 Physics Help Forum Mechanics II: Hamiltonian and Lagrangian of a relativistic free particle
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 Advanced Mechanics Advanced Mechanics Physics Help Forum Feb 12th 2018, 02:41 AM #1 Junior Member   Join Date: Feb 2018 Posts: 1 Mechanics II: Hamiltonian and Lagrangian of a relativistic free particle The Problem: I am given the Hamiltonian of the relativistic free particle. H(q,p)=sqrt(p^2c^2+m^2c^4) Assume c=1 1: Find Ham-1 and Ham-2 for m=0 2: Show L(q,q(dot))=-m*sqrt(1-(q(dot))^2/c^2) 3: Consider m=0, what does it mean? Equations Used: Ham-1: q(dot)=dH/dp Ham-2: p(dot)=-dH/dq L(q,q(dot))=pq(dot)-H(q,p) My attempt: 1: For m=0, c=1, H=p, Ham-1=d/dp*(p)=1 and Ham-2 -d/dq*(p)=0 2: We need to find p in terms of q and q(dot) to find L. From Ham-1 with m=/=0 q(dot)=p/sqrt(p^2+m^2)-> p=mq(dot)/sqrt(q(dot)^2-1) Using L(q,q(dot))=pq(dot)-H(q,p) and Ham-1=0 for m=0 L=sqrt(p^2+m^2)=-sqrt((m^2q(dot)^2)/(q(dot)^2-1)+m^2)=-m*sqrt((q(dot)^2)/(q(dot)^2-1)+1) I am given that L(q,q(dot)) should be -m*sqrt(1-q(dot)^2/c^2) but with c=1 L=-m*sqrt(1-q(dot)^2) Am I missing something simple algebraically or did I mess up a step earlier on? 3: I'm not sure what L=0 means. The value of H is the energy, so if the energy is 0 L=pq(dot). The momentum times the change in canonical position is 0? Thank you for the help!  Tags free, hamiltonian, lagrangian, mechanics, particle, relativistic Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post llawerif Quantum Physics 1 Oct 16th 2016 01:34 PM NicoleJS Quantum Physics 4 Sep 7th 2016 01:48 PM Kiwi_Dave Advanced Mechanics 1 Jun 14th 2013 01:07 PM suyoon Advanced Mechanics 1 Apr 19th 2010 04:35 AM nathanzzf Quantum Physics 4 Jan 4th 2010 04:07 AM