How to show that ##f_i(q_i, p_i)## is constant

[JD_PM]

Homework Statement

Consider a Hamiltonian system of $N$ degrees of freedom $q_i$, where $i = 1,......,N$. The momentum is $p_i$. Suppose the Hamiltonian has the following specific form:

$$H = H[f_1(q_1, p_1), f_2(q_2, p_2), ..., f_N(q_N, p_N)]$$

Where $f_i$ is some function of $q^i$ and $p_i$ alone.

Show that $f_i(q_i, p_i)$ is a constant of motion.

Relevant Equations

$$f(q_1, q_2, ...., q_N, p_1, p_2,...., p_N) = \text{constant}$$

Alright my idea is that, in order to show that $f_i(q_i, p_i)$ is a constant of motion, it would suffice to show that the Hamiltonian is equal to a constant.

Well, the Hamiltonian will be equal to a constant iff:

$$f(q_1, q_2, ..., q_N, p_1, p_2,..., p_N) = \text{constant}$$

Which is what we have to prove... So I am in a closed loop.

Could you give me a hint?

I am actually checking out Goldstein, section 2.6. He says:

'In many problems a number of first integrals of the equations of motion can be obtained immediately; by this we mean relations of the type (note that his $f$ is in function of $q_i$ and its derivatives wrt time alone, while mine is in function of $q_i$ and $p_i$ alone):

$$f(q_1, q_2, ..., \dot q_1, \dot q_2, ..., t) = \text{constant}$$

...These first integrals are of interest because they tell us something physically about the system.'

OK so we should be able to show $f(q_1, q_2, ..., p_1, p_2) = \text{constant}$ by Hamilton's principle then?

Thanks.

 

[PeroK]

Looks like a job for the Poisson bracket.

 

[JD_PM]

Mmm I saw it briefly in class. I see it is defined as follows:

$$[u, v] = \sum_{i=1}^{n} \Big( \frac{\partial u}{\partial q_i} \frac{\partial v}{\partial p_i} - \frac{\partial u}{\partial p_j} \frac{\partial v}{\partial q_j}\Big)$$

Where $u(q, p)$ and $v(q, p)$ are any two functions of position in the phase space $(q, p)$.

Thanks for the hint PeroK, I will investigate further.

 

[PeroK]

Mmm I saw it briefly in class. I see it is defined as follows:

$$[u, v] = \sum_{i=1}^{n} \Big( \frac{\partial u}{\partial q_i} \frac{\partial v}{\partial p_i} - \frac{\partial u}{\partial p_j} \frac{\partial v}{\partial q_j}\Big)$$

Where $u(q, p)$ and $v(q, p)$ are any two functions of position in the phase space $(q, p)$.

Thanks for the hint PeroK, I will investigate further.

Work out what $[f, H]$ says about $f$.

 

[JD_PM]

Work out what $[f, H]$ says about $f$.

And shouldn't $\frac{\partial f} {\partial t} = 0$ be involved somehow? That idea makes sense to me because we are looking for showing $f_i$ to be constant.

 

[PeroK]

And shouldn't $\frac{\partial f} {\partial t} = 0$ be involved somehow? That idea makes sense to me because we are looking for showing $f_i$ to be constant.

It's $\frac{df}{dt}$ you are interested in.

 

[JD_PM]

Oh I think I got it!

$$[f, H] = (\nabla_q f)(\nabla_p H) - (\nabla_p f)(\nabla_q H)$$

We identify the Hamilton equations:

$$\nabla_p H = \frac{\partial H}{\partial p_i} = \dot q_i$$

$$\nabla_q H = \frac{\partial H}{\partial q_i} = -\dot p_i$$

Thus:

$$[f, H] = (\nabla_q f)\dot q_i - (\nabla_p f)(-\dot p_i) \ \ \ \ \ \ \ \ (1)$$

But we know, because of the chain rule, that:

$$\frac{d}{dt} f(q, p) = \frac{df}{dq}\frac{dq}{dt} + \frac{df}{dp}\frac{dp}{dt} = \frac{df}{dq} \dot q + \frac{df}{dp} \dot p \ \ \ \ \ \ \ \ (2)$$

Note that (1) = (2).

$$[f, H] = \frac{d}{dt} f(q, p)$$

Thus, $f_i$ is a constant of motion iff $[f, H] = 0$