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## Main Question or Discussion Point

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 F

$$\sum_{i=1}^{N} \int_{t_A}^{t_B} \vec{F_i} \cdot \vec{v_i} dt = \sum_{i=1}^{N} (\phi_i(\vec{r_A}) - \phi_i(\vec{r_B})) $$

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 t

By Gregory, the total work done by all the external forces (that's the F

_{i}s ) 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 t

_{A}to t_{B}do the same sum as integrating from r_{A}to r_{B}?