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 Ham1 and Ham2 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:
Ham1: q(dot)=dH/dp
Ham2: p(dot)=dH/dq
L(q,q(dot))=pq(dot)H(q,p)
My attempt:
1: For m=0, c=1, H=p, Ham1=d/dp*(p)=1 and Ham2 d/dq*(p)=0
2: We need to find p in terms of q and q(dot) to find L. From Ham1 with m=/=0
q(dot)=p/sqrt(p^2+m^2)> p=mq(dot)/sqrt(q(dot)^21)
Using L(q,q(dot))=pq(dot)H(q,p) and Ham1=0 for m=0
L=sqrt(p^2+m^2)=sqrt((m^2q(dot)^2)/(q(dot)^21)+m^2)=m*sqrt((q(dot)^2)/(q(dot)^21)+1)
I am given that L(q,q(dot)) should be m*sqrt(1q(dot)^2/c^2) but with c=1 L=m*sqrt(1q(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!
