# Parameter for a saddle point

Tags:
1. Apr 17, 2015

### enot

Hello, this is my first post here, so if I do problems, please correct me and do not be upset = )
I have one small theoretical and one greater question.
small one first:
1. The problem statement, all variables and given/known data
I have a potential energy :
$$W(L)= -\frac{1}{4}k_4(L_x^4+L_y^4+L_z^4)$$
How can describe my potential, replacing L_z with L_x,L_y.
My question is: what the situation should be, that I can replace L_z with L_x, L_y
2. Relevant equations
$$W(L)= -\frac{1}{4}k_4(L_x^4+L_y^4+L_z^4)$$,
k_4 - constant

3. The attempt at a solution
I thought to write something like :
L_x_tilda^2 + L_y_tilda^2 + L_z_tilda^2 = constant(scalar)
L_x_tilda = (L_x + L_x(0))
L_y_tilda = (L_y + L_y(0))
L_z_tilda = (L_z + L_z(0))
if my special point is L(0) is (0,0,1) then:
L_x^2 + L_y^2 + L_z^2 + 2L_z L_z(0) = 1
1 + 2L_z L_z(0) = 1;
but what does it mean. I do not know. I stuck on this...

big one
1. The problem statement, all variables and given/known data
I have a potential energy again :
$$W(L)= -\frac{1}{4}k_4(L_x^4+L_y^4+L_z^4)$$
I have a system of diff. eq. that describe movement of a part(now without a z dimension)
$$\ddot{L_x} = - \omega_{af}^2 L_x - 2 \gamma_{af} \dot{L_x} - j L_y$$.

$$\ddot{L_y} = - \omega_{af}^2 L_y - 2 \gamma_{af} \dot{L_y} + j L_x$$.
where j - is the current ,
My question is: how can I find the value of a current, and other parameters like \omega_{af} , \gamma_{af} so that my system will go to the saddle point.

2. Relevant equations
?
3. The attempt at a solution
the idea is here the following I solve the system by naming
L_x = A exp(iwt)
L_y = Bexp(iwt)
then I gather
$$- A w^2 e^{iwt}= - \omega_{af}^2 A e^{iwt} - 2 \gamma_{af} A iw e^{iwt} - j B e^{iwt}$$

$$- B w^2 e^{iwt}= - \omega_{af}^2 B e^{iwt} - 2 \gamma_{af} B iw e^{iwt} + j A e^{iwt}$$

then I get the following :
$$(( w^2 - \omega_{af}^2 - 2 \gamma_{af} iw )^2 + ( j p_z )^2 )B= 0$$
and I solve such equations:
$$w^2 - \omega_{af}^2 - 2 \gamma_{af} iw = - i j$$
$$w^2 - \omega_{af}^2 - 2 \gamma_{af} iw = i j$$
So for the first eq. I get
$$w = \gamma_{af} i \pm \sqrt{ (- \gamma_{af}^2 + \omega_{af}^2 - i j)}$$
Then I use De Moivre's formula
$$z^n = |z| ^n \exp(i \phi n)$$
$$\sqrt { |z| } = \sqrt[4] { (- \gamma_{af}^2 + \omega_{af}^2)^2 + ( - j )^2 }$$
if current is small
$$\sqrt { |z| } = \sqrt {- \gamma_{af}^2 + \omega_{af}^2}$$
and
$$\phi = \arctan \frac {b}{a} = \arctan ( \frac { - j p_z} {- \gamma_{af}^2 + \omega_{af}^2} )$$.
and as $$\omega_{af}, \gamma_{af} > 0$$, $$\gamma_{af}^2 \ll \omega_{af}^2$$ we get:
$$\phi = \arctan ( -\frac { j } { \omega_{af}^2 } )$$
$$j \ll \omega_{af}^2$$ and $$\tan (\phi) = \frac {-j} {\omega_{af}^2} \approx \phi$$
$$\exp(\frac {i \phi} {2} ) \approx 1 + \left( \frac {i \phi} {2} \right) + O\left( \left( \frac {i \phi} {2} \right)^2\right)$$
$$\exp \left( - \frac {i j} {2\omega_{af}^2} \right ) \approx 1 + \left( - \frac {i j} {2\omega_{af}^2} \right) + O\left( \left( - \frac {i j} {2\omega_{af}^2} \right)^2 \right)$$

$$\sqrt z = \sqrt{ (- \gamma_{af}^2 + \omega_{af}^2 - i j p_z)} = \omega_{af} \left( 1 - \frac {i j} {2\omega_{af}^2} \right)$$

$$w = \pm \omega_{af} + i \left( \gamma_{af} \mp \frac { j } {2\omega_{af}} \right)$$
$$L_x = A \exp( iwt ) = A \exp \left ( \left ( -\gamma_{af} \pm \frac { j } {2\omega_{af}} \right) t \right) \times \exp \left ( \pm i \omega_{af} t \right )$$
FINALY, when
$$-\gamma_{af} \pm \frac { j } {2\omega_{af}} < 0$$.
we should get the stable situation.
But how to gather from here parameters I need to have a saddle point. People, help me please.

2. Apr 22, 2015

### Staff: Admin

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted