Consider a pairwise potential, such as the Lennard Jones potential, \[U(r) = 4 \epsilon \left[ \left(\frac{\sigma}{r}\right)^{12} - \left(\frac{\sigma}{r}\right)^{6} \right].\]Question: How can we get the force (magnitude and force) from such a potential?
For simplicity let us assume that we have only two particles A and B. Specifically, what is the force on particle A due to particle B?
We know that the force is the negative gradient of the potential; therefore, \[\begin{align*}
\mathbf{f}_{AB} & = -\dfrac{dU(r_{AB})}{d\mathbf{r}_A} \\
& = -\dfrac{dU(r_{AB})}{dr_{AB}} \color{blue} {\dfrac{dr_{AB}}{d\mathbf{r}_A}}
\end{align*}\] How do we evaluate the term in blue?
Let \(\mathbf{r}_A = x_A \mathbf{e}_x + y_A \mathbf{e}_y + z_A \mathbf{e}_z\), and \(\mathbf{r}_B = x_B \mathbf{e}_x + y_B \mathbf{e}_y + z_B \mathbf{e}_z\) be the positions of the two particles. Let \(\mathbf{r}_{AB} = \mathbf{r}_B - \mathbf{r}_A\) be a vector pointing in a direction from A to B. The distance between the two particles is:
\[r_{AB} = \sqrt{x_{AB}^2 + y_{AB}^2 + z_{AB}^2},\] where \(x_{AB}^2 = (x_B - x_A)^2\), etc.
We can now try to tackle the blue term: \[\begin{align*}
\dfrac{dr_{AB}}{d\mathbf{r}_A} & = \dfrac{dr_{AB}}{dx_A} \mathbf{e}_x + \dfrac{dr_{AB}}{dy_A} \mathbf{e}_y + \dfrac{dr_{AB}}{dz_A} \mathbf{e}_z\\
& = \dfrac{1}{2 r_{AB}} \left(\dfrac{dx_{AB}^2}{dx_A} \mathbf{e}_x + \dfrac{dy_{AB}^2}{dy_A} \mathbf{e}_y + \dfrac{dz_{AB}^2}{dz_A} \mathbf{e}_z \right) \\
& = \dfrac{1}{2 r_{AB}} \left(-2 x_{AB} \dfrac{dx_A}{dx_A} \mathbf{e}_x - 2 y_{AB} \dfrac{dy_{A}}{dy_A} \mathbf{e}_y - 2 z_{AB} \dfrac{dz_{A}}{dz_A} \mathbf{e}_z \right) \\
& = -\dfrac{\mathbf{r}_{AB}}{r_{AB}} \\
& = -\hat{\mathbf{r}}_{AB}
\end{align*}\] Thus, the force is, \[\mathbf{f}_{AB}(r) = \dfrac{dU(r)}{dr} \hat{\mathbf{r}}_{AB}.\] If \(U^{\prime}(r)\) is negative (repulsive regime of LJ, for instance), the direction of the force is along \( -\hat{\mathbf{r}}_{AB} = \hat{\mathbf{r}}_{AB}\); this force points in the direction of A from B, trying to push particle A away from particle B. If \(U^{\prime}(r)\) is positive (attractive part of LJ), the force acts along \(\hat{\mathbf{r}}_{AB}\).
For simplicity let us assume that we have only two particles A and B. Specifically, what is the force on particle A due to particle B?
We know that the force is the negative gradient of the potential; therefore, \[\begin{align*}
\mathbf{f}_{AB} & = -\dfrac{dU(r_{AB})}{d\mathbf{r}_A} \\
& = -\dfrac{dU(r_{AB})}{dr_{AB}} \color{blue} {\dfrac{dr_{AB}}{d\mathbf{r}_A}}
\end{align*}\] How do we evaluate the term in blue?
Let \(\mathbf{r}_A = x_A \mathbf{e}_x + y_A \mathbf{e}_y + z_A \mathbf{e}_z\), and \(\mathbf{r}_B = x_B \mathbf{e}_x + y_B \mathbf{e}_y + z_B \mathbf{e}_z\) be the positions of the two particles. Let \(\mathbf{r}_{AB} = \mathbf{r}_B - \mathbf{r}_A\) be a vector pointing in a direction from A to B. The distance between the two particles is:
\[r_{AB} = \sqrt{x_{AB}^2 + y_{AB}^2 + z_{AB}^2},\] where \(x_{AB}^2 = (x_B - x_A)^2\), etc.
We can now try to tackle the blue term: \[\begin{align*}
\dfrac{dr_{AB}}{d\mathbf{r}_A} & = \dfrac{dr_{AB}}{dx_A} \mathbf{e}_x + \dfrac{dr_{AB}}{dy_A} \mathbf{e}_y + \dfrac{dr_{AB}}{dz_A} \mathbf{e}_z\\
& = \dfrac{1}{2 r_{AB}} \left(\dfrac{dx_{AB}^2}{dx_A} \mathbf{e}_x + \dfrac{dy_{AB}^2}{dy_A} \mathbf{e}_y + \dfrac{dz_{AB}^2}{dz_A} \mathbf{e}_z \right) \\
& = \dfrac{1}{2 r_{AB}} \left(-2 x_{AB} \dfrac{dx_A}{dx_A} \mathbf{e}_x - 2 y_{AB} \dfrac{dy_{A}}{dy_A} \mathbf{e}_y - 2 z_{AB} \dfrac{dz_{A}}{dz_A} \mathbf{e}_z \right) \\
& = -\dfrac{\mathbf{r}_{AB}}{r_{AB}} \\
& = -\hat{\mathbf{r}}_{AB}
\end{align*}\] Thus, the force is, \[\mathbf{f}_{AB}(r) = \dfrac{dU(r)}{dr} \hat{\mathbf{r}}_{AB}.\] If \(U^{\prime}(r)\) is negative (repulsive regime of LJ, for instance), the direction of the force is along \( -\hat{\mathbf{r}}_{AB} = \hat{\mathbf{r}}_{AB}\); this force points in the direction of A from B, trying to push particle A away from particle B. If \(U^{\prime}(r)\) is positive (attractive part of LJ), the force acts along \(\hat{\mathbf{r}}_{AB}\).
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