Mechanics II: Hamiltonian and Lagrangian of a relativistic free particle

  • #1
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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!
 

Answers and Replies

  • #2
TSny
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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.
 
  • #3
dextercioby
Science Advisor
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Can the OP rewrite this in LaTex code?
 

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