June 2014 - mathfun@mailman.xmission.com

[math-fun] 17 Tesla magnetic field "trapped" in ball of type-II superconductor
by Warren D Smith 27 Jun '14

27 Jun '14

Amazingly enough (since we were just speaking about this) this paper just appeared http://iopscience.iop.org/0953-2048/27/8/082001/pdf/0953-2048_27_8_082001.p… "A trapped ﬁeld of 17.6T in melt-processed, bulk Gd-Ba-Cu-O reinforced with shrink-ﬁt steel" Superconductor Science and Technology 27 (2014) 082001 (5 pages). They have a golf-ball sized ball of GdBa2Cu3O7 superconductor surrounded by a cylinder of shrunk-on stainless steel for strength. They placed it, while warm, in the bore of an 18T electromagnet, cooled it to 26 kelvin, then gradually turned off the external field over about 1 day long turn-off time. Result: 17.6 Tesla of "trapped field." If warmed up to 50 kelvin, it still trapped 10 Tesla. If the "pinning" of the "flux tubes" by crystal defects had been perfect then this trapped field would have lasted forever at 26 kelvin. However, in fact, it gradually leaked away, falling to 17.4 Tesla after about 200 minutes. This is because the flux tubes were occasionally able to move because the pinning was not infinitely strong, hence migrated out of the superconductor and were lost. If a neutron star were a type-I superconductor, no magnetic field would be possible inside. If type-II then you could have flux tubes passing thru like in this experiment. However, I would expect, due to the claimed superfluid nature of neutron star stuff (i.e. not a crystal with defects), that the flux tubes would be mobile, in which case they, like in this experiment, would be expelled. However, I would guess the time scale for the expulsion would be seconds at most, not (as here) about 1 week. But this is just a guess. If the neutron star really were a crystal with defects, then the crucial question would be the pseudo-energy barrier that a flux tube would have to surmount to move from one path-of-defects to another, and whether that barrier was much larger than the thermal energy. (It perhaps also could quantum tunnel thru the barrier, an option which is of negligible importance for lab-accessible superconductors, but perhaps would not be negligible for neutron star stuff.) Trying to understand that purely by theory without experiment, seems to me very difficult and I would not believe anybody who claimed (as of year 2014) to be able to do it.

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Re: [math-fun] neutron star (electric field?)
by Bill Gosper 27 Jun '14

27 Jun '14

Would someone please explain to us GR and QM noobs why the concentric fields don't cancel by Gauss's law, like in the classical case? --rwg Richard E. Howard <rich(a)richardehoward.com>>Sounds like you are making a buoyancy argument that protons will sink and electrons float in a neutron star. I think that is a very complex argument, made even more so by the extreme gravity and quantum effects (they are both fermions with all that implies.) I suspect that saying protons are heavy and electrons light is just scratching the surface. Remember that collapsing P+e => n is reducing the number of particles in half and, thus, reducing the gravitational energy (they are fermions). A core collapse during a supernova is driven by the same energy considerations--fusion of nuclei that allows the diameter to be reduced (rather quickly) under gravitational force. In a neutron star, that collapse has happened with p+e => n already. Next step is when the gravitational force exceeds the exclusion principle energy and the neutrons combine--taking it all the way to a black hole... --R -----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Warren D Smith Sent: Thursday, June 26, 2014 11:29 AM To: math-fun(a)mailman.xmission.com Subject: [math-fun] neutron star (electric field?) oward: As has been mentioned in this thread, at the pressures in a neutron star a neutron is stable relative to a proton + electron. Electrostatic force pulls e + p => n together long before they can separate gravitationally. --R --also, this is refuted thus: the binding energy of electron & proton is 13.6 electron volts, which is tiny versus the thermal energies of order 10^9 celsius, far far more than enough to separate an e and p by infinite distance. --so I conclude there DEFINITELY will be large e & p "separation" both possible, and expected to, happen. So the "electrostatic force" Howard here is worried about, is basically neglectible on the microscale; the only effects worth discussing are macroscopic collaborative effects, which indeed is what I was speaking about, yielding a macroscale electric field.

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[math-fun] neutron star (electric field?)
by Warren D Smith 26 Jun '14

26 Jun '14

>oward: > As has been mentioned in this thread, at the pressures in a neutron star > a > neutron is stable relative to a proton + electron. > Electrostatic force pulls e + p => n together long before they can > separate gravitationally. > --R --also, this is refuted thus: the binding energy of electron & proton is 13.6 electron volts, which is tiny versus the thermal energies of order 10^9 celsius, far far more than enough to separate an e and p by infinite distance. --so I conclude there DEFINITELY will be large e & p "separation" both possible, and expected to, happen. So the "electrostatic force" Howard here is worried about, is basically neglectible on the microscale; the only effects worth discussing are macroscopic collaborative effects, which indeed is what I was speaking about, yielding a macroscale electric field.

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Re: [math-fun] Robert Smith's vector problem...
by Henry Baker 26 Jun '14

26 Jun '14

I didn't understand your "#" operation. Could you please try again? At 10:46 AM 6/26/2014, Robert Smith wrote: >Thanks for the info. > >I agree that a quaternion multiply is not sufficient, unless you >define quaternion-vector "multiplication" by the operator # above. > >Robert > >On Thu, Jun 26, 2014 at 6:07 AM, Henry Baker <hbaker1(a)pipeline.com> wrote: >> I did provide you with a single quaternion s.t. >> >> Q A (1/Q) = A' >> Q B (1/Q) = B' >> >> However, one normally (pun intended!) does the >> computation like this: >> >> (Q A conjugate(Q))/N(Q) = A' >> (Q B conjugate(Q))/N(Q) = B' >> >> where N(Q) = Q conjugate(Q) >> >> because it is cheaper/easier/more precise to >> normalize at the end. >> >> If you compute the _unit_ quaternion >> Q/sqrt(N(Q)), then in this case, >> >> (1/Q) = conjugate(Q) >> >> There is no way to do the rotation with only >> one quaternion multiply, because each quaternion >> multiply only computes a _reflection_, and you >> need 2 reflections to make a rotation. >> >> See Coxeter's wonderful paper about quaternion >> rotations & reflections here: >> >> www.math.utah.edu/~ptrapa/math-library/coxeter/Coxeter-Quaternions-and-refl… >> >> Coxeter shows that quaternions can even do >> all the rotations in 4D, but there you need >> a different quaternion for the pre- and post- >> multiplication in order to get enough degrees >> of freedom. >> >> At 12:13 AM 6/26/2014, Robert Smith wrote: >>>Last one for the night. (It's midnight here.) >>> >>>I got the above math to work numerically as well, but now I want to >>>see if it's possible to boil that down into a single quaternion Q, as >>>opposed to a function, so that for >>> >>> Q # v = Q * v * conj(Q), >>> >>>we have the simultaneously satisfied >>> >>> Q # A = A' >>> Q # B = B'. >>> >>>I am introducing all of these symbols (#, ., *, implicit) just to be >>>absolutely clear. I guess it's the programmer in me. >>> >>>Do you think there's a nice expression for this? I'm betting on "no", >>>but you've surprised me above. >>> >>>Robert >>> >>>On Wed, Jun 25, 2014 at 11:20 PM, Robert Smith <quad(a)symbo1ics.com> wrote: >>>> Sorry for the noise. I also meant >>>> >>>> Let R = (A' - A) x (B' - B) >>>> Let U = R x A >>>> Let U' = R x A' >>>> >>>> Robert >>>> >>>> On Wed, Jun 25, 2014 at 10:22 PM, Robert Smith <quad(a)symbo1ics.com> wrote: >>>>> Sorry, I guess what I said doesn't quite make sense in the last bit. >>>>> When I said Q * A, I meant more rotate(Q, A) = A', where >>>>> >>>>> rotate(Q, A) = Q * A * conj(Q). >>>>> >>>>> Robert >>>>> >>>>> On Wed, Jun 25, 2014 at 6:06 PM, Robert Smith <quad(a)symbo1ics.com> wrote: >>>>>> Henry: >>>>>> >>>>>> You've provided a function which maps a quaternion >>>>>> >>>>>> v = r + xi + yj + zk >>>>>> >>>>>> to a new quaternion >>>>>> >>>>>> v' = r' + x'i + y'j + z'k. >>>>>> >>>>>> If I followed along correctly, the expression can be rewritten as follows: >>>>>> >>>>>> A -> A' >>>>>> B -> B' >>>>>> >>>>>> Let S = (A' - A) x (B' - B) >>>>>> Let U = R x A >>>>>> Let U = R x A' >>>>>> >>>>>> Then the mapping sought is >>>>>> >>>>>> v -> >>>>>> (|U|^2 - U' * U) * v * (|U|^2 - U * U') >>>>>> --------------------------------------- >>>>>> 2(U . U)(U . (U + U')) >>>>>> >>>>>> which is equal to >>>>>> >>>>>> v -> >>>>>> (|U|^2 - U' * U) * v * (|U|^2 - U * U') >>>>>> --------------------------------------- >>>>>> 2 |U|^2 (|U|^2 + U . U') >>>>>> >>>>>> where implicit multiplication is scalar-by-scalar, the dot is vector >>>>>> dot product (which in all these cases, if U is interpreted as a >>>>>> quaternion, then it has a realpart of 0), and the asterisk * is >>>>>> quaternion multiply associating left to right. >>>>>> >>>>>> If the above interpretation is right, how can we recover a quaternion >>>>>> Q such that >>>>>> >>>>>> Q * A = A' >>>>>> Q * B = B' >>>>>> >>>>>> where a vector [x y z] is the quaternion xi + yj + ck? Just use v = 1? >>>>>> >>>>>> I haven't verified this correct yet because I'm trying to ensure all >>>>>> the types check out. :) >>>>>> >>>>>> Cheers, >>>>>> >>>>>> Robert >>>>>> >>>>>> On Wed, Jun 25, 2014 at 3:50 PM, Robert Smith <quad(a)symbo1ics.com> wrote: >>>>>>> Fantastic. I didn't expect it to be as nice as that, for certain! >>>>>>> >>>>>>> Robert >>>>>>> >>>>>>> On Fri, Jun 13, 2014 at 9:54 AM, Henry Baker <hbaker1(a)pipeline.com> wrote: >>>>>>>> I have confirmed via a number of numerical experiments that >>>>>>>> >>>>>>>> v->(|SxA|^2-(SxA')(SxA))v(|SxA|^2-(SxA)(SxA'))/Nq >>>>>>>> >>>>>>>> where S=(A'-A)x(B'-B) >>>>>>>> >>>>>>>> does indeed transform A->A' and B->B'. >>>>>>>> >>>>>>>> Here's a slightly simpler form for >>>>>>>> >>>>>>>> Nq = 2*|SxA|^2*(|SxA|^2+(SxA).(SxA')) >>>>>>>> = 2*[(SxA).(SxA)]*[(SxA).(Sx(A+A'))] >>>>>>>> >>>>>>>> At 09:24 PM 6/10/2014, Henry Baker wrote: >>>>>>>>>Another solution based on vectors & quaternions w/o matrices. >>>>>>>>> >>>>>>>>>If A,B are unit vectors (|A|=|B|=1), then rotating them >>>>>>>>>will preserve length (|A'|=|B'|=1). >>>>>>>>> >>>>>>>>>We need to warm up a bit before solving the full problem; >>>>>>>>>we first consider mapping A->A' without worrying about B. >>>>>>>>> >>>>>>>>>The most straightforward rotation that maps A->A' is >>>>>>>>>the quaternion q=sqrt((-A')A). Here (-A')A is >>>>>>>>>the quaternion product of the quaternion inverse of A' >>>>>>>>>and A. (For unit vector quaternions like A', (A')^-1=-A'.) >>>>>>>>>We use the standard mapping of vectors into >>>>>>>>>quaternions: A -> 0+A; i.e., the scalar part of (0+A) >>>>>>>>>is 0, and the vector part of (0+A) is A itself. >>>>>>>>> >>>>>>>>>The mapping >>>>>>>>> >>>>>>>>>v -> q v (q)^-1 = (q v q*)/(qq*) >>>>>>>>> >>>>>>>>>will transform A->A'. q* is conjugate(q). >>>>>>>>> >>>>>>>>>(The second form is preferred, since we don't have to >>>>>>>>>normalize q, and we don't have to compute absolute values.) >>>>>>>>> >>>>>>>>>The "sqrt()" above is the _quaternion_ square root, which >>>>>>>>>can be computed: >>>>>>>>> >>>>>>>>>sqrt(q) = (|q|+q)/sqrt(2*(|q|+S(q))) >>>>>>>>> >>>>>>>>>(Note that the denominator is a real scalar quantity.) >>>>>>>>> >>>>>>>>>Here S(q) is the "scalar part" of q, or >>>>>>>>> >>>>>>>>>S(q) = (q+q*)/2 >>>>>>>>> >>>>>>>>>But since we don't need a _normalized_ q, we can use a >>>>>>>>>cheap sqrt: >>>>>>>>> >>>>>>>>>elcheaposqrt(q) = |q|+q (!) >>>>>>>>> >>>>>>>>>Proof: >>>>>>>>> >>>>>>>>>(|q|+q)^2 = |q|^2+2|q|q+q^2 >>>>>>>>> = |q|^2+2|q|q+(Sq+Vq)^2 >>>>>>>>> = |q|^2+2|q|q+Sq^2+2SqVq+Vq^2 >>>>>>>>> = |q|^2+2|q|q+Sq^2+2SqVq-|Vq|^2 >>>>>>>>> = |q|^2+2|q|q+Sq^2+2SqVq-|Vq|^2-Sq^2+Sq^2 >>>>>>>>> = |q|^2+2|q|q+2Sq^2+2SqVq-|q|^2 >>>>>>>>> = 2|q|q+2Sq^2+2SqVq >>>>>>>>> = 2|q|q+2Sq(Sq+Vq) >>>>>>>>> = 2|q|q+2Sq q >>>>>>>>> = 2(|q|+Sq) q >>>>>>>>> >>>>>>>>>QED >>>>>>>>> >>>>>>>>>In particular, the function that takes A->A' is >>>>>>>>> >>>>>>>>>v -> (1-A'A)v(1-AA')/2/(1+A.A') >>>>>>>>> >>>>>>>>>(Here A.A' is the dot product of A & A'.) >>>>>>>>> >>>>>>>>>We now attack your actual question: you want a >>>>>>>>>rotation that _simultaneously_ transports >>>>>>>>>A->A' and B->B'. >>>>>>>>> >>>>>>>>>As Dan pointed out, the axis of rotation that we >>>>>>>>>want is (A'-A)x(B'-B), but we can't use (A'-A).(B'-B) >>>>>>>>>as the scalar part of a quaternion, because that >>>>>>>>>would rotate the projection of A into the projection >>>>>>>>>of B (or equivalently, the projection of A' into the >>>>>>>>>projection of B') -- not what we want. >>>>>>>>> >>>>>>>>>Let S=(A'-A)x(B'-B) be our axiS. If we are careful, >>>>>>>>>we won't have to normalize S into a unit vector. We >>>>>>>>>also no longer need B,B' since we only needed B,B' >>>>>>>>>to establish the axis S. >>>>>>>>> >>>>>>>>>If we compute SxA and SxA', we note that they are in >>>>>>>>>a plane orthogonal to S, and during the rotation that >>>>>>>>>maps A->A' and B->B', we will also map SxA into SxA'. >>>>>>>>>So the axis of this rotation is S itself. Also, >>>>>>>>>obviously, |SxA|=|SxA'|. >>>>>>>>> >>>>>>>>>So, we can now utilize the method of our warmup >>>>>>>>>exercise to map SxA into SxA' via the mapping: >>>>>>>>> >>>>>>>>>v->(|SxA|^2-(SxA')(SxA))v(|SxA|^2-(SxA)(SxA'))/Nq >>>>>>>>> >>>>>>>>>and Nq is the "absolute value squared" of >>>>>>>>>(|SxA|^2-(SxA')(SxA)) or of (|SxA|^2-(SxA)(SxA')). >>>>>>>>> >>>>>>>>>This is the mapping that we seek that will >>>>>>>>>simultaneously map A->A' and B->B'. >>>>>>>>> >>>>>>>>>Thus, although we talked about square roots and >>>>>>>>>absolute values in our development, we find that >>>>>>>>>the final result requires no square roots or >>>>>>>>>absolute values. >>>>>>>>> >>>>>>>>>I'd be surprised if there is a simpler expression >>>>>>>>>for computing this quaternion. >>>>>>>>> >>>>>>>>>At 04:03 PM 6/9/2014, Henry Baker wrote: >>>>>>>>>>We can take our matrix product below, which is >>>>>>>>>>an orthogonal matrix. Let's call it M. >>>>>>>>>> >>>>>>>>>>Then we can easily recover the quaternion from >>>>>>>>>>the orthogonal matrix M using the inverse _Cayley >>>>>>>>>>Transform_: >>>>>>>>>> >>>>>>>>>>S = (I-M)(I+M)^-1 >>>>>>>>>> >>>>>>>>>>The matrix I+M is invertible, because we have _rigid_ >>>>>>>>>>rotations (i.e., no reflections, hence no eigenvalues >>>>>>>>>>of -1). >>>>>>>>>> >>>>>>>>>>S is a _skew-symmetric_ matrix: >>>>>>>>>> >>>>>>>>>>[ 0 z -y] >>>>>>>>>>[-z 0 x] >>>>>>>>>>[ y -x 0] = S >>>>>>>>>> >>>>>>>>>>Then our (non-normalized) quaternion is [1,x,y,z]. >>>>>>>>>> >>>>>>>>>>http://en.wikipedia.org/wiki/Cayley_transform >>>>>>>>>> >>>>>>>>>>At 02:06 PM 6/7/2014, Henry Baker wrote: >>>>>>>>>>>We can easily do it with vectors & matrices: >>>>>>>>>>> >>>>>>>>>>>We are given A,B,A',B', where |A|=|B|=|A'|=|B'|. >>>>>>>>>>>R(A)=A', R(B)=B'. >>>>>>>>>>> >>>>>>>>>>>Because the rotation R is rigid, we know that >>>>>>>>>>> >>>>>>>>>>>R(AxB) = A'xB' = R(A)xR(B) >>>>>>>>>>> >>>>>>>>>>>Assuming that B is not a multiple of A, we can >>>>>>>>>>>form the 3x3 matrices R=matrix(A,B,AxB) and >>>>>>>>>>>R'=matrix(A',B',A'xB'). >>>>>>>>>>> >>>>>>>>>>>Our answer is then the matrix product >>>>>>>>>>> >>>>>>>>>>>R^-1 R' = >>>>>>>>>>> >>>>>>>>>>>matrix(A,B,AxB)^-1 matrix(A',B',A'xB') >>>>>>>>>>> >>>>>>>>>>>You are now free to convert this _orthogonal_ >>>>>>>>>>>matrix into a quaternion at your leisure. >>>>>>>>>>> >>>>>>>>>>>(Although AxB, A'xB' are not unit vectors, >>>>>>>>>>>hence R,R' are not orthogonal matrices, >>>>>>>>>>>nevertheless, R^-1 R' _is_ orthogonal.) >>>>>>>>>>> >>>>>>>>>>>(Contrary to Knuth, I did actually try this >>>>>>>>>>>on some numbers & it worked!) >>>>>>>>>>> >>>>>>>>>>>At 04:25 AM 6/7/2014, Dan Asimov wrote: >>>>>>>>>>>>In almost all cases, V := (A-A')x(B-B') will be a vector in the direction of the axis of rotation. Knowing V makes it easy to project say A and A' onto the perpendicular plane to C to determine the angle T. >>>>>>>>>>>> >>>>>>>>>>>>The remaining cases should be easy to exclude or deal with. >>>>>>>>>>>> >>>>>>>>>>>>--Dan >>>>>>>>>>>> >>>>>>>>>>>>On Jun 7, 2014, at 1:32 AM, Robert Smith <quad(a)symbo1ics.com> wrote: >>>>>>>>>>>> >>>>>>>>>>>>> Let A and B be unit vectors in R^3. Suppose they are rotated about >>>>>>>>>>>>> some vector V by an angle T, resulting in A' and B' respectively. What >>>>>>>>>>>>> are V and T? >>>>>>>>>>>>> >>>>>>>>>>>>> I set up a quadratic system using quaternions and got a result that >>>>>>>>>>>>> was 3 million terms large. Am I missing something?

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Re: [math-fun] Robert Smith's vector problem...
by Henry Baker 26 Jun '14

26 Jun '14

Smith's original question involved finding the unit quaternion q s.t. qAq*=A', qBq*=B', and |A|=|A'|=|B|=|B'|=1. The previous postings didn't fully explore the space of all quaternions which could rotate A->A'. The _smallest angle_ rotation is about the axis AxA', but these are not the only axes that rotate A->A'. Consider all q's s.t. qAq*=A'. But we can utilize any axis for q that lies in the plane bisecting the angle AOA', where O is the origin. Since |A|=|A'|, this bisecting plane is spanned by the vectors (A+A') and cross(A,A'). Indeed, calculations show us that for any real m,n, the quaternion q = m(A+A') + n(A.(A+A') + AxA') will rotate A->A'. Similarly, the quaternion q' = m'(B+B') + n'(B.(B+B') + BxB') rotates B->B'. Thus, we need to find the intersection of these two planes, one of which bisects AOA' and the other of which bisects BOB'; this will be our desired axis. The plane bisecting AOA' consists of all those vectors which are perpendicular to A'-A, while the lane bisecting BOB' consists of all those vectors which are perpendicular to B'-B. The only vectors perpendicular to both are multiples of (A'-A)x(B'-B), so this is the axis that we desire. Refer to the previous postings for the actual quaternion expressions for the desired quaternion. BTW, I finally broke down & wrote a quaternion expression simplifier for Maxima that enables me to check my results much more efficiently. I was surprised that no one else had apparently written such a simplifier, even though Maxima/Macsyma has been around for 40+ years.

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Re: [math-fun] Robert Smith's vector problem...
by Henry Baker 26 Jun '14

26 Jun '14

I did provide you with a single quaternion s.t. Q A (1/Q) = A' Q B (1/Q) = B' However, one normally (pun intended!) does the computation like this: (Q A conjugate(Q))/N(Q) = A' (Q B conjugate(Q))/N(Q) = B' where N(Q) = Q conjugate(Q) because it is cheaper/easier/more precise to normalize at the end. If you compute the _unit_ quaternion Q/sqrt(N(Q)), then in this case, (1/Q) = conjugate(Q) There is no way to do the rotation with only one quaternion multiply, because each quaternion multiply only computes a _reflection_, and you need 2 reflections to make a rotation. See Coxeter's wonderful paper about quaternion rotations & reflections here: www.math.utah.edu/~ptrapa/math-library/coxeter/Coxeter-Quaternions-and-refl… Coxeter shows that quaternions can even do all the rotations in 4D, but there you need a different quaternion for the pre- and post- multiplication in order to get enough degrees of freedom. At 12:13 AM 6/26/2014, Robert Smith wrote: >Last one for the night. (It's midnight here.) > >I got the above math to work numerically as well, but now I want to >see if it's possible to boil that down into a single quaternion Q, as >opposed to a function, so that for > > Q # v = Q * v * conj(Q), > >we have the simultaneously satisfied > > Q # A = A' > Q # B = B'. > >I am introducing all of these symbols (#, ., *, implicit) just to be >absolutely clear. I guess it's the programmer in me. > >Do you think there's a nice expression for this? I'm betting on "no", >but you've surprised me above. > >Robert > >On Wed, Jun 25, 2014 at 11:20 PM, Robert Smith <quad(a)symbo1ics.com> wrote: >> Sorry for the noise. I also meant >> >> Let R = (A' - A) x (B' - B) >> Let U = R x A >> Let U' = R x A' >> >> Robert >> >> On Wed, Jun 25, 2014 at 10:22 PM, Robert Smith <quad(a)symbo1ics.com> wrote: >>> Sorry, I guess what I said doesn't quite make sense in the last bit. >>> When I said Q * A, I meant more rotate(Q, A) = A', where >>> >>> rotate(Q, A) = Q * A * conj(Q). >>> >>> Robert >>> >>> On Wed, Jun 25, 2014 at 6:06 PM, Robert Smith <quad(a)symbo1ics.com> wrote: >>>> Henry: >>>> >>>> You've provided a function which maps a quaternion >>>> >>>> v = r + xi + yj + zk >>>> >>>> to a new quaternion >>>> >>>> v' = r' + x'i + y'j + z'k. >>>> >>>> If I followed along correctly, the expression can be rewritten as follows: >>>> >>>> A -> A' >>>> B -> B' >>>> >>>> Let S = (A' - A) x (B' - B) >>>> Let U = R x A >>>> Let U = R x A' >>>> >>>> Then the mapping sought is >>>> >>>> v -> >>>> (|U|^2 - U' * U) * v * (|U|^2 - U * U') >>>> --------------------------------------- >>>> 2(U . U)(U . (U + U')) >>>> >>>> which is equal to >>>> >>>> v -> >>>> (|U|^2 - U' * U) * v * (|U|^2 - U * U') >>>> --------------------------------------- >>>> 2 |U|^2 (|U|^2 + U . U') >>>> >>>> where implicit multiplication is scalar-by-scalar, the dot is vector >>>> dot product (which in all these cases, if U is interpreted as a >>>> quaternion, then it has a realpart of 0), and the asterisk * is >>>> quaternion multiply associating left to right. >>>> >>>> If the above interpretation is right, how can we recover a quaternion >>>> Q such that >>>> >>>> Q * A = A' >>>> Q * B = B' >>>> >>>> where a vector [x y z] is the quaternion xi + yj + ck? Just use v = 1? >>>> >>>> I haven't verified this correct yet because I'm trying to ensure all >>>> the types check out. :) >>>> >>>> Cheers, >>>> >>>> Robert >>>> >>>> On Wed, Jun 25, 2014 at 3:50 PM, Robert Smith <quad(a)symbo1ics.com> wrote: >>>>> Fantastic. I didn't expect it to be as nice as that, for certain! >>>>> >>>>> Robert >>>>> >>>>> On Fri, Jun 13, 2014 at 9:54 AM, Henry Baker <hbaker1(a)pipeline.com> wrote: >>>>>> I have confirmed via a number of numerical experiments that >>>>>> >>>>>> v->(|SxA|^2-(SxA')(SxA))v(|SxA|^2-(SxA)(SxA'))/Nq >>>>>> >>>>>> where S=(A'-A)x(B'-B) >>>>>> >>>>>> does indeed transform A->A' and B->B'. >>>>>> >>>>>> Here's a slightly simpler form for >>>>>> >>>>>> Nq = 2*|SxA|^2*(|SxA|^2+(SxA).(SxA')) >>>>>> = 2*[(SxA).(SxA)]*[(SxA).(Sx(A+A'))] >>>>>> >>>>>> At 09:24 PM 6/10/2014, Henry Baker wrote: >>>>>>>Another solution based on vectors & quaternions w/o matrices. >>>>>>> >>>>>>>If A,B are unit vectors (|A|=|B|=1), then rotating them >>>>>>>will preserve length (|A'|=|B'|=1). >>>>>>> >>>>>>>We need to warm up a bit before solving the full problem; >>>>>>>we first consider mapping A->A' without worrying about B. >>>>>>> >>>>>>>The most straightforward rotation that maps A->A' is >>>>>>>the quaternion q=sqrt((-A')A). Here (-A')A is >>>>>>>the quaternion product of the quaternion inverse of A' >>>>>>>and A. (For unit vector quaternions like A', (A')^-1=-A'.) >>>>>>>We use the standard mapping of vectors into >>>>>>>quaternions: A -> 0+A; i.e., the scalar part of (0+A) >>>>>>>is 0, and the vector part of (0+A) is A itself. >>>>>>> >>>>>>>The mapping >>>>>>> >>>>>>>v -> q v (q)^-1 = (q v q*)/(qq*) >>>>>>> >>>>>>>will transform A->A'. q* is conjugate(q). >>>>>>> >>>>>>>(The second form is preferred, since we don't have to >>>>>>>normalize q, and we don't have to compute absolute values.) >>>>>>> >>>>>>>The "sqrt()" above is the _quaternion_ square root, which >>>>>>>can be computed: >>>>>>> >>>>>>>sqrt(q) = (|q|+q)/sqrt(2*(|q|+S(q))) >>>>>>> >>>>>>>(Note that the denominator is a real scalar quantity.) >>>>>>> >>>>>>>Here S(q) is the "scalar part" of q, or >>>>>>> >>>>>>>S(q) = (q+q*)/2 >>>>>>> >>>>>>>But since we don't need a _normalized_ q, we can use a >>>>>>>cheap sqrt: >>>>>>> >>>>>>>elcheaposqrt(q) = |q|+q (!) >>>>>>> >>>>>>>Proof: >>>>>>> >>>>>>>(|q|+q)^2 = |q|^2+2|q|q+q^2 >>>>>>> = |q|^2+2|q|q+(Sq+Vq)^2 >>>>>>> = |q|^2+2|q|q+Sq^2+2SqVq+Vq^2 >>>>>>> = |q|^2+2|q|q+Sq^2+2SqVq-|Vq|^2 >>>>>>> = |q|^2+2|q|q+Sq^2+2SqVq-|Vq|^2-Sq^2+Sq^2 >>>>>>> = |q|^2+2|q|q+2Sq^2+2SqVq-|q|^2 >>>>>>> = 2|q|q+2Sq^2+2SqVq >>>>>>> = 2|q|q+2Sq(Sq+Vq) >>>>>>> = 2|q|q+2Sq q >>>>>>> = 2(|q|+Sq) q >>>>>>> >>>>>>>QED >>>>>>> >>>>>>>In particular, the function that takes A->A' is >>>>>>> >>>>>>>v -> (1-A'A)v(1-AA')/2/(1+A.A') >>>>>>> >>>>>>>(Here A.A' is the dot product of A & A'.) >>>>>>> >>>>>>>We now attack your actual question: you want a >>>>>>>rotation that _simultaneously_ transports >>>>>>>A->A' and B->B'. >>>>>>> >>>>>>>As Dan pointed out, the axis of rotation that we >>>>>>>want is (A'-A)x(B'-B), but we can't use (A'-A).(B'-B) >>>>>>>as the scalar part of a quaternion, because that >>>>>>>would rotate the projection of A into the projection >>>>>>>of B (or equivalently, the projection of A' into the >>>>>>>projection of B') -- not what we want. >>>>>>> >>>>>>>Let S=(A'-A)x(B'-B) be our axiS. If we are careful, >>>>>>>we won't have to normalize S into a unit vector. We >>>>>>>also no longer need B,B' since we only needed B,B' >>>>>>>to establish the axis S. >>>>>>> >>>>>>>If we compute SxA and SxA', we note that they are in >>>>>>>a plane orthogonal to S, and during the rotation that >>>>>>>maps A->A' and B->B', we will also map SxA into SxA'. >>>>>>>So the axis of this rotation is S itself. Also, >>>>>>>obviously, |SxA|=|SxA'|. >>>>>>> >>>>>>>So, we can now utilize the method of our warmup >>>>>>>exercise to map SxA into SxA' via the mapping: >>>>>>> >>>>>>>v->(|SxA|^2-(SxA')(SxA))v(|SxA|^2-(SxA)(SxA'))/Nq >>>>>>> >>>>>>>and Nq is the "absolute value squared" of >>>>>>>(|SxA|^2-(SxA')(SxA)) or of (|SxA|^2-(SxA)(SxA')). >>>>>>> >>>>>>>This is the mapping that we seek that will >>>>>>>simultaneously map A->A' and B->B'. >>>>>>> >>>>>>>Thus, although we talked about square roots and >>>>>>>absolute values in our development, we find that >>>>>>>the final result requires no square roots or >>>>>>>absolute values. >>>>>>> >>>>>>>I'd be surprised if there is a simpler expression >>>>>>>for computing this quaternion. >>>>>>> >>>>>>>At 04:03 PM 6/9/2014, Henry Baker wrote: >>>>>>>>We can take our matrix product below, which is >>>>>>>>an orthogonal matrix. Let's call it M. >>>>>>>> >>>>>>>>Then we can easily recover the quaternion from >>>>>>>>the orthogonal matrix M using the inverse _Cayley >>>>>>>>Transform_: >>>>>>>> >>>>>>>>S = (I-M)(I+M)^-1 >>>>>>>> >>>>>>>>The matrix I+M is invertible, because we have _rigid_ >>>>>>>>rotations (i.e., no reflections, hence no eigenvalues >>>>>>>>of -1). >>>>>>>> >>>>>>>>S is a _skew-symmetric_ matrix: >>>>>>>> >>>>>>>>[ 0 z -y] >>>>>>>>[-z 0 x] >>>>>>>>[ y -x 0] = S >>>>>>>> >>>>>>>>Then our (non-normalized) quaternion is [1,x,y,z]. >>>>>>>> >>>>>>>>http://en.wikipedia.org/wiki/Cayley_transform >>>>>>>> >>>>>>>>At 02:06 PM 6/7/2014, Henry Baker wrote: >>>>>>>>>We can easily do it with vectors & matrices: >>>>>>>>> >>>>>>>>>We are given A,B,A',B', where |A|=|B|=|A'|=|B'|. >>>>>>>>>R(A)=A', R(B)=B'. >>>>>>>>> >>>>>>>>>Because the rotation R is rigid, we know that >>>>>>>>> >>>>>>>>>R(AxB) = A'xB' = R(A)xR(B) >>>>>>>>> >>>>>>>>>Assuming that B is not a multiple of A, we can >>>>>>>>>form the 3x3 matrices R=matrix(A,B,AxB) and >>>>>>>>>R'=matrix(A',B',A'xB'). >>>>>>>>> >>>>>>>>>Our answer is then the matrix product >>>>>>>>> >>>>>>>>>R^-1 R' = >>>>>>>>> >>>>>>>>>matrix(A,B,AxB)^-1 matrix(A',B',A'xB') >>>>>>>>> >>>>>>>>>You are now free to convert this _orthogonal_ >>>>>>>>>matrix into a quaternion at your leisure. >>>>>>>>> >>>>>>>>>(Although AxB, A'xB' are not unit vectors, >>>>>>>>>hence R,R' are not orthogonal matrices, >>>>>>>>>nevertheless, R^-1 R' _is_ orthogonal.) >>>>>>>>> >>>>>>>>>(Contrary to Knuth, I did actually try this >>>>>>>>>on some numbers & it worked!) >>>>>>>>> >>>>>>>>>At 04:25 AM 6/7/2014, Dan Asimov wrote: >>>>>>>>>>In almost all cases, V := (A-A')x(B-B') will be a vector in the direction of the axis of rotation. Knowing V makes it easy to project say A and A' onto the perpendicular plane to C to determine the angle T. >>>>>>>>>> >>>>>>>>>>The remaining cases should be easy to exclude or deal with. >>>>>>>>>> >>>>>>>>>>--Dan >>>>>>>>>> >>>>>>>>>>On Jun 7, 2014, at 1:32 AM, Robert Smith <quad(a)symbo1ics.com> wrote: >>>>>>>>>> >>>>>>>>>>> Let A and B be unit vectors in R^3. Suppose they are rotated about >>>>>>>>>>> some vector V by an angle T, resulting in A' and B' respectively. What >>>>>>>>>>> are V and T? >>>>>>>>>>> >>>>>>>>>>> I set up a quadratic system using quaternions and got a result that >>>>>>>>>>> was 3 million terms large. Am I missing something?

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[math-fun] Need a distribution
by David Wilson 26 Jun '14

26 Jun '14

For a work project, I need a 1D continuous distribution that is like a Gaussian in that it has a "bulge" and "tails", but also has skewness (i.e. the two tails can independently drop off slowly or quickly). I would expect this distribution to have at least 3 parameters (e.g. mean, variance, and skewness). I need the equation for the density function and a way to choose parameters for a distribution to fit a set of data points. Any pointers from the more statistically fluent would be greatly appreciated. I am copying to seqfan, in hopes of possibly snagging some help there, apologies. Please reply to dwilson at ltk.com. Thanks in advance.

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Re: [math-fun] math-fun Digest, Vol 136, Issue 64
by Warren D Smith 26 Jun '14

26 Jun '14

> As has been mentioned in this thread, at the pressures in a neutron star a > neutron is stable relative to a proton + electron. > Electrostatic force pulls e + p => n together long before they can > separate gravitationally. > --R --I quote from "earlier in thread" post by E.Salamin: So the electron density is 4% of the neutron density, and by electrical neutrality the same holds for the proton density. Note that the electron density scales as the square of the neutron density, as long as the neutrons remain non-relativistic. So if we assume instead a neutron density of 10^44 m-3, the electron density will be 0.4% of the neutron density... WDS: If Salamin is anywhere near correct there will be plenty of free protons & electrons; and they presumably will be distributed in a near min-energy manner, which means separated with the protons nearer the bottom. The distribution will be such that the gravitational forces are exactly balanced by the coulombic forces. This will cause an electric field. If you deny the existence of free protons in the mix, then I guess I'd have to ask: why are you even talking about superconductivity at all? Wouldn't you then deny its existence too?

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[math-fun] neutron stars... electric field?
by Warren D Smith 26 Jun '14

26 Jun '14

And actually, now you've got me thinking about this, it occurs to me there should be a dramatic difference in internal structure between a superconducting and not neutron star. (Which the papers so far cited on this topic do not mention at all...) Let me explain. Assume that the interior is made of neutrons which are polluted to some extent by proton-electron mixture. Now here is the thing: protons are heavier and denser than electrons. Therefore, we would expect the protons to sink toward the center, while the electrons rise toward the outside. Which means, after that happens, there should be a very large static radial ELECTRIC field pointing out of the core, within the neutron star. The thing is a big capacitor. If there is no superconductivity, this seems to be something about which there should be no controversy. However, electric field is not possible inside a superconductor (whether type II or type I). So what do you physicists conclude from that? Crikey. The analogue in lab-accessible kinds of matter would be, I guess, if you had a bar of superconducting metal, and then you applied a magical force (analogue of gravity) to the electrons and nuclei inside, pulling them toward opposite ends of the bar. And then as a thought experiment (since there is no such magic force available) you ask "what would happen then?" My top guess, actually, is this would yield a state of matter that I think nobody has ever seen or conceived of before, namely: a superconductor containing a macroscopic permanent internal electric field. What. The. Hell. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] superconducting magnet confusion
by Warren D Smith 26 Jun '14

26 Jun '14

Richard E. Howard raised the point that type-II superconductors can permit magnetic field to penetrate thru them via "pinned flux tubes" which might permit a topologically-ball shaped permanent superconducting magnet to exist. This could overcome my "there's a huge magnetic field" objection to the claim neutron stars had superfluid+superconductive interior. --WDS: sorry, I should have pointed this out myself. However, I object that if the material is a "superfluid", not a "solid crystal lattice with crystal defects" then I find it hard to understand how we could have an "Abrikosov latttice of flux tubes pinned by crystal defects." On the contrary I would expect any flux tube to be mobile and hence be expelled. Therefore I would expect neutron star stuff, if really superfluid+superconductor, to be "type I" and hence to expel all magnetic field extremely quickly (result: exactly 0 field throughout the super-region). Also, the Earth's magnetic field is caused by convection currents in hot normal-conductive fluid forming a "dynamo." Ditto sun's field. But I would think there would be no such thing as convection currents in a superfluid. Instead in liquid helium heat is transmitted via "second sound", a wavelike rather than diffusive phenomenon quite unlike heat transmission in normal matter (including unlike it in liquid helium above the super-transition temperature). I suppose you might argue that a thin outer layer of non-super matter in a neutron star might be able to convect and might also be responsible for the magnetic field. But in that case we'd be speaking of an essentially TWO-dimensional "dynamo" quite unlike the Earth and Sun's 3D one and it seems dubious (?) such a thing could exist generating a macroscopic dipole field. Also, although I guess I should admit the logical possibility that a ball-shaped chunk of super-matter could be a permanent magnet if it is type-II, I am unaware of any actual example of this logical possibility. As far as I know, nobody has ever engineered a super-magnet that is topologically ball-shaped, and so it is not clear to me it really IS a logical possibility. Is it really? After a bit of thinking, I think yes, it really is possible as follows. Place your ball of type-II superconductor -- but heated well above Tc -- in a huge external field. Now cool it below Tc. Now the field presumably would be "frozen in" in the form of a flux-tube lattice even after we remove external field. So anyhow, at the very least I think one should concede these matters are subtle and worth analysis+discussion -- which I remind you, the "neutron stars have super interiors!" papers DID NOT DO AT ALL. At the worst, it might even be this all just kills the entire hypothesis.

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