The fields in Hamilton's principle for matter

Van

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nThe Field Theory of Fluids\n\n1) what are the correct fields to use in Hamilton\'s principle (for\nfluids or any continua)?\n\n2) what is the Lagrangian?\n\nWhen we know this, we can apply the formalism of classical field theory\nto get the conservation laws and eqs. of motion.\n\nThe fields and field theory for matter is just as important, and just\nas much\nof a classical field theory for matter as for the EM field.\n\nI see many papers talking about the Hamilton\'s principle for fluids,\nbut no mention is made of the following. I believe it is the starting\npoint for\nthe field theory of fluids.\n\nThe EM field F = dA since dF = 0 (Maxwell\'s homogeneous eqns.).\nThen A is the vector field in Hamilton\'s principle, and the Lagrangian\nL = - (F|F)/2 gives the rest of Maxwell\'s eqns.\n\nThe same can be done for matter--both for particles and continuous\nmatter.\n\nFor fluids we have the continuity eqn.\ndiv(J) = div(nu) = 0, which is best understood by writing it in\nterms of the 3-form *J = j on spacetime:\n\ndiv(J) = *d*J = *(dj) = 0 ==> dj = 0 so j is closed and thus exact. In\nfact\n\none can show that dj = 0 ==> j = dz^1 /\\ dz^2 /\\ dz^3\n\nwhere the z^i are the 3 scalar fields in Hamilton\'s principle, and they\nplay the role for fluids that A does for EM.\nThe z^i are the initial positions of the\nfluid elements or Lagrangian coords., and potentials for the\nmatter current J = nu and density n.\n\nThe action for a single particle of mass m is\n\nS = Int(L ds) ; where ds = - g(x\',x\') dt^2 , x\' = dx/dt, s = proper\ntime.\n\nThe Lagrangian is L = - sqrt[- (p|p)] = - m ; where p = mu is the\n4-momentum.\n\nFor cold fluids p --> J = nu and L = - sqrt[- (J|J)} = - sqrt(j|j)\n\n= - sqrt(dz^1 /\\ dz^2 /\\ dz^3|dz^1 /\\ dz^2 /\\ dz^3) = - n. We must add\nthe\ninternal energy when it is not 0; L = - n(1 + e) = - r = - (total\nenergy density). The action is S = Int(L d^4(x)), d^4(x) = 4D volume\nelement.\n\n>From here we can apply canonical field theory and get the\nenergy-momentum\ntensor\n\nT = gL - dz^i x dL/d(dz^i) = (r + p)uu + pg where p = pressure.\n\ndiv(T) = 0 then gives the motion.\n\nI have an unpublished paper on this if anyone wants it I would be glad\nto\nemail it.\n\nVan Jacques\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>The Field Theory of Fluids

1) what are the correct fields to use in Hamilton's principle (for
fluids or any continua)?

2) what is the Lagrangian?

When we know this, we can apply the formalism of classical field theory
to get the conservation laws and eqs. of motion.

The fields and field theory for matter is just as important, and just
as much
of a classical field theory for matter as for the EM field.

I see many papers talking about the Hamilton's principle for fluids,
but no mention is made of the following. I believe it is the starting
point for
the field theory of fluids.

The EM field [itex]F = dA[/itex] since [itex]dF =[/itex] (Maxwell's homogeneous eqns.).
Then A is the vector field in Hamilton's principle, and the Lagrangian
[itex]L = - (F|F)/2[/itex] gives the rest of Maxwell's eqns.

The same can be done for matter--both for particles and continuous
matter.

For fluids we have the continuity eqn.
[itex]div(J) = div(\nu) = 0,[/itex] which is best understood by writing it in
terms of the 3-form [itex]*J = j[/itex] on spacetime:

[tex]div(J) = *d*J = *(dj) =[/itex] ==> [itex]dj =[/itex] so j is closed and thus exact. In
fact

one can show that [itex]dj =[/itex] ==> [itex]j = dz^1 /\ dz^2 /\ dz^3[/tex]

where the [itex]z^i[/itex] are the 3 scalar fields in Hamilton's principle, and they
play the role for fluids that A does for EM.
The [itex]z^i[/itex] are the initial positions of the
fluid elements or Lagrangian coords., and potentials for the
matter current [itex]J = \nu[/itex] and density n.

The action for a single particle of mass m is

[itex]S = \Int(L ds) ;[/itex] where [itex]ds = - g(x',x') dt^2 , x' = dx/dt, s =[/itex] proper
time.

The Lagrangian is [itex]L = - \sqrt[- (p|p)] = - m ;[/itex] where [itex]p = \mu[/itex] is the
4-momentum.

For cold fluids p --> [itex]J = \nu[/itex] and [itex]L = - \sqrt[- (J|J)} = - \sqrt(j|j)= - \sqrt(dz^1 /\ dz^2 /\ dz^3|dz^1 /\ dz^2 /\ dz^3) = - n[/itex]. We must add
the
internal energy when it is not 0; [itex]L = - n(1 + e) = - r = -[/itex] (total
energy density). The action [itex]is S = \Int(L d^4(x)), d^4(x) = 4D[/itex] volume
element.

>From here we can apply canonical field theory and get the
energy-momentum
tensor

[itex]T = gL - dz^i x dL/d(dz^i) = (r + p)uu + pg[/itex] where p = pressure.

[itex]div(T) =[/itex] then gives the motion.

I have an unpublished paper on this if anyone wants it I would be glad
to
email it.

Van Jacques

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