Mass As Gravitational Charge & Derivation Of Geodesic Equation Of Motion

Alfred Einstead

<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>whopkins@csd.uwm.edu (Alfred Einstead) wrote:\n> Esa A E Peuha <esa.peuha@helsinki.fi> wrote:\n> > whopkins@csd.uwm.edu (Alfred Einstead) writes:\n> > > All of this should bother people, because General Relativity says\n> > > that mass is gravitational charge.\n> >\n> > This is [sic] _not_ what GR says. GR says that energy and momentum are\n> > the source of gravity.\n>\n> More correctly, mass = gravitational charge is exactly what the Principle\n> of Equivalence says, and that _therefore_ that energy and momentum are the\n> source of gravity.\n\nIn detail, restoring some of the context deleted from the original\narticle the quote was in reference to, the Lagrangian for a point\nsource is:\nL = 1/2 m g_{mn} dx^m/ds dx^n/ds.\n\nInserting the Lagrangian density\n*L = L delta^3(r - x(s))\ninto the Einstein-Hilbert action, you get the total action:\nS = integral [|g|^{1/2} (R + Lambda)/(16 pi G) + *L]\n[ |g| = |det(g_{mn})| ]\nwhich yields the corresponding field equations:\nG^{mn} |g|^{1/2} = 8 pi G m dx^m/ds dx^n/ds delta^3(r - x(s))\nthus identifying the charge as proportional to m.\n\n[The stress-tensor T^{mn}, defined generally in terms of the\nmatter Lagrangian by T^{mn} = 2 |g|^{-1/2} delta(L)/delta(g_{mn})\nyields the tensor density:\nT^{mn} |g|^{1/2} = m dx^m/ds dx^n/ds delta^3(r - x(s)).\n]\n\nFor a path-parametrized time-like motion this yields the\ngeneralized Poisson law:\n-1/2 R |g|^{1/2} = 4 pi G m delta^3(r - x(s)).\nthus identifying m as the gravitational charge of the point source.\n\n[In the special case g_{i0} = 0 = g_{0j}, g_{ij} = delta_{ij},\nthe generalized Poisson law reduces exactly to the Poisson equation\nfor |g|:\ndel^2(|g|^{1/2}) = 4 pi g m delta^3(r - x(s))\n].\n\nApplying the Bianchi identity for the Einstein tensor, one finds\nthat:\nd_m (G^{mn} |g|^{1/2}) + Gamma^n_{mp} G^{mp} |g|^{1/2} = 0\nwhere\nd_m denotes partial derivative,\nGamma^n_{mp} are the components of the connection\nwhich is the continuum version of the geodesic law.\n\nUnder the field law, this becomes (after dividing by 8 pi G):\nd_m (m dx^m/ds dx^n/ds delta^3(r-x(s)))\n+ m Gamma^n_{mp} dx^m/ds dx^p/ds delta^3(r - x(s)) = 0,\nwith\nd_m (m d(x^m)/ds d(x^n)/ds delta^3(r - x(s)))\n= m d(x^n)/ds (d(x^m)/ds d_m) delta^3(r - x(s))\n= m d(x^n)/ds (-d/ds delta^3(r - x(s)))\n= m d/ds (d(x^n)/ds) delta^3(r - x(s))\n= m d^2(x^n)/ds^2 delta^3(r - x(s)).\nThus\nm (d^2(x^n)/ds^2 + Gamma^n_{mp} dx^m/ds dx^p/ds) delta^3(r - x(s)) = 0\nwhich after integrating becomes the geodesic law for the point source:\nm (x^n\'\' + Gamma^n_{mp} x^m\' x^p\') = 0.\n\nThe Bianchi identity for point sources is the geodesic law, itself.\n\nThis is cast in first order form as the couple system:\np_m = m g_{mn} d(x^n)/ds\nd(p_m)/ds = dU(x,dx/ds)/ds\ndefining the gravitational potential\nU(x,v) = 1/2 g_{mn}(x) v^m v^n.\n[For slowly moving objects, U is just 1/2 g_{00}, itself].\n\nThe first equation identifies m as the inertia of the point source, the\nsecond equation identifies m as the coupling constant to the external\nfield U.\n\nTherefore, that\'s EXACTLY what Einstein\'s equations are indeed\nstating; namely, that:\n(1) m is the charge of the point source, as\na gravitating body\n(2) m is the charge of the point source, as\nthe coupling constant of the body to an external\nfield\nand\n(3) m is the inertia of the point source, thus identifying\nthe gravitational charge as being the inertial mass, itself.\n\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>whopkins@csd.uwm.edu (Alfred Einstead) wrote:
> Esa A E Peuha <esa.peuha@helsinki.fi> wrote:
> > whopkins@csd.uwm.edu (Alfred Einstead) writes:
> > > All of this should bother people, because General Relativity says
> > > that mass is gravitational charge.
> >
> > This is [sic] _not_ what GR says. GR says that energy and momentum are
> > the source of gravity.
>
> More correctly, mass = gravitational charge is exactly what the Principle
> of Equivalence says, and that _therefore_ that energy and momentum are the
> source of gravity.

In detail, restoring some of the context deleted from the original
article the quote was in reference to, the Lagrangian for a point
source is:
[itex]L = 1/2 m g_{mn} dx^m/ds dx^n/ds[/itex].

Inserting the Lagrangian density
[itex]*L = L \delta^3(r - x(s))[/itex]
into the Einstein-Hilbert action, you get the total action:
S = integral [itex][|g|^{1/2} (R + \Lambda)/(16 \pi G) + *L][ |g| = |det(g_{mn})| ][/itex]
which yields the corresponding field equations:
[itex]G^{mn} |g|^{1/2} = 8 \pi G m dx^m/ds dx^n/ds \delta^3(r - x(s))[/itex]
thus identifying the charge as proportional to m.

[The stress-tensor [itex]T^{mn},[/itex] defined generally in terms of the
matter Lagrangian by [itex]T^{mn} = 2 |g|^{-1/2} \delta(L)/\delta(g_{mn})[/itex]
yields the tensor density:
[itex]T^{mn} |g|^{1/2} = m dx^m/ds dx^n/ds \delta^3(r - x(s))[/itex].
]

For a path-parametrized time-like motion this yields the
generalized Poisson law:
[itex]-1/2 R |g|^{1/2} = 4 \pi G m \delta^3(r - x(s))[/itex].
thus identifying m as the gravitational charge of the point source.

[In the special case [itex]g_{i0} == g_{0j}, g_{ij} = \delta_{ij},[/itex]
the generalized Poisson law reduces exactly to the Poisson equation
for [itex]|g|:del^2(|g|^{1/2}) = 4 \pi g m \delta^3(r - x(s))[/itex]
].

Applying the Bianchi identity for the Einstein tensor, one finds
that:
[itex]d_m (G^{mn} |g|^{1/2}) + \Gamma^n_{mp} G^{mp} |g|^{1/2} =[/itex]
where
[itex]d_m[/itex] denotes partial derivative,
[itex]\Gamma^n_{mp}[/itex] are the components of the connection
which is the continuum version of the geodesic law.

Under the field law, this becomes (after dividing [itex]by 8 \pi G):d_m (m dx^m/ds dx^n/ds \delta^3(r-x(s)))+ m \Gamma^n_{mp} dx^m/ds dx^p/ds \delta^3(r - x(s)) = 0,[/itex]
with
[itex]d_m (m d(x^m)/ds d(x^n)/ds \delta^3(r - x(s)))= m d(x^n)/ds (d(x^m)/ds d_m) \delta^3(r - x(s))= m d(x^n)/ds (-d/ds \delta^3(r - x(s)))= m d/ds (d(x^n)/ds) \delta^3(r - x(s))= m d^2(x^n)/ds^2 \delta^3(r - x(s))[/itex].
Thus
[itex]m (d^2(x^n)/ds^2 + \Gamma^n_{mp} dx^m/ds dx^p/ds) \delta^3(r - x(s)) = [/itex]
which after integrating becomes the geodesic law for the point source:
[itex]m (x^n'' + \Gamma^n_{mp} x^m' x^p') = .[/itex]

The Bianchi identity for point sources is the geodesic law, itself.

This is cast in first order form as the couple system:
[itex]p_m = m g_{mn} d(x^n)/dsd(p_m)/ds = dU(x,dx/ds)/ds[/itex]
defining the gravitational potential
U(x,v) [itex]= 1/2 g_{mn}(x) v^m v^n[/itex].
[For slowly moving objects, U is just [itex]1/2 g_{00},[/itex] itself].

The first equation identifies m as the inertia of the point source, the
second equation identifies m as the coupling constant to the external
field U.

Therefore, that's EXACTLY what Einstein's equations are indeed
stating; namely, that:
(1) m is the charge of the point source, as
a gravitating body
(2) m is the charge of the point source, as
the coupling constant of the body to an external
field
and
(3) m is the inertia of the point source, thus identifying
the gravitational charge as being the inertial mass, itself.

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