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

## Homework Statement

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))=-msqrt(1-(q(dot))^2/c^2)
3: Consider m=0, what does it mean?

## Homework Equations

Ham-1: q(dot)=dH/dp
Ham-2: p(dot)=-dH/dq
L(q,q(dot))=pq(dot)-H(q,p)

## The Attempt at a Solution

1: For m=0, c=1, Ham-1=1 and Ham-2=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((m^2q(dot)^2)/(q(dot)^2-1)+m^2)=-m*sqrt((q(dot)^2)/(q(dot)^2-1)+1)

We are 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!

Related Advanced Physics Homework Help News on Phys.org
TSny
Homework Helper
Gold Member
The attempt at a solution
1: For m=0, c=1, Ham-1=1 and Ham-2=0
Ham-1 and Ham-2 are the equations ##\dot q=\partial H/ \partial p## and ##\dot p=-\partial H/ \partial q##, respectively. I don't understand the meaning of Ham-1=1 and Ham-2=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)
OK
-> p=mq(dot)/sqrt(q(dot)^2-1)
Note that you are taking the square root of a negative number. Check your derivation.
Using L(q,q(dot))=pq(dot)-H(q,p) and Ham-1=0 for m=0
L=-sqrt((m^2q(dot)^2)/(q(dot)^2-1)+m^2)=-m*sqrt((q(dot)^2)/(q(dot)^2-1)+1)
Again, the expression Ham-1 = 0 doesn't make sense to me.

dextercioby