we can simplify spin to your > > level of understanding, but if you're serious we can > > go to "spin connections" in Weinberg's QFT Vol.3, > > Eq.(31.5.17). > > Indeed. A spin connection is not a spin vector. > A spin connection is a connection in a spinor bundle. > It helps to know what the words mean, Ken. Yes, beware that a vector is a tensor. (rank 1). A spin vector may be generated in more than one way, that's the point, for example the primary Curl. Regards Ken S. Tucker |