# Parameter for a saddle point

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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