- #1
Rasalhague
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- 2
I have some questions about what I think is a fairly standard and common short-hand notation used in physics.
Today I watched lecture 2 in the nptelhrd series Classical Physics by Prof. V. Balakrishnan. In it, he models a kind of system called a simple harmonic oscillator, I think using [itex]TC = C \times \mathbb{R} = \mathbb{R}^2[/itex] for a state space (He calls it phase space, but I'll use the more general name, as phase space is said elsewhere to have a coordinate called "momentum" whereas he calls the corresponding coordinate "velocity".), where [itex]C[/itex] is the configuration space of the system, and [itex]TC[/itex] the tangent bundle thereon. He labels points in state space with [itex]q[/itex] and [itex]\dot{q}[/itex], thus [itex](q,\dot{q}) \in \mathbb{R}^2[/itex]. So far so good. Then he writes some equations:
[tex]\ddot{q}=-\omega q, \enspace\enspace\enspace V(q)=\frac{1}{2}m\omega^2q^2, \enspace\enspace\enspace m\ddot{q}=-\frac{\mathrm{d} V}{\mathrm{d} q}(q);[/tex]
[tex]\dot{q}=v, \enspace\enspace\enspace \dot{v}=-\frac{V'(q)}{m}, \enspace\enspace\enspace\frac{\mathrm{d} v}{\mathrm{d} q}=-\frac{\omega^2}{v}q.[/tex]
I'm not satisfied that I understand all of these symbols.
I think [itex]\omega = \sqrt{k/m}[/itex] and [itex]m[/itex] are constants (angular velocity and mass). I think [itex]\ddot{q}[/itex] should mean the value at [itex]t[/itex] of the second derivative of some function whose value at [itex]t[/itex] is labelled [itex]q[/itex]. I'm guessing this implicit function is the first component function, [itex]\gamma_1[/itex], of a curve function, [itex]\gamma : \mathbb{R} \rightarrow \mathbb{R}^2 \; |\; t \mapsto (\gamma_1(t),\gamma_2(t))[/itex], whose image is a trajectory in state space, and that this is an arbitrary element of the set of trajectories defined by the differential equation(s). I think [itex]V : \mathbb{R} \rightarrow \mathbb{R}[/itex] is a scalar field on the configuration space [itex]C = \mathbb{R}[/itex].
Does [itex]\dot{q}=v[/itex] mean [itex]\gamma_2(t)=f\circ\gamma_1(t)[/itex] for some unknown function [itex]f:\mathbb{R}\rightarrow \mathbb{R}[/itex]?
If so, does does [itex]\dot{v}[/itex] mean [itex](f\circ\gamma_1)'(t)[/itex] or [itex]f'\circ\gamma_1(t)[/itex]? I'm guessing the latter.
Is [itex]-\frac{V'(q)}{m}[/itex] to be read as [itex]-\frac{(V\circ\gamma_1)'(t)}{m}[/itex] or [itex]-\frac{V'\circ\gamma_1(t)}{m}[/itex]? Again, I'd guess the latter.
How about the final equation?
[tex]\frac{\mathrm{d} v}{\mathrm{d} q}=-\frac{\omega^2}{v}q[/tex]
Is it
[itex]f'\circ\gamma_1(t)=-\frac{\omega^2}{f\circ\gamma_1(t)}\gamma_1(t) \enspace ?[/itex]
Today I watched lecture 2 in the nptelhrd series Classical Physics by Prof. V. Balakrishnan. In it, he models a kind of system called a simple harmonic oscillator, I think using [itex]TC = C \times \mathbb{R} = \mathbb{R}^2[/itex] for a state space (He calls it phase space, but I'll use the more general name, as phase space is said elsewhere to have a coordinate called "momentum" whereas he calls the corresponding coordinate "velocity".), where [itex]C[/itex] is the configuration space of the system, and [itex]TC[/itex] the tangent bundle thereon. He labels points in state space with [itex]q[/itex] and [itex]\dot{q}[/itex], thus [itex](q,\dot{q}) \in \mathbb{R}^2[/itex]. So far so good. Then he writes some equations:
[tex]\ddot{q}=-\omega q, \enspace\enspace\enspace V(q)=\frac{1}{2}m\omega^2q^2, \enspace\enspace\enspace m\ddot{q}=-\frac{\mathrm{d} V}{\mathrm{d} q}(q);[/tex]
[tex]\dot{q}=v, \enspace\enspace\enspace \dot{v}=-\frac{V'(q)}{m}, \enspace\enspace\enspace\frac{\mathrm{d} v}{\mathrm{d} q}=-\frac{\omega^2}{v}q.[/tex]
I'm not satisfied that I understand all of these symbols.
I think [itex]\omega = \sqrt{k/m}[/itex] and [itex]m[/itex] are constants (angular velocity and mass). I think [itex]\ddot{q}[/itex] should mean the value at [itex]t[/itex] of the second derivative of some function whose value at [itex]t[/itex] is labelled [itex]q[/itex]. I'm guessing this implicit function is the first component function, [itex]\gamma_1[/itex], of a curve function, [itex]\gamma : \mathbb{R} \rightarrow \mathbb{R}^2 \; |\; t \mapsto (\gamma_1(t),\gamma_2(t))[/itex], whose image is a trajectory in state space, and that this is an arbitrary element of the set of trajectories defined by the differential equation(s). I think [itex]V : \mathbb{R} \rightarrow \mathbb{R}[/itex] is a scalar field on the configuration space [itex]C = \mathbb{R}[/itex].
Does [itex]\dot{q}=v[/itex] mean [itex]\gamma_2(t)=f\circ\gamma_1(t)[/itex] for some unknown function [itex]f:\mathbb{R}\rightarrow \mathbb{R}[/itex]?
If so, does does [itex]\dot{v}[/itex] mean [itex](f\circ\gamma_1)'(t)[/itex] or [itex]f'\circ\gamma_1(t)[/itex]? I'm guessing the latter.
Is [itex]-\frac{V'(q)}{m}[/itex] to be read as [itex]-\frac{(V\circ\gamma_1)'(t)}{m}[/itex] or [itex]-\frac{V'\circ\gamma_1(t)}{m}[/itex]? Again, I'd guess the latter.
How about the final equation?
[tex]\frac{\mathrm{d} v}{\mathrm{d} q}=-\frac{\omega^2}{v}q[/tex]
Is it
[itex]f'\circ\gamma_1(t)=-\frac{\omega^2}{f\circ\gamma_1(t)}\gamma_1(t) \enspace ?[/itex]