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I have to find a constant of motion associated to this Hamiltonian but I don't know how to proceed.

$$H=\frac{\mathbf{p_0}^2}{2m}+\frac{\mathbf{p_1}^2}{2m}+\frac{\mathbf{p_2}^2}{2m}-2V(\mathbf{r_1}- \mathbf{r_0})+V(\mathbf{r_2}-\mathbf{r_1})$$

where $$V(\mathbf x)=\frac {e^2}{|\mathbf x|}.$$

I don't know what $\mathbf x$ is.

This Hamiltonian refers to a system of 3 particles $(0,1,2)$ with mass $m$ and charge $e$.

The coordinates are $r^\alpha_i$ and conjugate momenta $p^\beta_j$ with $\alpha, \beta=0,1,2$.

I have written all the information that I have.

Qmechanic
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sunrise
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1 Answers1

3

Use Poisson brackets. A constant of motion is some function $F(q_i,p_i)$ of phase space that is independent of time, i.e. such that $\frac{dF}{dt}=0$. Now, consider that $$\frac{dF}{dt}=\frac{\partial{F}}{\partial{q}_1}\frac{d{q_1}}{d{t}}+\ldots+\frac{\partial{F}}{\partial{q}_n}\frac{dq_n}{dt}+\frac{\partial{F}}{\partial{p}_1}\frac{d{p_1}}{d{t}}+\ldots+\frac{\partial{F}}{\partial{p}_n}\frac{dp_n}{dt}+\frac{\partial{F}}{\partial{t}}\\=\sum_{j=1}^n\left(\frac{\partial{F}}{\partial{q}_j}\dot{q}_j+\frac{\partial{F}}{\partial{p}_j}\dot{p}_j\right)+\frac{\partial{F}}{\partial{t}}$$ but since we defined $F=F(q_i,p_i)$ ($F$ doesn't depend explicitly on time), $\partial{F}/\partial{t}=0$, so that you may just calculate $$\sum_{j=1}^n\left(\frac{\partial{F}}{\partial{q}_j}\dot{q}_j+\frac{\partial{F}}{\partial{p}_j}\dot{p}_j\right)=\sum_{j=1}^n\left(\frac{\partial{F}}{\partial{q}_j}\frac{\partial{\mathcal{H}}}{\partial{p}_j}-\frac{\partial{F}}{\partial{p}_j}\frac{\partial{\mathcal{H}}}{\partial{q}_j}\right)\equiv\left\{F,\mathcal{H}\right\}$$ which is the way the Poisson bracket $\{F,\mathcal{H}\}$ is defined. Note that your hamiltonian does not depend on time, so that immediately $\{\mathcal{H},\mathcal{H}\}=0$ and energy is conserved. This way you may look for other conserved quantities like linear momentum $p_k$ or angular momentum $L_k$, for example.