I am looking for a physics based proof for how one applies forces to each of the three atoms in a bond angle potential. I know the general principle that $F = -\nabla \Phi$ if $\Phi$ is your potential energy function. I am wondering how one goes from such a definition to the force on each atom and how to prove that those forces satisfy conservation of energy and momentum (or any other relevant physical laws).
For some more details, here is how LAMMPs applies forces on atoms from a simple harmonic bond angle potential:
$\Phi(\theta) = k(\theta - \theta_{0})^{2}$
Then $\Phi = k(\arccos(\frac{\mathbf{r_{1}\cdot r_{2}}}{r_{1}r_{2}})- \theta_{0})^{2}$ This leaves us with $$\frac{\partial \Phi}{\partial r_{1}} = 2k(\arccos(\frac{\mathbf{r_{1}\cdot r_{2}}}{r_{1}r_{2}})- \theta_{0})(-\frac{1}{\sqrt{1-(\frac{\mathbf{r_{1}\cdot r_{2}}}{r_{1}r_{2}})^{2}}})(-\frac{\mathbf{r_{1}\cdot r_{2}}}{r_{1}^{2}r_{2}})$$ $$\frac{\partial \Phi}{\partial r_{2}} = 2k(\arccos(\frac{\mathbf{r_{1}\cdot r_{2}}}{r_{1}r_{2}})- \theta_{0})(-\frac{1}{\sqrt{1-(\frac{\mathbf{r_{1}\cdot r_{2}}}{r_{1}r_{2}})^{2}}})(-\frac{\mathbf{r_{1}\cdot r_{2}}}{r_{1}r_{2}^{2}})$$ The forces that LAMMPs assigns to each atom are then $$\textbf{F}_{1} = (-\frac{\partial \Phi}{\partial r_{1}})(\frac{delx_{1}\hat{x}+dely_{1}\hat{y}+delz_{1}\hat{z}}{r_{1}}) + (-\frac{\partial \Phi}{\partial r_{2}})(\frac{delx_{2}\hat{x}+dely_{2}\hat{y}+delz_{2}\hat{z}}{\frac{\textbf{r}_{1}\cdot\textbf{r}_{2}}{r_{2}}})$$ $$\textbf{F}_{3} = (-\frac{\partial \Phi}{\partial r_{1}})(\frac{delx_{1}\hat{x}+dely_{1}\hat{y}+delz_{1}\hat{z}}{\frac{\textbf{r}_{1}\cdot\textbf{r}_{2}}{r_{1}}}) + (-\frac{\partial \Phi}{\partial r_{2}})(\frac{delx_{2}\hat{x}+dely_{2}\hat{y}+delz_{2}\hat{z}}{r_{2}})$$ $$\textbf{F}_{2} = -\textbf{F}_{1} - \textbf{F}_{3}$$
Any resources, proofs, suggestions on chapters from a textbook, etc. that would help to provide intuition on this problem would be very much appreciated.
EDIT:
I am going to post the derivation of the forces from this bond angle potential as presented by Swope and Ferguson 'Alternative Expressions for Energies and Forces Due to Angle Bending and Torsional Energy' 1992. Hopefully with this information plus the answer provided below by Shern Ren Tee, others will be able to use this post as a starting point for their own calculations.
Consider the following diagram as presented above.
$$\Phi(\theta) = k(\theta - \theta_{0})^{2}$$
We seek the following quantity to derive the forces on atom 1:
$$-\nabla_{1} \Phi(\theta)$$
where $\nabla_{1}$ is the derivative of the $\Phi$ potential holding the position of each other atom (atoms 2 and 3) constant. From here, we use a clever formulation of the chain rule to simplify much of the math for us:
$$-\nabla_{1} \Phi(\theta) = (-\frac{d\Phi}{d\theta})(\frac{d\theta}{d\cos(\theta)})(\nabla_{1}\cos(\theta))$$
The first two terms in this expression are trivial. $$(-\frac{d\Phi}{d\theta}) = 2k(\theta-\theta_{0}) = 2k(\arccos(\frac{\mathbf{r_{1}\cdot r_{2}}}{r_{1}r_{2}})- \theta_{0})$$ $$(\frac{d\theta}{d\cos(\theta)}) = \frac{1}{\frac{\cos(\theta)}{d\theta}} = -\frac{1}{\sin(\theta)} = -\frac{1}{\sqrt{1-\cos^{2}(\theta)}} = (-\frac{1}{\sqrt{1-(\frac{\mathbf{r_{1}\cdot r_{2}}}{r_{1}r_{2}})^{2}}})$$ The last term is slightly trickier. First we must realize the following: $$\nabla_{1} = \hat{x} \frac{\partial}{\partial x_{1}} + \hat{y} \frac{\partial}{\partial y_{1}} + \hat{z} \frac{\partial}{\partial z)}$$ Let us only consider the x component here for the following derivation. We can demonstrate that $$\hat{x}\nabla_{1}\cos(\theta) = \frac{\partial\cos(\theta)}{\partial x_{1}} = \frac{\partial\cos(\theta)}{\partial delx_{1}}\frac{\partial delx_{1}}{\partial x_{1}} + \frac{\partial\cos(\theta)}{\partial dely_{1}}\frac{\partial dely_{1}}{\partial x_{1}} + \frac{\partial\cos(\theta)}{\partial delz_{1}}\frac{\partial delz_{1}}{\partial x_{1}} + \frac{\partial\cos(\theta)}{\partial delx_{2}}\frac{\partial delx_{2}}{\partial x_{1}} + \frac{\partial\cos(\theta)}{\partial dely_{2}}\frac{\partial dely_{2}}{\partial x_{1}} + \frac{\partial\cos(\theta)}{\partial delz_{2}}\frac{\partial delz_{2}}{\partial x_{1}} = \frac{\cos(\theta)}{\partial delx_{1}}\hat{x}$$ Finally, we take that derivative to get the portion of the force in the x direction on atom 1 to see:
$$\frac{\partial\cos(\theta)}{\partial x_{1}} = \frac{\partial}{\partial delx_{1}}(\frac{\mathbf{r}_{1}\cdot\mathbf{r}_{2}}{r_{1}r_{2}}) = -\frac{1}{r_{1}r_{2}}(\frac{\mathbf{r}_{1}\cdot\mathbf{r}_{2}}{r_{1}^{2}}delx_{1}\hat{x} - delx_{2}\hat{x})$$ Or overall we get that the entire $\nabla_{1} \cos(\theta)$ term reduces to $$-\frac{1}{r_{1}r_{2}}(\frac{\mathbf{r}_{1}\cdot\mathbf{r}_{2}}{r_{1}^{2}}\mathbf{r}_{1} - \mathbf{r}_{2})$$ Which is identical to the result obtained from LAMMPs. I skipped over some of the more tedious steps, but I hope that this derivation may help others as a starting point.