Electron’s puzzles.
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From: PD on 12 May 2010 09:22 On May 11, 6:00 pm, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote: > 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). Ah, ok, then in that case we are agreement. > A spin vector may be generated in more than one way, > that's the point, for example the primary Curl. There is no curl that pertains to the quantum mechanical spin. > Regards > Ken S. Tucker
From: Ken S. Tucker on 12 May 2010 14:04 On May 12, 6:22 am, PD <thedraperfam...(a)gmail.com> wrote: > On May 11, 6:00 pm, "Ken S. Tucker" <dynam...(a)vianet.on.ca> wrote: > > > > > 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). > > Ah, ok, then in that case we are agreement. > > > A spin vector may be generated in more than one way, > > that's the point, for example the primary Curl. > > There is no curl that pertains to the quantum mechanical spin. Yes and no, depends on how you use it, this is a type of math that I find awkward, http://en.wikipedia.org/wiki/Stiefel-Whitney_class better to go over to tensors. Ken
From: glird on 12 May 2010 15:11 On May 9, 8:24 pm, waldofj <wald...(a)verizon.net> wrote: glird: >< In his 1905 SR paper Einstein wrote an equation, 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. > W: 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, > 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. True. As I said, Einstein never did define what _a_ denotes. Neither did you. > 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? ALMOST the same way you did when you wrote, "Constant acceleration (one is a constant, right?) means velocity is always increasing at a constant rate." Btw, how do you think that E got from "eta = a{c/sqrt(c^2-v^2)}y" to "eta = phi(v)y"? To be explicit, even if a = phi(v) allowed him to replace a with phi(v), how come c/sqrt(c^2-v^2) disappeared? >< he DELETED the symbol from his equations. > W: What's the point of writing one times something, you do realize multiplying by one doesn't change the expression, right? Do you realize that the only way for c/sqrt(c^2-v^2) to equal 1 is that v = 0? If so, then why bother with transformation equations at all? >< 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 seems 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. > W: 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? Thank you for taking the time to answer me. (I knew that my wording was misleading, and hoped that someone would correct it.) Perhaps it would have been clearer if I'd written it this way: Since eta and zeta are independent of v or any change in it, neither he nor anyone since seems to have realized that the latter requires that da/dv = 0; i.e. that there is NO change in the value of a as v changes. The point is this: If a = phi(v) means that a is a function of v, and if a(v) means the same thing, and if da/dv is another way to express that relation; then if phi(v) = da/dv = 1, and dv = 4 or 3 or 6 then da = 4 or 3 or 6 also. But if there is no change in the relation between eta and y as v changes, as is the case in the LTE, then that relation is NOT a function of v! Btw, E's "proof" that "phi(v) = 1" was defective anyhow. glird
From: glird on 12 May 2010 15:19 On May 9, 8:24 pm, waldofj <wald...(a)verizon.net> wrote: > Or is it you don't know what phi(v) = 0 means? Perhaps I do. Especially wrt E's actual sentence, "a is a function phi(v)" Otoh, how do you write "a is a function phi of the velocity v" in calculus? glird
From: waldofj on 14 May 2010 10:39
On May 12, 3:11 pm, glird <gl...(a)aol.com> wrote: > On May 9, 8:24 pm, waldofj <wald...(a)verizon.net> wrote: > > glird: >< In his 1905 SR paper Einstein wrote an equation, > 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. > > > W: 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, > > > 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. > > True. As I said, Einstein never did define what _a_ denotes. > Neither did you. that's correct. As I said above it turns out that _a_ doesn't denote anything so there's nothing to define. btw we have both been making a mistake when we write a = phi(v). that implies that a is a variable that is being set to the value returned by phi(v). That's not what it is. We should be using the symbol for "is the same as" (I can't reproduce it here with ascii text) instead of equals. In other words, _a_ is just a typographical substitution for phi(v), easier to type _a_ than phi(v). As to what _a_ is (here we go again) _a_ is a mathematical artifact that arises from the method used to derive the equations, nothing more, nothing less. It requires a subsequent analysis of the problem to determine if it denotes anything or not. As it turns out, _a_ doesn't denote anything. > > > 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? > > ALMOST the same way you did when you wrote, "Constant acceleration > (one is a constant, right?) means velocity is always increasing at a > constant rate." that doesn't make any sense. > Btw, how do you think that E got from > "eta = a{c/sqrt(c^2-v^2)}y" to "eta = phi(v)y"? > To be explicit, even if a = phi(v) allowed him to replace a with > phi(v), how come c/sqrt(c^2-v^2) disappeared? good question. I have commented on this before but I don't mind repeating myself. "I ain't proud, or tired" first of all c/sqrt(c^2-v^2) is just beta so I'll just write beta if you don't mind. (actually nowadays it's called gamma but I'll stick to the old terminology for this discussion). If you look at the equations on page 45 in the dover publication and then look at the equations at the top of page 46 you will see that beta has been divided out of all four equations. This step has been skipped and no explanation is given for it. That's why I say a page is missing from the dover publication. The short answer: this is done to make the equations compatible with the principle of relativity while maintaining compatibility with the constancy of the speed of light. If you want the long answer, ask me in a different thread. > > >< he DELETED the symbol from his equations. > > > W: What's the point of writing one times something, you do realize > multiplying by one doesn't change the expression, right? > > Do you realize that the only way for c/sqrt(c^2-v^2) to equal 1 is > that v = 0? If so, then why bother with transformation equations at > all? who says beta should equal one? > > >< 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 seems 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. > > > W: 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? > > Thank you for taking the time to answer me. (I knew that my wording > was misleading, and hoped that someone would correct it.) Perhaps it > would have been clearer if I'd written it this way: > Since eta and zeta are independent of v or any change in it, neither > he nor anyone since seems to have realized that the latter requires > that da/dv = 0; i.e. that there is NO change in the value of a as v > changes. > > The point is this: If a = phi(v) means that a is a function of v, it does > and if a(v) means the same thing, it does > and if da/dv is another way to > express that relation; stop right there, it most definatly is not > then if phi(v) = da/dv = 1, and dv = 4 or 3 or > 6 then da = 4 or 3 or 6 also. But if there is no change in the > relation between eta and y as v changes, as is the case in the LTE, > then that relation is NOT a function of v! phi(v) = 1 means phi(v), or _a_ , or a(v) (all the same thing) is a constant function such as: (0 * V) + 1 no matter what value I put in for v it always returns 1 you said above: > Since eta and zeta are independent of v or any change in it, neither > he nor anyone since seems to have realized that the latter requires > that da/dv = 0; i.e. that there is NO change in the value of a as v > changes. of course everyone realized that, that's the behavior of a constant function > > Btw, E's "proof" that "phi(v) = 1" was defective anyhow. no, only your assumptions about it are defective. > > glird |