Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

From Lagrangian I got two primary constraint $\phi_i$ and $\phi$. And my Hamiltonian in presence of the constraints becomes- $$H_p=p\dot q-L+\lambda_i\phi_i+\lambda\phi$$ here the $\lambda_i$ and $\lambda$ are Lagrange undetermined multiplier. Now from $\dot \phi_i=[\phi_i,H_p]$ I got secondary constraint $\Sigma_k$ and from $\dot \phi=[\phi,H_p]$ I got another secondary constraint $\Sigma$ . To satisfy the consistency condition I calculated the $\dot \Sigma_k=[\Sigma_k,H_p]$ and $\dot\Sigma=[\Sigma,H_p]$. $$$$ From the relation I have $\dot \Sigma\approx0$. But the $\dot \Sigma_k$ gives the value of of $\lambda_i$. Now can anyone help me how can I further analyze the constraints in this case? Do I have to put the value of $\lambda_i$ in the equation of $H_p$ and calculate the commutation again? An example would be lovely.

share|improve this question

1 Answer 1

As I know, from the time consistency of $\Sigma_i$ you will not get another constraint. You have to substitute the Lagrange multiplier in the Hamiltonian by what you got. But I can not understand what you want to say by $\dot{\Sigma}=0$. If you mean 0=0, it is the end! But if it gives you an expression you have to continue and consider $\dot{\Sigma}=0$ as a new constraint. Check whether this new expression does not be a linear combination of other constraints. Generally It is not necessary to add secondary constraints to the complete Hamiltonian ($H_p$).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.