Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Classical Physics
Quantum Physics
Quantum Interpretations
Special and General Relativity
Atomic and Condensed Matter
Nuclear and Particle Physics
Beyond the Standard Model
Cosmology
Astronomy and Astrophysics
Other Physics Topics
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Classical Physics
Quantum Physics
Quantum Interpretations
Special and General Relativity
Atomic and Condensed Matter
Nuclear and Particle Physics
Beyond the Standard Model
Cosmology
Astronomy and Astrophysics
Other Physics Topics
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Physics
Classical Physics
Understanding Hamiltonian Conservation Laws
Reply to thread
Message
[QUOTE="vanhees71, post: 5399406, member: 260864"] 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. [/QUOTE]
Insert quotes…
Post reply
Forums
Physics
Classical Physics
Understanding Hamiltonian Conservation Laws
Back
Top