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Hamilton's equations of motion?

  1. Oct 25, 2009 #1
    Hi,
    This is on my electrodynamics homework and I haven't been able to get anywhere with it. Here it is:

    The Hamiltonian of a particle of mass m, charge q, position r, momentum p, in an external field defined by a vector potential A(r,t) and scalar potential [tex]\phi[/tex](r,t) is given below:

    H(r,p) = (1/2m)[p - qA(r,t)]2 + q[tex]\phi[/tex](r,t) = (1/2m)(pjpj - 2qpjAj +q2AjAj) + q[tex]\phi[/tex]

    Calculate Hamilton's equations of motion. You can use the relations

    [tex]\frac{\partial p_{j}}{\partial p_{i}} = \delta _{ij}[/tex]

    and

    [tex]\frac{\partial p_{j}}{\partial r_{i}} = 0.[/tex]

    So, attempted solution...
    according to my notes, these equations of motion are

    [tex]\frac{\partial H}{\partial p_{i}} = \frac{dr_{i}}{dt}[/tex]

    and

    [tex]\frac{\partial H}{\partial r_{i}} = -\frac{dp_{i}}{dt}[/tex]

    What I tried to do is stick the Hamiltonian into the two equations, but I am a bit confused about what the j and i subscripts are all about. I assumed they meant j = x,y,z, so I put H into those equations to get expressions for j = x, y and z.

    I got

    [tex]\frac{dH}{dp_{x}} = 1/m(p_{x} - qA_{x})[/tex] and equivalent for y and z,

    and

    [tex]\frac{dH}{dr_{x}} = 1/2m(-2qp_{x}(d/dr_{x})A_{x} + q^{2}(d/dr_{x})A^{2}_{x}[/tex] and equivalent for y and z.

    This is pretty much as far as I have got with this (ie. nowhere). I don't really understand what I'm supposed to be doing (ie. what the point of putting H into these equations is), and I dont see what use the 2 relations I was given in the question are. I have given all the information the question has. Can anyone shed any light on what the hell this question is on about?
    Is there some way of rewriting the vector and scalar potentials in terms of momentum or something?
    I am totally lost here and my notes barely even touch on this stuff.

    Thanks.
     
  2. jcsd
  3. Oct 25, 2009 #2
    I'm certain you are correct about the subscripts, thats pretty standard notation.

    The relations:
    ∂pj/∂pi = δij
    &
    ∂pj/∂ri = 0

    I BELIEVE are essentially just stating the independence of each component of p & r, that is... pi does not depend on pj ∀ i ≠j and pi has no dependence on position... your end results for pi
    should not contain any rx ry or rz in them,
    UP TO: this point i could be wrong about something.

    beyond that the equations of motion are as you already listed them (according to your notes) so I believe you're supposed to be solving the differential equations:
    aka eliminate the time derivative part of the equations to get equations momentum equations ∈ pᵢ and position equations ∈ rᵢ .. the "relations" I believe are there to make all this easier.
    Its been a while since I've taken any electrodynamics (even then it was more introductory-ish) and i'm not really familiar with the hamilton's equations (maybe i didn't use them in my class) but it seems to me the "point of this" may be to show you how using the hamiltonian can make your life, in general, a lot easier .... sorry if i'm wrong about anything! I hope it helps
     
  4. Oct 25, 2009 #3
    You're missing a term in one of your equations of motion above.

    You have a particle moving about in some vector & scalar fields that just happen to be electromagnetic in nature and you're using Hamilton's equations of motion to determine how the particle would move in the field. This sounds an awful lot like normal Classical Mechanics, it just happens that you're looking at particles in an E&M field. I believe the exercise is to show you the wide use of Hamilton's EOM.
     
  5. Oct 29, 2009 #4
    I'm in this class and i'm really stuck too. I tried doing the same but i have no idea how to progress from there.
     
  6. Oct 29, 2009 #5
    I've pretty much got this stuff done now actually, I was trying to make things more complicated than they needed to be.
    All this question wanted was to get simplified expressions of dH/dP and dH/dr, for use in the later questions, but you have to really be careful with the i and j subscripts.

    Also the dH/dr that I got on here is wrong.

    this is useful for the later questions, if you really are trying to do the same homework as me:

    http://quantummechanics.ucsd.edu/ph130a/130_notes/node302.html
     
  7. Oct 29, 2009 #6
    For this thing [tex]
    \frac{\partial p_{j}}{\partial p_{i}} = \delta _{ij}
    [/tex]

    does it mean any component over any different component is equal to the same thing?
     
  8. Oct 29, 2009 #7
    i took that to mean that it is equal to 1 if i=j, and it is equal to zero if i=/= j. think its called a kronecker delta.
    it helps simplify the dH/dp equation of motion.


    look on that link I pasted, its slightly different because it starts off with q as a momentum vector, and uses e as the charge, but you can follow it through to the final lorentz force derivation.
     
  9. Oct 29, 2009 #8
    That's correct.
     
  10. Oct 29, 2009 #9
    oh ok. thanks! maybe that will help. kronecker delta rings a bell

    I'll have a look through that site too.

    I got same for dH/dr but with the the + q dI(r,t) / dr on the end
     
  11. Oct 29, 2009 #10
    what I currently have is:

    [tex]\ \frac{\partial H}{\partial r_{i}} = (\frac{q^2A_{j}}{m} - \frac{qp_{j}}{m})\frac{\partial A_{j}}{\partial r_{i}} + q\frac{\partial \Phi}{\partial r_{i}} [/tex]

    im 95% sure thats right
     
  12. Oct 29, 2009 #11

    I think you may be a little off there...I get:

    [tex]
    \frac{\partial H}{\partial r_i}=\frac{1}{2m}\left[q^2\frac{\partial\left(A_iA_j\right)}{\partial r_i}-2qp_i\frac{\partial A_i}{\partial r_i}\right]+q\frac{\partial\phi}{\partial r_i}
    [/tex]

    Just because [tex]\partial_{r_i}p_j=0[/tex] doesn't mean that [tex]\partial_{r_i}A_j=0[/tex] as well.
     
  13. Oct 29, 2009 #12
    where have you got the [tex]\ A_{i} [/tex] parts from? all I have is 2 j's.
    I have used what I got in the last part of the question, and I'm almost finished it, it seems to agree with what is on that link too.

    actually you might be making the same mistake I did.

    what I was doing was just using one subscript, like I changed the equations of motion from the notes to have j's instead of i's, so I was differentiating with respect to j. I've changed that now and differentiated the hamilton with respect to i... the subscripts seem to be important... you will see at the end why.
     
  14. Oct 29, 2009 #13
    I get this


    dH/dri = 1/2m (q^2*dAj^2/dri - 2*q*pj*dAj/dri) + q*dI/dri

    dH/dpi = 1/2m ( dpj^2/dpi - q*Aj*dpj/dpi)


    (sorry im too confused about how to use the latex code!)
     
  15. Oct 29, 2009 #14
    ah, I havent differentiated Aj2, ive chain ruled AjAj which i think gives a different result, other than that I think we get the same?
     
  16. Oct 29, 2009 #15
    It looks like you've done dAjAj/dri = 2* dAj/dri which i don't think works


    I'll try see if i can get the same though!
     
  17. Oct 29, 2009 #16
    i said (d/dri)AjAj = Aj(d/dri)Aj + Aj(d/dri)Aj = 2Aj(d/dri)Aj
     
  18. Oct 29, 2009 #17
    Either way, you should have [tex]\partial_{r_i}(A_jA_j)[/tex] or [tex]\partial_{r_i}(A_iA_i)[/tex] or whatever indices you should choose. What you did is let one of those two A's not be acted on by [tex]\partial_{r_i}[/tex] and I don't believe you can do that.
     
  19. Oct 29, 2009 #18

    ah ok, that makes sense. can't see anything wrong then.
     
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