[ 9 posts ] 
 Exercise [20.08] b 
Author Message

Joined: 03 Jun 2010, 15:18
Posts: 136
Post Exercise [20.08] b
Attached my solution.
It is not really different from the one by Gandalf http://www.roadtoreality.info/viewtopic.php?f=16&t=1427&start=0(that also covers other topics) but is somehow more elementary and may help to clarify some aspects not obvious (e.g. the "mixed term" problem raised in the discussion).


Attachments:
Exercise_20_08.pdf [90.37 KiB]
Downloaded 142 times
14 Sep 2010, 17:28

Joined: 12 Jul 2010, 07:44
Posts: 154
Post Re: Exercise [20.08] b
There are lots of different comments about this exercise all over the board.
I thought they all ought to be referred to from one post.

As well as Roberto's solution (above) and gandalf's solution (previous topic), there is also a discussion about the "mixed term" here:

http://roadtoreality.info/viewtopic.php?f=5&t=1623


Laura prepared a comment about this for Prof Penrose, discussed here:

http://roadtoreality.info/viewtopic.php?f=20&t=59

The actual comment can be downloaded here (as a PDF): http://roadtoreality.info/download/file.php?id=21

And another similar comment by Laura is posted on her website here (also a PDF):

http://camoo.freeshell.org/20.8.pdf


Last edited by deant on 16 May 2012, 00:05, edited 3 times in total.

13 May 2012, 08:35

Joined: 12 Jul 2010, 07:44
Posts: 154
Post Re: Exercise [20.08] b
Roberto, I think the last line of your solution should be:

"V has a minimum if and only if Q_{hk} is non-negative definite".


It's entirely possible to have a Q_{hk} such that Q_{hk}q^h q^k=0 but q^h \neq 0. Just consider a potential field that depends (say) only on the z coordinate, but is unaffected by x or y.

The result is small oscillations in the z direction (assuming stable equilibrium), but constant-velocity motion in the x-y plane; the matrix W is degenerate in this case, so for some q's (namely linear combinations of x and y) the corresponding eigenvalue is zero.

V still has a minimum in this case, it's just that it depends only on the value of z.


Last edited by deant on 15 May 2012, 22:13, edited 1 time in total.

13 May 2012, 09:46

Joined: 03 Jun 2010, 15:18
Posts: 136
Post Re: Exercise [20.08] b
I cant' agree with you. A point P is a minimum for V(P) if in every point Q of a neighbour of P it is V(Q)>V(P), strictly greater. This does not happen in your example, because a near point in x direction has the same value of V, so it is just V(Q)>= V(P). It is just a stationary point, not a minimum.
In physical terms, that is an equilibrium point but the equilibrium is unstable: as yourself noted, a small perturbation to momentum in x direction causes a constant velocity motion that moves the system away from the equilibrium point; this means that equilibrium is unstable.

Of course there are situations (e.g. V_1= x^4 + y^4 + z^4 and V_2= -V_1) in which behaviour of V cannot be recognised by the analisys of linearised system: in both cases matrix Q vanishes, but first case is a clearly minimum and latter case is a maximum; so in first case equilibrium is stable and in second case unstable. This kind of systems cannot be analysed by linearisation, and you have to continue the Taylor development up to higher orders.


14 May 2012, 14:03

Joined: 12 Jul 2010, 07:44
Posts: 154
Post Re: Exercise [20.08] b
I don't disagree with anything you said, except perhaps the semantics. I take "V is a minimum" to mean "there is no point with a lower value of V (perhaps only locally)". So it's just a matter of definition.

Also, in the case with constant motion in the x-y plane, I don't believe this equilibrium situation should be called unstable, I believe the correct description for this kind of equilibrium condition is "meta-stable". "Unstable" refers to the case where perturbations can grow exponentially.


15 May 2012, 11:20

Joined: 03 Jun 2010, 15:18
Posts: 136
Post Re: Exercise [20.08] b
Of course is a matter of definition, however I think that definition of minimum for V as "there is no point with a lower or equal value of V (perhaps only locally)" is quite standard. Otherwise the function V(x)=5 will have a minimum at every x, that is quite counter-intuitive.

About stability: there are many definitions of stability (see e.g.
http://en.wikipedia.org/wiki/Stability_theory and http://en.wikipedia.org/wiki/Lyapunov_stability) but if a system, after a perturbation, moves away from the equilibrium point and the distance increases without any limit with time (exponentially or linearly or in any other way), the point of equilibrium is defined as unstable.

Metastability is quite a different concept ( see http://en.wikipedia.org/wiki/Metastability ); it is normally used with reference to thermodynamical systems but in the context of Hamiltonian dynamics can be considered metastable a point that is a local minimum but not a global minimum. When the perturbation is small enough, the system remains near to the metastability point (so it behaves as a stability point) but when the perturbation is made larger enough it moves away, going to a global minimum (if any) or to infinity; so it behaves as an instability point.


15 May 2012, 17:22

Joined: 12 Jul 2010, 07:44
Posts: 154
Post Re: Exercise [20.08] b
Hmmm... I think that it's perfectly appropriate to say that V(x)=5 has a minimum; Its minimum is "5" (at every x). It just doesn't have a (unique) minimum point, which is a stronger condition then merely having a minimum (value).

However, in the context of your solution, I'll grant you that "V has a minimum", although technically ambiguous, is most naturally interpreted as "V has a minimum point at equilibrium".

However, the actual issue here is that V can have a minimum point at equilibrium, while Q_{hk} might nonetheless not be positive definite - as you already demonstrated with your V_1=x^4+y^4+z^4 example. So the "if and only if" part of your solution's final sentence is incorrect. (As is my first suggested correction to it, in my second post from the top of the thread).

A corrected version of it would be:

V has a minimum point at equilibrium if Q_{hk} is positive definite;
Q_{hk} is non-negative definite if V has a minimum point at equilibrium.



---------

About metastability you are correct - I apologise. I'm afraid I had the term "metastable" confused with "neutrally stable" (or "critically stable"/"marginally stable"), which may sometimes be used to describe situations such as the example I gave, for linear systems. (It does entirely depend on the definition of "stability", though, like you said).

Also this would generally only apply to the linearised system, not the actual system, because (like you already pointed out) the actual stability of the system when Q_{hk} is non-negative definite with (at least one) zero eigenvalue, cannot be determined from the linearisation alone but must be determined by taking higher order terms of the Taylor expansion into account.


Last edited by deant on 17 May 2012, 17:53, edited 1 time in total.

15 May 2012, 23:35

Joined: 03 Jun 2010, 15:18
Posts: 136
Post Re: Exercise [20.08] b
deant wrote:
Hmmm... I think that it's perfectly appropriate to say that V(x)=5 has a minimum; Its minimum is "5" (at every x). It just doesn't have a (unique) minimum point, which is a stronger condition then merely having a minimum (value).

However, in the context of your solution, I'll grant you that "V has a minimum", although technically ambiguous, is most naturally interpreted as "V has a minimum point at equilibrium".


Yes, in this context I used “minimum” with the meaning of “minimum point” not of “minimum value of the function".

deant wrote:
A corrected version of it would be:

V has a minimum point at equilibrium if Q_{hk} is positive definite;
Q_{hk} is non-negative definite if V has a minimum point at equilibrium.



I think we can summarise the results of whole discussion as follows:

a) If at equilibrium the matrix Q is positive definite, this is a minimum point for V.
b) If at equilibrium the matrix Q is negative definite or is not definite, this cannot be a minimum point for V.
c) If at equilibrium the matrix Q is non-negative definite, behavior of V at that point cannot be determined just by Q (higher terms of Taylor expansions are needed).

A) If the equilibrium point is minimum for V, Q is non-negative definite. (This is equivalent to b)
B) If the equilibrium point is not minimum for V, Q is not positive definite (may be negative definite, non-negative definite, not definite). (This is equivalent to a)


17 May 2012, 16:54

Joined: 12 Jul 2010, 07:44
Posts: 154
Post Re: Exercise [20.08] b
Yes, I think that about sums it up!


17 May 2012, 17:42
   [ 9 posts ] 


cron