Find the stable equilibrium points

[evinda]

Hello! (Wave)

Let a particle be forced to move over the sphere $x^2+y^2+z^2=1$, that is subject to gravitational forces while also to an additional "dynamic" $V(x,y,z)=x+y$. Find the stable equilibrium points , if they exist.I have found a similar example:

Let $\vec{F}$ be the gravity field near the surface of the earth, i.e. let $\vec{F}=(F_x, F_y, F_z)$ where $F_x=0, F_y=0$ and $F_z=-mg$, where $g$ is the acceleration du to the gravity. Which are the equilibrium points, if a point particle of mass $m$ is restricted to the sphere $\phi(x,y,z)=x^2+y^2+z^2-r^2=0(r>0)$ ? Which of these are stable?

The solution is the following:We notice that $\vec{F}$ is a gradient field with "dynamic" $V=mgz$. Using the method of Lagrange multipliers to get the possible extrema , we have the equations

$$\nabla V= \lambda \nabla \phi \\ \phi=0$$

or equivalently$$0=2 \lambda x \\ 0= 2 \lambda y \\mg=2 \lambda z \\ x^2+y^2+z^2-r^2=0$$

The solution of the system of equations is $x=0, y=0, z = \pm r, \lambda= \pm \frac{mg}{2r}$.

From a theorem we get that $P_1=(0,0,-r)$ and $P_2=(0,0,r)$ are equilibrium points. Checking $V=mgz$ we get that $P_1$ is stable, but $P_2$ isn't.Do we consider also in our case a gravity field $\vec{F}$ ? If so what will this represent? (Sweating)If so then $\vec{F}=-\nabla V=-(1,1,0)$, right?

Then using the method of Lagrange multipliers we get the following extrema:

$$\left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right), \left( -\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$$Right?

 

[I like Serena]

Hello! (Wave)

Let a particle be forced to move over the sphere $x^2+y^2+z^2=1$, that is subject to gravitational forces while also to an additional "dynamic" $V(x,y,z)=x+y$. Find the stable equilibrium points , if they exist.I have found a similar example:

Let $\vec{F}$ be the gravity field near the surface of the earth, i.e. let $\vec{F}=(F_x, F_y, F_z)$ where $F_x=0, F_y=0$ and $F_z=-mg$, where $g$ is the acceleration du to the gravity. Which are the equilibrium points, if a point particle of mass $m$ is restricted to the sphere $\phi(x,y,z)=x^2+y^2+z^2-r^2=0(r>0)$ ? Which of these are stable?

The solution is the following:We notice that $\vec{F}$ is a gradient field with "dynamic" $V=mgz$. Using the method of Lagrange multipliers to get the possible extrema , we have the equations

$$\nabla V= \lambda \nabla \phi \\ \phi=0$$

or equivalently$$0=2 \lambda x \\ 0= 2 \lambda y \\mg=2 \lambda z \\ x^2+y^2+z^2-r^2=0$$

The solution of the system of equations is $x=0, y=0, z = \pm r, \lambda= \pm \frac{mg}{2r}$.

From a theorem we get that $P_1=(0,0,-r)$ and $P_2=(0,0,r)$ are equilibrium points. Checking $V=mgz$ we get that $P_1$ is stable, but $P_2$ isn't.Do we consider also in our case a gravity field $\vec{F}$ ? If so what will this represent? (Sweating)If so then $\vec{F}=-\nabla V=-(1,1,0)$, right?

Then using the method of Lagrange multipliers we get the following extrema:

$$\left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right), \left( -\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$$Right?

Hey evinda! (Smile)

It says that $V$ is additional to the force of gravity.
I interpret that to mean that we have the dynamic $V=x+y$ and additionally $V_g=mgz$.
In physics those correspond to contributions to the potential $V_{tot}$ of the particle, which is then:
$$V_{tot} = V + V_g = x + y + mgz$$
Or alternatively:
$$\vec F_{tot} = -\nabla V_{tot} = - \nabla V + \vec F_g = (-1,-1,-mg)$$

Applying the Lagrange multipliers, we then get:
$$\nabla V_{tot} = \lambda (x^2+y^2+z^2-1) \\ x^2+y^2+z^2-1 = 0$$
(Thinking)

 

[evinda]

Hey evinda! (Smile)

It says that $V$ is additional to the force of gravity.
I interpret that to mean that we have the dynamic $V=x+y$ and additionally $V_g=mgz$.
In physics those correspond to contributions to the potential $V_{tot}$ of the particle, which is then:
$$V_{tot} = V + V_g = x + y + mgz$$
Or alternatively:
$$\vec F_{tot} = -\nabla V_{tot} = - \nabla V + \vec F = (-1,-1,-mg)$$

What is $\vec{F}$ in this case?
Is $V_g$ always in the form $V_g=mgz$? If so, why?

 

[I like Serena]

What is $\vec{F}$ in this case?
Is $V_g$ always in the form $V_g=mgz$? If so, why?

That should be $\vec F_g$, which represents the gravitational force.

The actual form of the gravitational force is:
$$\vec F_g = -\frac{GMm}{r^2}\mathbf{\hat r}$$
where $r$ is the distance of a particle with mass $m$ to another mass $M$.
Close to the surface of the earth, it is usually approximated as:
$$\vec F_g = -mg\mathbf{\hat z}$$
as it is in your example. (Thinking)

 

[evinda]

That should be $\vec F_g$, which represents the gravitational force.

$\vec F_g$ is the force of gravity that is exerted to the particle?

The actual form of the gravitational force is:
$$\vec F_g = -\frac{GMm}{r^2}\mathbf{\hat r}$$
where $r$ is the distance of a particle with mass $m$ to another mass $M$.
Close to the surface of the earth, it is usually approximated as:
$$\vec F_g = -mg\mathbf{\hat z}$$
as it is in your example. (Thinking)

So this is known, right?Also what does $V_{tot}$ represent? What is meant with dynamic?

 

[I like Serena]

$\vec F_g$ is the force of gravity that is exerted to the particle?

So this is known, right?

Also what does $V_{tot}$ represent? What is meant with dynamic?

Yes and yes.

I don't know what is meant with dynamic. That's what you came up with.
For me it is the potential energy or the energy in a conservative field of a particle. (Nerd)

Hmm... doesn't the word dynamic come from δύναμις, which (also) means energy? (Wondering)

 

[evinda]

Yes and yes.

I don't know what is meant with dynamic. That's what you came up with.
For me it is the potential energy or the energy in a conservative field of a particle. (Nerd)

Hmm... doesn't the word dynamic come from δύναμις, which (also) means energy? (Wondering)

δύναμις means force.

 

[I like Serena]

δύναμις means force.

Okay, so the derivative of this 'dynamic' is δύναμις. (Angel)

 

[evinda]

Okay, so the derivative of this 'dynamic' is δύναμις. (Angel)

Yes, right. (Bigsmile)