Is the phase diagram for an oscilliator always clockwise?

[AlonsoMcLaren]

Is the phase diagram for an oscilliator (all sorts of oscilliations) always clockwise? If it is so, why?

 

[Jolb]

Clockwise or counterclockwise depends on how you draw your axes. You might choose p to the right and x up, or x to the right and p up. These two configurations would obviously change the orientation.

However, I think I know what you're referring to. Let's restrict our attention to a simple harmonic (F=-kx). Let's draw our phase space diagram with x to the right and p upwards.

Now let's say our system happens to be at a point on the x axis, in the positive region. (x is positive, p=0.) The force -kx would be to the left, meaning the particle's momentum would be decreasing at that point. So the path of the particle through phase space would be down from its initial point on the x axis, i.e. clockwise. (A similar argument applies for p=0 but x negative, yeilding again clockwise.) Similarly, if x=0 but p is positive, there is no force, but the particle is moving in the positive x direction, so this would also be clockwise.

You should see that because an oscillator is a restoring force, and since p is proportional to dx/dt, p always has the same sign as dx/dt but x always has the opposite sign as dp/dt. This translates to the fact that phase space points on the positive x-axis always evolve towards the negative p axis, then go to the negative x-axis, then to the positive p axis, and back to the starting point (given no damping/forcing/dissipation). This could be clockwise or counterclockwise depending on how you draw your axes.

 

[Redbelly98]

Short answer:

A positive momentum means that the position coordinate is increasing; a negative momentum means that position is decreasing.

It is conventional to draw oscillator diagrams with positive momentum in the +y direction, and positive position in the +x direction. For that way of drawing the axes, the motion has to be clockwise.