Rasalhague
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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 TC = C \times \mathbb{R} = \mathbb{R}^2 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 C is the configuration space of the system, and TC the tangent bundle thereon. He labels points in state space with q and \dot{q}, thus (q,\dot{q}) \in \mathbb{R}^2. So far so good. Then he writes some equations:
\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);
\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.
I'm not satisfied that I understand all of these symbols.
I think \omega = \sqrt{k/m} and m are constants (angular velocity and mass). I think \ddot{q} should mean the value at t of the second derivative of some function whose value at t is labelled q. I'm guessing this implicit function is the first component function, \gamma_1, of a curve function, \gamma : \mathbb{R} \rightarrow \mathbb{R}^2 \; |\; t \mapsto (\gamma_1(t),\gamma_2(t)), 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 V : \mathbb{R} \rightarrow \mathbb{R} is a scalar field on the configuration space C = \mathbb{R}.
Does \dot{q}=v mean \gamma_2(t)=f\circ\gamma_1(t) for some unknown function f:\mathbb{R}\rightarrow \mathbb{R}?
If so, does does \dot{v} mean (f\circ\gamma_1)'(t) or f'\circ\gamma_1(t)? I'm guessing the latter.
Is -\frac{V'(q)}{m} to be read as -\frac{(V\circ\gamma_1)'(t)}{m} or -\frac{V'\circ\gamma_1(t)}{m}? Again, I'd guess the latter.
How about the final equation?
\frac{\mathrm{d} v}{\mathrm{d} q}=-\frac{\omega^2}{v}q
Is it
f'\circ\gamma_1(t)=-\frac{\omega^2}{f\circ\gamma_1(t)}\gamma_1(t) \enspace ?
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 TC = C \times \mathbb{R} = \mathbb{R}^2 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 C is the configuration space of the system, and TC the tangent bundle thereon. He labels points in state space with q and \dot{q}, thus (q,\dot{q}) \in \mathbb{R}^2. So far so good. Then he writes some equations:
\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);
\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.
I'm not satisfied that I understand all of these symbols.
I think \omega = \sqrt{k/m} and m are constants (angular velocity and mass). I think \ddot{q} should mean the value at t of the second derivative of some function whose value at t is labelled q. I'm guessing this implicit function is the first component function, \gamma_1, of a curve function, \gamma : \mathbb{R} \rightarrow \mathbb{R}^2 \; |\; t \mapsto (\gamma_1(t),\gamma_2(t)), 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 V : \mathbb{R} \rightarrow \mathbb{R} is a scalar field on the configuration space C = \mathbb{R}.
Does \dot{q}=v mean \gamma_2(t)=f\circ\gamma_1(t) for some unknown function f:\mathbb{R}\rightarrow \mathbb{R}?
If so, does does \dot{v} mean (f\circ\gamma_1)'(t) or f'\circ\gamma_1(t)? I'm guessing the latter.
Is -\frac{V'(q)}{m} to be read as -\frac{(V\circ\gamma_1)'(t)}{m} or -\frac{V'\circ\gamma_1(t)}{m}? Again, I'd guess the latter.
How about the final equation?
\frac{\mathrm{d} v}{\mathrm{d} q}=-\frac{\omega^2}{v}q
Is it
f'\circ\gamma_1(t)=-\frac{\omega^2}{f\circ\gamma_1(t)}\gamma_1(t) \enspace ?