Mixing
Hektic Oscillator
It's fascinating that whereas the solution of the quartic oscillator (restoring force $propto r^3$) problem can be expressed as the distance-from-origin of a point moving with unit speed along a lemniscate $r=sqrt{cos(2theta)}$ (made a silly mistake there at first - fixed it now), the solution for an ordinary quadratic oscillator (restoring force $propto r$) can likewise be expressed as the distance-from-origin of a point moving with unit speed along a fattened quasi-lemniscate $r=costheta$ ... or two congruent circles just fitting within the unit circle.
Unlike as the case of the quartic oscillator, which can infact be elemenarily realised, a prototype realisation being transverse oscillations of a mass at the centre of a taught elastic string, I can't offhand think of how a hektic oscillator (restoring force $propto r^5$) might simply be physically realised. Would that have its quasi-lemniscate?
I suppose it must exist ... it's the solubility of it, really, that raises the questions. I think the lobes would be rather teardrop -shaped ... on the grounds that as the exponent increases, the scenario becomes more like that of the particle simply bouncing off 'sheer walls' of force - so the lemniscate would tend more & more to be two increasingly flattened 'blobs' each on the end of an increasingly thin 'stalk' lying diametrically opposed to the other.
I think it might be one of those situations in which you have two figures, one for each of two scenarios one of which is one step further further along in a progression, that have a simple & beautiful relationship to each other; but in which when you try to extend it indefinitely it just starts getting kind of ... ugly. Quaternions - octonions, etc; exponentiation - tetration, etc ... it seems to happen with such regularity. And sometimes you just so want the progression to be extensible indefinitely!
Update
See answer.
differential-equations dynamical-systems elliptic-functions
asked Nov 21 '18 at 7:22
AmbretteOrrisey
57410
|
show 2 more comments
It's fascinating that whereas the solution of the quartic oscillator (restoring force $propto r^3$) problem can be expressed as the distance-from-origin of a point moving with unit speed along a lemniscate $r=sqrt{cos(2theta)}$ (made a silly mistake there at first - fixed it now), the solution for an ordinary quadratic oscillator (restoring force $propto r$) can likewise be expressed as the distance-from-origin of a point moving with unit speed along a fattened quasi-lemniscate $r=costheta$ ... or two congruent circles just fitting within the unit circle.
Unlike as the case of the quartic oscillator, which can infact be elemenarily realised, a prototype realisation being transverse oscillations of a mass at the centre of a taught elastic string, I can't offhand think of how a hektic oscillator (restoring force $propto r^5$) might simply be physically realised. Would that have its quasi-lemniscate?
I suppose it must exist ... it's the solubility of it, really, that raises the questions. I think the lobes would be rather teardrop -shaped ... on the grounds that as the exponent increases, the scenario becomes more like that of the particle simply bouncing off 'sheer walls' of force - so the lemniscate would tend more & more to be two increasingly flattened 'blobs' each on the end of an increasingly thin 'stalk' lying diametrically opposed to the other.
I think it might be one of those situations in which you have two figures, one for each of two scenarios one of which is one step further further along in a progression, that have a simple & beautiful relationship to each other; but in which when you try to extend it indefinitely it just starts getting kind of ... ugly. Quaternions - octonions, etc; exponentiation - tetration, etc ... it seems to happen with such regularity. And sometimes you just so want the progression to be extensible indefinitely!
Update
See answer.
differential-equations dynamical-systems elliptic-functions
asked Nov 21 '18 at 7:22
AmbretteOrrisey
57410
1
I don't think 'Hektic' is the right word... maybe you mean 'quintic'?
– Yuriy S
Nov 21 '18 at 13:51
Convention of denoting the kind of oscillator according to the shape of the potential well rather than the shape of the restoring force is actually one that is used ..
– AmbretteOrrisey
Nov 21 '18 at 13:55
certainly "quartic oscillator" is used for an oscillator having a cubic restoring force. Physicists do seem to prefer to frame such things in terms of energy rather than force. It translates better into quantum mechanics, without a doubt ... which is of-a-piece with the way that in QM there no longer really is anymore a thing with a trajectory. And preceding that, there were Lagrangian & Hamiltionian forms of mechanics. And in fluid mechanics you have velocity potentials, which end up-being what is solved-for. I think as a general rule physicists find the integral form easier.
– AmbretteOrrisey
Nov 21 '18 at 14:02
But now you mention it, it usually is Latin rather than Greek prefices that are used for things like that. Afterall it's "quartic" rather than "tetric" isn't it, & "quintic" rather that "pentic".
– AmbretteOrrisey
Nov 21 '18 at 14:10
However, if it were derived from Greek rather than Latin, it's not atall unlikely that it would be rendered "hektic" rather that "hextic", as in the old Greek from which these particles are taken there is a liberal use of inflexion to 'relieve' difficult concatenation of consonants - and the substitution of 'κ' for 'ξ' in such a word would be a quite typical instance of such practice.
– AmbretteOrrisey
Nov 21 '18 at 14:22
|
show 2 more comments
It's fascinating that whereas the solution of the quartic oscillator (restoring force $propto r^3$) problem can be expressed as the distance-from-origin of a point moving with unit speed along a lemniscate $r=sqrt{cos(2theta)}$ (made a silly mistake there at first - fixed it now), the solution for an ordinary quadratic oscillator (restoring force $propto r$) can likewise be expressed as the distance-from-origin of a point moving with unit speed along a fattened quasi-lemniscate $r=costheta$ ... or two congruent circles just fitting within the unit circle.
Unlike as the case of the quartic oscillator, which can infact be elemenarily realised, a prototype realisation being transverse oscillations of a mass at the centre of a taught elastic string, I can't offhand think of how a hektic oscillator (restoring force $propto r^5$) might simply be physically realised. Would that have its quasi-lemniscate?
I suppose it must exist ... it's the solubility of it, really, that raises the questions. I think the lobes would be rather teardrop -shaped ... on the grounds that as the exponent increases, the scenario becomes more like that of the particle simply bouncing off 'sheer walls' of force - so the lemniscate would tend more & more to be two increasingly flattened 'blobs' each on the end of an increasingly thin 'stalk' lying diametrically opposed to the other.
I think it might be one of those situations in which you have two figures, one for each of two scenarios one of which is one step further further along in a progression, that have a simple & beautiful relationship to each other; but in which when you try to extend it indefinitely it just starts getting kind of ... ugly. Quaternions - octonions, etc; exponentiation - tetration, etc ... it seems to happen with such regularity. And sometimes you just so want the progression to be extensible indefinitely!
Update
See answer.
differential-equations dynamical-systems elliptic-functions
asked Nov 21 '18 at 7:22
AmbretteOrrisey
57410
It's fascinating that whereas the solution of the quartic oscillator (restoring force $propto r^3$) problem can be expressed as the distance-from-origin of a point moving with unit speed along a lemniscate $r=sqrt{cos(2theta)}$ (made a silly mistake there at first - fixed it now), the solution for an ordinary quadratic oscillator (restoring force $propto r$) can likewise be expressed as the distance-from-origin of a point moving with unit speed along a fattened quasi-lemniscate $r=costheta$ ... or two congruent circles just fitting within the unit circle.
Unlike as the case of the quartic oscillator, which can infact be elemenarily realised, a prototype realisation being transverse oscillations of a mass at the centre of a taught elastic string, I can't offhand think of how a hektic oscillator (restoring force $propto r^5$) might simply be physically realised. Would that have its quasi-lemniscate?
I suppose it must exist ... it's the solubility of it, really, that raises the questions. I think the lobes would be rather teardrop -shaped ... on the grounds that as the exponent increases, the scenario becomes more like that of the particle simply bouncing off 'sheer walls' of force - so the lemniscate would tend more & more to be two increasingly flattened 'blobs' each on the end of an increasingly thin 'stalk' lying diametrically opposed to the other.
I think it might be one of those situations in which you have two figures, one for each of two scenarios one of which is one step further further along in a progression, that have a simple & beautiful relationship to each other; but in which when you try to extend it indefinitely it just starts getting kind of ... ugly. Quaternions - octonions, etc; exponentiation - tetration, etc ... it seems to happen with such regularity. And sometimes you just so want the progression to be extensible indefinitely!
Update
See answer.
differential-equations dynamical-systems elliptic-functions
differential-equations dynamical-systems elliptic-functions
asked Nov 21 '18 at 7:22
AmbretteOrrisey
57410
asked Nov 21 '18 at 7:22
AmbretteOrrisey
57410
edited Nov 21 '18 at 12:30
asked Nov 21 '18 at 7:22
AmbretteOrrisey
57410
asked Nov 21 '18 at 7:22
AmbretteOrrisey
57410
asked Nov 21 '18 at 7:22
AmbretteOrrisey
57410
57410
1
I don't think 'Hektic' is the right word... maybe you mean 'quintic'?
– Yuriy S
Nov 21 '18 at 13:51
Convention of denoting the kind of oscillator according to the shape of the potential well rather than the shape of the restoring force is actually one that is used ..
– AmbretteOrrisey
Nov 21 '18 at 13:55
certainly "quartic oscillator" is used for an oscillator having a cubic restoring force. Physicists do seem to prefer to frame such things in terms of energy rather than force. It translates better into quantum mechanics, without a doubt ... which is of-a-piece with the way that in QM there no longer really is anymore a thing with a trajectory. And preceding that, there were Lagrangian & Hamiltionian forms of mechanics. And in fluid mechanics you have velocity potentials, which end up-being what is solved-for. I think as a general rule physicists find the integral form easier.
– AmbretteOrrisey
Nov 21 '18 at 14:02
But now you mention it, it usually is Latin rather than Greek prefices that are used for things like that. Afterall it's "quartic" rather than "tetric" isn't it, & "quintic" rather that "pentic".
– AmbretteOrrisey
Nov 21 '18 at 14:10
However, if it were derived from Greek rather than Latin, it's not atall unlikely that it would be rendered "hektic" rather that "hextic", as in the old Greek from which these particles are taken there is a liberal use of inflexion to 'relieve' difficult concatenation of consonants - and the substitution of 'κ' for 'ξ' in such a word would be a quite typical instance of such practice.
– AmbretteOrrisey
Nov 21 '18 at 14:22
|
show 2 more comments
1
I don't think 'Hektic' is the right word... maybe you mean 'quintic'?
– Yuriy S
Nov 21 '18 at 13:51
Convention of denoting the kind of oscillator according to the shape of the potential well rather than the shape of the restoring force is actually one that is used ..
– AmbretteOrrisey
Nov 21 '18 at 13:55
certainly "quartic oscillator" is used for an oscillator having a cubic restoring force. Physicists do seem to prefer to frame such things in terms of energy rather than force. It translates better into quantum mechanics, without a doubt ... which is of-a-piece with the way that in QM there no longer really is anymore a thing with a trajectory. And preceding that, there were Lagrangian & Hamiltionian forms of mechanics. And in fluid mechanics you have velocity potentials, which end up-being what is solved-for. I think as a general rule physicists find the integral form easier.
– AmbretteOrrisey
Nov 21 '18 at 14:02
But now you mention it, it usually is Latin rather than Greek prefices that are used for things like that. Afterall it's "quartic" rather than "tetric" isn't it, & "quintic" rather that "pentic".
– AmbretteOrrisey
Nov 21 '18 at 14:10
However, if it were derived from Greek rather than Latin, it's not atall unlikely that it would be rendered "hektic" rather that "hextic", as in the old Greek from which these particles are taken there is a liberal use of inflexion to 'relieve' difficult concatenation of consonants - and the substitution of 'κ' for 'ξ' in such a word would be a quite typical instance of such practice.
– AmbretteOrrisey
Nov 21 '18 at 14:22
1
1
I don't think 'Hektic' is the right word... maybe you mean 'quintic'?
– Yuriy S
Nov 21 '18 at 13:51
I don't think 'Hektic' is the right word... maybe you mean 'quintic'?
– Yuriy S
Nov 21 '18 at 13:51
Convention of denoting the kind of oscillator according to the shape of the potential well rather than the shape of the restoring force is actually one that is used ..
– AmbretteOrrisey
Nov 21 '18 at 13:55
Convention of denoting the kind of oscillator according to the shape of the potential well rather than the shape of the restoring force is actually one that is used ..
– AmbretteOrrisey
Nov 21 '18 at 13:55
certainly "quartic oscillator" is used for an oscillator having a cubic restoring force. Physicists do seem to prefer to frame such things in terms of energy rather than force. It translates better into quantum mechanics, without a doubt ... which is of-a-piece with the way that in QM there no longer really is anymore a thing with a trajectory. And preceding that, there were Lagrangian & Hamiltionian forms of mechanics. And in fluid mechanics you have velocity potentials, which end up-being what is solved-for. I think as a general rule physicists find the integral form easier.
– AmbretteOrrisey
Nov 21 '18 at 14:02
certainly "quartic oscillator" is used for an oscillator having a cubic restoring force. Physicists do seem to prefer to frame such things in terms of energy rather than force. It translates better into quantum mechanics, without a doubt ... which is of-a-piece with the way that in QM there no longer really is anymore a thing with a trajectory. And preceding that, there were Lagrangian & Hamiltionian forms of mechanics. And in fluid mechanics you have velocity potentials, which end up-being what is solved-for. I think as a general rule physicists find the integral form easier.
– AmbretteOrrisey
Nov 21 '18 at 14:02
But now you mention it, it usually is Latin rather than Greek prefices that are used for things like that. Afterall it's "quartic" rather than "tetric" isn't it, & "quintic" rather that "pentic".
– AmbretteOrrisey
Nov 21 '18 at 14:10
But now you mention it, it usually is Latin rather than Greek prefices that are used for things like that. Afterall it's "quartic" rather than "tetric" isn't it, & "quintic" rather that "pentic".
– AmbretteOrrisey
Nov 21 '18 at 14:10
However, if it were derived from Greek rather than Latin, it's not atall unlikely that it would be rendered "hektic" rather that "hextic", as in the old Greek from which these particles are taken there is a liberal use of inflexion to 'relieve' difficult concatenation of consonants - and the substitution of 'κ' for 'ξ' in such a word would be a quite typical instance of such practice.
– AmbretteOrrisey
Nov 21 '18 at 14:22
However, if it were derived from Greek rather than Latin, it's not atall unlikely that it would be rendered "hektic" rather that "hextic", as in the old Greek from which these particles are taken there is a liberal use of inflexion to 'relieve' difficult concatenation of consonants - and the substitution of 'κ' for 'ξ' in such a word would be a quite typical instance of such practice.
– AmbretteOrrisey
Nov 21 '18 at 14:22
|
show 2 more comments
1 Answer
1
active
oldest
votes
Now I think about it more carefully, I see that the lemniscate clearly exists: it can be made to exist by the expedient of setting (I'll just show case of an abstract dimensionless quasilemniscate)
$$frac{1}{sqrt{1-r^{2n}}}=sqrt{1+r^2left(frac{dtheta}{dr}right)^2}$$
whence
$$frac{dtheta}{dr}=frac{r^{n-1}}{sqrt{1-r^{2n}}}$$
whence (setting $u=r^n$)
$$ntheta=operatorname{acos}u$$
whence $$r=(cos(ntheta))^{1/n} .$$
So clearly there is always a solution-as-quasilemniscate for any positive integer value of $n$ - the problem of solving for it can be done very nicely 'short-circuiting' solving for the differential equation for $r$ in terms of time that results in evermore complex elliptic-type functions with increasing $n$. Relating length along the quasilemniscate to time is then just a matter of fetching in the characteristic length & time from the physical differential equation. And solving for $r$ in terms of time is then one with getting an explicit expression for $r$ in terms of arclength along the lemniscate, rather than in terms of $theta$.
It's very pleasant to see that there is a beautifully simple answer to this question afterall! Wasn't expecting it. I feel again I ought to have seen it sooner!
It doesn't result in those teardrop-shaped lobes that I was expecting, though, does it: the lobes are just narrower & nearlier sharp-cornered. In fact, come to think of it, the idea of teardrop-shaped lobes is extremely silly, really, coz you'd need to have $theta$ going backwards for part of its course!! In the limit of high $n$ they would tend to become just two line-segments, depicting the particle's behaviour tending towards bouncing backwards & forwards between elastic collisions. Which reminds me: I at first described the elliptic function solution of the cubic|quartic oscillator as being like a 'fattened' sine function: that's inaccurate, isn't it, as it's going to be somewhat morphing towards a triangular sawtooth waveform - the limit waveform, as just said.
I find it really quite astonishing, actually, just how strangely fitting the generalised lemniscate (or quasi -lemniscate, or super -lemniscate (by analogy with superellipse)) is for expressing - or encapsulating - the solution of the generalised (generalised in the sense of the restoring-force being proportional to any odd power of displacement) harmonic oscillator. And it's because the differential of arc length in polar co-ordinates, and the differential of time with respect to radius in terms of radius in the differential equation of the harmonic oscillator, meld together so conveniently to give an elementary integral in $r^n$ ... $n$ being by definition such that restoring force $propto r^{2n-1}$ or shape of oscillator's potential well is of the form $r^{2n}$.
UPDATE
I've also just thought of how a hektic|pentic||sectic|quintic oscillator could be realised - and I don't know how it is I missed it: as in the quartic|cubic oscillator, affix the oscillating mass at the centre of a string ... but this time it doesn't need to be elastic, because instead of fixing the ends of the string to fixed points, each of them is in turn fixed to the centre of a elastic string the ends of each of which are fixed to fixed points! Obviously this process can be continued indefinitely ... but! ... it would probably become so fragile you would need a gravitiless environment in which to set it up! But I've finally clocked how an oscillator of any degree could in principle be set-up!
answered Nov 21 '18 at 12:31
AmbretteOrrisey
57410
I don't think there's much occasion or scope for the actual setting-up of such an oscillator - in fact designing it & optimising the positioning of the 2^(n-2)-1 strings would be an interesting mathematical question in itself ... but I have heard advanced now & then the idea that if a machine or settup emonstrating some physical principle is at least plausible then more likely than not there is some operation in nature somewhere that is isomorphic to the operation of it. I can't remember whether a name was cited in connection with any of these mentionings.
– AmbretteOrrisey
Nov 22 '18 at 10:56
2^(n-1) fixed fastening points & 2^(n-1)-1 strings, rather.
– AmbretteOrrisey
Nov 22 '18 at 11:21
Actually, doing that wouldn't yield an index of the restoring force of 2n-1, would it? It would rather be 2×3^(n-2)+1. Oh!! I give up!!
– AmbretteOrrisey
Nov 22 '18 at 17:24
No ... 2^n-1. I'm fairly sure about that now.
– AmbretteOrrisey
Nov 22 '18 at 21:24
add a comment |
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Now I think about it more carefully, I see that the lemniscate clearly exists: it can be made to exist by the expedient of setting (I'll just show case of an abstract dimensionless quasilemniscate)
$$frac{1}{sqrt{1-r^{2n}}}=sqrt{1+r^2left(frac{dtheta}{dr}right)^2}$$
whence
$$frac{dtheta}{dr}=frac{r^{n-1}}{sqrt{1-r^{2n}}}$$
whence (setting $u=r^n$)
$$ntheta=operatorname{acos}u$$
whence $$r=(cos(ntheta))^{1/n} .$$
So clearly there is always a solution-as-quasilemniscate for any positive integer value of $n$ - the problem of solving for it can be done very nicely 'short-circuiting' solving for the differential equation for $r$ in terms of time that results in evermore complex elliptic-type functions with increasing $n$. Relating length along the quasilemniscate to time is then just a matter of fetching in the characteristic length & time from the physical differential equation. And solving for $r$ in terms of time is then one with getting an explicit expression for $r$ in terms of arclength along the lemniscate, rather than in terms of $theta$.
It's very pleasant to see that there is a beautifully simple answer to this question afterall! Wasn't expecting it. I feel again I ought to have seen it sooner!
It doesn't result in those teardrop-shaped lobes that I was expecting, though, does it: the lobes are just narrower & nearlier sharp-cornered. In fact, come to think of it, the idea of teardrop-shaped lobes is extremely silly, really, coz you'd need to have $theta$ going backwards for part of its course!! In the limit of high $n$ they would tend to become just two line-segments, depicting the particle's behaviour tending towards bouncing backwards & forwards between elastic collisions. Which reminds me: I at first described the elliptic function solution of the cubic|quartic oscillator as being like a 'fattened' sine function: that's inaccurate, isn't it, as it's going to be somewhat morphing towards a triangular sawtooth waveform - the limit waveform, as just said.
I find it really quite astonishing, actually, just how strangely fitting the generalised lemniscate (or quasi -lemniscate, or super -lemniscate (by analogy with superellipse)) is for expressing - or encapsulating - the solution of the generalised (generalised in the sense of the restoring-force being proportional to any odd power of displacement) harmonic oscillator. And it's because the differential of arc length in polar co-ordinates, and the differential of time with respect to radius in terms of radius in the differential equation of the harmonic oscillator, meld together so conveniently to give an elementary integral in $r^n$ ... $n$ being by definition such that restoring force $propto r^{2n-1}$ or shape of oscillator's potential well is of the form $r^{2n}$.
UPDATE
I've also just thought of how a hektic|pentic||sectic|quintic oscillator could be realised - and I don't know how it is I missed it: as in the quartic|cubic oscillator, affix the oscillating mass at the centre of a string ... but this time it doesn't need to be elastic, because instead of fixing the ends of the string to fixed points, each of them is in turn fixed to the centre of a elastic string the ends of each of which are fixed to fixed points! Obviously this process can be continued indefinitely ... but! ... it would probably become so fragile you would need a gravitiless environment in which to set it up! But I've finally clocked how an oscillator of any degree could in principle be set-up!
answered Nov 21 '18 at 12:31
AmbretteOrrisey
57410
I don't think there's much occasion or scope for the actual setting-up of such an oscillator - in fact designing it & optimising the positioning of the 2^(n-2)-1 strings would be an interesting mathematical question in itself ... but I have heard advanced now & then the idea that if a machine or settup emonstrating some physical principle is at least plausible then more likely than not there is some operation in nature somewhere that is isomorphic to the operation of it. I can't remember whether a name was cited in connection with any of these mentionings.
– AmbretteOrrisey
Nov 22 '18 at 10:56
2^(n-1) fixed fastening points & 2^(n-1)-1 strings, rather.
– AmbretteOrrisey
Nov 22 '18 at 11:21
Actually, doing that wouldn't yield an index of the restoring force of 2n-1, would it? It would rather be 2×3^(n-2)+1. Oh!! I give up!!
– AmbretteOrrisey
Nov 22 '18 at 17:24
No ... 2^n-1. I'm fairly sure about that now.
– AmbretteOrrisey
Nov 22 '18 at 21:24
add a comment |
Now I think about it more carefully, I see that the lemniscate clearly exists: it can be made to exist by the expedient of setting (I'll just show case of an abstract dimensionless quasilemniscate)
$$frac{1}{sqrt{1-r^{2n}}}=sqrt{1+r^2left(frac{dtheta}{dr}right)^2}$$
whence
$$frac{dtheta}{dr}=frac{r^{n-1}}{sqrt{1-r^{2n}}}$$
whence (setting $u=r^n$)
$$ntheta=operatorname{acos}u$$
whence $$r=(cos(ntheta))^{1/n} .$$
So clearly there is always a solution-as-quasilemniscate for any positive integer value of $n$ - the problem of solving for it can be done very nicely 'short-circuiting' solving for the differential equation for $r$ in terms of time that results in evermore complex elliptic-type functions with increasing $n$. Relating length along the quasilemniscate to time is then just a matter of fetching in the characteristic length & time from the physical differential equation. And solving for $r$ in terms of time is then one with getting an explicit expression for $r$ in terms of arclength along the lemniscate, rather than in terms of $theta$.
It's very pleasant to see that there is a beautifully simple answer to this question afterall! Wasn't expecting it. I feel again I ought to have seen it sooner!
It doesn't result in those teardrop-shaped lobes that I was expecting, though, does it: the lobes are just narrower & nearlier sharp-cornered. In fact, come to think of it, the idea of teardrop-shaped lobes is extremely silly, really, coz you'd need to have $theta$ going backwards for part of its course!! In the limit of high $n$ they would tend to become just two line-segments, depicting the particle's behaviour tending towards bouncing backwards & forwards between elastic collisions. Which reminds me: I at first described the elliptic function solution of the cubic|quartic oscillator as being like a 'fattened' sine function: that's inaccurate, isn't it, as it's going to be somewhat morphing towards a triangular sawtooth waveform - the limit waveform, as just said.
I find it really quite astonishing, actually, just how strangely fitting the generalised lemniscate (or quasi -lemniscate, or super -lemniscate (by analogy with superellipse)) is for expressing - or encapsulating - the solution of the generalised (generalised in the sense of the restoring-force being proportional to any odd power of displacement) harmonic oscillator. And it's because the differential of arc length in polar co-ordinates, and the differential of time with respect to radius in terms of radius in the differential equation of the harmonic oscillator, meld together so conveniently to give an elementary integral in $r^n$ ... $n$ being by definition such that restoring force $propto r^{2n-1}$ or shape of oscillator's potential well is of the form $r^{2n}$.
UPDATE
I've also just thought of how a hektic|pentic||sectic|quintic oscillator could be realised - and I don't know how it is I missed it: as in the quartic|cubic oscillator, affix the oscillating mass at the centre of a string ... but this time it doesn't need to be elastic, because instead of fixing the ends of the string to fixed points, each of them is in turn fixed to the centre of a elastic string the ends of each of which are fixed to fixed points! Obviously this process can be continued indefinitely ... but! ... it would probably become so fragile you would need a gravitiless environment in which to set it up! But I've finally clocked how an oscillator of any degree could in principle be set-up!
answered Nov 21 '18 at 12:31
AmbretteOrrisey
57410
I don't think there's much occasion or scope for the actual setting-up of such an oscillator - in fact designing it & optimising the positioning of the 2^(n-2)-1 strings would be an interesting mathematical question in itself ... but I have heard advanced now & then the idea that if a machine or settup emonstrating some physical principle is at least plausible then more likely than not there is some operation in nature somewhere that is isomorphic to the operation of it. I can't remember whether a name was cited in connection with any of these mentionings.
– AmbretteOrrisey
Nov 22 '18 at 10:56
2^(n-1) fixed fastening points & 2^(n-1)-1 strings, rather.
– AmbretteOrrisey
Nov 22 '18 at 11:21
Actually, doing that wouldn't yield an index of the restoring force of 2n-1, would it? It would rather be 2×3^(n-2)+1. Oh!! I give up!!
– AmbretteOrrisey
Nov 22 '18 at 17:24
No ... 2^n-1. I'm fairly sure about that now.
– AmbretteOrrisey
Nov 22 '18 at 21:24
add a comment |
Now I think about it more carefully, I see that the lemniscate clearly exists: it can be made to exist by the expedient of setting (I'll just show case of an abstract dimensionless quasilemniscate)
$$frac{1}{sqrt{1-r^{2n}}}=sqrt{1+r^2left(frac{dtheta}{dr}right)^2}$$
whence
$$frac{dtheta}{dr}=frac{r^{n-1}}{sqrt{1-r^{2n}}}$$
whence (setting $u=r^n$)
$$ntheta=operatorname{acos}u$$
whence $$r=(cos(ntheta))^{1/n} .$$
So clearly there is always a solution-as-quasilemniscate for any positive integer value of $n$ - the problem of solving for it can be done very nicely 'short-circuiting' solving for the differential equation for $r$ in terms of time that results in evermore complex elliptic-type functions with increasing $n$. Relating length along the quasilemniscate to time is then just a matter of fetching in the characteristic length & time from the physical differential equation. And solving for $r$ in terms of time is then one with getting an explicit expression for $r$ in terms of arclength along the lemniscate, rather than in terms of $theta$.
It's very pleasant to see that there is a beautifully simple answer to this question afterall! Wasn't expecting it. I feel again I ought to have seen it sooner!
It doesn't result in those teardrop-shaped lobes that I was expecting, though, does it: the lobes are just narrower & nearlier sharp-cornered. In fact, come to think of it, the idea of teardrop-shaped lobes is extremely silly, really, coz you'd need to have $theta$ going backwards for part of its course!! In the limit of high $n$ they would tend to become just two line-segments, depicting the particle's behaviour tending towards bouncing backwards & forwards between elastic collisions. Which reminds me: I at first described the elliptic function solution of the cubic|quartic oscillator as being like a 'fattened' sine function: that's inaccurate, isn't it, as it's going to be somewhat morphing towards a triangular sawtooth waveform - the limit waveform, as just said.
I find it really quite astonishing, actually, just how strangely fitting the generalised lemniscate (or quasi -lemniscate, or super -lemniscate (by analogy with superellipse)) is for expressing - or encapsulating - the solution of the generalised (generalised in the sense of the restoring-force being proportional to any odd power of displacement) harmonic oscillator. And it's because the differential of arc length in polar co-ordinates, and the differential of time with respect to radius in terms of radius in the differential equation of the harmonic oscillator, meld together so conveniently to give an elementary integral in $r^n$ ... $n$ being by definition such that restoring force $propto r^{2n-1}$ or shape of oscillator's potential well is of the form $r^{2n}$.
UPDATE
I've also just thought of how a hektic|pentic||sectic|quintic oscillator could be realised - and I don't know how it is I missed it: as in the quartic|cubic oscillator, affix the oscillating mass at the centre of a string ... but this time it doesn't need to be elastic, because instead of fixing the ends of the string to fixed points, each of them is in turn fixed to the centre of a elastic string the ends of each of which are fixed to fixed points! Obviously this process can be continued indefinitely ... but! ... it would probably become so fragile you would need a gravitiless environment in which to set it up! But I've finally clocked how an oscillator of any degree could in principle be set-up!
answered Nov 21 '18 at 12:31
AmbretteOrrisey
57410
Now I think about it more carefully, I see that the lemniscate clearly exists: it can be made to exist by the expedient of setting (I'll just show case of an abstract dimensionless quasilemniscate)
$$frac{1}{sqrt{1-r^{2n}}}=sqrt{1+r^2left(frac{dtheta}{dr}right)^2}$$
whence
$$frac{dtheta}{dr}=frac{r^{n-1}}{sqrt{1-r^{2n}}}$$
whence (setting $u=r^n$)
$$ntheta=operatorname{acos}u$$
whence $$r=(cos(ntheta))^{1/n} .$$
So clearly there is always a solution-as-quasilemniscate for any positive integer value of $n$ - the problem of solving for it can be done very nicely 'short-circuiting' solving for the differential equation for $r$ in terms of time that results in evermore complex elliptic-type functions with increasing $n$. Relating length along the quasilemniscate to time is then just a matter of fetching in the characteristic length & time from the physical differential equation. And solving for $r$ in terms of time is then one with getting an explicit expression for $r$ in terms of arclength along the lemniscate, rather than in terms of $theta$.
It's very pleasant to see that there is a beautifully simple answer to this question afterall! Wasn't expecting it. I feel again I ought to have seen it sooner!
It doesn't result in those teardrop-shaped lobes that I was expecting, though, does it: the lobes are just narrower & nearlier sharp-cornered. In fact, come to think of it, the idea of teardrop-shaped lobes is extremely silly, really, coz you'd need to have $theta$ going backwards for part of its course!! In the limit of high $n$ they would tend to become just two line-segments, depicting the particle's behaviour tending towards bouncing backwards & forwards between elastic collisions. Which reminds me: I at first described the elliptic function solution of the cubic|quartic oscillator as being like a 'fattened' sine function: that's inaccurate, isn't it, as it's going to be somewhat morphing towards a triangular sawtooth waveform - the limit waveform, as just said.
I find it really quite astonishing, actually, just how strangely fitting the generalised lemniscate (or quasi -lemniscate, or super -lemniscate (by analogy with superellipse)) is for expressing - or encapsulating - the solution of the generalised (generalised in the sense of the restoring-force being proportional to any odd power of displacement) harmonic oscillator. And it's because the differential of arc length in polar co-ordinates, and the differential of time with respect to radius in terms of radius in the differential equation of the harmonic oscillator, meld together so conveniently to give an elementary integral in $r^n$ ... $n$ being by definition such that restoring force $propto r^{2n-1}$ or shape of oscillator's potential well is of the form $r^{2n}$.
UPDATE
I've also just thought of how a hektic|pentic||sectic|quintic oscillator could be realised - and I don't know how it is I missed it: as in the quartic|cubic oscillator, affix the oscillating mass at the centre of a string ... but this time it doesn't need to be elastic, because instead of fixing the ends of the string to fixed points, each of them is in turn fixed to the centre of a elastic string the ends of each of which are fixed to fixed points! Obviously this process can be continued indefinitely ... but! ... it would probably become so fragile you would need a gravitiless environment in which to set it up! But I've finally clocked how an oscillator of any degree could in principle be set-up!
answered Nov 21 '18 at 12:31
AmbretteOrrisey
57410
edited Nov 22 '18 at 10:11
answered Nov 21 '18 at 12:31
AmbretteOrrisey
57410
answered Nov 21 '18 at 12:31
AmbretteOrrisey
57410
answered Nov 21 '18 at 12:31
AmbretteOrrisey
57410
57410
I don't think there's much occasion or scope for the actual setting-up of such an oscillator - in fact designing it & optimising the positioning of the 2^(n-2)-1 strings would be an interesting mathematical question in itself ... but I have heard advanced now & then the idea that if a machine or settup emonstrating some physical principle is at least plausible then more likely than not there is some operation in nature somewhere that is isomorphic to the operation of it. I can't remember whether a name was cited in connection with any of these mentionings.
– AmbretteOrrisey
Nov 22 '18 at 10:56
2^(n-1) fixed fastening points & 2^(n-1)-1 strings, rather.
– AmbretteOrrisey
Nov 22 '18 at 11:21
Actually, doing that wouldn't yield an index of the restoring force of 2n-1, would it? It would rather be 2×3^(n-2)+1. Oh!! I give up!!
– AmbretteOrrisey
Nov 22 '18 at 17:24
No ... 2^n-1. I'm fairly sure about that now.
– AmbretteOrrisey
Nov 22 '18 at 21:24
add a comment |
I don't think there's much occasion or scope for the actual setting-up of such an oscillator - in fact designing it & optimising the positioning of the 2^(n-2)-1 strings would be an interesting mathematical question in itself ... but I have heard advanced now & then the idea that if a machine or settup emonstrating some physical principle is at least plausible then more likely than not there is some operation in nature somewhere that is isomorphic to the operation of it. I can't remember whether a name was cited in connection with any of these mentionings.
– AmbretteOrrisey
Nov 22 '18 at 10:56
2^(n-1) fixed fastening points & 2^(n-1)-1 strings, rather.
– AmbretteOrrisey
Nov 22 '18 at 11:21
Actually, doing that wouldn't yield an index of the restoring force of 2n-1, would it? It would rather be 2×3^(n-2)+1. Oh!! I give up!!
– AmbretteOrrisey
Nov 22 '18 at 17:24
No ... 2^n-1. I'm fairly sure about that now.
– AmbretteOrrisey
Nov 22 '18 at 21:24
I don't think there's much occasion or scope for the actual setting-up of such an oscillator - in fact designing it & optimising the positioning of the 2^(n-2)-1 strings would be an interesting mathematical question in itself ... but I have heard advanced now & then the idea that if a machine or settup emonstrating some physical principle is at least plausible then more likely than not there is some operation in nature somewhere that is isomorphic to the operation of it. I can't remember whether a name was cited in connection with any of these mentionings.
– AmbretteOrrisey
Nov 22 '18 at 10:56
I don't think there's much occasion or scope for the actual setting-up of such an oscillator - in fact designing it & optimising the positioning of the 2^(n-2)-1 strings would be an interesting mathematical question in itself ... but I have heard advanced now & then the idea that if a machine or settup emonstrating some physical principle is at least plausible then more likely than not there is some operation in nature somewhere that is isomorphic to the operation of it. I can't remember whether a name was cited in connection with any of these mentionings.
– AmbretteOrrisey
Nov 22 '18 at 10:56
2^(n-1) fixed fastening points & 2^(n-1)-1 strings, rather.
– AmbretteOrrisey
Nov 22 '18 at 11:21
2^(n-1) fixed fastening points & 2^(n-1)-1 strings, rather.
– AmbretteOrrisey
Nov 22 '18 at 11:21
Actually, doing that wouldn't yield an index of the restoring force of 2n-1, would it? It would rather be 2×3^(n-2)+1. Oh!! I give up!!
– AmbretteOrrisey
Nov 22 '18 at 17:24
Actually, doing that wouldn't yield an index of the restoring force of 2n-1, would it? It would rather be 2×3^(n-2)+1. Oh!! I give up!!
– AmbretteOrrisey
Nov 22 '18 at 17:24
No ... 2^n-1. I'm fairly sure about that now.
– AmbretteOrrisey
Nov 22 '18 at 21:24
No ... 2^n-1. I'm fairly sure about that now.
– AmbretteOrrisey
Nov 22 '18 at 21:24
add a comment |
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I don't think 'Hektic' is the right word... maybe you mean 'quintic'?
– Yuriy S
Nov 21 '18 at 13:51
Convention of denoting the kind of oscillator according to the shape of the potential well rather than the shape of the restoring force is actually one that is used ..
– AmbretteOrrisey
Nov 21 '18 at 13:55
certainly "quartic oscillator" is used for an oscillator having a cubic restoring force. Physicists do seem to prefer to frame such things in terms of energy rather than force. It translates better into quantum mechanics, without a doubt ... which is of-a-piece with the way that in QM there no longer really is anymore a thing with a trajectory. And preceding that, there were Lagrangian & Hamiltionian forms of mechanics. And in fluid mechanics you have velocity potentials, which end up-being what is solved-for. I think as a general rule physicists find the integral form easier.
– AmbretteOrrisey
Nov 21 '18 at 14:02
But now you mention it, it usually is Latin rather than Greek prefices that are used for things like that. Afterall it's "quartic" rather than "tetric" isn't it, & "quintic" rather that "pentic".
– AmbretteOrrisey
Nov 21 '18 at 14:10
However, if it were derived from Greek rather than Latin, it's not atall unlikely that it would be rendered "hektic" rather that "hextic", as in the old Greek from which these particles are taken there is a liberal use of inflexion to 'relieve' difficult concatenation of consonants - and the substitution of 'κ' for 'ξ' in such a word would be a quite typical instance of such practice.
– AmbretteOrrisey
Nov 21 '18 at 14:22