# Mechanics II: Hamiltonian and Lagrangian of a relativistic free particle

#### thoover

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!