zackiechan
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I'm working through Gregory's Classical Mechanics and came across his derivation of energy conservation for a system of N particles that is unconstrained. We get to assume all the external forces are conservative, so we can write them as the gradient of a potential energy. There's a step he makes in the derivation that has me confused.
By Gregory, the total work done by all the external forces (that's the Fis ) is:
$$\sum_{i=1}^{N} \int_{t_A}^{t_B} \vec{F_i} \cdot \vec{v_i} dt = \sum_{i=1}^{N} (\phi_i(\vec{r_A}) - \phi_i(\vec{r_B})) $$
What I don't understand is how to go from the integral:
$$\int_{t_A}^{t_B} \vec{F_i} \cdot \vec{v_i} dt$$ to the potentials.
My idea is:
$$\int_{t_A}^{t_B} \vec{F_i} \cdot \vec{v_i} dt = \int_{t_A}^{t_B} -\nabla \phi_i \cdot \vec{v_i} dt = \int_{\vec{r_A}}^{\vec{r_B}} -\nabla \phi_i \cdot \vec{dr} = \phi_i(\vec{r_A}) - \phi_i(\vec{r_B})$$
My questions are :
Can we go from a time integral to a position integral without messing with the potentials? I know they are path independent, but are they time independent?
Does integrating from tA to tB do the same sum as integrating from rA to rB?
By Gregory, the total work done by all the external forces (that's the Fis ) is:
$$\sum_{i=1}^{N} \int_{t_A}^{t_B} \vec{F_i} \cdot \vec{v_i} dt = \sum_{i=1}^{N} (\phi_i(\vec{r_A}) - \phi_i(\vec{r_B})) $$
What I don't understand is how to go from the integral:
$$\int_{t_A}^{t_B} \vec{F_i} \cdot \vec{v_i} dt$$ to the potentials.
My idea is:
$$\int_{t_A}^{t_B} \vec{F_i} \cdot \vec{v_i} dt = \int_{t_A}^{t_B} -\nabla \phi_i \cdot \vec{v_i} dt = \int_{\vec{r_A}}^{\vec{r_B}} -\nabla \phi_i \cdot \vec{dr} = \phi_i(\vec{r_A}) - \phi_i(\vec{r_B})$$
My questions are :
Can we go from a time integral to a position integral without messing with the potentials? I know they are path independent, but are they time independent?
Does integrating from tA to tB do the same sum as integrating from rA to rB?