The Hamiltonian

[Physgeek64]

I'm a little confused about the hamiltonian.

Once you have the hamiltonian how can you find conserved quantities. I understand that if it has no explicit dependence on time then the hamiltonian itself is conserved, but how would you get specific conservation laws from this?

Many thanks

 

[vanhees71]

Unfortunately you do not tell us about your level. Do you know Poisson brackets? If so, you question is quite easy to answer. Suppose you have an arbitrary phase-space function $f(t,q^k,p_j)$ ($k,j \in \{1,\ldots,f \}$) then the total time derivative is
$$\frac{\mathrm{d}}{ \mathrm{d} t} f=\dot{q}^k \frac{\partial f}{\partial q^k}+\dot{p}_j \frac{\partial f}{\partial p_j} + \partial_t f,$$
where the latter partial time derivative refers to the explicit time dependence of $f$ only. Now use the Hamilton equations of motion
$$\dot{q}^k=\frac{\partial H}{\partial p_k}, \quad \dot{p}_j=-\frac{\partial H}{\partial q^j}.$$
Plugging this in the time derivative you get
$$\frac{\mathrm{d}}{ \mathrm{d} t} f=\frac{\partial f}{\partial q^k} \frac{\partial H}{\partial p_k} - \frac{\partial f}{\partial p_j} \frac{\partial H}{\partial q^j} + \partial_t f=\{f,H \}+\partial_t f.$$
A quantity is thus obviously conserved by definition if this expression vanishes.

Applying this to the Hamiltonian itself you get
$$\frac{\mathrm{d}}{\mathrm{d} t} H=\{H,H\}+\partial_t H=\partial_t H,$$
i.e., $H$ is conserved (along the trajectory of the system) if and only if it is not explicitly time dependent.

Now an infinitesimal canonical transformation is generated by an arbitrary phase-space distribution function $G$,
$$\delta q^k=\frac{\partial G}{\partial p_k}\delta \alpha=\{q^k,G\} \delta \alpha, \quad \delta p_j=-\frac{\partial G}{\partial q^j} \delta \alpha =\{p_j,G \} \delta \alpha, \quad \delta H=\partial_t G \delta \alpha.$$
From this it is easy to show that
$$H'(t,q+\delta q,p+\delta p) = H(t,q,p),$$
i.e., that the infinitesimal canonical transformation is a symmetry of the Hamiltonian, if and only if
$$\{H,G \}+\partial_t G=0,$$
but that means that
$$\frac{\mathrm{d}}{\mathrm{d} t} G=0$$
along the trajectory of the system, i.e., the generator of a symmetry transformation is a conserved quantity, and also any conserved quantity is the generator of a symmetry transformation. That means that there's a one-to-one relation between the generators of symmetries and conserved quantities, which is one of Noether's famous theorems.

 

[Physgeek64]

Unfortunately you do not tell us about your level. Do you know Poisson brackets? If so, you question is quite easy to answer. Suppose you have an arbitrary phase-space function $f(t,q^k,p_j)$ ($k,j \in \{1,\ldots,f \}$) then the total time derivative is
$$\frac{\mathrm{d}}{ \mathrm{d} t} f=\dot{q}^k \frac{\partial f}{\partial q^k}+\dot{p}_j \frac{\partial f}{\partial p_j} + \partial_t f,$$
where the latter partial time derivative refers to the explicit time dependence of $f$ only. Now use the Hamilton equations of motion
$$\dot{q}^k=\frac{\partial H}{\partial p_k}, \quad \dot{p}_j=-\frac{\partial H}{\partial q^j}.$$
Plugging this in the time derivative you get
$$\frac{\mathrm{d}}{ \mathrm{d} t} f=\frac{\partial f}{\partial q^k} \frac{\partial H}{\partial p_k} - \frac{\partial f}{\partial p_j} \frac{\partial H}{\partial q^j} + \partial_t f=\{f,H \}+\partial_t f.$$
A quantity is thus obviously conserved by definition if this expression vanishes.

Applying this to the Hamiltonian itself you get
$$\frac{\mathrm{d}}{\mathrm{d} t} H=\{H,H\}+\partial_t H=\partial_t H,$$
i.e., $H$ is conserved (along the trajectory of the system) if and only if it is not explicitly time dependent.

Now an infinitesimal canonical transformation is generated by an arbitrary phase-space distribution function $G$,
$$\delta q^k=\frac{\partial G}{\partial p_k}\delta \alpha=\{q^k,G\} \delta \alpha, \quad \delta p_j=-\frac{\partial G}{\partial q^j} \delta \alpha =\{p_j,G \} \delta \alpha, \quad \delta H=\partial_t G \delta \alpha.$$
From this it is easy to show that
$$H'(t,q+\delta q,p+\delta p) = H(t,q,p),$$
i.e., that the infinitesimal canonical transformation is a symmetry of the Hamiltonian, if and only if
$$\{H,G \}+\partial_t G=0,$$
but that means that
$$\frac{\mathrm{d}}{\mathrm{d} t} G=0$$
along the trajectory of the system, i.e., the generator of a symmetry transformation is a conserved quantity, and also any conserved quantity is the generator of a symmetry transformation. That means that there's a one-to-one relation between the generators of symmetries and conserved quantities, which is one of Noether's famous theorems.

Sorry- no I don't know about Poisson brackets- I'm a complete novice. Haven't encountered hamiltonians before, nor do I know much about them.

Thank you for your response though! Unfortunately I can't see any of the maths you've included- For some reason my computer thinks its an error

 

[Edgardo]

Unfortunately I can't see any of the maths you've included- For some reason my computer thinks its an error

Right-click on the math error, go to Math Settings -> Math Renderer and try e.g. HTML-CSS (or some other renderer).