# Energy of trajectories

• Dec 20th 2012, 12:03 PM
darren86
Energy of trajectories
Consider the system:

xdot=y
ydot=-ax+x^3, a=constant>0

derived from the equation xdotdot=F(x)=-ax+x^3

There are 3 fixed points, 2 of which are saddle points

The energy E=y^2/2 + U (U=potential which is ax^2/2-x^4/4))

Note that U was found by setting F=-dU/dx and solving

I'm quite new to energy, am I right to assume that E=constant in the system? and if so can this technique not just be applied to all systems, hence all systems have constant energy (this sounds totally wrong)?

Next, I'm asked to find E along the trajectories that join the saddle points. Does anyone have any pointers how to do this?
• Dec 20th 2012, 01:47 PM
HallsofIvy
Re: Energy of trajectories
Not all systems are "conservative" so there does not necessarily exist a "potential" function (what you are calling "energy").

Those are, by the way, physics terms. I would say that "not all systems are 'exact' and so there does not necessarily exist an "anti-derivative".
• Dec 21st 2012, 08:46 AM
darren86
Re: Energy of trajectories
Quote:

Originally Posted by HallsofIvy
Not all systems are "conservative" so there does not necessarily exist a "potential" function (what you are calling "energy").

Those are, by the way, physics terms. I would say that "not all systems are 'exact' and so there does not necessarily exist an "anti-derivative".

thanks, any ideas on the second part of the question?
• Dec 21st 2012, 09:32 AM
HallsofIvy
Re: Energy of trajectories
Integrate the force vector "dot" the differential of arclength for the path.
• Dec 21st 2012, 10:41 AM
darren86
Re: Energy of trajectories
Quote:

Originally Posted by HallsofIvy
Integrate the force vector "dot" the differential of arclength for the path.

Is there any way to do this using the expression in my original post for energy?

I mean if E is constant along trajectories, I'm assuming at all points on the trajectory that join the 2 saddle points, E is some constant, so how do I find the value of E at any point on the trajectory?
• Dec 21st 2012, 04:19 PM
ILikeSerena
Re: Energy of trajectories
Hi darren86! :)

The total energy of your system is the kinetic energy plus the potential energy, which must be constant.
With a mass m moving along an x-axis, it takes the form:
E = (1/2)m xdot^2 + V(x) = constant

If you take the derivative with respect to t, you get:
m xdot xdotdot + V'(x) xdot = 0

which simplifies to:
m xdotdot = -V'(x)

You have already deduced which form V(x) takes.
To be a little more in line with physics, it is actually:
V(x) = m(ax^2/2-x^4/4) + V0

where V0 some constant, or more specifically the potential energy when x=0.

The stationary points occur when xdot = 0.
When this happens we have:
E = (1/2)m xdot^2 + V(x) = 0 + V(x)
In other words:
V(x) = m(ax^2/2-x^4/4) + V0 = E
Solve for x and you find your stationary points.

Can you analyze the graph of V(x) versus x?
And perhaps draw it?

You should find the V(x) takes certain maxima somewhere.
These are interesting points.
Can you find them?

In particular you'll have 2 maxima that have the same potential energy.
So if we release a test mass at one of these maxima with initial speed zero, it will "just" remain between these maxima.
And moreover the total energy E of the system will be equal to the potential energy in these maxima.

These maxima are what you are asked for.
• Dec 27th 2012, 10:43 AM
darren86
Re: Energy of trajectories
Quote:

Originally Posted by ILikeSerena
Hi darren86! :)

The total energy of your system is the kinetic energy plus the potential energy, which must be constant.
With a mass m moving along an x-axis, it takes the form:
E = (1/2)m xdot^2 + V(x) = constant

If you take the derivative with respect to t, you get:
m xdot xdotdot + V'(x) xdot = 0

which simplifies to:
m xdotdot = -V'(x)

You have already deduced which form V(x) takes.
To be a little more in line with physics, it is actually:
V(x) = m(ax^2/2-x^4/4) + V0

where V0 some constant, or more specifically the potential energy when x=0.

The stationary points occur when xdot = 0.
When this happens we have:
E = (1/2)m xdot^2 + V(x) = 0 + V(x)
In other words:
V(x) = m(ax^2/2-x^4/4) + V0 = E
Solve for x and you find your stationary points.

Can you analyze the graph of V(x) versus x?
And perhaps draw it?

You should find the V(x) takes certain maxima somewhere.
These are interesting points.
Can you find them?

In particular you'll have 2 maxima that have the same potential energy.
So if we release a test mass at one of these maxima with initial speed zero, it will "just" remain between these maxima.
And moreover the total energy E of the system will be equal to the potential energy in these maxima.

These maxima are what you are asked for.

when you say the stationary points occur when xdot=0, are we interested in these because they correspond to the fixed points?

I kind of lost what you were saying when you were talking about the maxima of V(x). I have plotted the function and seen the maxima but am confused how this relates to the question.

If we take our expression for E and find the value of it at the fixed points (which is equal), does this mean that at all points on the trajectory between the fixed points, the trajectory will will have this energy? (as we have shown that E is constant)?

however at the exact fixed point the trajectory will not move at all, does this matter?
• Dec 27th 2012, 01:42 PM
ILikeSerena
Re: Energy of trajectories
Quote:

I kind of lost what you were saying when you were talking about the maxima of V(x). I have plotted the function and seen the maxima but am confused how this relates to the question.
So?
At which x positions does V(x) take its maxima?
And how high are these maxima?

Quote:

however at the exact fixed point the trajectory will not move at all, does this matter?
The point of the problem is that you find the behavior of the system.

http://serc.carleton.edu/images/intr.../Stability.jpg

This picture shows the relevant equilibrium types.

If the system starts between the 2 maxima of V(x) with an energy that is less than the V(x) in those 2 maxima, it will oscillate between the corresponding 2 positions (stable).
If the system starts between the 2 maxima with a greater energy, it will be unstable and move to infinity.
If the system starts either left or right of those 2 maxima, regardless of the energy, the system is unstable and will move to infinity.

So the interesting question is what the threshold energy is, such that the system is on the verge of becoming unstable.
As a side note, the system will be in an unstable equilibrium at this point.